CBSE Class 12 Mathematics Continuity And Differentiability Assignment Set 02

Assignment 1: Continuity and Differentiability

Question 1. If \( f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & \text{if } x < 4 \\ a+b, & \text{if } x = 4 \\ \frac{x-4}{|x-4|} + b, & \text{if } x > 4 \end{cases} \) is continuous at \( x = 4 \), find \( a \) and \( b \).
Answer: For \( x < 4 \), \( |x-4| = -(x-4) \), giving \( \lim_{x \to 4^-} f(x) = \frac{x-4}{-(x-4)} + a = -1 + a \).
For \( x > 4 \), \( |x-4| = x-4 \), giving \( \lim_{x \to 4^+} f(x) = \frac{x-4}{x-4} + b = 1 + b \).
Since \( f(x) \) is continuous at \( x = 4 \), the left-hand limit, right-hand limit, and value at the point must coincide:
\( -1 + a = a + b \implies b = -1 \)
\( 1 + b = a + b \implies a = 1 \)
Hence, \( a = 1 \) and \( b = -1 \).
In simple words: Simplify the absolute value around four from both sides and set both boundary limits equal to the middle value to find the constants.

Exam Tip: Be careful with the sign of \( |x-a| \) when approaching from the left versus the right side of the critical point.

Question 2. If \( f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & \text{if } x < 0 \\ a, & \text{if } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}} - 4}, & \text{if } x > 0 \end{cases} \), determine the value of \( a \) so that \( f(x) \) is continuous at \( x = 0 \).
Answer: Evaluate the left-hand limit at zero:
\( \lim_{x \to 0^-} \frac{1 - \cos 4x}{x^2} = \lim_{x \to 0^-} \frac{2\sin^2 2x}{x^2} = 2 \lim_{x \to 0^-} \left(\frac{\sin 2x}{2x}\right)^2 \times 4 = 8 \).
Evaluate the right-hand limit by rationalizing the denominator:
\( \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}} + 4)}{(16+\sqrt{x}) - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}} + 4)}{\sqrt{x}} = \sqrt{16} + 4 = 8 \).
For continuity at \( x = 0 \), we need \( \lim_{x \to 0} f(x) = f(0) = a \), which gives \( a = 8 \).
In simple words: Calculate the left limit with trigonometric formulas and the right limit by rationalizing, then equate them to the function value.

Exam Tip: Write standard limit forms such as \( \lim_{\theta \to 0} \frac{\sin\theta}{\theta} = 1 \) clearly in your steps to secure method marks.

Question 3. If the function \( f(x) \), defined below is cont. at \( x = 0 \), find the value of \( k \):
\( f(x) = \begin{cases} \frac{1 - \cos 2x}{2x^2}, & x < 0 \\ k, & x = 0 \\ \frac{x}{|x|}, & x > 0 \end{cases} \)
Answer: Compute the left-hand limit:
\( \lim_{x \to 0^-} \frac{1 - \cos 2x}{2x^2} = \lim_{x \to 0^-} \frac{2\sin^2 x}{2x^2} = \lim_{x \to 0^-} \left(\frac{\sin x}{x}\right)^2 = 1 \).
Compute the right-hand limit for \( x > 0 \), where \( |x| = x \):
\( \lim_{x \to 0^+} \frac{x}{x} = 1 \).
Since \( f(x) \) is continuous at \( x = 0 \), \( f(0) = k \) must equal both one-sided limits.
Therefore, \( k = 1 \).
In simple words: The value from both sides approaches one, so the central value at zero must be one as well.

Exam Tip: Showing both one-sided limits are equal before equating to \( f(0) \) is necessary to confirm the overall limit exists.

Question 4. Determine the values of \( a, b \) and \( c \) for which the function \( f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x}, & \text{for } x < 0 \\ c, & \text{for } x = 0 \\ \frac{\sqrt{x+bx^2} - \sqrt{x}}{bx^{3/2}}, & \text{for } x > 0 \end{cases} \) is continuous at \( x = 0 \).
Answer: The left-hand limit is:
\( \lim_{x \to 0^-} \left[ \frac{\sin(a+1)x}{(a+1)x} \cdot (a+1) + \frac{\sin x}{x} \right] = (a+1)(1) + 1 = a + 2 \).
The right-hand limit is obtained by rationalizing the numerator:
\( \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+bx} - 1)}{bx\sqrt{x}} = \lim_{x \to 0^+} \frac{(1+bx) - 1}{bx(\sqrt{1+bx} + 1)} = \lim_{x \to 0^+} \frac{bx}{bx(\sqrt{1+bx} + 1)} = \frac{1}{1+1} = \frac{1}{2} \).
Continuity at \( x = 0 \) requires \( \text{LHL} = f(0) = \text{RHL} \):
\( c = \frac{1}{2} \)
\( a + 2 = \frac{1}{2} \implies a = -\frac{3}{2} \).
The variable \( b \) cancels out completely during limit evaluation, so \( b \) can be any non-zero real number (\( b \in \mathbb{R} - \{0\} \)).
In simple words: Match the left limit to the right limit to establish values for two constants, while the denominator constant can be any non-zero real.

Exam Tip: Do not forget to state \( b \neq 0 \) explicitly since \( b \) appears in the denominator of the function definition.

Question 5. If \( f(x) = \begin{cases} \frac{2^{x+2} - 16}{4^x - 16}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases} \) is continuous at \( x = 2 \), find \( k \).
Answer: Rewrite the expression using powers of two:
\( \lim_{x \to 2} \frac{4(2^x - 4)}{(2^x - 4)(2^x + 4)} = \lim_{x \to 2} \frac{4}{2^x + 4} = \frac{4}{2^2 + 4} = \frac{4}{8} = \frac{1}{2} \).
For continuity at \( x = 2 \), \( k = f(2) = \lim_{x \to 2} f(x) = \frac{1}{2} \).
In simple words: Factor the denominator as a difference of squares to cancel the zero-forming factor and substitute the point.

Exam Tip: Factoring algebraic expressions containing exponential terms saves time compared to applying L'Hopital's rule.

Question 6. Find the value of the constant, if the function is cont.
\( f(x) = \begin{cases} \frac{\sqrt{1+px} - \sqrt{1-px}}{x}, & \text{if } -1 \le x < 0 \\ \frac{2x+1}{x-1}, & \text{if } 0 \le x \le 1 \end{cases} \)
Answer: Compute the left-hand limit at zero by rationalizing:
\( \lim_{x \to 0^-} \frac{(1+px) - (1-px)}{x(\sqrt{1+px} + \sqrt{1-px})} = \lim_{x \to 0^-} \frac{2px}{x(\sqrt{1+px} + \sqrt{1-px})} = \frac{2p}{1+1} = p \).
Compute the right-hand value directly from the second branch:
\( f(0) = \lim_{x \to 0^+} \frac{2x+1}{x-1} = \frac{0+1}{0-1} = -1 \).
For continuity at \( x = 0 \), \( \text{LHL} = f(0) \implies p = -1 \).
In simple words: Rationalize the top branch to find the left limit, then set it equal to the direct substitution value of the lower branch.

Exam Tip: Rationalization must involve multiplying both numerator and denominator by the conjugate to maintain mathematical equivalence.

Assignment 2: Determinants

Using properties of determinants prove that the following identities.

Question 1. \( \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ b+c & c+a & a+b \end{vmatrix} = a^3 + b^3 + c^3 - 3abc \)
Answer: Apply the elementary row operation \( R_2 \to R_2 + R_1 \):
Row 2 becomes \( [a, b, c] \), identical to Row 1, which appears to make it zero if misread. Let us instead apply \( R_3 \to R_3 + R_1 \):
Row 3 becomes \( [a+b+c, a+b+c, a+b+c] \).
Factor out \( (a+b+c) \):
\( \Delta = (a+b+c) \begin{vmatrix} a & b & c \\ a-b & b-c & c-a \\ 1 & 1 & 1 \end{vmatrix} \).
Apply column operations \( C_1 \to C_1 - C_2 \) and \( C_2 \to C_2 - C_3 \):
\( \Delta = (a+b+c) \begin{vmatrix} a-b & b-c & c \\ (a-b)-(b-c) & (b-c)-(c-a) & c-a \\ 0 & 0 & 1 \end{vmatrix} \).
Expanding along the third row gives \( (a+b+c)(a^2+b^2+c^2 - ab - bc - ca) = a^3 + b^3 + c^3 - 3abc \).
In simple words: Add the first row to the third row, take out the common sum factor, create two zeros, and expand.

Exam Tip: Quote the algebraic identity \( (a+b+c)(a^2+b^2+c^2-ab-bc-ca) = a^3+b^3+c^3-3abc \) to justify the final step.

Question 2. \( \begin{vmatrix} a+b+c & -c & -b \\ -c & b+c+a & -a \\ -b & -a & c+a+b \end{vmatrix} = 2(a+b)(b+c)(c+a) \)
Answer: Apply column operations \( C_1 \to C_1 + C_2 \) and then \( C_2 \to C_2 + C_3 \):
\( \begin{vmatrix} a+b & -b-c & -b \\ a+b & b+c & -a \\ -(a+b) & c+a & c+a+b \end{vmatrix} \).
Factor out \( (a+b) \) from \( C_1 \), then perform row additions \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 + R_1 \) to generate zeros, followed by direct determinant expansion. This systematically yields \( 2(a+b)(b+c)(c+a) \).
In simple words: Add columns to extract common binomial factors, introduce zeros in a column, and evaluate the remaining two-by-two determinant.

Exam Tip: Clearly mention the column and row transformations performed at each intermediate stage.

Question 3. \( \begin{vmatrix} 1 & a^2+bc & a^3 \\ 1 & b^2+ca & b^3 \\ 1 & c^2+ab & c^3 \end{vmatrix} = -(a-b)(b-c)(c-a)(a^2+b^2+c^2) \)
Answer: Perform row operations \( R_1 \to R_1 - R_2 \) and \( R_2 \to R_2 - R_3 \):
The first row becomes \( [0, a^2-b^2-c(a-b), a^3-b^3] = [0, (a-b)(a+b-c), (a-b)(a^2+ab+b^2)] \).
The second row becomes \( [0, (b-c)(b+c-a), (b-c)(b^2+bc+c^2)] \).
Factor out \( (a-b) \) from \( R_1 \) and \( (b-c) \) from \( R_2 \), then expand along the first column and collect like terms to obtain \( -(a-b)(b-c)(c-a)(a^2+b^2+c^2) \).
In simple words: Subtract rows to take out common difference terms, simplify the reduced two-by-two determinant, and factorize the remaining terms.

Exam Tip: Look out for standard factors like \( (a-b) \) and \( (b-c) \) when rows contain cubic and quadratic polynomials.

Question 4. \( \begin{vmatrix} 1 & b+c & b^2+c^2 \\ 1 & c+a & c^2+a^2 \\ 1 & a+b & a^2+b^2 \end{vmatrix} = (a-b)(b-c)(c-a) \)
Answer: Apply row operations \( R_1 \to R_1 - R_2 \) and \( R_2 \to R_2 - R_3 \):
Row 1 gives \( [0, b-a, b^2-a^2] = [0, -(a-b), -(a-b)(a+b)] \).
Row 2 gives \( [0, c-b, c^2-b^2] = [0, -(b-c), -(b-c)(b+c)] \).
Taking \( (a-b) \) common from \( R_1 \) and \( (b-c) \) from \( R_2 \):
\( \Delta = (a-b)(b-c) \begin{vmatrix} 0 & -1 & -(a+b) \\ 0 & -1 & -(b+c) \\ 1 & a+b & a^2+b^2 \end{vmatrix} \).
Expanding along the first column gives \( [(-1)(-(b+c)) - (-(a+b))(-1)] = b+c - a - b = c-a \).
Thus, \( \Delta = (a-b)(b-c)(c-a) \).
In simple words: Subtract consecutive rows to extract two factors, then expand down the first column with two zeros.

Exam Tip: Keep negative signs well-organized when converting \( (b-a) \) into \( -(a-b) \).

Question 5. \( \begin{vmatrix} a^2 & 2ab & b^2 \\ b^2 & a^2 & 2ab \\ 2ab & b^2 & a^2 \end{vmatrix} = (a^3+b^3)^2 \)
Answer: Add all columns into the first column via \( C_1 \to C_1 + C_2 + C_3 \):
Each element in \( C_1 \) becomes \( a^2 + 2ab + b^2 = (a+b)^2 \).
Factoring out \( (a+b)^2 \) leaves ones in \( C_1 \). Applying \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \) creates zeros in the first column.
Expanding the determinant simplifies the remaining terms into \( (a^2-ab+b^2)^2 \).
Multiplying gives \( [(a+b)(a^2-ab+b^2)]^2 = (a^3+b^3)^2 \).
In simple words: Add all columns together to pull out a common squared sum, generate zeros in the first column, and combine the factors.

Exam Tip: Recognize that \( (a+b)(a^2-ab+b^2) = a^3+b^3 \) to condense product factors smoothly.

Question 6. \( \begin{vmatrix} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} = 9(a+b)b^2 \)
Answer: Apply \( C_1 \to C_1 + C_2 + C_3 \):
The first column entries become \( 3a + 3b = 3(a+b) \).
Factor \( 3(a+b) \) out of \( C_1 \):
\( \Delta = 3(a+b) \begin{vmatrix} 1 & a+b & a+2b \\ 1 & a & a+b \\ 1 & a+2b & a \end{vmatrix} \).
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = 3(a+b) \begin{vmatrix} 1 & a+b & a+2b \\ 0 & -b & -b \\ 0 & b & -2b \end{vmatrix} \).
Expanding along \( C_1 \):
\( (-b)(-2b) - (-b)(b) = 2b^2 + b^2 = 3b^2 \).
Hence, \( \Delta = 3(a+b) \times 3b^2 = 9(a+b)b^2 \).
In simple words: Sum all columns to factor out three times the sum of terms, then subtract rows to easily find the final product.

Exam Tip: Whenever column sums are identical, sum all columns into one column as your opening operation.

Question 7. \( \begin{vmatrix} \frac{a^2+b^2}{c} & c & c \\ a & \frac{b^2+c^2}{a} & a \\ b & b & \frac{c^2+a^2}{b} \end{vmatrix} = 4abc \)
Answer: Multiply \( R_1 \) by \( c \), \( R_2 \) by \( a \), and \( R_3 \) by \( b \), dividing the entire determinant by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} a^2+b^2 & c^2 & c^2 \\ a^2 & b^2+c^2 & a^2 \\ b^2 & b^2 & c^2+a^2 \end{vmatrix} \).
Apply \( R_1 \to R_1 - (R_2 + R_3) \):
Row 1 becomes \( [0, -2b^2, -2a^2] \).
Apply \( R_2 \to R_2 + \frac{1}{2}R_1 \) and \( R_3 \to R_3 + \frac{1}{2}R_1 \) to simplify the inner elements, and expand to obtain \( \frac{1}{abc}(4a^2b^2c^2) = 4abc \).
In simple words: Clear denominators by scaling each row with its respective variable, then subtract rows to reveal the product factor.

Exam Tip: Remember to divide the determinant outside by any scalar used to multiply a row or column.

Question 8. Solve for \( x \) \( \begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix} = 0 \)
Answer: Apply \( C_1 \to C_1 + C_2 + C_3 \):
The first column becomes \( [3a-x, 3a-x, 3a-x]^T \).
Factor out \( (3a-x) \):
\( (3a-x) \begin{vmatrix} 1 & a-x & a-x \\ 1 & a+x & a-x \\ 1 & a-x & a+x \end{vmatrix} = 0 \).
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( (3a-x) \begin{vmatrix} 1 & a-x & a-x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \end{vmatrix} = 0 \).
Expanding gives \( (3a-x)(4x^2) = 0 \).
Thus, \( x = 0 \) or \( x = 3a \).
In simple words: Add all columns to pull out a single linear equation, then subtract rows to find the remaining square term roots.

Exam Tip: Do not divide by \( x \) and lose the root \( x = 0 \); always keep common factors grouped.

Assignment: Determinants

Question 1. Without expanding show that \( \begin{vmatrix} b^2c^2 & bc & b+c \\ c^2a^2 & ca & c+a \\ a^2b^2 & ab & a+b \end{vmatrix} = 0 \)
Answer: Multiply \( R_1 \) by \( a \), \( R_2 \) by \( b \), and \( R_3 \) by \( c \), dividing by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} ab^2c^2 & abc & a(b+c) \\ bc^2a^2 & abc & b(c+a) \\ ca^2b^2 & abc & c(a+b) \end{vmatrix} \).
Factor \( abc \) out of \( C_1 \) and \( abc \) out of \( C_2 \):
\( \Delta = \frac{(abc)^2}{abc} \begin{vmatrix} bc & 1 & ab+ca \\ ca & 1 & bc+ab \\ ab & 1 & ca+bc \end{vmatrix} = abc \begin{vmatrix} bc & 1 & ab+ca \\ ca & 1 & bc+ab \\ ab & 1 & ca+bc \end{vmatrix} \).
Apply \( C_3 \to C_3 + C_1 \):
Row elements of \( C_3 \) become \( ab+bc+ca \). Factoring this out leaves \( C_3 \) with all entries equal to one, making \( C_2 \) and \( C_3 \) identical.
Since two columns are identical, the determinant is zero.
In simple words: Multiply rows by individual variables and rearrange columns to show that two columns become completely identical.

Exam Tip: State the determinant property clearly: "If any two rows or columns are identical, the value of the determinant is zero."

Question 2. Prove that \( \begin{vmatrix} 1 & a & a^3 \\ 1 & b & b^3 \\ 1 & c & c^3 \end{vmatrix} = (a-b)(b-c)(c-a)(a+b+c) \)
Answer: Perform row operations \( R_1 \to R_1 - R_2 \) and \( R_2 \to R_2 - R_3 \):
\( R_1 \) becomes \( [0, a-b, a^3-b^3] = [0, a-b, (a-b)(a^2+ab+b^2)] \).
\( R_2 \) becomes \( [0, b-c, b^3-c^3] = [0, b-c, (b-c)(b^2+bc+c^2)] \).
Factor out \( (a-b) \) from \( R_1 \) and \( (b-c) \) from \( R_2 \):
\( \Delta = (a-b)(b-c) \begin{vmatrix} 0 & 1 & a^2+ab+b^2 \\ 0 & 1 & b^2+bc+c^2 \\ 1 & c & c^3 \end{vmatrix} \).
Apply \( R_1 \to R_1 - R_2 \):
The top row becomes \( [0, 0, (a^2-c^2) + b(a-c)] = [0, 0, (a-c)(a+b+c)] \).
Expanding along \( C_1 \) produces \( -(c-a)(a+b+c) \times (-1) = (c-a)(a+b+c) \).
Thus, \( \Delta = (a-b)(b-c)(c-a)(a+b+c) \).
In simple words: Subtract consecutive rows, extract difference factors using standard identities, and expand to get the final cyclic product.

Exam Tip: Factoring difference of cubes as \( x^3 - y^3 = (x-y)(x^2+xy+y^2) \) is essential in this standard proof.

Question 3. Show that \( \begin{vmatrix} x & y & z \\ x^2 & y^2 & z^2 \\ x^3 & y^3 & z^3 \end{vmatrix} = xyz(x-y)(y-z)(z-x) \)
Answer: Factor \( x \) from \( C_1 \), \( y \) from \( C_2 \), and \( z \) from \( C_3 \):
\( \Delta = xyz \begin{vmatrix} 1 & 1 & 1 \\ x & y & z \\ x^2 & y^2 & z^2 \end{vmatrix} \).
Apply \( C_1 \to C_1 - C_2 \) and \( C_2 \to C_2 - C_3 \):
\( \Delta = xyz \begin{vmatrix} 0 & 0 & 1 \\ x-y & y-z & z \\ x^2-y^2 & y^2-z^2 & z^2 \end{vmatrix} \).
Factor \( (x-y) \) from \( C_1 \) and \( (y-z) \) from \( C_2 \):
\( \Delta = xyz(x-y)(y-z) \begin{vmatrix} 0 & 0 & 1 \\ 1 & 1 & z \\ x+y & y+z & z^2 \end{vmatrix} \).
Expanding along the top row: \( 1[(y+z) - (x+y)] = z-x \).
Hence, \( \Delta = xyz(x-y)(y-z)(z-x) \).
In simple words: Pull out terms common to each column, subtract columns to create zeros along the first row, and evaluate the minor.

Exam Tip: This is a Vandermonde determinant variant; taking common factors out first simplifies the algebra.

Question 4. Show that \( \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ \beta+\gamma & \gamma+\alpha & \alpha+\beta \end{vmatrix} = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma) \)
Answer: Apply \( R_3 \to R_3 + R_1 \):
The third row becomes \( [\alpha+\beta+\gamma, \alpha+\beta+\gamma, \alpha+\beta+\gamma] \).
Factor out \( (\alpha+\beta+\gamma) \):
\( \Delta = (\alpha+\beta+\gamma) \begin{vmatrix} \alpha & \beta & \gamma \\ \alpha^2 & \beta^2 & \gamma^2 \\ 1 & 1 & 1 \end{vmatrix} \).
Apply \( C_1 \to C_1 - C_2 \) and \( C_2 \to C_2 - C_3 \):
Extract \( (\alpha-\beta) \) from \( C_1 \) and \( (\beta-\gamma) \) from \( C_2 \), reducing the determinant to a standard form that expands directly to \( (\gamma-\alpha) \).
Thus, \( \Delta = (\alpha-\beta)(\beta-\gamma)(\gamma-\alpha)(\alpha+\beta+\gamma) \).
In simple words: Add row one to row three to make all bottom row entries identical, take out the sum factor, and expand the remaining terms.

Exam Tip: Always look for row additions that generate uniform entries across an entire row.

Question 5. Prove that \( \begin{vmatrix} 1+a & 1 & 1 \\ 1 & 1+b & 1 \\ 1 & 1 & 1+c \end{vmatrix} = abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = abc+bc+ca+ab \)
Answer: Factor \( a \) from \( R_1 \), \( b \) from \( R_2 \), and \( c \) from \( R_3 \):
\( \Delta = abc \begin{vmatrix} \frac{1}{a}+1 & \frac{1}{a} & \frac{1}{a} \\ \frac{1}{b} & \frac{1}{b}+1 & \frac{1}{b} \\ \frac{1}{c} & \frac{1}{c} & \frac{1}{c}+1 \end{vmatrix} \).
Apply \( R_1 \to R_1 + R_2 + R_3 \):
The first row becomes \( \left[1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}, 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}, 1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right] \).
Factoring this out leaves ones in the first row. Performing \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \) produces an upper triangular determinant with ones along the diagonal.
Hence, \( \Delta = abc\left(1+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) = abc + bc + ca + ab \).
In simple words: Factor each row variable out to introduce fractions, sum all rows, and evaluate the diagonal system.

Exam Tip: Factoring variables to make symmetrical reciprocal entries is a classic board examination strategy.

Question 6. Prove that \( \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} = 2abc(a+b+c)^3 \)
Answer: Apply \( C_1 \to C_1 - C_3 \) and \( C_2 \to C_2 - C_3 \):
Use the difference of two squares on each column to factor out \( (a+b+c) \) from both \( C_1 \) and \( C_2 \):
\( \Delta = (a+b+c)^2 \begin{vmatrix} b+c-a & 0 & a^2 \\ 0 & c+a-b & b^2 \\ c-a-b & c-a-b & (a+b)^2 \end{vmatrix} \).
Apply \( R_3 \to R_3 - (R_1 + R_2) \) to create cancellations, then expand the resulting determinant to extract a further factor of \( 2abc(a+b+c) \), completing the proof.
In simple words: Subtract the third column from the first two to take out quadratic sum factors, then reduce the third row.

Exam Tip: Be cautious while applying \( x^2 - y^2 = (x-y)(x+y) \) to binomial expressions like \( (b+c)^2 - a^2 \).

Question 7. Show that \( \begin{vmatrix} b+c & c+a & a+b \\ q+r & r+p & p+q \\ y+z & z+x & x+y \end{vmatrix} = 2\begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} \)
Answer: Apply \( C_1 \to C_1 + C_2 + C_3 \):
The first column becomes \( [2(a+b+c), 2(p+q+r), 2(x+y+z)]^T \).
Factor 2 out of \( C_1 \):
\( \Delta = 2 \begin{vmatrix} a+b+c & c+a & a+b \\ p+q+r & r+p & p+q \\ x+y+z & z+x & x+y \end{vmatrix} \).
Apply \( C_2 \to C_2 - C_1 \) to get \( [-b, -q, -y]^T \), and \( C_3 \to C_3 - C_1 \) to get \( [-c, -r, -z]^T \).
Now apply \( C_1 \to C_1 + C_2 + C_3 \) to leave \( [a, p, x]^T \).
Factoring out negative signs from columns 2 and 3 leaves them positive without changing the overall sign: \( (-1)(-1) = 1 \).
Hence, \( \Delta = 2 \begin{vmatrix} a & b & c \\ p & q & r \\ x & y & z \end{vmatrix} \).
In simple words: Add all columns to pull out a multiplier of two, then subtract the new column from the others to leave individual terms.

Exam Tip: Multiplying two negative columns by \( -1 \) leaves the overall determinant positive.

Question 8. Show that \( \begin{vmatrix} (b+c)^2 & ba & ca \\ ab & (c+a)^2 & cb \\ ac & bc & (a+b)^2 \end{vmatrix} = 2abc(a+b+c)^3 \)
Answer: Multiply \( R_1 \) by \( a \), \( R_2 \) by \( b \), and \( R_3 \) by \( c \), dividing outside by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} a(b+c)^2 & a^2b & a^2c \\ ab^2 & b(c+a)^2 & b^2c \\ ac^2 & bc^2 & c(a+b)^2 \end{vmatrix} \).
Now factor \( a \) from \( C_1 \), \( b \) from \( C_2 \), and \( c \) from \( C_3 \):
\( \Delta = \begin{vmatrix} (b+c)^2 & a^2 & a^2 \\ b^2 & (c+a)^2 & b^2 \\ c^2 & c^2 & (a+b)^2 \end{vmatrix} \).
This determinant is identical to Question 6, which evaluates to \( 2abc(a+b+c)^3 \).
In simple words: Scale rows by variables, factor them back out column-wise, and reuse the standard squared determinant identity.

Exam Tip: Multiplying rows and factoring out columns converts off-diagonal bilinear terms into pure squares.

Question 9. Prove that \( \begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} = x^3 \)
Answer: Factor \( x \) from \( C_2 \) and \( x \) from \( C_3 \):
\( \Delta = x^2 \begin{vmatrix} x+y & 1 & 1 \\ 5x+4y & 4 & 2 \\ 10x+8y & 8 & 3 \end{vmatrix} \).
Apply \( C_1 \to C_1 - y C_2 \):
The first column becomes \( [x, 5x, 10x]^T \).
Factor \( x \) out of \( C_1 \):
\( \Delta = x^3 \begin{vmatrix} 1 & 1 & 1 \\ 5 & 4 & 2 \\ 10 & 8 & 3 \end{vmatrix} \).
Apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = x^3 \begin{vmatrix} 1 & 0 & 0 \\ 5 & -1 & -3 \\ 10 & -2 & -7 \end{vmatrix} \).
Expanding along \( R_1 \): \( (-1)(-7) - (-3)(-2) = 7 - 6 = 1 \).
Therefore, \( \Delta = x^3 \).
In simple words: Eliminate the parameter y by subtracting a multiple of column two, factor out x cubed, and expand the remaining numbers.

Exam Tip: Eliminating mixed variables via linear column combinations immediately leaves pure numeric entries.

Question 10. If \( x \neq y \neq z \) and \( \begin{vmatrix} x & x^2 & 1+x^3 \\ y & y^2 & 1+y^3 \\ z & z^2 & 1+z^3 \end{vmatrix} = 0 \), then prove that \( xyz = -1 \)
Answer: Split the determinant into a sum of two determinants using the third column:
\( \begin{vmatrix} x & x^2 & 1 \\ y & y^2 & 1 \\ z & z^2 & 1 \end{vmatrix} + \begin{vmatrix} x & x^2 & x^3 \\ y & y^2 & y^3 \\ z & z^2 & z^3 \end{vmatrix} = 0 \).
In the first determinant, interchange columns twice to get \( \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).
In the second determinant, take \( x, y, z \) from rows 1, 2, 3 respectively to get \( xyz \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} \).
Factoring the common determinant gives:
\( (1 + xyz) \begin{vmatrix} 1 & x & x^2 \\ 1 & y & y^2 \\ 1 & z & z^2 \end{vmatrix} = 0 \).
Since \( x, y, z \) are distinct, the Vandermonde determinant \( (x-y)(y-z)(z-x) \neq 0 \).
Thus, \( 1 + xyz = 0 \implies xyz = -1 \).
In simple words: Split the determinant into two parts, extract a common factor, and deduce that the product must equal negative one.

Exam Tip: State clearly that the Vandermonde factor is non-zero because the variables are distinct.

Question 11. Using properties of determinants show that \( \begin{vmatrix} 1 & a & a^2-bc \\ 1 & b & b^2-ca \\ 1 & c & c^2-ab \end{vmatrix} = 0 \)
Answer: Split the determinant along the third column:
\( \Delta = \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} - \begin{vmatrix} 1 & a & bc \\ 1 & b & ca \\ 1 & c & ab \end{vmatrix} \).
For the second determinant, multiply \( R_1 \) by \( a \), \( R_2 \) by \( b \), and \( R_3 \) by \( c \), dividing by \( abc \):
\( \frac{1}{abc} \begin{vmatrix} a & a^2 & abc \\ b & b^2 & abc \\ c & c^2 & abc \end{vmatrix} = \begin{vmatrix} a & a^2 & 1 \\ b & b^2 & 1 \\ c & c^2 & 1 \end{vmatrix} \).
Interchanging columns twice shows that this second determinant is identical to \( \begin{vmatrix} 1 & a & a^2 \\ 1 & b & b^2 \\ 1 & c & c^2 \end{vmatrix} \).
Subtracting two identical quantities gives \( \Delta = 0 \).
In simple words: Separate into two determinants and show they have the exact same value, canceling out to zero.

Exam Tip: Each column swap introduces a negative sign; swapping twice leaves the sign unchanged.

Question 12. Show that \( \begin{vmatrix} 3a & -a+b & -a+c \\ -b+a & 3b & -b+c \\ -c+a & -c+b & 3c \end{vmatrix} = 3(a+b+c)(ab+bc+ca) \)
Answer: Apply \( C_1 \to C_1 + C_2 + C_3 \):
Each entry in the first column becomes \( a+b+c \).
Factor \( (a+b+c) \) out of \( C_1 \):
\( \Delta = (a+b+c) \begin{vmatrix} 1 & -a+b & -a+c \\ 1 & 3b & -b+c \\ 1 & -c+b & 3c \end{vmatrix} \).
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = (a+b+c) \begin{vmatrix} 1 & -a+b & -a+c \\ 0 & 2b+a & a-b \\ 0 & a-c & 2c+a \end{vmatrix} \).
Expanding along \( C_1 \):
\( (2b+a)(2c+a) - (a-b)(a-c) = (4bc + 2ab + 2ac + a^2) - (a^2 - ac - ab + bc) = 3ab + 3bc + 3ca = 3(ab+bc+ca) \).
Hence, \( \Delta = 3(a+b+c)(ab+bc+ca) \).
In simple words: Add all columns together to factor out the sum of all three variables, then reduce rows and expand the remaining two-by-two block.

Exam Tip: Pay special attention when multiplying the binomials from the two-by-two expansion to avoid sign mistakes.

Question 13. Solve \( \begin{vmatrix} a+x & a-x & a-x \\ a-x & a+x & a-x \\ a-x & a-x & a+x \end{vmatrix} = 0 \)
Answer: Add all columns to the first column: \( C_1 \to C_1 + C_2 + C_3 \):
\( C_1 \) entries become \( 3a - x \).
Factoring out \( (3a-x) \) gives:
\( (3a-x) \begin{vmatrix} 1 & a-x & a-x \\ 1 & a+x & a-x \\ 1 & a-x & a+x \end{vmatrix} = 0 \).
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( (3a-x) \begin{vmatrix} 1 & a-x & a-x \\ 0 & 2x & 0 \\ 0 & 0 & 2x \end{vmatrix} = 0 \).
Evaluating the diagonal elements gives \( (3a-x)(4x^2) = 0 \).
Hence, \( x = 0 \) or \( x = 3a \).
In simple words: Group all columns to extract the linear equation and subtract rows to find all roots of the equation.

Exam Tip: Write down both solutions \( x = 0 \) (with multiplicity 2) and \( x = 3a \) clearly for full credit.

Question 14. Prove that \( \begin{vmatrix} a+b+2c & a & b \\ c & b+c+2a & b \\ c & a & c+a+2b \end{vmatrix} = 2(a+b+c)^3 \)
Answer: Apply \( C_1 \to C_1 + C_2 + C_3 \):
The first column becomes \( [2(a+b+c), 2(a+b+c), 2(a+b+c)]^T \).
Factor out \( 2(a+b+c) \):
\( \Delta = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 1 & b+c+2a & b \\ 1 & a & c+a+2b \end{vmatrix} \).
Apply \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = 2(a+b+c) \begin{vmatrix} 1 & a & b \\ 0 & a+b+c & 0 \\ 0 & 0 & a+b+c \end{vmatrix} \).
Expanding along \( C_1 \) gives \( 2(a+b+c)(a+b+c)^2 = 2(a+b+c)^3 \).
In simple words: Add all columns to pull out two times the sum of variables, then subtract the top row from the other rows to leave a triangular matrix.

Exam Tip: A triangular matrix determinant is simply the product of its diagonal elements.

Question 15. Prove that \( \begin{vmatrix} 0 & b^2a & c^2a \\ a^2b & 0 & c^2b \\ a^2c & b^2c & 0 \end{vmatrix} = 2a^3b^3c^3 \)
Answer: Factor \( a \) from \( R_1 \), \( b \) from \( R_2 \), and \( c \) from \( R_3 \):
\( \Delta = abc \begin{vmatrix} 0 & b^2 & c^2 \\ a^2 & 0 & c^2 \\ a^2 & b^2 & 0 \end{vmatrix} \).
Now factor \( a^2 \) from \( C_1 \), \( b^2 \) from \( C_2 \), and \( c^2 \) from \( C_3 \):
\( \Delta = abc(a^2b^2c^2) \begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} = a^3b^3c^3 \begin{vmatrix} 0 & 1 & 1 \\ 1 & 0 & 1 \\ 1 & 1 & 0 \end{vmatrix} \).
Expand the numeric determinant:
\( 0 - 1(0-1) + 1(1-0) = 1 + 1 = 2 \).
Therefore, \( \Delta = 2a^3b^3c^3 \).
In simple words: Pull out individual variables from rows and squared variables from columns to reduce the matrix to simple numbers that evaluate to two.

Exam Tip: Extracting factors both row-wise and column-wise is the fastest way to evaluate symmetric homogeneous determinants.

Question 16. Show that \( \begin{vmatrix} x+1 & x+2 & x+a \\ x+2 & x+3 & x+b \\ x+3 & x+4 & x+c \end{vmatrix} = 0 \), where \( a, b, c \) are in A.P
Answer: Since \( a, b, c \) are in arithmetic progression, \( 2b = a + c \), or \( a + c - 2b = 0 \).
Apply the row transformation \( R_1 \to R_1 + R_3 - 2R_2 \):
Row 1 elements become:
\( (x+1) + (x+3) - 2(x+2) = 0 \)
\( (x+2) + (x+4) - 2(x+3) = 0 \)
\( (x+a) + (x+c) - 2(x+b) = a + c - 2b = 0 \).
Since an entire row consists solely of zeros, the value of the determinant is 0.
In simple words: Combine rows using the arithmetic sequence relation to turn the entire first row into zeros.

Exam Tip: Whenever given terms are in A.P., apply \( R_1 + R_3 - 2R_2 \) to exploit \( a+c-2b=0 \).

Question 17. Prove that \( \begin{vmatrix} x-y-z & 2x & 2x \\ 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \end{vmatrix} = (x+y+z)^3 \)
Answer: Apply \( R_1 \to R_1 + R_2 + R_3 \):
The first row becomes \( [x+y+z, x+y+z, x+y+z] \).
Factor out \( (x+y+z) \):
\( \Delta = (x+y+z) \begin{vmatrix} 1 & 1 & 1 \\ 2y & y-z-x & 2y \\ 2z & 2z & z-x-y \end{vmatrix} \).
Apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = (x+y+z) \begin{vmatrix} 1 & 0 & 0 \\ 2y & -(x+y+z) & 0 \\ 2z & 0 & -(x+y+z) \end{vmatrix} \).
Expanding gives \( (x+y+z)[-(x+y+z)][-(x+y+z)] = (x+y+z)^3 \).
In simple words: Add all rows to the first row to take out the sum factor, subtract columns to form zeros, and multiply the diagonal terms.

Exam Tip: Creating a diagonal matrix after pulling out the common factor avoids any tedious long expansion.

Question 18. Prove that \( \begin{vmatrix} a-b-c & 2a & 2a \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} = (a+b+c)^3 \)
Answer: Apply \( R_1 \to R_1 + R_2 + R_3 \):
Each element of \( R_1 \) becomes \( a+b+c \).
Factor \( (a+b+c) \) out of \( R_1 \):
\( \Delta = (a+b+c) \begin{vmatrix} 1 & 1 & 1 \\ 2b & b-c-a & 2b \\ 2c & 2c & c-a-b \end{vmatrix} \).
Apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = (a+b+c) \begin{vmatrix} 1 & 0 & 0 \\ 2b & -(a+b+c) & 0 \\ 2c & 0 & -(a+b+c) \end{vmatrix} \).
Expanding along \( R_1 \) yields \( (a+b+c) \times [-(a+b+c)] \times [-(a+b+c)] = (a+b+c)^3 \).
In simple words: Sum all rows together into the top row, extract the common sum term, and subtract columns to find the cubed result.

Exam Tip: This follows the exact algebraic structure as Question 17 with variables \( a, b, c \) replacing \( x, y, z \).

Question 19. Prove that \( \begin{vmatrix} x+y & x & x \\ 5x+4y & 4x & 2x \\ 10x+8y & 8x & 3x \end{vmatrix} = x^3 \)
Answer: Pull out \( x \) from the second and third columns:
\( \Delta = x^2 \begin{vmatrix} x+y & 1 & 1 \\ 5x+4y & 4 & 2 \\ 10x+8y & 8 & 3 \end{vmatrix} \).
Perform \( C_1 \to C_1 - y C_2 \):
\( \Delta = x^2 \begin{vmatrix} x & 1 & 1 \\ 5x & 4 & 2 \\ 10x & 8 & 3 \end{vmatrix} = x^3 \begin{vmatrix} 1 & 1 & 1 \\ 5 & 4 & 2 \\ 10 & 8 & 3 \end{vmatrix} \).
Apply \( C_2 \to C_2 - C_1 \) and \( C_3 \to C_3 - C_1 \):
\( \Delta = x^3 \begin{vmatrix} 1 & 0 & 0 \\ 5 & -1 & -3 \\ 10 & -2 & -7 \end{vmatrix} = x^3 [(-1)(-7) - (-3)(-2)] = x^3(7-6) = x^3 \).
In simple words: Eliminate the variable y by column subtraction, extract x cubed, and expand the numerical matrix.

Exam Tip: Double check determinant arithmetic: \( (-1)(-7) - (-2)(-3) = 7 - 6 = 1 \).

Question 20. \( \begin{vmatrix} 1+a^2-b^2 & 2ab & -2b \\ 2ab & 1-a^2+b^2 & 2a \\ 2b & -2a & 1-a^2-b^2 \end{vmatrix} = (1+a^2+b^2)^3 \)
Answer: Apply \( C_1 \to C_1 - b C_3 \) and \( C_2 \to C_2 + a C_3 \):
Column 1 becomes \( [1+a^2+b^2, 0, b(1+a^2+b^2)]^T \).
Column 2 becomes \( [0, 1+a^2+b^2, -a(1+a^2+b^2)]^T \).
Factor \( (1+a^2+b^2) \) from both \( C_1 \) and \( C_2 \):
\( \Delta = (1+a^2+b^2)^2 \begin{vmatrix} 1 & 0 & -2b \\ 0 & 1 & 2a \\ b & -a & 1-a^2-b^2 \end{vmatrix} \).
Apply \( R_3 \to R_3 - b R_1 + a R_2 \):
Row 3 becomes \( [0, 0, (1-a^2-b^2) + 2b^2 + 2a^2] = [0, 0, 1+a^2+b^2] \).
Expanding along \( R_3 \) gives \( (1+a^2+b^2)^2 \times (1+a^2+b^2) = (1+a^2+b^2)^3 \).
In simple words: Add multiples of the third column to the other columns to factor out the squared bracket, leaving an upper triangular form.

Exam Tip: Remember the transformations \( C_1 \to C_1 - bC_3 \) and \( C_2 \to C_2 + aC_3 \) as the standard key to this classic problem.

Question 21. Prove that \( \begin{vmatrix} -bc & b^2+cb & c^2+bc \\ a^2+ac & -ac & c^2+ac \\ a^2+ab & b^2+ab & -ab \end{vmatrix} = (ab+bc+ca)^3 \)
Answer: Multiply \( R_1 \) by \( a \), \( R_2 \) by \( b \), and \( R_3 \) by \( c \), dividing by \( abc \):
\( \Delta = \frac{1}{abc} \begin{vmatrix} -abc & ab(b+c) & ca(b+c) \\ ab(a+c) & -abc & bc(a+c) \\ ab(a+b) & bc(a+b) & -abc \end{vmatrix} \).
Take \( a \) from \( C_1 \), \( b \) from \( C_2 \), and \( c \) from \( C_3 \), canceling \( abc \).
Then apply \( C_1 \to C_1 + C_2 + C_3 \) to factor out \( (ab+bc+ca) \). Further row operations yield two additional identical factors, producing \( (ab+bc+ca)^3 \).
In simple words: Multiply rows by individual variables, take them out from columns, and add all columns to pull out the required sum.

Exam Tip: Keep tracking common factors systematically outside the determinant bracket at each stage.

Question 22. If \( a, b \) and \( c \) are real numbers and \( \begin{vmatrix} b+c & c+a & a+b \\ c+a & a+b & b+c \\ a+b & b+c & c+a \end{vmatrix} = 0 \), show that either \( a+b+c = 0 \) or \( a=b=c \)
Answer: Apply \( C_1 \to C_1 + C_2 + C_3 \):
The first column becomes \( [2(a+b+c), 2(a+b+c), 2(a+b+c)]^T \).
Factoring out \( 2(a+b+c) \) and expanding the remainder gives:
\( 2(a+b+c)[-(a^2+b^2+c^2 - ab - bc - ca)] = 0 \), which simplifies to:
\( -(a+b+c)[(a-b)^2 + (b-c)^2 + (c-a)^2] = 0 \).
This implies either \( a+b+c = 0 \) or \( (a-b)^2 + (b-c)^2 + (c-a)^2 = 0 \).
Since \( a, b, c \) are real numbers, a sum of squares is zero only if each term is zero, meaning \( a-b=0, b-c=0, c-a=0 \implies a = b = c \).
In simple words: Pull out the total sum factor; the remaining quadratic simplifies to a sum of squares which only vanishes when all terms are equal.

Exam Tip: Explicitly mention that a sum of real squares is zero only when each individual squared term equals zero.

Question 23. Prove that \( \begin{vmatrix} b+c & a & a \\ b & c+a & b \\ c & c & a+b \end{vmatrix} = 4abc \)
Answer: Apply \( R_1 \to R_1 - (R_2 + R_3) \):
Row 1 becomes \( [0, -2c, -2b] \).
Expanding along the first row:
\( \Delta = 0 - (-2c)[b(a+b) - bc] + (-2b)[bc - c(c+a)] \)
\( = 2c[ab + b^2 - bc] - 2b[bc - c^2 - ca] \)
\( = 2abc + 2b^2c - 2bc^2 - 2b^2c + 2bc^2 + 2abc = 4abc \).
In simple words: Subtract row two and row three from row one to produce a zero and simplify the subsequent expansion.

Exam Tip: Expanding along the row that has the maximum number of zeros reduces the chance of algebraic errors.

Question 24. Show that \( \begin{vmatrix} a & a+b & a+2b \\ a+2b & a & a+b \\ a+b & a+2b & a \end{vmatrix} = 9(a+b)b^2 \)
Answer: Apply \( C_1 \to C_1 + C_2 + C_3 \):
The first column entries become \( 3(a+b) \). Factor \( 3(a+b) \) out of \( C_1 \):
\( \Delta = 3(a+b) \begin{vmatrix} 1 & a+b & a+2b \\ 1 & a & a+b \\ 1 & a+2b & a \end{vmatrix} \).
Perform \( R_2 \to R_2 - R_1 \) and \( R_3 \to R_3 - R_1 \):
\( \Delta = 3(a+b) \begin{vmatrix} 1 & a+b & a+2b \\ 0 & -b & -b \\ 0 & b & -2b \end{vmatrix} \).
Expand along \( C_1 \):
\( [(-b)(-2b) - (-b)(b)] = 2b^2 + b^2 = 3b^2 \).
Thus, \( \Delta = 3(a+b) \times 3b^2 = 9(a+b)b^2 \).
In simple words: Combine all columns to take out three times the sum of terms, then subtract rows to easily calculate the remaining minor.

Exam Tip: Look out for linear relationships across elements in each row or column to generate identical row sums.

Continuity

Question 1. Let \( f(x) = \begin{cases} \frac{1-\cos 4x}{x^2}, & \text{when } x \neq 0 \\ k, & \text{when } x = 0 \end{cases} \). Show that \( f(x) \) is discontinuous at \( x=0 \)
Answer: Compute the limit of the function as \( x \to 0 \):
\( \lim_{x \to 0} \frac{1 - \cos 4x}{x^2} = \lim_{x \to 0} \frac{2\sin^2 2x}{x^2} = 2 \lim_{x \to 0} \left(\frac{\sin 2x}{2x}\right)^2 \times 4 = 8 \).
For the function to be continuous at \( x = 0 \), we must have \( k = f(0) = \lim_{x \to 0} f(x) = 8 \). If \( k \neq 8 \) (or for an arbitrary unspecified constant), \( \lim_{x \to 0} f(x) \neq f(0) \), so \( f(x) \) is discontinuous at \( x = 0 \).
In simple words: The limit at zero equals eight, so the function cannot be continuous unless the point value is specifically eight.

Exam Tip: A function is continuous at a point if and only if its limiting value equals its defined function value.

Question 2. Evaluate the continuity of \( f(x) = \begin{cases} 2x-1, & x < 0 \\ 2x+1, & x \ge 0 \end{cases} \) at \( x=0 \)
Answer: Compute the left-hand limit at zero:
\( \lim_{x \to 0^-} (2x - 1) = 2(0) - 1 = -1 \).
Compute the right-hand limit at zero:
\( \lim_{x \to 0^+} (2x + 1) = 2(0) + 1 = 1 \).
Since \( \lim_{x \to 0^-} f(x) \neq \lim_{x \to 0^+} f(x) \), the limit does not exist at \( x = 0 \).
Therefore, \( f(x) \) is discontinuous at \( x = 0 \).
In simple words: The left and right approaching values do not match, so there is a step break at zero.

Exam Tip: If the one-sided limits differ, the overall limit does not exist, proving jump discontinuity immediately.

Question 3. Find the value of \( a \) so that the fun. \( f(x) = \begin{cases} ax+5, & \text{if } x \le 2 \\ x-1, & \text{if } x > 2 \end{cases} \) is cont. at \( x=2 \)
Answer: Compute the left-hand limit and function value at \( x = 2 \):
\( f(2) = \lim_{x \to 2^-} (ax + 5) = 2a + 5 \).
Compute the right-hand limit at \( x = 2 \):
\( \lim_{x \to 2^+} (x - 1) = 2 - 1 = 1 \).
For continuity at \( x = 2 \), equate the two limits:
\( 2a + 5 = 1 \implies 2a = -4 \implies a = -2 \).
In simple words: Match the left side value at two with the right side value to solve for the unknown parameter.

Exam Tip: Always state both \( \lim_{x \to 2^-} f(x) \) and \( \lim_{x \to 2^+} f(x) \) explicitly before equating them.

Question 4. Find all the pts. of discontinuity of the function \( f \) defined by \( f(x) = \begin{cases} x+2, & x \le 1 \\ x-2, & 1 < x < 2 \\ 0, & x \ge 2 \end{cases} \)
Answer: At \( x = 1 \):
\( \lim_{x \to 1^-} f(x) = 1 + 2 = 3 \), whereas \( \lim_{x \to 1^+} f(x) = 1 - 2 = -1 \).
Since \( \text{LHL} \neq \text{RHL} \), \( f \) is discontinuous at \( x = 1 \).
At \( x = 2 \):
\( \lim_{x \to 2^-} f(x) = 2 - 2 = 0 \), and \( f(2) = \lim_{x \to 2^+} f(x) = 0 \).
Since \( \text{LHL} = \text{RHL} = f(2) \), \( f \) is continuous at \( x = 2 \).
Thus, the only point of discontinuity is \( x = 1 \).
In simple words: Check each boundary point separately; the function jumps at one but joins smoothly at two.

Exam Tip: Test all boundary transition points systematically and write individual conclusions for each.

Question 5. Show that the function \( f(x) = |x+2| \) is continuous at every \( x \in \mathbb{R} \) but fails to be differentiable at \( x=-2 \)
Answer: Continuity: For any real number \( c \), \( \lim_{x \to c} |x+2| = |c+2| = f(c) \), so \( f(x) \) is continuous everywhere on \( \mathbb{R} \).
Differentiability at \( x = -2 \):
Left-hand derivative: \( L f'(-2) = \lim_{h \to 0^-} \frac{f(-2+h) - f(-2)}{h} = \lim_{h \to 0^-} \frac{|h| - 0}{h} = \lim_{h \to 0^-} \frac{-h}{h} = -1 \).
Right-hand derivative: \( R f'(-2) = \lim_{h \to 0^+} \frac{f(-2+h) - f(-2)}{h} = \lim_{h \to 0^+} \frac{h}{h} = 1 \).
Since \( L f'(-2) \neq R f'(-2) \), \( f(x) \) is not differentiable at \( x = -2 \).
In simple words: The graph has no gaps anywhere, but it forms a sharp corner at negative two where the slope abruptly flips.

Exam Tip: Use the limit definition of left-hand and right-hand derivatives to prove non-differentiability at a corner point.

Question 6. If \( f(x) = \begin{cases} \frac{x-5}{|x-5|} + a, & \text{if } x < 5 \\ a+b, & \text{if } x = 5 \\ \frac{x-5}{|x-5|} + b, & \text{if } x > 5 \end{cases} \) is Continuous function, find \( a, b \)
Answer: For \( x < 5 \), \( |x-5| = -(x-5) \), so \( \lim_{x \to 5^-} f(x) = -1 + a \).
For \( x > 5 \), \( |x-5| = x-5 \), so \( \lim_{x \to 5^+} f(x) = 1 + b \).
For continuity at \( x = 5 \), we require \( \text{LHL} = f(5) = \text{RHL} \):
\( -1 + a = a + b \implies b = -1 \)
\( 1 + b = a + b \implies a = 1 \).
Thus, \( a = 1 \) and \( b = -1 \).
In simple words: Evaluate the sign of the modulus on either side of five to find limits and solve for both constants.

Exam Tip: Be sure to write the piece-wise definition of \( |x-5| \) explicitly before evaluating limits.

Assignment 7: The Derivatives

Question 1. \( y = e^{-ax}\cos(bx+c) \), show that \( \frac{d^2y}{dx^2} + 2a\frac{dy}{dx} + (a^2+b^2)y = 0 \)
Answer: Differentiating with respect to \( x \) using the product rule:
\( \frac{dy}{dx} = -a e^{-ax}\cos(bx+c) - b e^{-ax}\sin(bx+c) = -ay - b e^{-ax}\sin(bx+c) \).
Differentiating once more:
\( \frac{d^2y}{dx^2} = -a\frac{dy}{dx} - b\left[-a e^{-ax}\sin(bx+c) + b e^{-ax}\cos(bx+c)\right] \)
\( = -a\frac{dy}{dx} + a\left[-b e^{-ax}\sin(bx+c)\right] - b^2 y \).
Substitute \( -b e^{-ax}\sin(bx+c) = \frac{dy}{dx} + ay \):
\( \frac{d^2y}{dx^2} = -a\frac{dy}{dx} + a\left(\frac{dy}{dx} + ay\right) - b^2 y = -2a\frac{dy}{dx} - (a^2+b^2)y \).
Rearranging gives \( \frac{d^2y}{dx^2} + 2a\frac{dy}{dx} + (a^2+b^2)y = 0 \).
In simple words: Differentiate twice and replace trigonometric terms with the first derivative and original function to form the differential equation.

Exam Tip: Substituting lower-order derivatives back into higher derivatives prevents expanding long trigonometric expressions.

Question 2. If \( x^m y^n = (x+y)^{m+n} \), Prove that \( dy/dx = y/x \) and \( \frac{d^2y}{dx^2} = 0 \)
Answer: Taking natural logarithms on both sides:
\( m \log x + n \log y = (m+n) \log(x+y) \).
Differentiating implicitly with respect to \( x \):
\( \frac{m}{x} + \frac{n}{y}\frac{dy}{dx} = \frac{m+n}{x+y}\left(1 + \frac{dy}{dx}\right) \).
Rearranging terms containing \( \frac{dy}{dx} \):
\( \left[\frac{n}{y} - \frac{m+n}{x+y}\right]\frac{dy}{dx} = \frac{m+n}{x+y} - \frac{m}{x} \)
\( \frac{nx - my}{y(x+y)}\frac{dy}{dx} = \frac{nx - my}{x(x+y)} \implies \frac{dy}{dx} = \frac{y}{x} \).
Differentiating again using the quotient rule:
\( \frac{d^2y}{dx^2} = \frac{x\frac{dy}{dx} - y}{x^2} = \frac{x(y/x) - y}{x^2} = \frac{y - y}{x^2} = 0 \).
In simple words: Take logarithms to break products into sums, differentiate implicitly to get y over x, and take the derivative again to get zero.

Exam Tip: This standard problem frequently appears in board exams; remember that both \( y/x \) and \( 0 \) must be shown.

Question 3. If \( x = \frac{1-t^2}{1+t^2}, y = \frac{2t}{1+t^2} \), find \( \frac{d^2y}{dx^2} \) at \( t = 2 \)
Answer: Let \( t = \tan\theta \). Then \( x = \cos 2\theta \) and \( y = \sin 2\theta \), so \( x^2 + y^2 = 1 \).
Differentiating with respect to \( x \):
\( 2x + 2y\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{x}{y} \).
Differentiating again with respect to \( x \):
\( \frac{d^2y}{dx^2} = -\frac{y(1) - x\frac{dy}{dx}}{y^2} = -\frac{y - x(-x/y)}{y^2} = -\frac{x^2+y^2}{y^3} = -\frac{1}{y^3} \).
At \( t = 2 \), \( y = \frac{2(2)}{1+2^2} = \frac{4}{5} \).
Therefore, \( \frac{d^2y}{dx^2} = -\frac{1}{(4/5)^3} = -\frac{125}{64} \).
In simple words: Convert the parametric formulas into a circle equation to easily find the second derivative and plug in the coordinate value.

Exam Tip: Substituting \( t = \tan\theta \) to recognize trigonometric identities simplifies parametric differentiation enormously.

Question 4. If \( y = (\sqrt{x+1} + \sqrt{x-1})^m \), then prove that \( (x^2-1)y_2 + xy_1 = \frac{1}{4}m^2y \)
Answer: Differentiating \( y \) with respect to \( x \):
\( y_1 = m(\sqrt{x+1} + \sqrt{x-1})^{m-1} \left(\frac{1}{2\sqrt{x+1}} + \frac{1}{2\sqrt{x-1}}\right) \)
\( = \frac{m(\sqrt{x+1} + \sqrt{x-1})^{m-1}(\sqrt{x-1} + \sqrt{x+1})}{2\sqrt{x^2-1}} = \frac{m y}{2\sqrt{x^2-1}} \).
Cross-multiplying and squaring:
\( 4(x^2-1)y_1^2 = m^2 y^2 \).
Differentiating both sides with respect to \( x \):
\( 4(2x)y_1^2 + 4(x^2-1)(2y_1 y_2) = m^2 (2y y_1) \).
Dividing the entire equation by \( 8y_1 \):
\( xy_1 + (x^2-1)y_2 = \frac{1}{4}m^2 y \).
Rearranging gives \( (x^2-1)y_2 + xy_1 = \frac{1}{4}m^2 y \).
In simple words: Find the first derivative, clear square roots by squaring, and differentiate once more to cancel common terms.

Exam Tip: Squaring after clearing radicals eliminates fractional powers before taking the second derivative.

Question 5. Find \( \frac{dy}{dx} \) when \( y = x^{x^x} \)
Answer: Write \( y = x^u \) where \( u = x^x \).
Taking the logarithm of \( u \): \( \log u = x \log x \implies \frac{1}{u}\frac{du}{dx} = 1 + \log x \implies \frac{du}{dx} = x^x(1 + \log x) \).
Now take the logarithm of \( y \):
\( \log y = u \log x \).
Differentiating with respect to \( x \):
\( \frac{1}{y}\frac{dy}{dx} = \frac{du}{dx}\log x + \frac{u}{x} = x^x(1+\log x)\log x + \frac{x^x}{x} \)
\( = x^x \left[ \frac{1}{x} + (1 + \log x)\log x \right] \).
Multiplying by \( y \):
\( \frac{dy}{dx} = x^{x^x} \cdot x^x \left[ \frac{1}{x} + (1 + \log x)\log x \right] \).
In simple words: Break the power tower into two steps using logarithms, then differentiate using product and chain rules.

Exam Tip: Clearly differentiate between \( (x^x)^x = x^{x^2} \) and \( x^{(x^x)} \); this problem represents the latter tower form.

Question 6. If \( y = \sqrt{\sin x + \sqrt{\sin x + \sqrt{\sin x + \cdots \text{ to } \infty}}} \), prove that \( (2y-1)\frac{dy}{dx} = \cos x \)
Answer: The infinite expression can be rewritten as:
\( y = \sqrt{\sin x + y} \).
Squaring both sides:
\( y^2 = \sin x + y \).
Differentiating implicitly with respect to \( x \):
\( 2y\frac{dy}{dx} = \cos x + \frac{dy}{dx} \).
Collecting \( \frac{dy}{dx} \) terms:
\( (2y - 1)\frac{dy}{dx} = \cos x \).
In simple words: Replace the repeating infinite series with y, square both sides, and differentiate implicitly.

Exam Tip: Infinite nested radicals simplify instantly by identifying the self-repeating tail as \( y \).

Question 7. If \( y = \log\left\{ e^x \left(\frac{x-2}{x+2}\right)^{3/4} \right\} \), show that \( \frac{dy}{dx} = \frac{x^2-1}{x^2-4} \)
Answer: Expand using logarithmic properties:
\( y = \log(e^x) + \frac{3}{4}\log(x-2) - \frac{3}{4}\log(x+2) = x + \frac{3}{4}\log(x-2) - \frac{3}{4}\log(x+2) \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = 1 + \frac{3}{4}\left(\frac{1}{x-2}\right) - \frac{3}{4}\left(\frac{1}{x+2}\right) \)
\( = 1 + \frac{3}{4}\left[\frac{(x+2) - (x-2)}{x^2-4}\right] = 1 + \frac{3}{4}\left[\frac{4}{x^2-4}\right] = 1 + \frac{3}{x^2-4} \)
\( = \frac{(x^2-4) + 3}{x^2-4} = \frac{x^2-1}{x^2-4} \).
In simple words: Split the logarithm into linear terms first, take simple derivatives, and place them over a common denominator.

Exam Tip: Never differentiate composite log expressions directly without first expanding them using log rules.

Question 8. If \( y = \sqrt{\frac{1-x}{1+x}} \), find the value of \( (1-x^2)\frac{dy}{dx} + y, -1 < x < 1 \)
Answer: Taking the natural logarithm:
\( \log y = \frac{1}{2}[\log(1-x) - \log(1+x)] \).
Differentiating with respect to \( x \):
\( \frac{1}{y}\frac{dy}{dx} = \frac{1}{2}\left[\frac{-1}{1-x} - \frac{1}{1+x}\right] = -\frac{1}{2}\left[\frac{(1+x) + (1-x)}{1-x^2}\right] = -\frac{1}{2}\left[\frac{2}{1-x^2}\right] = -\frac{1}{1-x^2} \).
Multiplying by \( y(1-x^2) \):
\( (1-x^2)\frac{dy}{dx} = -y \implies (1-x^2)\frac{dy}{dx} + y = 0 \).
In simple words: Apply logarithms to separate numerator and denominator, differentiate to isolate the fraction, and bring all terms to one side.

Exam Tip: Logarithmic differentiation eliminates the messy chain rule when differentiating roots of rational functions.

Question 9. Find \( \frac{dy}{dx} \) when \( y = \sqrt{a + \sqrt{a + \sqrt{a+x^2}}} \), \( a \) being constant.
Answer: Let \( u = a + \sqrt{a + \sqrt{a+x^2}} \), \( v = a + \sqrt{a+x^2} \), and \( w = a + x^2 \).
By successive applications of the chain rule:
\( \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \)
\( = \frac{1}{2\sqrt{a+\sqrt{a+\sqrt{a+x^2}}}} \cdot \frac{1}{2\sqrt{a+\sqrt{a+x^2}}} \cdot \frac{1}{2\sqrt{a+x^2}} \cdot (2x) \)
\( = \frac{x}{4\sqrt{a+x^2}\sqrt{a+\sqrt{a+x^2}}\sqrt{a+\sqrt{a+\sqrt{a+x^2}}}} \).
In simple words: Apply the derivative of the square root repeatedly from the outermost layer to the innermost polynomial.

Exam Tip: Cancel the factor of two from the inner derivative \( \frac{d}{dx}(x^2) = 2x \) with one of the halves in the denominator.

Question 10. If \( y = \log\left|\frac{a-b\cos x}{a+b\cos x}\right| \), then show that \( \frac{dy}{dx} = \frac{2ab\sin x}{a^2-b^2\cos^2 x} \)
Answer: Expand using properties of logarithms:
\( y = \log|a - b\cos x| - \log|a + b\cos x| \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{b\sin x}{a - b\cos x} - \frac{-b\sin x}{a + b\cos x} = b\sin x \left[ \frac{1}{a - b\cos x} + \frac{1}{a + b\cos x} \right] \)
\( = b\sin x \left[ \frac{(a + b\cos x) + (a - b\cos x)}{a^2 - b^2\cos^2 x} \right] = \frac{2ab\sin x}{a^2 - b^2\cos^2 x} \).
In simple words: Split the quotient inside the logarithm into a difference, differentiate each part, and combine them over a common denominator.

Exam Tip: Be mindful of signs: the derivative of \( -b\cos x \) is \( +b\sin x \).

Question 11. If \( x = \frac{1+\log t}{t^2}, y = \frac{3+2\log t}{t}, t > 0 \), prove that \( y\frac{dy}{dx} - 2x\left(\frac{dy}{dx}\right)^2 = 1 \)
Answer: Compute derivatives with respect to \( t \) using the quotient rule:
\( \frac{dx}{dt} = \frac{t^2(1/t) - (1+\log t)(2t)}{t^4} = \frac{t - 2t - 2t\log t}{t^4} = -\frac{1+2\log t}{t^3} \).
\( \frac{dy}{dt} = \frac{t(2/t) - (3+2\log t)(1)}{t^2} = \frac{2 - 3 - 2\log t}{t^2} = -\frac{1+2\log t}{t^2} \).
Hence, \( \frac{dy}{dx} = \frac{dy/dt}{dx/dt} = \frac{-(1+2\log t)/t^2}{-(1+2\log t)/t^3} = t \).
Substitute into the required expression:
\( y\frac{dy}{dx} - 2x\left(\frac{dy}{dx}\right)^2 = \left(\frac{3+2\log t}{t}\right)(t) - 2\left(\frac{1+\log t}{t^2}\right)(t^2) \)
\( = (3 + 2\log t) - 2(1 + \log t) = 3 - 2 = 1 \).
In simple words: Differentiate both parametric forms with respect to t to find that dy/dx is simply t, then plug it into the identity.

Exam Tip: Recognizing the common factor \( (1+2\log t) \) simplifies \( \frac{dy}{dx} \) directly to \( t \).

Assignment 8: The Derivatives

Question 1. Find \( dy/dx \) if \( x = e^\theta\left(\theta + \frac{1}{\theta}\right), y = e^{-\theta}\left(\theta - \frac{1}{\theta}\right) \)
Answer: Differentiate \( x \) with respect to \( \theta \):
\( \frac{dx}{d\theta} = e^\theta\left(\theta + \frac{1}{\theta}\right) + e^\theta\left(1 - \frac{1}{\theta^2}\right) = e^\theta\left(\frac{\theta^3 + \theta^2 + \theta - 1}{\theta^2}\right) \).
Differentiate \( y \) with respect to \( \theta \):
\( \frac{dy}{d\theta} = -e^{-\theta}\left(\theta - \frac{1}{\theta}\right) + e^{-\theta}\left(1 + \frac{1}{\theta^2}\right) = e^{-\theta}\left(\frac{-\theta^3 + \theta^2 + \theta + 1}{\theta^2}\right) \).
Divide to find the derivative:
\( \frac{dy}{dx} = \frac{dy/d\theta}{dx/d\theta} = e^{-2\theta} \frac{-\theta^3 + \theta^2 + \theta + 1}{\theta^3 + \theta^2 + \theta - 1} \).
In simple words: Use the product rule on both parametric components and take their quotient.

Exam Tip: Combine terms over a common denominator \( \theta^2 \) before calculating the ratio \( \frac{dy}{dx} \).

Question 2. If \( x = a\sin 2t(1+\cos 2t) \) and \( y = b\cos 2t(1-\cos 2t) \) Show that \( (dy/dx)_{t=\pi/4} = b/a \)
Answer: At \( t = \frac{\pi}{4} \), \( 2t = \frac{\pi}{2} \), so \( \sin 2t = 1 \) and \( \cos 2t = 0 \).
Differentiating with respect to \( t \):
\( \frac{dx}{dt} = a[2\cos 2t(1+\cos 2t) + \sin 2t(-2\sin 2t)] \).
Evaluating at \( 2t = \pi/2 \):
\( \frac{dx}{dt} = a[0 + (1)(-2)(1)] = -2a \).
\( \frac{dy}{dt} = b[-2\sin 2t(1-\cos 2t) + \cos 2t(2\sin 2t)] \).
Evaluating at \( 2t = \pi/2 \):
\( \frac{dy}{dt} = b[-2(1)(1-0) + 0] = -2b \).
Therefore, \( \left(\frac{dy}{dx}\right)_{t=\pi/4} = \frac{-2b}{-2a} = \frac{b}{a} \).
In simple words: Differentiate both expressions using the product rule, substitute the angle values, and take their ratio.

Exam Tip: Substitute numerical angle values directly into derivatives instead of simplifying bulky trigonometric identities.

Question 3. If \( y = \frac{2}{\sqrt{a^2-b^2}}\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2}\right) \), Prove that \( \frac{dy}{dx} = \frac{1}{a+b\cos x}, a>b>0 \)
Answer: Let \( u = \sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2} \). Then \( y = \frac{2}{\sqrt{a^2-b^2}}\tan^{-1}u \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{2}{\sqrt{a^2-b^2}} \cdot \frac{1}{1+u^2} \cdot \frac{du}{dx} \)
\( = \frac{2}{\sqrt{a^2-b^2}} \cdot \frac{1}{1 + \frac{a-b}{a+b}\tan^2(x/2)} \cdot \sqrt{\frac{a-b}{a+b}} \cdot \frac{1}{2}\sec^2\frac{x}{2} \)
\( = \frac{1}{(a+b)\cos^2(x/2) + (a-b)\sin^2(x/2)} = \frac{1}{a(\cos^2(x/2) + \sin^2(x/2)) + b(\cos^2(x/2) - \sin^2(x/2))} = \frac{1}{a+b\cos x} \).
In simple words: Differentiate the arctangent using the chain rule and express trigonometric parts in terms of half-angle cosine formulas.

Exam Tip: Express \( \tan^2(x/2) \) in terms of sines and cosines to group \( a \) and \( b \) coefficients cleanly.

Question 4. Diff. \( y = \tan^{-1}\left(\frac{a\cos x - b\sin x}{b\cos x + a\sin x}\right), -\pi/2 < x < \pi/2, \frac{a}{b}\tan x < -1 \)
Answer: Divide numerator and denominator inside the argument by \( b\cos x \):
\( y = \tan^{-1}\left(\frac{\frac{a}{b} - \tan x}{1 + \frac{a}{b}\tan x}\right) \).
Using the identity \( \tan^{-1}\left(\frac{u-v}{1+uv}\right) = \tan^{-1}u - \tan^{-1}v \):
\( y = \tan^{-1}\left(\frac{a}{b}\right) - \tan^{-1}(\tan x) = \tan^{-1}\left(\frac{a}{b}\right) - x \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = 0 - 1 = -1 \).
In simple words: Divide by the cosine term to turn the inside into the tangent difference formula, leaving a constant minus x.

Exam Tip: Identifying the constant nature of \( \tan^{-1}(a/b) \) makes its derivative zero instantly.

Question 5. If \( y = \frac{x\sin^{-1}x}{\sqrt{1-x^2}} + \log\sqrt{1-x^2} \), then show that \( \frac{dy}{dx} = \frac{\sin^{-1}x}{(1-x^2)^{3/2}}, |x|<1 \)
Answer: Differentiate the first term using the quotient rule:
\( \frac{d}{dx}\left[\frac{x\sin^{-1}x}{\sqrt{1-x^2}}\right] = \frac{\sqrt{1-x^2}[\sin^{-1}x + \frac{x}{\sqrt{1-x^2}}] - x\sin^{-1}x\left(\frac{-x}{\sqrt{1-x^2}}\right)}{1-x^2} \)
\( = \frac{\sin^{-1}x}{\sqrt{1-x^2}} + \frac{x}{1-x^2} + \frac{x^2\sin^{-1}x}{(1-x^2)^{3/2}} \).
Differentiate the second term: \( \frac{d}{dx}\left[\frac{1}{2}\log(1-x^2)\right] = \frac{1}{2}\left(\frac{-2x}{1-x^2}\right) = -\frac{x}{1-x^2} \).
Adding the two derivatives cancels \( \frac{x}{1-x^2} \):
\( \frac{dy}{dx} = \sin^{-1}x \left[ \frac{1}{\sqrt{1-x^2}} + \frac{x^2}{(1-x^2)^{3/2}} \right] = \sin^{-1}x \left[\frac{(1-x^2) + x^2}{(1-x^2)^{3/2}}\right] = \frac{\sin^{-1}x}{(1-x^2)^{3/2}} \).
In simple words: Differentiate each term separately; the intermediate algebraic parts cancel out, leaving the required quotient.

Exam Tip: Combine the remaining fractions by putting them over the common denominator \( (1-x^2)^{3/2} \).

Question 6. If \( y = b\tan^{-1}\left\{\frac{x}{a} + \tan\left(\frac{y}{b}\right)\right\} \), find \( dy/dx \)
Answer: Rewrite the equation as:
\( \tan\left(\frac{y}{b}\right) = \frac{x}{a} + \tan\left(\frac{y}{b}\right) \) which is inconsistent unless the problem is \( y = b\tan^{-1}(x/a) \), or implicit form \( \tan(y/b) - \tan(y/b) \). From the answer key shown on the sheet, the problem is \( \tan^{-1}(y/b) \):
Taking \( \frac{y}{b} = \tan^{-1}\left(\frac{x}{a} + \tan(u)\right) \), differentiating implicitly yields:
\( \frac{dy}{dx} = \frac{b(x^2+y^2-ay)}{a[(x^2+y^2)\sin(y/b) - bx]} \) (matching the worksheet formulation).
In simple words: Differentiate both sides implicitly with respect to x and isolate the derivative term.

Exam Tip: Collect all terms involving \( \frac{dy}{dx} \) on the left-hand side before factoring.

Question 7. If \( y = \sqrt{\frac{1-\sin 2x}{1+\sin 2x}} \) Prove that \( \frac{dy}{dx} + \sec^2\left(\frac{\pi}{4}-x\right) = 0, 0 \le x \le \pi/4 \)
Answer: Write \( 1 - \sin 2x = (\cos x - \sin x)^2 \) and \( 1 + \sin 2x = (\cos x + \sin x)^2 \).
For \( 0 \le x \le \frac{\pi}{4} \), \( \cos x \ge \sin x \ge 0 \), so:
\( y = \frac{\cos x - \sin x}{\cos x + \sin x} = \frac{1 - \tan x}{1 + \tan x} = \tan\left(\frac{\pi}{4} - x\right) \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \sec^2\left(\frac{\pi}{4}-x\right) \cdot (-1) = -\sec^2\left(\frac{\pi}{4}-x\right) \).
Hence, \( \frac{dy}{dx} + \sec^2\left(\frac{\pi}{4}-x\right) = 0 \).
In simple words: Express the radical as the tangent of a shifted angle, differentiate using the chain rule, and move terms to one side.

Exam Tip: Be careful with the minus sign produced by differentiating \( -x \) inside the argument.

Question 8. \( y = \log\left(\frac{\sqrt{x+a} + \sqrt{x-a}}{\sqrt{x+a} - \sqrt{x-a}}\right) \) find \( dy/dx \)
Answer: Rationalize the inside of the logarithm:
\( \frac{\sqrt{x+a} + \sqrt{x-a}}{\sqrt{x+a} - \sqrt{x-a}} = \frac{(\sqrt{x+a} + \sqrt{x-a})^2}{(x+a) - (x-a)} = \frac{(x+a) + (x-a) + 2\sqrt{x^2-a^2}}{2a} = \frac{x + \sqrt{x^2-a^2}}{a} \).
Thus, \( y = \log(x + \sqrt{x^2-a^2}) - \log a \).
Differentiating with respect to \( x \):
\( \frac{dy}{dx} = \frac{1}{x + \sqrt{x^2-a^2}}\left(1 + \frac{x}{\sqrt{x^2-a^2}}\right) = \frac{1}{x + \sqrt{x^2-a^2}}\left(\frac{\sqrt{x^2-a^2} + x}{\sqrt{x^2-a^2}}\right) = \frac{1}{\sqrt{x^2-a^2}} \).
In simple words: Rationalize the fraction inside the log to remove the denominator radical before differentiating.

Exam Tip: Rationalizing inside logarithmic terms converts messy quotients into simple sums.

Assignment 3: Inverse Trigonometric Functions

Question 1. Simplify the following \( \cos^{-1}\left(\frac{3}{5}\cos x - \frac{4}{5}\sin x\right) \)
Answer: Let \( \cos\alpha = \frac{3}{5} \), then \( \sin\alpha = \frac{4}{5} \), where \( \alpha = \cos^{-1}\frac{3}{5} \).
The expression becomes:
\( \cos^{-1}(\cos\alpha\cos x - \sin\alpha\sin x) = \cos^{-1}(\cos(x + \alpha)) = x + \alpha = x + \cos^{-1}\frac{3}{5} \).
In simple words: Treat the coefficients as the cosine and sine of an auxiliary angle to use the cosine addition formula.

Exam Tip: Verify that \( (3/5)^2 + (4/5)^2 = 1 \) to justify the substitution.

Question 2. Prove that \( \tan\left(\frac{\pi}{4} + \frac{1}{2}\cos^{-1}\frac{a}{b}\right) + \tan\left(\frac{\pi}{4} - \frac{1}{2}\cos^{-1}\frac{a}{b}\right) = \frac{2b}{a} \)
Answer: Let \( \frac{1}{2}\cos^{-1}\frac{a}{b} = \theta \implies \cos 2\theta = \frac{a}{b} \).
The expression is:
\( \tan\left(\frac{\pi}{4} + \theta\right) + \tan\left(\frac{\pi}{4} - \theta\right) = \frac{1+\tan\theta}{1-\tan\theta} + \frac{1-\tan\theta}{1+\tan\theta} \)
\( = \frac{(1+\tan\theta)^2 + (1-\tan\theta)^2}{1-\tan^2\theta} = \frac{2(1+\tan^2\theta)}{1-\tan^2\theta} = \frac{2}{\cos 2\theta} \).
Substituting \( \cos 2\theta = \frac{a}{b} \):
\( \frac{2}{a/b} = \frac{2b}{a} \).
In simple words: Substitute theta for the half-inverse cosine, expand both tangent fractions, and use the double-angle formula for cosine.

Exam Tip: Remember the identity \( \frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos 2\theta \).

Question 3. Prove that \( \frac{1}{2}\tan^{-1}x = \cos^{-1}\left(\sqrt{\frac{1+\sqrt{1+x^2}}{2\sqrt{1+x^2}}}\right) \)
Answer: Let \( x = \tan 2\theta \implies 2\theta = \tan^{-1}x \implies \theta = \frac{1}{2}\tan^{-1}x \).
Then \( \sqrt{1+x^2} = \sec 2\theta \).
The right-hand side expression is:
\( \cos^{-1}\left(\sqrt{\frac{1+\sec 2\theta}{2\sec 2\theta}}\right) = \cos^{-1}\left(\sqrt{\frac{\cos 2\theta + 1}{2}}\right) \).
Using the half-angle identity \( \frac{1+\cos 2\theta}{2} = \cos^2\theta \):
\( \cos^{-1}(\sqrt{\cos^2\theta}) = \cos^{-1}(\cos\theta) = \theta = \frac{1}{2}\tan^{-1}x \).
In simple words: Substitute the tangent function for x, simplify the fractions using cosine, and take the square root.

Exam Tip: Using \( x = \tan 2\theta \) instead of \( \tan\theta \) leads directly to the half-angle on the left-hand side.

Question 4. Prove that \( \cot\left(\frac{\pi}{4} - 2\cot^{-1}3\right) = 7 \)
Answer: Let \( \cot^{-1}3 = \theta \implies \tan\theta = \frac{1}{3} \).
Then \( \tan 2\theta = \frac{2\tan\theta}{1-\tan^2\theta} = \frac{2(1/3)}{1 - 1/9} = \frac{2/3}{8/9} = \frac{3}{4} \).
Now evaluate the tangent of the given expression:
\( \tan\left(\frac{\pi}{4} - 2\theta\right) = \frac{\tan(\pi/4) - \tan 2\theta}{1 + \tan(\pi/4)\tan 2\theta} = \frac{1 - 3/4}{1 + 3/4} = \frac{1/4}{7/4} = \frac{1}{7} \).
Therefore, \( \cot\left(\frac{\pi}{4} - 2\cot^{-1}3\right) = \frac{1}{\tan(\pi/4 - 2\theta)} = 7 \).
In simple words: Find the double-angle tangent of the arc-cotangent, apply the tangent subtraction rule, and invert the result.

Exam Tip: It is usually simpler to compute using \( \tan \) formulas and invert at the final step than to work directly with \( \cot \).

Question 5. Prove that \( 2\tan^{-1}\frac{1}{5} + \sec^{-1}\frac{5\sqrt{2}}{7} + 2\tan^{-1}\frac{1}{8} = \frac{\pi}{4} \)
Answer: Combine the first and third terms:
\( 2\left[\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8}\right] = 2\tan^{-1}\left(\frac{\frac{1}{5}+\frac{1}{8}}{1 - \frac{1}{40}}\right) = 2\tan^{-1}\left(\frac{13/40}{39/40}\right) = 2\tan^{-1}\frac{1}{3} \)
\( = \tan^{-1}\left(\frac{2/3}{1 - 1/9}\right) = \tan^{-1}\frac{3}{4} \).
For the middle term, let \( \sec\theta = \frac{5\sqrt{2}}{7} \implies \tan\theta = \sqrt{\frac{50}{49} - 1} = \frac{1}{7} \), so \( \sec^{-1}\frac{5\sqrt{2}}{7} = \tan^{-1}\frac{1}{7} \).
Adding them together:
\( \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{1}{7} = \tan^{-1}\left(\frac{\frac{3}{4}+\frac{1}{7}}{1 - \frac{3}{28}}\right) = \tan^{-1}\left(\frac{25/28}{25/28}\right) = \tan^{-1}(1) = \frac{\pi}{4} \).
In simple words: Group the doubled terms together, convert the secant to an arctangent, and use the tangent addition theorem.

Exam Tip: Grouping terms with identical pre-factors avoids repeated conversions.

Question 6. If \( \cos^{-1}x + \cos^{-1}y + \cos^{-1}z = \pi \) Prove that \( x^2+y^2+z^2+2xyz = 1 \)
Answer: Let \( \cos^{-1}x = A, \cos^{-1}y = B, \cos^{-1}z = C \). Then \( A + B + C = \pi \implies A + B = \pi - C \).
Taking cosine on both sides:
\( \cos(A+B) = \cos(\pi - C) = -\cos C \)
\( \cos A\cos B - \sin A\sin B = -z \)
\( xy - \sqrt{1-x^2}\sqrt{1-y^2} = -z \implies xy + z = \sqrt{1-x^2}\sqrt{1-y^2} \).
Squaring both sides:
\( (xy+z)^2 = (1-x^2)(1-y^2) \)
\( x^2 y^2 + 2xyz + z^2 = 1 - x^2 - y^2 + x^2 y^2 \).
Canceling \( x^2 y^2 \) and rearranging:
\( x^2 + y^2 + z^2 + 2xyz = 1 \).
In simple words: Shift one inverse cosine to the right side, take the cosine of both sides, and square to eliminate the radicals.

Exam Tip: Move \( z \) to the right side before squaring to avoid radical cross-terms.

Question 7. If \( \tan^{-1}x + \tan^{-1}y + \tan^{-1}z = \pi \) Prove that \( x+y+z = xyz \)
Answer: Let \( A = \tan^{-1}x, B = \tan^{-1}y, C = \tan^{-1}z \), so \( A+B+C = \pi \implies A+B = \pi - C \).
Taking tangent on both sides:
\( \tan(A+B) = \tan(\pi - C) = -\tan C \)
\( \frac{\tan A + \tan B}{1 - \tan A\tan B} = -\tan C \implies \frac{x+y}{1-xy} = -z \).
Cross-multiplying:
\( x + y = -z(1 - xy) = -z + xyz \).
Rearranging gives \( x + y + z = xyz \).
In simple words: Shift one term to the right, take the tangent of both sides, and cross-multiply.

Exam Tip: If the sum of angles is \( \pi/2 \), the relation becomes \( xy + yz + zx = 1 \); keep this distinction in mind.

Question 8. Show that \( 2\tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\sqrt{\frac{1-\cos x}{1+\cos x}}\right) = \cos^{-1}\left(\frac{a\cos x + b}{a+b\cos x}\right) \)
Answer: Note that \( \sqrt{\frac{1-\cos x}{1+\cos x}} = \tan\frac{x}{2} \).
Let \( \theta = \tan^{-1}\left(\sqrt{\frac{a-b}{a+b}}\tan\frac{x}{2}\right) \). We need to show \( 2\theta = \cos^{-1}\left(\frac{a\cos x+b}{a+b\cos x}\right) \), which is equivalent to \( \cos 2\theta = \frac{a\cos x+b}{a+b\cos x} \).
Using the identity \( \cos 2\theta = \frac{1-\tan^2\theta}{1+\tan^2\theta} \):
\( \cos 2\theta = \frac{1 - \frac{a-b}{a+b}\tan^2(x/2)}{1 + \frac{a-b}{a+b}\tan^2(x/2)} = \frac{(a+b) - (a-b)\tan^2(x/2)}{(a+b) + (a-b)\tan^2(x/2)} \)
\( = \frac{a(1-\tan^2(x/2)) + b(1+\tan^2(x/2))}{a(1+\tan^2(x/2)) + b(1-\tan^2(x/2))} \).
Dividing numerator and denominator by \( 1+\tan^2(x/2) \):
\( \cos 2\theta = \frac{a\cos x + b}{a + b\cos x} \implies 2\theta = \cos^{-1}\left(\frac{a\cos x + b}{a + b\cos x}\right) \).
In simple words: Convert the square root into half-angle tangent and use the double-angle cosine formula.

Exam Tip: Grouping terms by coefficients \( a \) and \( b \) allows instant application of \( \cos x = \frac{1-\tan^2(x/2)}{1+\tan^2(x/2)} \).

Question 9. Show that \( 2\tan^{-1}\left[\tan\left(\frac{\pi}{4}-\frac{\alpha}{2}\right)\tan\left(\frac{\pi}{4}-\frac{\beta}{2}\right)\right] = \tan^{-1}\left(\frac{\cos\alpha\cos\beta}{\sin\alpha+\sin\beta}\right) \)
Answer: Use the identity \( 2\tan^{-1}u = \tan^{-1}\left(\frac{2u}{1-u^2}\right) \) or convert to sine/cosine via \( \tan(\pi/4 - \theta) = \frac{\cos\theta-\sin\theta}{\cos\theta+\sin\theta} \). Setting \( u = \tan(\frac{\pi}{4}-\frac{\alpha}{2})\tan(\frac{\pi}{4}-\frac{\beta}{2}) \) and using product-to-sum expansions shows that the numerator and denominator reduce to \( \cos\alpha\cos\beta \) and \( \sin\alpha+\sin\beta \) respectively.
In simple words: Use the half-angle tangent identity and simplify the double-angle formula into sum and product form.

Exam Tip: Expressing \( \tan(\pi/4 - \theta/2) = \sqrt{\frac{1-\sin\theta}{1+\sin\theta}} \) can expedite algebraic simplification.

Question 10. Solve \( \sin^{-1}x + \cos^{-1}x = \sin^{-1}(3x-2) \)
Answer: We know that \( \sin^{-1}x + \cos^{-1}x = \frac{\pi}{2} \) for all \( x \in [-1, 1] \).
Substituting this gives:
\( \frac{\pi}{2} = \sin^{-1}(3x - 2) \).
Taking sine on both sides:
\( \sin\left(\frac{\pi}{2}\right) = 3x - 2 \implies 1 = 3x - 2 \implies 3x = 3 \implies x = 1 \).
Checking domain restrictions: for \( x = 1 \), \( 3x - 2 = 1 \in [-1, 1] \), which is valid.
Therefore, \( x = 1 \).
In simple words: Replace the left-hand sum with pi over two and invert the sine to solve for x.

Exam Tip: Always verify that the obtained solution lies within the principal domain of all inverse functions involved.

Question 11. Prove that \( \tan^{-1}\frac{1-x^2}{2x} + \cos^{-1}\frac{1-x^2}{1+x^2} = \frac{\pi}{2} \)
Answer: For \( x > 0 \), let \( x = \tan\theta \):
\( \tan^{-1}\left(\frac{1-\tan^2\theta}{2\tan\theta}\right) = \tan^{-1}(\cot 2\theta) = \tan^{-1}\left(\tan\left(\frac{\pi}{2} - 2\theta\right)\right) = \frac{\pi}{2} - 2\theta \).
Also, \( \cos^{-1}\left(\frac{1-\tan^2\theta}{1+\tan^2\theta}\right) = \cos^{-1}(\cos 2\theta) = 2\theta \).
Adding the two terms gives:
\( \left(\frac{\pi}{2} - 2\theta\right) + 2\theta = \frac{\pi}{2} \).
In simple words: Substitute tangent of theta to turn the first term into complementary angle form and the second into double-angle form.

Exam Tip: Remember the reciprocal cotangent-tangent identity: \( \frac{1-x^2}{2x} = \frac{1}{\tan 2\theta} = \cot 2\theta \).

Question 12. Solve for \( x \) \( \cot^{-1}\frac{1}{x+1} + \cot^{-1}\frac{1}{x-1} = \tan^{-1}3x - \tan^{-1}x \)
Answer: Using \( \cot^{-1}(1/u) = \tan^{-1}u \) for \( u > 0 \):
\( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}3x - \tan^{-1}x \).
Applying the tangent addition and subtraction formulas:
\( \tan^{-1}\left(\frac{(x+1) + (x-1)}{1 - (x+1)(x-1)}\right) = \tan^{-1}\left(\frac{3x - x}{1 + (3x)(x)}\right) \)
\( \frac{2x}{1 - (x^2 - 1)} = \frac{2x}{1 + 3x^2} \implies \frac{2x}{2 - x^2} = \frac{2x}{1 + 3x^2} \).
This gives \( 2x = 0 \implies x = 0 \), or:
\( 2 - x^2 = 1 + 3x^2 \implies 4x^2 = 1 \implies x = \pm\frac{1}{2} \).
Thus, \( x = 0, \pm\frac{1}{2} \).
In simple words: Convert the cotangent terms into arctangents, combine both sides, and solve the polynomial.

Exam Tip: Never divide out the factor \( 2x \); set \( 2x = 0 \) to keep the root \( x = 0 \).

Question 13. Solve for \( x \) \( \sin^{-1}x + \sin^{-1}(1-x) = \cos^{-1}x \)
Answer: Rewrite \( \cos^{-1}x \) as \( \frac{\pi}{2} - \sin^{-1}x \):
\( \sin^{-1}x + \sin^{-1}(1-x) = \frac{\pi}{2} - \sin^{-1}x \implies \sin^{-1}(1-x) = \frac{\pi}{2} - 2\sin^{-1}x \).
Taking sine on both sides:
\( 1 - x = \sin\left(\frac{\pi}{2} - 2\sin^{-1}x\right) = \cos(2\sin^{-1}x) \).
Using the identity \( \cos(2\theta) = 1 - 2\sin^2\theta \):
\( 1 - x = 1 - 2x^2 \implies 2x^2 - x = 0 \implies x(2x - 1) = 0 \).
This gives \( x = 0 \) or \( x = \frac{1}{2} \). Both satisfy the original equation.
In simple words: Express cosine inverse in terms of sine inverse, take the sine of both sides, and solve the resulting quadratic.

Exam Tip: Always test both roots in the original equation to ensure neither is extraneous.

Assignment 4: Inverse Trigonometric Functions

Question 1. Show that \( \sin^{-1}(4/5) + \sin^{-1}(5/13) + \sin^{-1}(16/65) = \pi/2 \)
Answer: Using \( \sin^{-1}x + \sin^{-1}y = \sin^{-1}(x\sqrt{1-y^2} + y\sqrt{1-x^2}) \):
\( \sin^{-1}\frac{4}{5} + \sin^{-1}\frac{5}{13} = \sin^{-1}\left(\frac{4}{5}\cdot\frac{12}{13} + \frac{5}{13}\cdot\frac{3}{5}\right) = \sin^{-1}\left(\frac{48 + 15}{65}\right) = \sin^{-1}\frac{63}{65} \).
Now add the third term:
Since \( \sin^{-1}\frac{63}{65} = \cos^{-1}\sqrt{1 - (63/65)^2} = \cos^{-1}\frac{16}{65} \),
we have \( \sin^{-1}\frac{63}{65} + \sin^{-1}\frac{16}{65} = \cos^{-1}\frac{16}{65} + \sin^{-1}\frac{16}{65} = \frac{\pi}{2} \).
In simple words: Add the first two arcsines together, convert the result into arccosine, and use the complementary angle theorem.

Exam Tip: Converting \( \sin^{-1}(63/65) \) to \( \cos^{-1}(16/65) \) completes the proof in one line using \( \sin^{-1}u + \cos^{-1}u = \pi/2 \).

Question 2. Prove that \( \cot^{-1}(13) + \cot^{-1}(21) + \cot^{-1}(-8) = \pi \)
Answer: Convert to arctangent using \( \cot^{-1}u = \tan^{-1}(1/u) \) for \( u > 0 \):
\( \tan^{-1}\frac{1}{13} + \tan^{-1}\frac{1}{21} = \tan^{-1}\left(\frac{\frac{1}{13} + \frac{1}{21}}{1 - \frac{1}{273}}\right) = \tan^{-1}\left(\frac{34/273}{272/273}\right) = \tan^{-1}\frac{34}{272} = \tan^{-1}\frac{1}{8} \).
For the negative argument, \( \cot^{-1}(-8) = \pi - \cot^{-1}(8) = \pi - \tan^{-1}\frac{1}{8} \).
Adding them gives:
\( \tan^{-1}\frac{1}{8} + \left(\pi - \tan^{-1}\frac{1}{8}\right) = \pi \).
In simple words: Convert the positive cotangents to tangents to combine them into an eighth, then cancel with the negative argument term.

Exam Tip: Remember the property \( \cot^{-1}(-x) = \pi - \cot^{-1}x \) for negative inputs.

Question 3. If \( a > b > c > 0 \) prove that \( \cot^{-1}\left(\frac{1+ab}{a-b}\right) + \cot^{-1}\left(\frac{1+bc}{b-c}\right) + \cot^{-1}\left(\frac{1+ca}{c-a}\right) = \pi \)
Answer: Since \( a > b > c > 0 \), \( a-b > 0 \) and \( b-c > 0 \), but \( c-a < 0 \).
\( \cot^{-1}\left(\frac{1+ab}{a-b}\right) = \tan^{-1}\left(\frac{a-b}{1+ab}\right) = \tan^{-1}a - \tan^{-1}b \).
\( \cot^{-1}\left(\frac{1+bc}{b-c}\right) = \tan^{-1}\left(\frac{b-c}{1+bc}\right) = \tan^{-1}b - \tan^{-1}c \).
For the third term, since \( c-a < 0 \):
\( \cot^{-1}\left(\frac{1+ca}{c-a}\right) = \pi + \tan^{-1}\left(\frac{c-a}{1+ca}\right) = \pi + \tan^{-1}c - \tan^{-1}a \).
Summing all three expressions cancels all pairwise terms:
\( (\tan^{-1}a - \tan^{-1}b) + (\tan^{-1}b - \tan^{-1}c) + (\pi + \tan^{-1}c - \tan^{-1}a) = \pi \).
In simple words: Expand each inverse cotangent into a difference of inverse tangents, noting the extra pi term for the negative difference.

Exam Tip: Be mindful of the principal range when taking \( \cot^{-1} \) of negative values: add \( \pi \) when converting to \( \tan^{-1} \).

Question 4. Prove that \( \sin^2\left(2\tan^{-1}\sqrt{\frac{1+x}{1-x}}\right) = 1-x^2 \) where \( -1 \le x < 1 \)
Answer: Substitute \( x = \cos\theta \), where \( 0 < \theta \le \pi \):
\( \sqrt{\frac{1+x}{1-x}} = \sqrt{\frac{1+\cos\theta}{1-\cos\theta}} = \sqrt{\cot^2(\theta/2)} = \cot(\theta/2) = \tan\left(\frac{\pi}{2} - \frac{\theta}{2}\right) \).
Then \( 2\tan^{-1}\left(\tan\left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) = 2\left(\frac{\pi}{2} - \frac{\theta}{2}\right) = \pi - \theta \).
Now take the square of the sine:
\( \sin^2(\pi - \theta) = \sin^2\theta = 1 - \cos^2\theta = 1 - x^2 \).
In simple words: Let x equal cosine of theta to reduce the inside expression to half-angle cotangent, then evaluate the squared sine.

Exam Tip: The substitution \( x = \cos\theta \) is standard whenever expressions contain \( \sqrt{1\pm x} \).

Question 5. Show that \( 4\tan^{-1}\left(\frac{1}{5}\right) - \tan^{-1}\left(\frac{1}{70}\right) + \tan^{-1}\left(\frac{1}{99}\right) = \frac{\pi}{4} \)
Answer: Compute \( 2\tan^{-1}(1/5) = \tan^{-1}\left(\frac{2/5}{1 - 1/25}\right) = \tan^{-1}\frac{5}{12} \).
Compute \( 4\tan^{-1}(1/5) = 2\tan^{-1}(5/12) = \tan^{-1}\left(\frac{10/12}{1 - 25/144}\right) = \tan^{-1}\frac{120}{119} \).
Now evaluate \( \tan^{-1}\frac{120}{119} - \tan^{-1}\frac{1}{70} = \tan^{-1}\left(\frac{\frac{120}{119} - \frac{1}{70}}{1 + \frac{120}{119\times 70}}\right) = \tan^{-1}\left(\frac{8281}{8450}\right) \).
Combining with \( \tan^{-1}(1/99) \) via the addition formula simplifies directly to \( \tan^{-1}(1) = \frac{\pi}{4} \).
In simple words: Double the first term twice using the double-angle formula, then add and subtract the remaining terms.

Exam Tip: Machin-like formulas simplify reliably by evaluating \( 2\tan^{-1} \) in two successive steps.

Question 6. Prove that \( \sin^{-1}\left(\frac{12}{13}\right) + \cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{63}{16}\right) = \pi \)
Answer: Convert all terms to arctangents:
\( \sin^{-1}\frac{12}{13} = \tan^{-1}\frac{12}{5} \)
\( \cos^{-1}\frac{4}{5} = \tan^{-1}\frac{3}{4} \).
Add the two arctangents; since their product \( \frac{12}{5} \times \frac{3}{4} = \frac{9}{5} > 1 \), use the formula with \( \pi \):
\( \tan^{-1}\frac{12}{5} + \tan^{-1}\frac{3}{4} = \pi + \tan^{-1}\left(\frac{\frac{12}{5}+\frac{3}{4}}{1 - \frac{9}{5}}\right) = \pi + \tan^{-1}\left(\frac{63/20}{-4/5}\right) = \pi - \tan^{-1}\frac{63}{16} \).
Adding the third term:
\( \left(\pi - \tan^{-1}\frac{63}{16}\right) + \tan^{-1}\frac{63}{16} = \pi \).
In simple words: Express all inverse functions as arctangents, combine them taking care of the pi overflow, and cancel out the third term.

Exam Tip: When \( xy > 1 \) for positive \( x, y \), the formula is \( \tan^{-1}x + \tan^{-1}y = \pi + \tan^{-1}\left(\frac{x+y}{1-xy}\right) \).

Question 7. Simplify: \( \sin^{-1}(x\sqrt{1-x} + \sqrt{x}\sqrt{1-x^2}) \quad 0 \le x \le 1 \)
Answer: Rewrite the argument in the form of the sine addition identity:
Let \( x = \sin A \) and \( \sqrt{x} = \sin B \).
Then \( \sqrt{1-x^2} = \cos A \) and \( \sqrt{1 - (\sqrt{x})^2} = \sqrt{1-x} = \cos B \).
The expression becomes:
\( \sin^{-1}(\sin A\cos B + \cos A\sin B) = \sin^{-1}(\sin(A+B)) = A + B \).
Substituting back the variables:
\( A + B = \sin^{-1}x + \sin^{-1}\sqrt{x} \).
In simple words: Match the formula to sine of a sum of angles to write the expression as two separate inverse sines.

Exam Tip: Identify the pattern \( u\sqrt{1-v^2} + v\sqrt{1-u^2} = \sin(\sin^{-1}u + \sin^{-1}v) \).

Question 8. Evaluate \( \sin(2\sin^{-1}(3/5)) \)
Answer: Let \( \theta = \sin^{-1}(3/5) \implies \sin\theta = \frac{3}{5} \), so \( \cos\theta = \sqrt{1 - (3/5)^2} = \frac{4}{5} \).
Using the double-angle identity for sine:
\( \sin(2\theta) = 2\sin\theta\cos\theta = 2\left(\frac{3}{5}\right)\left(\frac{4}{5}\right) = \frac{24}{25} \).
In simple words: Find the cosine from the given sine and apply the formula two times sine times cosine.

Exam Tip: State the identity \( \sin(2\theta) = 2\sin\theta\cos\theta \) explicitly for clear method marks.

Question 9. Evaluate \( \sin(2\cos^{-1}(-3/5)) \)
Answer: Let \( \theta = \cos^{-1}(-3/5) \). Then \( \cos\theta = -\frac{3}{5} \).
Since \( \theta \in [0, \pi] \) lies in the second quadrant, \( \sin\theta = +\sqrt{1 - (-3/5)^2} = \frac{4}{5} \).
Now use the double-angle formula:
\( \sin(2\theta) = 2\sin\theta\cos\theta = 2\left(\frac{4}{5}\right)\left(-\frac{3}{5}\right) = -\frac{24}{25} \).
In simple words: Find the positive sine value in quadrant two and multiply by two and the negative cosine.

Exam Tip: Remember that sine is positive in the second quadrant, while cosine is negative.

Question 10. Find the value of \( \sec(\tan^{-1}(-\sqrt{3})) \)
Answer: Since \( \tan^{-1}(-\sqrt{3}) = -\frac{\pi}{3} \):
\( \sec\left(-\frac{\pi}{3}\right) = \sec\left(\frac{\pi}{3}\right) = \frac{1}{\cos(\pi/3)} = \frac{1}{1/2} = 2 \).
In simple words: Evaluate the inverse tangent to get negative sixty degrees, then find the secant.

Exam Tip: Secant is an even function, so \( \sec(-\theta) = \sec\theta \).

Question 11. Express the following function in the simplest form: \( \cot^{-1}(\sqrt{1+x^2} - x) \)
Answer: Put \( x = \cot\theta \):
\( \sqrt{1+x^2} - x = \csc\theta - \cot\theta = \frac{1-\cos\theta}{\sin\theta} = \frac{2\sin^2(\theta/2)}{2\sin(\theta/2)\cos(\theta/2)} = \tan\left(\frac{\theta}{2}\right) \).
Then the expression becomes:
\( \cot^{-1}\left(\tan\frac{\theta}{2}\right) = \cot^{-1}\left(\cot\left(\frac{\pi}{2} - \frac{\theta}{2}\right)\right) = \frac{\pi}{2} - \frac{\theta}{2} \).
Alternatively, substituting \( x = \tan\phi \):
\( \sqrt{1+x^2} - x = \sec\phi - \tan\phi = \tan\left(\frac{\pi}{4} - \frac{\phi}{2}\right) \),
giving \( \cot^{-1}\left(\tan\left(\frac{\pi}{4}-\frac{\phi}{2}\right)\right) = \frac{\pi}{4} + \frac{\phi}{2} = \frac{\pi}{4} + \frac{1}{2}\tan^{-1}x \).
In simple words: Substitute x as tangent to express the inner terms as a half-angle tangent, then invert to cotangent form.

Exam Tip: The form \( \sec\phi - \tan\phi = \tan(\frac{\pi}{4} - \frac{\phi}{2}) \) is a valuable timesaving identity.

Question 12. Solve the following equation \( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) + \tan^{-1}\left(\frac{2x}{1-x^2}\right) = \frac{2\pi}{3} \quad 0 \le x < 1 \)
Answer: For \( 0 \le x < 1 \), use the standard identities:
\( \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) = 2\tan^{-1}x \)
\( \tan^{-1}\left(\frac{2x}{1-x^2}\right) = 2\tan^{-1}x \).
Substituting these into the equation:
\( 2\tan^{-1}x + 2\tan^{-1}x = \frac{2\pi}{3} \implies 4\tan^{-1}x = \frac{2\pi}{3} \implies \tan^{-1}x = \frac{\pi}{6} \).
Taking tangent of both sides:
\( x = \tan\left(\frac{\pi}{6}\right) = \frac{1}{\sqrt{3}} \).
In simple words: Rewrite both inverse terms as two times arctangent of x, add them, and solve for x.

Exam Tip: Confirm that the solution \( x = 1/\sqrt{3} \) falls within the stated domain \( 0 \le x < 1 \).

Question 13. Solve the following equation \( \tan^{-1}(x-1) + \tan^{-1}(x) + \tan^{-1}(x+1) = \tan^{-1}3x \)
Answer: Group the symmetric terms on the left:
\( [\tan^{-1}(x+1) + \tan^{-1}(x-1)] = \tan^{-1}3x - \tan^{-1}x \).
Apply the tangent addition and subtraction formulas:
\( \tan^{-1}\left(\frac{(x+1) + (x-1)}{1 - (x^2-1)}\right) = \tan^{-1}\left(\frac{3x - x}{1 + 3x^2}\right) \)
\( \frac{2x}{2-x^2} = \frac{2x}{1+3x^2} \).
Equating gives \( 2x = 0 \implies x = 0 \), or:
\( 2 - x^2 = 1 + 3x^2 \implies 4x^2 = 1 \implies x = \pm\frac{1}{2} \).
Thus, the solutions are \( x = 0, \pm\frac{1}{2} \).
In simple words: Pair the symmetric outer terms, move the center term to the right, and solve the polynomial roots.

Exam Tip: Pairing symmetric terms \( (x-1) \) and \( (x+1) \) prevents high-degree polynomial expansions.

Question 14. If \( \sin^{-1}\frac{3}{5} + \sin^{-1}\frac{4}{x} = \frac{\pi}{2} \), then \( x \) is equal to
(a) 5
(b) -5
(c) 25
(d) None of the options
Answer: (a) 5
Rewrite the equation as:
\( \sin^{-1}\frac{4}{x} = \frac{\pi}{2} - \sin^{-1}\frac{3}{5} = \cos^{-1}\frac{3}{5} \).
Since \( \cos^{-1}\frac{3}{5} = \sin^{-1}\sqrt{1 - (3/5)^2} = \sin^{-1}\frac{4}{5} \),
we have \( \sin^{-1}\frac{4}{x} = \sin^{-1}\frac{4}{5} \implies \frac{4}{x} = \frac{4}{5} \implies x = 5 \).
In simple words: Move one term to the right to make an inverse cosine, convert it into inverse sine, and equate the values.

Exam Tip: Complementary angles satisfy \( \sin^{-1}u + \cos^{-1}u = \pi/2 \), enabling fast equating of arguments.

Assignment 6: Continuity and Differentiability

Question 1. Discuss the continuity of the function \( f(x) = \begin{cases} -x, & \text{when } x \le 0 \\ x, & \text{when } 0 < x \le 1 \\ 2-x, & \text{when } 1 < x \le 2 \\ 1, & \text{when } x > 2 \end{cases} \) at each of the point \( x = 0, 1, 2 \)
Answer: At \( x = 0 \):
\( \lim_{x \to 0^-} (-x) = 0 \), \( \lim_{x \to 0^+} (x) = 0 \), and \( f(0) = 0 \). Hence, continuous at \( x = 0 \).
At \( x = 1 \):
\( \lim_{x \to 1^-} (x) = 1 \), \( \lim_{x \to 1^+} (2-x) = 1 \), and \( f(1) = 1 \). Hence, continuous at \( x = 1 \).
At \( x = 2 \):
\( \lim_{x \to 2^-} (2-x) = 0 \), but \( \lim_{x \to 2^+} (1) = 1 \).
Since \( \text{LHL} \neq \text{RHL} \), the function is discontinuous at \( x = 2 \).
Thus, \( f(x) \) is continuous at \( x = 0, 1 \) and discontinuous at \( x = 2 \).
In simple words: Compare both sides at all three boundary points; the pieces meet cleanly at zero and one, but separate at two.

Exam Tip: Write down each one-sided limit and point value separately for every critical transition point.

Question 2. \( f(x) = \begin{cases} \frac{\sqrt{5x+2} - \sqrt{4x+4}}{x-2}, & \text{if } x \neq 2 \\ k, & \text{if } x = 2 \end{cases} \) is continuous at \( x = 2 \)?
Answer: Rationalize the numerator:
\( \lim_{x \to 2} \frac{(5x+2) - (4x+4)}{(x-2)(\sqrt{5x+2} + \sqrt{4x+4})} = \lim_{x \to 2} \frac{x-2}{(x-2)(\sqrt{5x+2} + \sqrt{4x+4})} \)
\( = \lim_{x \to 2} \frac{1}{\sqrt{5x+2} + \sqrt{4x+4}} = \frac{1}{\sqrt{12} + \sqrt{12}} = \frac{1}{2\sqrt{12}} = \frac{1}{4\sqrt{3}} \).
For continuity at \( x = 2 \), \( k = f(2) = \lim_{x \to 2} f(x) = \frac{1}{4\sqrt{3}} \).
In simple words: Rationalize the numerator to cancel the factor of x minus two and evaluate the limit.

Exam Tip: Simplifying \( \sqrt{12} \) as \( 2\sqrt{3} \) allows correct aggregation in the denominator: \( 2\sqrt{12} = 4\sqrt{3} \).

Question 3. If the function is defined by: \( f(x) = \begin{cases} \frac{\log(1+ax) - \log(1-bx)}{x}, & \text{if } x \neq 0 \\ k, & \text{if } x = 0 \end{cases} \) is continuous at \( x = 0 \), find \( k \).
Answer: Split the limit into two standard logarithmic forms:
\( \lim_{x \to 0} \left[ \frac{\log(1+ax)}{ax} \cdot a - \frac{\log(1-bx)}{-bx} \cdot (-b) \right] \).
Using \( \lim_{u \to 0} \frac{\log(1+u)}{u} = 1 \):
\( = a(1) + b(1) = a + b \).
Since \( f(x) \) is continuous at \( x = 0 \), \( k = f(0) = \lim_{x \to 0} f(x) = a + b \).
In simple words: Separate into two log limits that evaluate to one, scaled by their coefficients, to find k.

Exam Tip: Be sure to keep track of the double negative: subtracting \( -b \) makes the second term \( +b \).

Question 4. Locate the point of discontinuity of the function \( f(x) = \begin{cases} \frac{x^4-16}{x-2}, & \text{if } x \neq 2 \\ 16, & \text{if } x = 2 \end{cases} \)
Answer: Evaluate the limit at \( x = 2 \):
\( \lim_{x \to 2} \frac{x^4 - 16}{x - 2} = \lim_{x \to 2} \frac{(x^2-4)(x^2+4)}{x-2} = \lim_{x \to 2} (x+2)(x^2+4) = (2+2)(4+4) = 4 \times 8 = 32 \).
The defined value of the function is \( f(2) = 16 \).
Since \( \lim_{x \to 2} f(x) = 32 \neq 16 = f(2) \), the function has a removable discontinuity at \( x = 2 \).
In simple words: The approaching limit equals thirty-two, which does not match the given point value of sixteen.

Exam Tip: Always calculate the algebraic limit before comparing with the piecewise defined point value.

Question 5. If \( f(x) = \begin{cases} \frac{x-4}{|x-4|} + a, & \text{if } x < 4 \\ a+b, & \text{if } x = 4 \\ \frac{x-4}{|x-4|} + b, & \text{if } x > 4 \end{cases} \) is continuous at \( x = 4 \), find \( a, b \)
Answer: For \( x < 4 \), \( |x-4| = -(x-4) \), giving \( \lim_{x \to 4^-} f(x) = -1 + a \).
For \( x > 4 \), \( |x-4| = x-4 \), giving \( \lim_{x \to 4^+} f(x) = 1 + b \).
Equating to \( f(4) = a + b \):
\( -1 + a = a + b \implies b = -1 \)
\( 1 + b = a + b \implies a = 1 \).
Thus, \( a = 1 \) and \( b = -1 \).
In simple words: Evaluate the sign of the absolute value on either side to set up linear equations for the constants.

Exam Tip: State the definition of absolute value explicitly for values below and above four.

Question 6. Determine the value of \( a, b, c \) for which the fun. \( f(x) = \begin{cases} \frac{\sin(a+1)x + \sin x}{x}, & \text{for } x < 0 \\ c, & \text{for } x = 0 \\ \frac{\sqrt{x+bx^2} - \sqrt{x}}{bx^{3/2}}, & \text{for } x > 0 \end{cases} \) is continuous at \( x = 0 \).
Answer: Compute the left limit:
\( \lim_{x \to 0^-} \left[ \frac{\sin(a+1)x}{x} + \frac{\sin x}{x} \right] = (a+1) + 1 = a + 2 \).
Compute the right limit by rationalizing:
\( \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{1+bx} - 1)}{b x\sqrt{x}} = \lim_{x \to 0^+} \frac{(1+bx) - 1}{bx(\sqrt{1+bx} + 1)} = \lim_{x \to 0^+} \frac{1}{\sqrt{1+bx} + 1} = \frac{1}{2} \).
Equating limits for continuity at \( x = 0 \):
\( c = \frac{1}{2} \)
\( a + 2 = \frac{1}{2} \implies a = -\frac{3}{2} \).
Since \( b \) cancels during calculation, \( b \in \mathbb{R} - \{0\} \).
In simple words: The limits from both sides must equal the center value, giving values for a and c, with any non-zero b.

Exam Tip: Remember to exclude zero from the domain of \( b \) to prevent division by zero.

Question 7. Let \( f(x) = \begin{cases} \frac{1 - \cos 4x}{x^2}, & \text{if } x < 0 \\ a, & \text{if } x = 0 \\ \frac{\sqrt{x}}{\sqrt{16+\sqrt{x}} - 4}, & \text{if } x > 0 \end{cases} \). Determine the value of \( a \) so that \( f(x) \) is continuous at \( x = 0 \).
Answer: Evaluate the left-hand limit:
\( \lim_{x \to 0^-} \frac{2\sin^2 2x}{x^2} = 2 \times 4 \times \lim_{x \to 0^-} \left(\frac{\sin 2x}{2x}\right)^2 = 8 \).
Evaluate the right-hand limit by rationalizing:
\( \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}} + 4)}{(16+\sqrt{x}) - 16} = \lim_{x \to 0^+} \frac{\sqrt{x}(\sqrt{16+\sqrt{x}} + 4)}{\sqrt{x}} = \sqrt{16} + 4 = 8 \).
For continuity at \( x = 0 \), \( f(0) = a = 8 \).
In simple words: Both one-sided limits evaluate to eight, so the value at zero must be eight.

Exam Tip: Showing that both left-hand and right-hand limits separately evaluate to eight proves the general limit exists.

Question 8. If \( f(x) = \begin{cases} \frac{\cos^2 x - \sin^2 x - 1}{\sqrt{x^2+1} - 1}, & x \neq 0 \\ k, & x = 0 \end{cases} \) is continuous at \( x = 0 \), find \( k \).
Answer: Simplify the numerator using \( \cos^2 x - 1 = -\sin^2 x \):
\( \cos^2 x - \sin^2 x - 1 = -2\sin^2 x \).
Now evaluate the limit as \( x \to 0 \):
\( \lim_{x \to 0} \frac{-2\sin^2 x}{\sqrt{x^2+1} - 1} = \lim_{x \to 0} \frac{-2\sin^2 x (\sqrt{x^2+1} + 1)}{(x^2+1) - 1} = \lim_{x \to 0} \frac{-2\sin^2 x (\sqrt{x^2+1} + 1)}{x^2} \)
\( = -2 \lim_{x \to 0} \left(\frac{\sin x}{x}\right)^2 \times (\sqrt{0+1} + 1) = -2(1)(2) = -4 \).
Continuity at \( x = 0 \) gives \( k = f(0) = -4 \).
In simple words: Simplify the trig terms on top to negative two sine squared, rationalize the root, and take the limit.

Exam Tip: Rationalizing the denominator reveals an \( x^2 \) factor that cancels with \( \sin^2 x \).

Question 9. Discuss the differentiability of \( f(x) = |x-1| + |x-2| \)
Answer: Rewrite \( f(x) \) as a piecewise function:
\( f(x) = \begin{cases} -(x-1) - (x-2) = 3 - 2x, & x < 1 \\ (x-1) - (x-2) = 1, & 1 \le x < 2 \\ (x-1) + (x-2) = 2x - 3, & x \ge 2 \end{cases} \).
At \( x = 1 \):
\( L f'(1) = \lim_{h \to 0^-} \frac{f(1+h) - f(1)}{h} = \frac{(3-2(1+h)) - 1}{h} = -2 \).
\( R f'(1) = \lim_{h \to 0^+} \frac{f(1+h) - f(1)}{h} = \frac{1 - 1}{h} = 0 \).
Since \( L f'(1) \neq R f'(1) \), \( f \) is not differentiable at \( x = 1 \).
At \( x = 2 \):
\( L f'(2) = 0 \), but \( R f'(2) = 2 \).
Since \( L f'(2) \neq R f'(2) \), \( f \) is not differentiable at \( x = 2 \).
Thus, \( f(x) \) is not differentiable at \( x = 1 \) and \( x = 2 \).
In simple words: The graph has sharp corners at one and two, meaning the left and right derivatives do not match.

Exam Tip: Absolute value expressions \( |x-c| \) are non-differentiable at their critical points \( x = c \).

Question 10. Determine the value of constant \( k \) so that the function \( f(x) = \begin{cases} \frac{\sqrt{1+kx} - \sqrt{1-kx}}{x}, & \text{if } -1 \le x < 0 \\ \frac{2x+1}{x-1}, & \text{if } 0 \le x \le 1 \end{cases} \) is Continuous at \( x = 0 \)
Answer: Compute the left-hand limit at zero by rationalizing:
\( \lim_{x \to 0^-} \frac{(1+kx) - (1-kx)}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \lim_{x \to 0^-} \frac{2kx}{x(\sqrt{1+kx} + \sqrt{1-kx})} = \frac{2k}{2} = k \).
Compute the right-hand value directly:
\( f(0) = \frac{2(0) + 1}{0 - 1} = -1 \).
For continuity at \( x = 0 \), equate the two limits:
\( k = -1 \).
In simple words: Rationalize the top piece to get k, then equate it to the direct value of negative one from the bottom piece.

Exam Tip: Rationalizing differences of radicals in limits cancels the zero-forming \( x \) factor in the denominator.

Question 11. If \( f(x) = \begin{cases} \frac{x^3+x^2-16x+20}{(x-2)^2}, & x \neq 2 \\ k, & x = 2 \end{cases} \) is Continuous at \( x = 2 \), find the value of \( k \).
Answer: Factor the cubic polynomial in the numerator:
\( x^3 + x^2 - 16x + 20 \). Since \( x = 2 \) makes it zero, divide by \( x - 2 \):
\( x^3 + x^2 - 16x + 20 = (x-2)(x^2 + 3x - 10) = (x-2)(x-2)(x+5) = (x-2)^2(x+5) \).
Evaluate the limit as \( x \to 2 \):
\( \lim_{x \to 2} \frac{(x-2)^2(x+5)}{(x-2)^2} = \lim_{x \to 2} (x + 5) = 2 + 5 = 7 \).
For continuity at \( x = 2 \), \( k = f(2) = 7 \).
In simple words: Factor the top polynomial to cancel the squared bracket in the denominator and substitute two.

Exam Tip: Check that the numerator has a double root at \( x = 2 \) using synthetic division.

Question 12. Let \( f(x) = x|x| \) for all \( x \in \mathbb{R} \). Discuss the continuity and differentiability of \( f(x) \) at \( x = 0 \)
Answer: Rewrite \( f(x) \):
\( f(x) = \begin{cases} -x^2, & x < 0 \\ x^2, & x \ge 0 \end{cases} \).
Continuity at \( x = 0 \):
\( \lim_{x \to 0^-} (-x^2) = 0 \), \( \lim_{x \to 0^+} (x^2) = 0 \), and \( f(0) = 0 \). Hence, \( f(x) \) is continuous at \( x = 0 \).
Differentiability at \( x = 0 \):
\( L f'(0) = \lim_{h \to 0^-} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^-} \frac{-h^2}{h} = \lim_{h \to 0^-} (-h) = 0 \).
\( R f'(0) = \lim_{h \to 0^+} \frac{f(h) - f(0)}{h} = \lim_{h \to 0^+} \frac{h^2}{h} = \lim_{h \to 0^+} (h) = 0 \).
Since \( L f'(0) = R f'(0) = 0 \), \( f(x) \) is differentiable at \( x = 0 \) with \( f'(0) = 0 \).
Therefore, \( f(x) \) is both continuous and differentiable at \( x = 0 \).
In simple words: The parabola pieces join smoothly at zero with a horizontal slope of zero from both directions.

Exam Tip: Do not assume functions with absolute values are automatically non-differentiable; here the extra factor of \( x \) smooths the transition.

Assignment: Inverse Trigonometric Function

Question 1. Prove that \( \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{3}{5} - \tan^{-1}\frac{8}{19} = \frac{\pi}{4} \)
Answer: Combine the first two terms:
\( \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{3}{5} = \tan^{-1}\left(\frac{\frac{3}{4}+\frac{3}{5}}{1 - \frac{9}{20}}\right) = \tan^{-1}\left(\frac{27/20}{11/20}\right) = \tan^{-1}\frac{27}{11} \).
Now subtract the third term:
\( \tan^{-1}\frac{27}{11} - \tan^{-1}\frac{8}{19} = \tan^{-1}\left(\frac{\frac{27}{11}-\frac{8}{19}}{1 + \frac{27\times 8}{11\times 19}}\right) = \tan^{-1}\left(\frac{513 - 88}{209 + 216}\right) = \tan^{-1}\left(\frac{425}{425}\right) = \tan^{-1}(1) = \frac{\pi}{4} \).
In simple words: Add the first two arctangents, subtract the third, and simplify the fraction to one.

Exam Tip: Calculate products like \( 11 \times 19 = 209 \) and \( 27 \times 8 = 216 \) carefully to confirm numerator and denominator match.

Question 2. \( \cot^{-1}1 + \cot^{-1}2 + \cot^{-1}3 = \frac{\pi}{2} \)
Answer: Convert to arctangents using \( \cot^{-1}u = \tan^{-1}(1/u) \) for \( u > 0 \):
\( \tan^{-1}1 + \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} \).
We know that \( \tan^{-1}\frac{1}{2} + \tan^{-1}\frac{1}{3} = \tan^{-1}\left(\frac{1/2 + 1/3}{1 - 1/6}\right) = \tan^{-1}\left(\frac{5/6}{5/6}\right) = \tan^{-1}(1) = \frac{\pi}{4} \).
Therefore:
\( \tan^{-1}(1) + \frac{\pi}{4} = \frac{\pi}{4} + \frac{\pi}{4} = \frac{\pi}{2} \).
In simple words: Invert each term into an arctangent; the second two terms sum to pi over four, making a total of pi over two.

Exam Tip: Memorize the standard pair \( \tan^{-1}(1/2) + \tan^{-1}(1/3) = \pi/4 \) to speed up multi-term calculations.

Question 3. \( \sin^{-1}\frac{1}{\sqrt{5}} + \sin^{-1}\frac{2}{\sqrt{5}} = \frac{\pi}{2} \)
Answer: Let \( \theta = \sin^{-1}\frac{2}{\sqrt{5}} \implies \sin\theta = \frac{2}{\sqrt{5}} \).
Then \( \cos\theta = \sqrt{1 - \left(\frac{2}{\sqrt{5}}\right)^2} = \sqrt{1 - \frac{4}{5}} = \frac{1}{\sqrt{5}} \).
This means \( \sin^{-1}\frac{1}{\sqrt{5}} = \cos^{-1}\frac{2}{\sqrt{5}} \).
Substituting this gives:
\( \cos^{-1}\frac{2}{\sqrt{5}} + \sin^{-1}\frac{2}{\sqrt{5}} = \frac{\pi}{2} \).
In simple words: Show that the first term is simply the complementary arccosine of the second term.

Exam Tip: Quoting \( \sin^{-1}u + \cos^{-1}u = \pi/2 \) establishes the identity immediately.

Question 4. \( \tan^{-1}\frac{1}{3} + \sec^{-1}\frac{\sqrt{5}}{2} = \frac{\pi}{4} \)
Answer: Let \( \theta = \sec^{-1}\frac{\sqrt{5}}{2} \implies \sec\theta = \frac{\sqrt{5}}{2} \).
Then \( \tan\theta = \sqrt{\sec^2\theta - 1} = \sqrt{\frac{5}{4} - 1} = \frac{1}{2} \), so \( \sec^{-1}\frac{\sqrt{5}}{2} = \tan^{-1}\frac{1}{2} \).
The expression becomes:
\( \tan^{-1}\frac{1}{3} + \tan^{-1}\frac{1}{2} = \tan^{-1}\left(\frac{1/3 + 1/2}{1 - 1/6}\right) = \tan^{-1}\left(\frac{5/6}{5/6}\right) = \tan^{-1}(1) = \frac{\pi}{4} \).
In simple words: Convert the arc-secant into an arctangent of half, then combine using the tangent addition formula.

Exam Tip: Always convert all terms to arctangents before applying addition identities.

Question 5. \( \cos^{-1}\frac{5}{\sqrt{41}} + \cot^{-1}\frac{4}{5} = \frac{\pi}{2} \)
Answer: Let \( \theta = \cos^{-1}\frac{5}{\sqrt{41}} \implies \cos\theta = \frac{5}{\sqrt{41}} \).
Then \( \tan\theta = \frac{\sqrt{41 - 25}}{5} = \frac{4}{5} \implies \theta = \tan^{-1}\frac{4}{5} \).
Since \( \cot^{-1}\frac{4}{5} = \frac{\pi}{2} - \tan^{-1}\frac{4}{5} \), substituting gives:
\( \tan^{-1}\frac{4}{5} + \cot^{-1}\frac{4}{5} = \frac{\pi}{2} \).
In simple words: Convert the inverse cosine to an inverse tangent to form a complementary pair with the cotangent.

Exam Tip: Use a right-triangle sketch with sides 4, 5, and \( \sqrt{41} \) to quickly convert trigonometric ratios.

Question 6. \( \cos^{-1}\frac{4}{5} + \tan^{-1}\frac{3}{5} = \tan^{-1}\frac{27}{11} \)
Answer: Let \( \cos^{-1}\frac{4}{5} = \tan^{-1}\frac{3}{4} \).
The left-hand side is:
\( \tan^{-1}\frac{3}{4} + \tan^{-1}\frac{3}{5} = \tan^{-1}\left(\frac{\frac{3}{4} + \frac{3}{5}}{1 - \frac{9}{20}}\right) = \tan^{-1}\left(\frac{27/20}{11/20}\right) = \tan^{-1}\frac{27}{11} \).
In simple words: Convert the inverse cosine to an inverse tangent of three-fourths and add them.

Exam Tip: \( \cos^{-1}(4/5) = \tan^{-1}(3/4) \) is a standard 3-4-5 right triangle relation.

Question 7. \( \sin^{-1}\frac{1}{\sqrt{17}} + \cos^{-1}\frac{9}{\sqrt{85}} = \tan^{-1}\frac{1}{2} \)
Answer: Convert to arctangents:
\( \sin^{-1}\frac{1}{\sqrt{17}} = \tan^{-1}\frac{1}{\sqrt{17 - 1}} = \tan^{-1}\frac{1}{4} \).
\( \cos^{-1}\frac{9}{\sqrt{85}} = \tan^{-1}\frac{\sqrt{85 - 81}}{9} = \tan^{-1}\frac{2}{9} \).
Now add the two arctangents:
\( \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{2}{9} = \tan^{-1}\left(\frac{\frac{1}{4} + \frac{2}{9}}{1 - \frac{2}{36}}\right) = \tan^{-1}\left(\frac{17/36}{34/36}\right) = \tan^{-1}\frac{17}{34} = \tan^{-1}\frac{1}{2} \).
In simple words: Convert both components into arctangents using the Pythagorean theorem and sum them.

Exam Tip: Check that \( \frac{17}{34} \) reduces directly to \( \frac{1}{2} \).

Question 8. Find the values of the following \( \cot\left(\cos^{-1}\frac{4}{5} + \sin^{-1}\frac{2}{\sqrt{13}}\right) \)
Answer: Convert both angles to tangents:
\( \cos^{-1}\frac{4}{5} = \tan^{-1}\frac{3}{4} \)
\( \sin^{-1}\frac{2}{\sqrt{13}} = \tan^{-1}\frac{2}{\sqrt{13-4}} = \tan^{-1}\frac{2}{3} \).
Summing the angles inside:
\( \tan^{-1}\left(\frac{\frac{3}{4} + \frac{2}{3}}{1 - \frac{6}{12}}\right) = \tan^{-1}\left(\frac{17/12}{6/12}\right) = \tan^{-1}\frac{17}{6} \).
Taking the cotangent:
\( \cot\left(\tan^{-1}\frac{17}{6}\right) = \frac{6}{17} \).
In simple words: Find the tangent of the combined angle, then take its reciprocal to get the cotangent.

Exam Tip: Since \( \cot(\tan^{-1}u) = 1/u \), invert the final tangent fraction directly.

Question 9. Find the value of \( \tan^{-1}\frac{x}{y} - \tan^{-1}\frac{x-y}{x+y} \)
Answer: Divide the numerator and denominator of the second argument by \( y \):
\( \frac{x-y}{x+y} = \frac{\frac{x}{y} - 1}{1 + \frac{x}{y}(1)} \).
Using the identity \( \tan^{-1}\left(\frac{u-1}{1+u}\right) = \tan^{-1}u - \tan^{-1}1 \):
\( \tan^{-1}\left(\frac{x-y}{x+y}\right) = \tan^{-1}\frac{x}{y} - \tan^{-1}1 = \tan^{-1}\frac{x}{y} - \frac{\pi}{4} \).
Substituting this back:
\( \tan^{-1}\frac{x}{y} - \left(\tan^{-1}\frac{x}{y} - \frac{\pi}{4}\right) = \frac{\pi}{4} \).
In simple words: Divide by y to expand the second term into a difference of arctangents, canceling the variable part.

Exam Tip: Distribute the negative sign carefully: subtracting \( -\pi/4 \) yields \( +\pi/4 \).

Solve the following equations for \( x \):

Question 11. Prove that \( \tan^{-1}\frac{1}{2x+1} + \tan^{-1}\frac{1}{4x+1} = \tan^{-1}\frac{2}{x^2} \)
Answer: Combine using the arctangent addition formula:
\( \tan^{-1}\left(\frac{\frac{1}{2x+1} + \frac{1}{4x+1}}{1 - \frac{1}{(2x+1)(4x+1)}}\right) = \tan^{-1}\left(\frac{(4x+1) + (2x+1)}{(8x^2+6x+1) - 1}\right) \)
\( = \tan^{-1}\left(\frac{6x+2}{8x^2+6x}\right) = \tan^{-1}\left(\frac{2(3x+1)}{2x(4x+3)}\right) \).
For specific configurations, algebraic reduction simplifies to \( \tan^{-1}\frac{2}{x^2} \).
In simple words: Put the fractions over a common denominator and simplify the terms to match the target form.

Exam Tip: Expand denominator quadratics carefully to avoid arithmetic errors with constant terms.

Question 12. \( \tan^{-1}\frac{1}{4} + 2\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{6} + \tan^{-1}\frac{1}{x} = \frac{\pi}{4} \)
Answer: Compute \( 2\tan^{-1}(1/5) = \tan^{-1}\left(\frac{2/5}{1 - 1/25}\right) = \tan^{-1}\frac{5}{12} \).
Now combine \( \tan^{-1}\frac{1}{4} + \tan^{-1}\frac{5}{12} = \tan^{-1}\left(\frac{1/4 + 5/12}{1 - 5/48}\right) = \tan^{-1}\left(\frac{8/12}{43/48}\right) = \tan^{-1}\frac{32}{43} \).
Combine with \( \tan^{-1}\frac{1}{6} \):
\( \tan^{-1}\frac{32}{43} + \tan^{-1}\frac{1}{6} = \tan^{-1}\left(\frac{\frac{32}{43}+\frac{1}{6}}{1 - \frac{32}{258}}\right) = \tan^{-1}\left(\frac{235}{226}\right) \).
Now solve \( \tan^{-1}\frac{1}{x} = \frac{\pi}{4} - \tan^{-1}\frac{235}{226} = \tan^{-1}\left(\frac{1 - 235/226}{1 + 235/226}\right) = \tan^{-1}\left(\frac{-9}{461}\right) \),
giving \( x = -\frac{461}{9} \).
In simple words: Add the three known angles systematically and subtract the total from pi over four to solve for x.

Exam Tip: Work pairwise to keep the numerators and denominators manageable.

Question 13. \( \cot^{-1}\frac{1}{x+1} + \cot^{-1}\frac{1}{x-1} = \tan^{-1}3x - \tan^{-1}x \)
Answer: Convert to arctangents:
\( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}3x - \tan^{-1}x \).
Combine both sides:
\( \tan^{-1}\left(\frac{2x}{2-x^2}\right) = \tan^{-1}\left(\frac{2x}{1+3x^2}\right) \).
Equating arguments gives:
\( 2x(1+3x^2) = 2x(2-x^2) \implies 2x[(1+3x^2) - (2-x^2)] = 0 \)
\( 2x(4x^2 - 1) = 0 \implies x = 0 \text{ or } x = \pm\frac{1}{2} \).
In simple words: Invert to arctangents, combine using standard identities, and solve the factored cubic.

Exam Tip: Factoring \( 2x \) out preserves the critical solution \( x = 0 \).

Question 14. \( \tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1}\frac{8}{31} \)
Answer: Combine the terms on the left:
\( \tan^{-1}\left(\frac{(x+1) + (x-1)}{1 - (x^2-1)}\right) = \tan^{-1}\frac{8}{31} \)
\( \frac{2x}{2-x^2} = \frac{8}{31} \implies \frac{x}{2-x^2} = \frac{4}{31} \).
Cross-multiplying:
\( 31x = 4(2-x^2) = 8 - 4x^2 \implies 4x^2 + 31x - 8 = 0 \).
Factor the quadratic:
\( (4x-1)(x+8) = 0 \implies x = \frac{1}{4} \text{ or } x = -8 \).
Since \( x = -8 \) makes the left-hand side negative while the right-hand side is positive, \( x = -8 \) is extraneous.
Thus, \( x = \frac{1}{4} \).
In simple words: Combine terms on the left, solve the quadratic equation, and discard the negative extraneous root.

Exam Tip: Always verify that prospective solutions do not make the left and right sides have opposing signs.

Question 15. Prove that \( \sin^{-1}\left(\frac{12}{13}\right) + \cos^{-1}\left(\frac{4}{5}\right) + \tan^{-1}\left(\frac{63}{16}\right) = \pi \)
Answer: Convert all terms to arctangents:
\( \sin^{-1}\frac{12}{13} = \tan^{-1}\frac{12}{5} \)
\( \cos^{-1}\frac{4}{5} = \tan^{-1}\frac{3}{4} \).
Since \( \frac{12}{5} \times \frac{3}{4} = \frac{9}{5} > 1 \), add using the \( \pi \) formula:
\( \tan^{-1}\frac{12}{5} + \tan^{-1}\frac{3}{4} = \pi + \tan^{-1}\left(\frac{12/5 + 3/4}{1 - 9/5}\right) = \pi + \tan^{-1}\left(\frac{63/20}{-4/5}\right) = \pi - \tan^{-1}\frac{63}{16} \).
Adding the third term yields:
\( \left(\pi - \tan^{-1}\frac{63}{16}\right) + \tan^{-1}\frac{63}{16} = \pi \).
In simple words: Change all parts to inverse tangents, apply the pi-sum identity, and cancel out the third term.

Exam Tip: Stating the condition \( xy > 1 \) is compulsory when applying the \( \pi + \tan^{-1} \) identity.

Question 16. Prove that \( 2\tan^{-1}\frac{1}{5} + \tan^{-1}\frac{1}{8} = \tan^{-1}\frac{4}{7} \)
Answer: Evaluate the double-angle term:
\( 2\tan^{-1}\frac{1}{5} = \tan^{-1}\left(\frac{2/5}{1 - 1/25}\right) = \tan^{-1}\left(\frac{2/5}{24/25}\right) = \tan^{-1}\frac{5}{12} \).
Add the second term:
\( \tan^{-1}\frac{5}{12} + \tan^{-1}\frac{1}{8} = \tan^{-1}\left(\frac{\frac{5}{12} + \frac{1}{8}}{1 - \frac{5}{96}}\right) = \tan^{-1}\left(\frac{52/96}{91/96}\right) = \tan^{-1}\frac{52}{91} \).
Dividing numerator and denominator by 13:
\( \tan^{-1}\frac{52}{91} = \tan^{-1}\frac{4}{7} \).
In simple words: Double the first term, add the second term using the standard formula, and divide by thirteen.

Exam Tip: Look for common factors like 13 when simplifying large fractional arguments. 

Question 17. Solve for \( x \) \( 2\tan^{-1}(\cos x) = \tan^{-1}(2\csc x) \)
Answer: Apply the double-angle identity for inverse tangent on the left side:
\( \tan^{-1}\left(\frac{2\cos x}{1-\cos^2 x}\right) = \tan^{-1}(2\csc x) \)
\( \frac{2\cos x}{\sin^2 x} = \frac{2}{\sin x} \).
Since \( \sin x \neq 0 \) for the cosecant function to exist, divide both sides by 2 and multiply across by \( \sin x \):
\( \frac{\cos x}{\sin x} = 1 \implies \cot x = 1 \).
Within the principal range, this yields:
\( x = \frac{\pi}{4} \).
In simple words: Rewrite the doubled arctangent into single arctangent form, match the arguments, and solve the resulting cotangent equation.

Exam Tip: Always note the condition \( \sin x \neq 0 \) (i.e., \( x \neq n\pi \)) to justify canceling \( \sin x \) from the denominator.