CBSE Class 11 Mathematics Limits And Derivatives Worksheet Set 01

Q1. Evaluate the following :-

(i) \(\lim_{x \to 1} \left( \frac{2}{x^2 - 1} + \frac{1}{x - 1} \right)\)
Answer:
Given expression = \(\lim_{x \to 1} \left( \frac{2}{(x-1)(x+1)} + \frac{1}{x-1} \right)\)
Taking LCM = \(\lim_{x \to 1} \left( \frac{2 + x + 1}{(x-1)(x+1)} \right)\)
= \(\lim_{x \to 1} \frac{x + 3}{(x-1)(x+1)}\)
Substituting \(x = 1\):
= \(\frac{1 + 3}{(1-1)(1+1)} = \frac{4}{0} = \infty\) (Not defined / approaches infinity)
In simple words: Simplify the fraction by finding a common denominator and then evaluate the limit.

Exam Tip: Always factorize denominators before combining rational terms.

(ii) \(\lim_{x \to 2} \frac{x^3 - 3x^2 + 4}{x^4 - 8x^2 + 16}\)
Answer:
Put \(x = 2\), we get \(\frac{0}{0}\) form.
Numerator: \(x^3 - 3x^2 + 4 = (x-2)(x^2 - x - 2) = (x-2)^2(x+1)\)
Denominator: \(x^4 - 8x^2 + 16 = (x^2 - 4)^2 = ((x-2)(x+2))^2 = (x-2)^2(x+2)^2\)
Now, \(\lim_{x \to 2} \frac{(x-2)^2(x+1)}{(x-2)^2(x+2)^2} = \lim_{x \to 2} \frac{x+1}{(x+2)^2}\)
= \(\frac{2+1}{(2+2)^2} = \frac{3}{16}\)
In simple words: Factorize both numerator and denominator completely to cancel out the common zero-making factor.

Exam Tip: Use synthetic division or factor theorem to factorize cubic polynomials.

(iii) \(\lim_{x \to \sqrt{2}} \frac{x^4 - 4}{x^2 + 3\sqrt{2}x - 8}\)
Answer:
Numerator: \(x^4 - 4 = (x^2 - 2)(x^2 + 2) = (x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)\)
Denominator: \(x^2 + 3\sqrt{2}x - 8 = x^2 + 4\sqrt{2}x - \sqrt{2}x - 8 = (x + 4\sqrt{2})(x - \sqrt{2})\)
Now, \(\lim_{x \to \sqrt{2}} \frac{(x - \sqrt{2})(x + \sqrt{2})(x^2 + 2)}{(x + 4\sqrt{2})(x - \sqrt{2})} = \lim_{x \to \sqrt{2}} \frac{(x + \sqrt{2})(x^2 + 2)}{x + 4\sqrt{2}}\)
Substitute \(x = \sqrt{2}\):
\(= \frac{(\sqrt{2} + \sqrt{2})((\sqrt{2})^2 + 2)}{\sqrt{2} + 4\sqrt{2}} = \frac{(2\sqrt{2})(2 + 2)}{5\sqrt{2}} = \frac{2\sqrt{2} \times 4}{5\sqrt{2}} = \frac{8}{5}\)
In simple words: Factorize using algebraic identities and cancel the term causing zero in the denominator.

Exam Tip: Splitting the middle term with radical coefficients requires careful grouping.

(iv) \(\lim_{x \to 3} (x^2 - 9)\left(\frac{1}{x+3} + \frac{1}{x-3}\right)\)
Answer:
Simplify the bracket term:
\(\frac{1}{x+3} + \frac{1}{x-3} = \frac{(x-3) + (x+3)}{(x+3)(x-3)} = \frac{2x}{x^2 - 9}\)
Now substitute back into the limit:
\(\lim_{x \to 3} (x^2 - 9) \cdot \frac{2x}{x^2 - 9} = \lim_{x \to 3} 2x\)
= \(2(3) = 6\)
In simple words: Combine the fractions inside the bracket first, which neatly cancels out the outer term.

Exam Tip: Simplify complex algebraic expressions before substituting limit values.

(v) \(\lim_{x \to 0} \frac{\sqrt{2 + x} - \sqrt{2}}{x}\)
Answer:
Rationalize the numerator:
\(= \lim_{x \to 0} \frac{(\sqrt{2+x} - \sqrt{2})(\sqrt{2+x} + \sqrt{2})}{x(\sqrt{2+x} + \sqrt{2})}\)
\(= \lim_{x \to 0} \frac{(2 + x) - 2}{x(\sqrt{2+x} + \sqrt{2})}\)
\(= \lim_{x \to 0} \frac{x}{x(\sqrt{2+x} + \sqrt{2})} = \lim_{x \to 0} \frac{1}{\sqrt{2+x} + \sqrt{2}}\)
Substitute \(x = 0\):
\(= \frac{1}{\sqrt{2+0} + \sqrt{2}} = \frac{1}{2\sqrt{2}}\)
In simple words: Multiply numerator and denominator by the conjugate of the numerator to remove the square root.

Exam Tip: Conjugate multiplication is the standard technique for limits involving square roots.

(vi) \(\lim_{x \to a} \frac{\sqrt{a^2 + x^2} - \sqrt{a^2 - x^2}}{x^2}\)
Answer:
Rationalize the numerator:
\(= \lim_{x \to a} \frac{(\sqrt{a^2+x^2} - \sqrt{a^2-x^2})(\sqrt{a^2+x^2} + \sqrt{a^2-x^2})}{x^2(\sqrt{a^2+x^2} + \sqrt{a^2-x^2})}\)
\(= \lim_{x \to a} \frac{(a^2 + x^2) - (a^2 - x^2)}{x^2(\sqrt{a^2+x^2} + \sqrt{a^2-x^2})}\)
\(= \lim_{x \to a} \frac{2x^2}{x^2(\sqrt{a^2+x^2} + \sqrt{a^2-x^2})} = \lim_{x \to a} \frac{2}{\sqrt{a^2+x^2} + \sqrt{a^2-x^2}}\)
Substitute \(x = a\):
\(= \frac{2}{\sqrt{a^2+a^2} + \sqrt{a^2-a^2}} = \frac{2}{\sqrt{2a^2} + 0} = \frac{2}{\sqrt{2}a} = \frac{\sqrt{2}}{a}\)
In simple words: Rationalize the numerator, simplify terms, and cancel \(x^2\).

Exam Tip: Be careful with signs when subtracting radical expressions.

(vii) \(\lim_{x \to 0} \frac{1 - \cos 2mx}{1 - \cos 2nx}\)
Answer:
Using trigonometric identity \(1 - \cos 2\theta = 2\sin^2\theta\):
\(= \lim_{x \to 0} \frac{2\sin^2(mx)}{2\sin^2(nx)} = \lim_{x \to 0} \left( \frac{\sin mx}{\sin nx} \right)^2\)
Multiply and divide by appropriate powers of \(x\):
\(= \lim_{x \to 0} \left( \frac{\frac{\sin mx}{mx} \cdot mx}{\frac{\sin nx}{nx} \cdot nx} \right)^2 = \left( \frac{m}{n} \right)^2 = \frac{m^2}{n^2}\)
In simple words: Convert cosine to sine using double angle formulas and use the standard limit \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\).

Exam Tip: Memorize standard trigonometric limit forms.

(viii) \(\lim_{x \to 0} \frac{1 - \cos 2x}{x^2}\)
Answer:
Using \(1 - \cos 2x = 2\sin^2 x\):
\(= \lim_{x \to 0} \frac{2\sin^2 x}{x^2} = 2 \lim_{x \to 0} \left( \frac{\sin x}{x} \right)^2\)
Since \(\lim_{x \to 0} \frac{\sin x}{x} = 1\):
\(= 2(1)^2 = 2\)
In simple words: Apply the half/double angle formula for cosine and use the fundamental limit.

Exam Tip: This is a standard limit result worth remembering for quick checks.

(ix) \(\lim_{x \to 0} \frac{\sin 2x + \sin 3x}{2x + \sin 3x}\)
Answer:
Divide numerator and denominator by \(x\):
\(= \lim_{x \to 0} \frac{\frac{\sin 2x}{x} + \frac{\sin 3x}{x}}{\frac{2x}{x} + \frac{\sin 3x}{x}}\)
\(= \lim_{x \to 0} \frac{2\left(\frac{\sin 2x}{2x}\right) + 3\left(\frac{\sin 3x}{3x}\right)}{2 + 3\left(\frac{\sin 3x}{3x}\right)}\)
Applying \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= \frac{2(1) + 3(1)}{2 + 3(1)} = \frac{5}{5} = 1\)
In simple words: Divide every term by \(x\) to form standard sine limits.

Exam Tip: Dividing by \(x\) is a powerful technique for trigonometric limits at zero.

(x) \(\lim_{x \to 0} \frac{\tan x - \sin x}{\sin^3 x}\)
Answer:
Rewrite \(\tan x = \frac{\sin x}{\cos x}\):
\(= \lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - \sin x}{\sin^3 x} = \lim_{x \to 0} \frac{\sin x \left(\frac{1}{\cos x} - 1\right)}{\sin^3 x}\)
\(= \lim_{x \to 0} \frac{1 - \cos x}{\cos x \sin^2 x} = \lim_{x \to 0} \frac{1 - \cos x}{\cos x (1 - \cos^2 x)}\)
\(= \lim_{x \to 0} \frac{1 - \cos x}{\cos x (1 - \cos x)(1 + \cos x)} = \lim_{x \to 0} \frac{1}{\cos x(1 + \cos x)}\)
Substitute \(x = 0\):
\(= \frac{1}{\cos 0(1 + \cos 0)} = \frac{1}{1(1 + 1)} = \frac{1}{2}\)
In simple words: Express tangent in terms of sine and cosine, factor out terms, and simplify using trigonometric identities.

Exam Tip: Convert all trigonometric functions to sine and cosine if simplification is unclear.

(xi) \(\lim_{x \to 0} \frac{\sec 4x - \sec 2x}{\sec 3x - \sec x}\)
Answer:
Rewrite secants as cosines:
\(= \lim_{x \to 0} \frac{\frac{1}{\cos 4x} - \frac{1}{\cos 2x}}{\frac{1}{\cos 3x} - \frac{1}{\cos x}} = \lim_{x \to 0} \frac{\frac{\cos 2x - \cos 4x}{\cos 4x \cos 2x}}{\frac{\cos x - \cos 3x}{\cos 3x \cos x}}\)
Using \(\cos A - \cos B = 2\sin\left(\frac{B+A}{2}\right)\sin\left(\frac{B-A}{2}\right)\):
\(\cos 2x - \cos 4x = 2\sin 3x \sin x\)
\(\cos x - \cos 3x = 2\sin 2x \sin x\)
Substitute into the limit:
\(= \lim_{x \to 0} \frac{\frac{2\sin 3x \sin x}{\cos 4x \cos 2x}}{\frac{2\sin 2x \sin x}{\cos 3x \cos x}} = \lim_{x \to 0} \left( \frac{\sin 3x}{\sin 2x} \cdot \frac{\cos 3x \cos x}{\cos 4x \cos 2x} \right)\)
\(= \left( \lim_{x \to 0} \frac{\frac{\sin 3x}{3x} \cdot 3}{\frac{\sin 2x}{2x} \cdot 2} \right) \cdot \left( \frac{\cos 0 \cos 0}{\cos 0 \cos 0} \right) = \frac{3}{2} \cdot 1 = \frac{3}{2}\)
In simple words: Convert to cosine, apply sum-to-product formulas, and use standard sine limits.

Exam Tip: Sum-to-product trigonometric formulas turn difficult subtractions into manageable products.

(xii) \(\lim_{x \to a} \frac{(x + 2)^{5/3} - (a + 2)^{5/3}}{x - a}\)
Answer:
Using standard formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\):
Let \(y = x + 2\) and \(b = a + 2\). As \(x \to a\), \(y \to b\).
\(= \lim_{y \to b} \frac{y^{5/3} - b^{5/3}}{y - b}\)
\(= \frac{5}{3} (a + 2)^{\frac{5}{3} - 1} = \frac{5}{3} (a + 2)^{2/3}\)
In simple words: Substitute variables to match the standard algebraic limit formula for powers.

Exam Tip: Substitution helps recognize standard formulas instantly.

(xiii) \(\lim_{x \to 2} \frac{x^5 - 32}{x^2 - 4}\)
Answer:
Rewrite \(32 = 2^5\) and \(4 = 2^2\):
\(= \lim_{x \to 2} \frac{x^5 - 2^5}{x^2 - 2^2} = \lim_{x \to 2} \left( \frac{\frac{x^5 - 2^5}{x - 2}}{\frac{x^2 - 2^2}{x - 2}} \right)\)
Applying \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\):
\(= \frac{5(2^{5-1})}{2(2^{2-1})} = \frac{5(16)}{2(2)} = \frac{80}{4} = 20\)
In simple words: Divide both numerator and denominator by \(x - 2\) and apply the standard power limit theorem.

Exam Tip: Splitting fractions by \(x - a\) is the quickest way to evaluate \(\frac{x^n - a^n}{x^m - b^m}\) limits.

(xiv) \(\lim_{x \to 1} \frac{x^{16} - 1}{x^5 - 1}\)
Answer:
Divide numerator and denominator by \(x - 1\):
\(= \lim_{x \to 1} \frac{\frac{x^{16} - 1^16}{x - 1}}{\frac{x^5 - 1^5}{x - 1}}\)
Applying \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\):
\(= \frac{16(1)^{15}}{5(1)^4} = \frac{16}{5}\)
In simple words: Divide by \(x - 1\) above and below to apply the standard formula.

Exam Tip: Powers of 1 are always 1, simplifying coefficient calculations.

(xv) \(\lim_{x \to 0} \frac{e^{5x} - 1}{x}\)
Answer:
Using the standard exponential limit \(\lim_{\theta \to 0} \frac{e^\theta - 1}{\theta} = 1\):
\(= \lim_{x \to 0} \frac{e^{5x} - 1}{5x} \times 5 = 1 \times 5 = 5\)
In simple words: Adjust the denominator to match the exponent and apply the standard exponential limit.

Exam Tip: Remember \(\lim_{x \to 0} \frac{e^{kx} - 1}{x} = k\).

(xvi) \(\lim_{x \to 0} \frac{e^{5x} - 1}{e^{8x} - 1}\)
Answer:
Divide numerator and denominator by \(x\):
\(= \lim_{x \to 0} \frac{\frac{e^{5x} - 1}{x}}{\frac{e^{8x} - 1}{x}} = \lim_{x \to 0} \frac{5 \left(\frac{e^{5x} - 1}{5x}\right)}{8 \left(\frac{e^{8x} - 1}{8x}\right)}\)
\(= \frac{5(1)}{8(1)} = \frac{5}{8}\)
In simple words: Divide both parts by \(x\) and use the standard exponential limit.

Exam Tip: The answer is simply the ratio of the coefficients of \(x\) in the exponents.

(xvii) \(\lim_{x \to 0} \frac{\log(1 + 5x)}{x}\)
Answer:
Using the standard logarithmic limit \(\lim_{\theta \to 0} \frac{\log(1+\theta)}{\theta} = 1\):
\(= \lim_{x \to 0} 5 \cdot \frac{\log(1 + 5x)}{5x} = 5 \times 1 = 5\)
In simple words: Multiply and divide by 5 to match the logarithmic standard limit form.

Exam Tip: Remember \(\lim_{x \to 0} \frac{\log(1+kx)}{x} = k\).

(xviii) \(\lim_{x \to 0} \frac{\log(1 + 5x)}{\log(1 + 10x)}\)
Answer:
Divide numerator and denominator by \(x\):
\(= \lim_{x \to 0} \frac{\frac{\log(1+5x)}{x}}{\frac{\log(1+10x)}{x}} = \lim_{x \to 0} \frac{5 \left(\frac{\log(1+5x)}{5x}\right)}{10 \left(\frac{\log(1+10x)}{10x}\right)}\)
\(= \frac{5(1)}{10(1)} = \frac{5}{10} = \frac{1}{2}\)
In simple words: Divide by \(x\) and apply the standard log limit.

Exam Tip: The limit equals the ratio of the linear coefficients inside the log arguments.

(xix) \(\lim_{x \to 3} \frac{e^x - e^3}{x - 3}\)
Answer:
Using the standard derivative definition / limit formula \(\lim_{x \to a} \frac{e^x - e^a}{x - a} = e^a\):
Here \(a = 3\), so the limit is \(e^3\).
In simple words: This expression is precisely the derivative of \(e^x\) at \(x = 3\).

Exam Tip: Recognize derivative definitions disguised as limits.

(xx) \(\lim_{x \to 0} \frac{1 - \cos 2x}{\sin^2 x}\)
Answer:
Using \(1 - \cos 2x = 2\sin^2 x\):
\(= \lim_{x \to 0} \frac{2\sin^2 x}{\sin^2 x} = \lim_{x \to 0} 2 = 2\)
In simple words: Substitute the double angle formula for cosine, and \(\sin^2 x\) cancels out directly.

Exam Tip: Always look for direct trigonometric cancellation before applying complex theorems.

Q2. If \(\lim_{x \to a} \frac{x^3 - a^3}{x - a} = \lim_{x \to 1} \frac{x^4 - 1}{x - 1}\), then find the value of a.
Answer:
Apply the standard formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\) on both sides:
LHS: \(3 a^{3-1} = 3a^2\)
RHS: \(4 (1)^{4-1} = 4(1)^3 = 4\)
Equating LHS and RHS:
\(3a^2 = 4\)
\(a^2 = \frac{4}{3}\)
\(a = \pm \frac{2}{\sqrt{3}}\)
In simple words: Evaluate both sides using the standard algebraic limit formula and solve for \(a\).

Exam Tip: Do not forget both positive and negative square roots.

Q3. If \(\lim_{x \to 2} \frac{x^n - 2^n}{x - 2} = 80\) and if n is a +ve integer, find n.
Answer:
Using the standard formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\):
\(n (2)^{n-1} = 80\)
Test integer values for \(n\):
If \(n = 5\): \(5 \cdot (2)^{5-1} = 5 \cdot 2^4 = 5 \times 16 = 80\).
Thus, \(n = 5\).
In simple words: Apply the power limit formula and test integer values for \(n\) that satisfy the equation.

Exam Tip: Prime factorization of the constant on the RHS often reveals the exponent quickly.

Q4. If \(f(x) = \begin{cases} 5x - 4, & \text{if } x < 1 \\ 4x^3 - 3x, & \text{if } x > 1 \end{cases}\) Find (i) \(\lim_{x \to 1} f(x)\) (ii) \(\lim_{x \to 4} f(x)\) (iii) \(\lim_{x \to 3} f(x)\) (iv) \(\lim_{x \to 2} f(x)\)
Answer:
(i) For \(\lim_{x \to 1} f(x)\):
Left-hand limit (LHL as \(x \to 1^-\)): \(\lim_{x \to 1} (5x - 4) = 5(1) - 4 = 1\)
Right-hand limit (RHL as \(x \to 1^+\)): \(\lim_{x \to 1} (4x^3 - 3x) = 4(1)^3 - 3(1) = 1\)
Since LHL = RHL = 1, \(\lim_{x \to 1} f(x) = 1\).

(ii) For \(\lim_{x \to 4} f(x)\):
Since \(4 > 1\), use function \(f(x) = 4x^3 - 3x\):
\(= 4(4)^3 - 3(4) = 4(64) - 12 = 256 - 12 = 244\)

(iii) For \(\lim_{x \to 3} f(x)\):
Since \(3 > 1\), use function \(f(x) = 4x^3 - 3x\):
\(= 4(3)^3 - 3(3) = 4(27) - 9 = 108 - 9 = 99\)

(iv) For \(\lim_{x \to 2} f(x)\):
Since \(2 > 1\), use function \(f(x) = 4x^3 - 3x\):
\(= 4(2)^3 - 3(2) = 4(8) - 6 = 32 - 6 = 26\)
In simple words: Check which branch of the piecewise function applies for the given point of limit.

Exam Tip: For points where the function definition changes (like \(x = 1\)), always check both LHL and RHL.

Q5. If \(f(x) = \begin{cases} 4, & \text{if } x \ge 3 \\ Kx + 1, & \text{if } x < 3 \end{cases}\) and \(\lim_{x \to 3} f(x)\) exists, find K.
Answer:
For the limit to exist at \(x = 3\), Left-Hand Limit must equal Right-Hand Limit:
\(\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x)\)
LHL as \(x \to 3^-\): \(K(3) + 1 = 3K + 1\)
RHL as \(x \to 3^+\): \(4\)
Equating them:
\(3K + 1 = 4\)
\(3K = 3 \implies K = 1\)
In simple words: Equate the left-hand limit and right-hand limit at the boundary point 3 to solve for \(K\).

Exam Tip: Limit existence at a point requires LHL = RHL.

Q6. If \(f(x) = \begin{cases} \cos x, & \text{if } x \ge 0 \\ x + k, & \text{if } x < 0 \end{cases}\) and \(\lim_{x \to 0} f(x)\) exists, find K.
Answer:
For the limit to exist at \(x = 0\):
\(\lim_{x \to 0^-} f(x) = \lim_{x \to 0^+} f(x)\)
LHL as \(x \to 0^-\): \(0 + k = k\)
RHL as \(x \to 0^+\): \(\cos 0 = 1\)
Equating both:
\(k = 1\)
In simple words: Set the limit from the left equal to the limit from the right at \(x = 0\).

Exam Tip: Remember \(\cos 0 = 1\).

Q7. If \(f(x) = \begin{cases} (x - 1)K, & \text{if } x \ge 3 \\ x + 1, & \text{if } x < 3 \end{cases}\) and \(\lim_{x \to 3} f(x)\) exists, then find the value of K.
Answer:
For the limit to exist at \(x = 3\):
\(\lim_{x \to 3^-} f(x) = \lim_{x \to 3^+} f(x)\)
LHL as \(x \to 3^-\): \(3 + 1 = 4\)
RHL as \(x \to 3^+\): \((3 - 1)K = 2K\)
Equating both:
\(2K = 4 \implies K = 2\)
In simple words: Equate the expressions obtained from both sides of 3.

Exam Tip: Carefully substitute the boundary value into each branch.

Q8. Find the derivative of the following functions from first principle / delta Method / ab-initio method.
Answer:
The first principle formula for derivative is:
\(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\)

(i) \(f(x) = \sin x\)
\(f'(x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}\)
Using sum-to-product formula \(\sin A - \sin B = 2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\):
\(= \lim_{h \to 0} \frac{2\cos\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}\)
\(= \lim_{h \to 0} \cos\left(x + \frac{h}{2}\right) \cdot \lim_{h \to 0} \frac{\sin(h/2)}{h/2}\)
\(= \cos(x + 0) \cdot 1 = \cos x\)

(ii) \(f(x) = \sqrt{x}\)
\(f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h}\)
Rationalize the numerator:
\(= \lim_{h \to 0} \frac{(\sqrt{x+h} - \sqrt{x})(\sqrt{x+h} + \sqrt{x})}{h(\sqrt{x+h} + \sqrt{x})} = \lim_{h \to 0} \frac{x+h - x}{h(\sqrt{x+h} + \sqrt{x})}\)
\(= \lim_{h \to 0} \frac{1}{\sqrt{x+h} + \sqrt{x}} = \frac{1}{2\sqrt{x}}\)

(iii) \(f(x) = \sin(x + a)\)
\(f'(x) = \lim_{h \to 0} \frac{\sin(x+h+a) - \sin(x+a)}{h}\)
Using factor formula:
\(= \lim_{h \to 0} \frac{2\cos\left(x + a + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h} = \cos(x + a) \cdot 1 = \cos(x + a)\)

(iv) \(f(x) = \cos x\)
\(f'(x) = \lim_{h \to 0} \frac{\cos(x+h) - \cos x}{h}\)
Using \(\cos A - \cos B = -2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\):
\(= \lim_{h \to 0} \frac{-2\sin\left(x + \frac{h}{2}\right)\sin(h/2)}{h} = -\sin x \cdot 1 = -\sin x\)

(v) \(f(x) = \tan(2x + 1)\)
\(f'(x) = \lim_{h \to 0} \frac{\tan(2(x+h)+1) - \tan(2x+1)}{h}\)
Convert to sine and cosine:
\(= \lim_{h \to 0} \frac{\frac{\sin(2x+2h+1)}{\cos(2x+2h+1)} - \frac{\sin(2x+1)}{\cos(2x+1)}}{h}\)
\(= \lim_{h \to 0} \frac{\sin(2x+2h+1)\cos(2x+1) - \cos(2x+2h+1)\sin(2x+1)}{h \cos(2x+2h+1)\cos(2x+1)}\)
Using \(\sin(A-B)\):
\(= \lim_{h \to 0} \frac{\sin(2h)}{h \cos(2x+2h+1)\cos(2x+1)} = \frac{2}{\cos^2(2x+1)} = 2\sec^2(2x+1)\)

(vi) \(f(x) = \frac{x+1}{x-1}\)
\(f'(x) = \lim_{h \to 0} \frac{\frac{x+h+1}{x+h-1} - \frac{x+1}{x-1}}{h}\)
\(= \lim_{x \to 0} \frac{(x+h+1)(x-1) - (x+1)(x+h-1)}{h(x+h-1)(x-1)}\)
Expanding numerator: \((x^2 - x + hx - h + x - 1) - (x^2 + hx - x + x + h - 1) = -2h\)
\(= \lim_{h \to 0} \frac{-2h}{h(x+h-1)(x-1)} = \frac{-2}{(x-1)^2}\)

(vii) \(f(x) = \sin(ax + b)\)
\(f'(x) = \lim_{h \to 0} \frac{\sin(a(x+h)+b) - \sin(ax+b)}{h}\)
Using sum-to-product formula:
\(= \lim_{h \to 0} \frac{2\cos\left(ax + b + \frac{ah}{2}\right)\sin\left(\frac{ah}{2}\right)}{h}\)
Multiply and divide by \(a / 2\):
\(= \cos(ax + b) \cdot a = a\cos(ax + b)\)

(ix) \(f(x) = (ax + b)^2\)
\(f'(x) = \lim_{h \to 0} \frac{(a(x+h)+b)^2 - (ax+b)^2}{h}\)
Using \(A^2 - B^2 = (A-B)(A+B)\):
\(= \lim_{h \to 0} \frac{[ (ax+ah+b) - (ax+b) ][ (ax+ah+b) + (ax+b) ]}{h}\)
\(= \lim_{h \to 0} \frac{ah(2ax + ah + 2b)}{h} = \lim_{h \to 0} a(2ax + ah + 2b) = 2a(ax + b)\)
In simple words: Apply the limit definition of derivative \(\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\) combined with algebraic or trigonometric expansions.

Exam Tip: Practice first-principle derivations for standard functions thoroughly as they frequently appear in long-answer questions.

Q9. If \(y = x^5\) show that \(x \frac{dy}{dx} - 5y = 0\).
Answer:
Given \(y = x^5\).
Differentiating w.r.t. \(x\):
\(\frac{dy}{dx} = 5x^4\)
Multiply both sides by \(x\):
\(x \frac{dy}{dx} = x(5x^4) = 5x^5\)
Since \(y = x^5\):
\(x \frac{dy}{dx} = 5y \implies x \frac{dy}{dx} - 5y = 0\). Hence proved.
In simple words: Find the derivative and substitute it into the required differential relation.

Exam Tip: Substitute the original function value to establish the final identity.

Q10. If \(y = \left(\sin \frac{x}{2} + \cos \frac{x}{2}\right)^2\), find \(\frac{dy}{dx}\) at \(x = \frac{\pi}{6}\).
Answer:
Expand the function:
\(y = \sin^2\left(\frac{x}{2}\right) + \cos^2\left(\frac{x}{2}\right) + 2\sin\left(\frac{x}{2}\right)\cos\left(\frac{x}{2}\right)\)
Using \(\sin^2\theta + \cos^2\theta = 1\) and \(2\sin\theta\cos\theta = \sin 2\theta\):
\(y = 1 + \sin\left(2 \cdot \frac{x}{2}\right) = 1 + \sin x\)
Differentiating w.r.t. \(x\):
\(\frac{dy}{dx} = \cos x\)
At \(x = \frac{\pi}{6}\):
\(\frac{dy}{dx} = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\)
In simple words: Simplify the squared trigonometric expression using identities before differentiating.

Exam Tip: Trigonometric simplification before differentiation saves considerable time and prevents errors.

Q11. Differentiate w.r. to x
Answer:

(i) \(\frac{x^2 + 3}{x^2 - 5}\)
Using quotient rule \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\):
\(\frac{d}{dx}\left(\frac{x^2+3}{x^2-5}\right) = \frac{2x(x^2 - 5) - (x^2 + 3)(2x)}{(x^2 - 5)^2}\)
\(= \frac{2x^3 - 10x - 2x^3 - 6x}{(x^2 - 5)^2} = \frac{-16x}{(x^2 - 5)^2}\)

(ii) \(\frac{1 + \tan x}{1 - \tan x}\)
Note that \(\frac{1 + \tan x}{1 - \tan x} = \tan\left(\frac{\pi}{4} + x\right)\)
Derivative = \(\sec^2\left(\frac{\pi}{4} + x\right)\)
Alternatively using quotient rule:
\(= \frac{\sec^2 x(1 - \tan x) - (1 + \tan x)(-\sec^2 x)}{(1 - \tan x)^2} = \frac{2\sec^2 x}{(1 - \tan x)^2}\)

(iii) \(\frac{\sec x + \tan x}{\sec x - \tan x}\)
Simplify the fraction:
\(\frac{\frac{1}{\sin x} + \dots}{\dots} \text{ or multiply numerator and denominator by } (\sec x + \tan x)\):
\(= (\sec x + \tan x)^2\)
Derivative = \(2(\sec x + \tan x)(\sec x \tan x + \sec^2 x) = 2\sec x(\sec x + \tan x)^2\)

(iv) \(\frac{x + \cos x}{\tan x}\)
Using quotient rule:
\(= \frac{(1 - \sin x)\tan x - (x + \cos x)\sec^2 x}{\tan^2 x}\)

(v) \(x^4 - 2\sin x + 3\cos x\)
Derivative = \(4x^3 - 2\cos x - 3\sin x\)

(vi) \(\left(x + \frac{1}{x}\right)\left(x + \frac{1}{x}\right)\) (Assuming product or standard expansion)
Let \(y = \left(x + \frac{1}{x}\right)^2 = x^2 + 2 + \frac{1}{x^2}\)
Derivative = \(2x - 2x^{-3} = 2x - \frac{2}{x^3}\)

(viii) \((x + \cos x)(x - \tan x)\)
Using product rule \((uv)' = u'v + uv'\):
\(= (1 - \sin x)(x - \tan x) + (x + \cos x)(1 - \sec^2 x)\)

(viii [second]) \(\frac{1 + x}{1 + x}\) (Assuming \(\frac{1+\sqrt{x}}{1-x}\) or similar; let us evaluate \(\frac{1+x}{1+x} = 1\), derivative is 0. Assuming OCR typo for \(\frac{1+\sqrt{x}}{1-\sqrt{x}}\)):
Let \(y = \frac{1+\sqrt{x}}{1-\sqrt{x}}\)
Quotient rule: \(\frac{\frac{1}{2\sqrt{x}}(1-\sqrt{x}) - (1+\sqrt{x})\left(-\frac{1}{2\sqrt{x}}\right)}{(1-\sqrt{x})^2} = \frac{1}{\sqrt{x}(1-\sqrt{x})^2}\)

(ix) \(\frac{a + x}{a - x}\)
Using quotient rule:
\(= \frac{(0 + 1)(a - x) - (a + x)(0 - 1)}{(a - x)^2} = \frac{a - x + a + x}{(a - x)^2} = \frac{2a}{(a - x)^2}\)

(x) \(\frac{x^2(1 + \cos x)}{\sin x}\)
Using quotient rule or product rule after splitting:
Let \(u = x^2(1+\cos x)\), \(v = \sin x\)
\(u' = 2x(1+\cos x) + x^2(-\sin x)\)
\(v' = \cos x\)
\(\frac{dy}{dx} = \frac{[2x(1+\cos x) - x^2\sin x]\sin x - x^2(1+\cos x)\cos x}{\sin^2 x}\)

(xi) \((x^2 - 3x + 2)(x + 2)\)
Expand first: \(x^3 + 2x^2 - 3x^2 - 6x + 2x + 4 = x^3 - x^2 - 4x + 4\)
Derivative = \(3x^2 - 2x - 4\)
In simple words: Apply appropriate derivative rules: product rule, quotient rule, or chain rule.

Exam Tip: Expanding polynomials before differentiating often saves time over the product rule.

Q12. If \(y = \frac{x}{x + 4}\), prove that \(x \frac{dy}{dx} = y(1 - y)\).
Answer:
Given \(y = \frac{x}{x + 4}\).
Find \(\frac{dy}{dx}\) using quotient rule:
\(\frac{dy}{dx} = \frac{1(x + 4) - x(1)}{(x + 4)^2} = \frac{4}{(x + 4)^2}\)
Now consider LHS: \(x \frac{dy}{dx} = x \cdot \frac{4}{(x + 4)^2} = \frac{4x}{(x + 4)^2}\)
Now consider RHS: \(y(1 - y) = \frac{x}{x + 4} \left(1 - \frac{x}{x + 4}\right) = \frac{x}{x + 4} \left(\frac{x + 4 - x}{x + 4}\right) = \frac{x}{x + 4} \cdot \frac{4}{x + 4} = \frac{4x}{(x + 4)^2}\)
Since LHS = RHS, \(x \frac{dy}{dx} = y(1 - y)\). Hence proved.
In simple words: Calculate the derivative and independently simplify both sides of the identity to show they match.

Exam Tip: Working on LHS and RHS separately is a great strategy for prove-that derivative questions.

Q13. If \(y = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \dots\), show that \(\frac{dy}{dx} = y\).
Answer:
Note that the given series is the expansion of \(e^x\), so \(y = e^x\).
Differentiating w.r.t. \(x\):
\(\frac{dy}{dx} = \frac{d}{dx}(e^x) = e^x = y\). Hence proved.
Alternatively, differentiate term by term:
\(\frac{dy}{dx} = 0 + \frac{1}{1!} + \frac{2x}{2!} + \frac{3x^2}{3!} + \dots\)
\(= 1 + \frac{x}{1!} + \frac{x^2}{2!} + \dots = y\).
In simple words: Recognize the series as \(e^x\) or differentiate term-by-term to recover the exact same series.

Exam Tip: Memorize standard Maclaurin series expansions like \(e^x\), \(\sin x\), and \(\cos x\).

Q14. If \(y = \sqrt{\frac{1 - \cos 2x}{1 + \cos 2x}}\), find \(\frac{dy}{dx}\).
Answer:
Using trigonometric identities \(1 - \cos 2x = 2\sin^2 x\) and \(1 + \cos 2x = 2\cos^2 x\):
\(y = \sqrt{\frac{2\sin^2 x}{2\cos^2 x}} = \sqrt{\tan^2 x} = \tan x\)
Differentiating w.r.t. \(x\):
\(\frac{dy}{dx} = \sec^2 x\)
In simple words: Simplify the radical expression using double angle formulas to get \(\tan x\), then differentiate.

Exam Tip: Simplifying inside the square root before differentiation avoids complex chain rules.

Q15. Differentiate \(\frac{4x - 1}{(5 - 2x)^3}\) and then find its value at \(x = 2\).
Answer:
Let \(y = \frac{4x - 1}{(5 - 2x)^3}\).
Using quotient rule:
\(\frac{dy}{dx} = \frac{4(5 - 2x)^3 - (4x - 1) \cdot 3(5 - 2x)^2(-2)}{(5 - 2x)^6}\)
Factor out \((5 - 2x)^2\) from the numerator:
\(= \frac{(5 - 2x)^2 [ 4(5 - 2x) + 6(4x - 1) ]}{(5 - 2x)^6}\)
\(= \frac{20 - 8x + 24x - 6}{(5 - 2x)^4} = \frac{16x + 14}{(5 - 2x)^4}\)
Now, evaluate at \(x = 2\):
\(\left.\frac{dy}{dx}\right|_{x=2} = \frac{16(2) + 14}{(5 - 2(2))^4} = \frac{32 + 14}{(5 - 4)^4} = \frac{46}{1^4} = 46\)
In simple words: Apply the quotient rule, simplify the algebraic expression in the numerator, and substitute \(x = 2\).

Exam Tip: Factoring out common terms in numerator derivatives simplifies algebraic substitution.

Q16. If \(ax^2 + 2hxy + by^2 + 2gx + 2fy + C = 0\), find \(\frac{dy}{dx}\).
Answer:
Differentiating implicitly w.r.t. \(x\):
\(\frac{d}{dx}(ax^2) + \frac{d}{dx}(2hxy) + \frac{d}{dx}(by^2) + \frac{d}{dx}(2gx) + \frac{d}{dx}(2fy) + \frac{d}{dx}(C) = 0\)
\(2ax + 2h\left(y + x \frac{dy}{dx}\right) + 2by \frac{dy}{dx} + 2g + 2f \frac{dy}{dx} = 0\)
Group terms containing \(\frac{dy}{dx}\):
\(2\left(hx + by + f\right) \frac{dy}{dx} = -2ax - 2hy - 2g\)
\(\frac{dy}{dx} = -\frac{ax + hy + g}{hx + by + f}\)
In simple words: Differentiate every term implicitly with respect to \(x\), treating \(y\) as a function of \(x\), then solve for \(\frac{dy}{dx}\).

Exam Tip: Remember to use the product rule for the mixed term \(xy\): \(\frac{d}{dx}(xy) = y + x\frac{dy}{dx}\).

Q17. Find the derivative of \(f(x) = (1 - x)(1 + x)(1 + x^2)(1 + x^4)(1 + x^8)\) & hence find \(f'(1)\).
Answer:
Simplify the expression by combining terms using difference of squares \((a-b)(a+b) = a^2 - b^2\):
\(f(x) = [(1 - x)(1 + x)](1 + x^2)(1 + x^4)(1 + x^8)\)
\(f(x) = (1 - x^2)(1 + x^2)(1 + x^4)(1 + x^8)\)
\(f(x) = (1 - x^4)(1 + x^4)(1 + x^8)\)
\(f(x) = (1 - x^8)(1 + x^8) = 1 - x^{16}\)
Now, differentiate w.r.t. \(x\):
\(f'(x) = \frac{d}{dx}(1 - x^{16}) = -16x^{15}\)
Now find \(f'(1)\):
\(f'(1) = -16(1)^{15} = -16\)
In simple words: Use algebraic identity repeatedly to collapse the product into a simple binomial before differentiating.

Exam Tip: Spotting algebraic patterns saves you from applying the product rule 5 times.