CBSE Class 11 Mathematics Probability Assignment Set 02

Q1. Evaluate each of the following limits:

(i) \(\lim_{x \to 0} \frac{\sqrt{1+x} - \sqrt{1-x}}{x}\)
Answer: Rationalize the numerator by multiplying and dividing by \((\sqrt{1+x} + \sqrt{1-x})\):
\(= \lim_{x \to 0} \frac{(\sqrt{1+x} - \sqrt{1-x})(\sqrt{1+x} + \sqrt{1-x})}{x(\sqrt{1+x} + \sqrt{1-x})}\)
\(= \lim_{x \to 0} \frac{(1+x) - (1-x)}{x(\sqrt{1+x} + \sqrt{1-x})}\)
\(= \lim_{x \to 0} \frac{2x}{x(\sqrt{1+x} + \sqrt{1-x})}\)
\(= \lim_{x \to 0} \frac{2}{\sqrt{1+x} + \sqrt{1-x}}\)
\(= \frac{2}{\sqrt{1+0} + \sqrt{1-0}} = \frac{2}{1 + 1} = 1\)
In simple words: Multiply top and bottom by the conjugate to remove the square roots in the numerator.

Exam Tip: Rationalization is the standard technique when encountering limits with square roots yielding the 0/0 indeterminate form.

(ii) \(\lim_{x \to \frac{\pi}{2}} \frac{1 + \cos 2x}{(x - \frac{\pi}{2})^2}\)
Answer: Put \(x - \frac{\pi}{2} = t\), so as \(x \to \frac{\pi}{2}\), \(t \to 0\). Then \(x = t + \frac{\pi}{2}\):
\(\cos 2x = \cos 2\left(t + \frac{\pi}{2}\right) = \cos(2t + \pi) = -\cos 2t\)
Substitute into the limit:
\(\lim_{t \to 0} \frac{1 - \cos 2t}{t^2}\)
Since \(1 - \cos 2t = 2 \sin^2 t\):
\(= \lim_{t \to 0} \frac{2 \sin^2 t}{t^2} = 2 \left( \lim_{t \to 0} \frac{\sin t}{t} \right)^2 = 2(1)^2 = 2\) (Wait, reviewing standard substitution: using standard formula gives Ans. 1/2 as given in problem, let us re-evaluate via L'Hopital's rule or standard limits).
Actually, \(1 + \cos 2x = 2 \cos^2 x\). Let \(x - \frac{\pi}{2} = t\), then \(\cos x = \cos\left(t + \frac{\pi}{2}\right) = -\sin t\).
So \(1 + \cos 2x = 2(-\sin t)^2 = 2 \sin^2 t\). Wait, \(1 + \cos 2x = 2 \cos^2 x = 2(-\sin t)^2 = 2 \sin^2 t\).
Thus \(\lim_{t \to 0} \frac{2 \sin^2 t}{t^2} = 2\). Wait, the source answer is 1/2. Let us re-verify: \(\lim_{x \to \frac{\pi}{2}} \frac{1 + \cos 2x}{(x - \frac{\pi}{2})^2}\).
Using L'Hopital's Rule twice:
Numerator derivative: \(-2 \sin 2x\), Second derivative: \(-4 \cos 2x\)
Denominator derivative: \(2(x - \frac{\pi}{2})\), Second derivative: \(2\)
At \(x = \frac{\pi}{2}\), \(\frac{-4 \cos \pi}{2} = \frac{-4(-1)}{2} = 2\).
In simple words: Substitute the variable or apply L'Hopital's rule twice to resolve the indeterminate form.

Exam Tip: L'Hopital's rule is very powerful for trigonometric limits of 0/0 form.

(iii) \(\lim_{x \to 0} \frac{\sqrt[3]{8+x} - 2}{x}\)
Answer: Let \(8+x = y^3\), so as \(x \to 0\), \(y \to 2\):
\(\lim_{y \to 2} \frac{y - 2}{y^3 - 8}\)
Using formula \(y^3 - 8 = (y-2)(y^2 + 2y + 4)\):
\(= \lim_{y \to 2} \frac{y - 2}{(y-2)(y^2 + 2y + 4)}\)
\(= \lim_{y \to 2} \frac{1}{y^2 + 2y + 4}\)
\(= \frac{1}{2^2 + 2(2) + 4} = \frac{1}{4 + 4 + 4} = \frac{1}{12}\)
In simple words: Substitute a new variable to clear the cube root, then factor the denominator.

Exam Tip: Substitution simplifies radical expressions into standard algebraic polynomials.

(iv) \(\lim_{x \to 1} \frac{x^4 - 3x^3 + 2}{x^3 - 5x^2 + 3x + 1}\)
Answer: Check value at \(x = 1\):
Numerator: \(1^4 - 3(1)^3 + 2 = 1 - 3 + 2 = 0\)
Denominator: \(1^3 - 5(1)^2 + 3(1) + 1 = 1 - 5 + 3 + 1 = 0\)
Since it is 0/0 form, apply L'Hopital's Rule:
Numerator derivative: \(4x^3 - 9x^2\)
Denominator derivative: \(3x^2 - 10x + 3\)
Substitute \(x = 1\):
\(\frac{4(1)^3 - 9(1)^2}{3(1)^2 - 10(1) + 3} = \frac{4 - 9}{3 - 10 + 3} = \frac{-5}{-4} = \frac{5}{4}\)
In simple words: Differentiate top and bottom separately when direct substitution gives zero over zero.

Exam Tip: L'Hopital's rule saves time compared to polynomial long division for 0/0 rational functions.

(v) \(\lim_{x \to 0} \frac{\sqrt{1+2x} - \sqrt{1-2x}}{\sin x}\)
Answer: Rationalize the numerator:
\(= \lim_{x \to 0} \frac{(\sqrt{1+2x} - \sqrt{1-2x})(\sqrt{1+2x} + \sqrt{1-2x})}{\sin x (\sqrt{1+2x} + \sqrt{1-2x})}\)
\(= \lim_{x \to 0} \frac{(1+2x) - (1-2x)}{\sin x (\sqrt{1+2x} + \sqrt{1-2x})}\)
\(= \lim_{x \to 0} \frac{4x}{\sin x (\sqrt{1+2x} + \sqrt{1-2x})}\)
Split the limit using \(\lim_{x \to 0} \frac{x}{\sin x} = 1\):
\(= 4 \times \left( \lim_{x \to 0} \frac{x}{\sin x} \right) \times \left( \lim_{x \to 0} \frac{1}{\sqrt{1+2x} + \sqrt{1-2x}} \right)\)
\(= 4 \times 1 \times \frac{1}{\sqrt{1} + \sqrt{1}} = 4 \times \frac{1}{2} = 2\)
In simple words: Rationalize numerator and separate the standard limit term \(\frac{\sin x}{x}\).

Exam Tip: Combine standard trigonometric limits with algebraic rationalization.

(vi) \(\lim_{x \to 0} \frac{1 - \cos x \cos 2x}{x^2}\)
Answer: Add and subtract \(\cos x\) in the numerator:
\(= \lim_{x \to 0} \frac{1 - \cos x + \cos x - \cos x \cos 2x}{x^2}\)
\(= \lim_{x \to 0} \left[ \frac{1 - \cos x}{x^2} + \frac{\cos x(1 - \cos 2x)}{x^2} \right]\)
We know \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\), and \(1 - \cos 2x = 2 \sin^2 x\):
\(= \frac{1}{2} + \lim_{x \to 0} \left[ \cos x \times \frac{2 \sin^2 x}{x^2} \right]\)
\(= \frac{1}{2} + 1 \times 2 \times (1)^2 = \frac{1}{2} + 2 = \frac{3}{2}\)
In simple words: Split the expression into two standard trigonometric limit components.

Exam Tip: Memorize standard limits like \(\lim_{x \to 0} \frac{1 - \cos x}{x^2} = \frac{1}{2}\).

(vii) \(\lim_{x \to \frac{\pi}{6}} \frac{\sqrt{3} \sin x - \cos x}{x - \frac{\pi}{6}}\)
Answer: Divide and multiply numerator by 2:
\(= \lim_{x \to \frac{\pi}{6}} \frac{2 \left(\frac{\sqrt{3}}{2} \sin x - \frac{1}{2} \cos x\right)}{x - \frac{\pi}{6}}\)
Since \(\frac{\sqrt{3}}{2} = \cos \frac{\pi}{6}\) and \(\frac{1}{2} = \sin \frac{\pi}{6}\):
\(= 2 \lim_{x \to \frac{\pi}{6}} \frac{\sin\left(x - \frac{\pi}{6}\right)}{x - \frac{\pi}{6}}\)
Let \(x - \frac{\pi}{6} = t\), then as \(x \to \frac{\pi}{6}\), \(t \to 0\):
\(= 2 \lim_{t \to 0} \frac{\sin t}{t} = 2(1) = 2\)
In simple words: Use trigonometric compound angle formulas to simplify the expression into a standard limit.

Exam Tip: Introducing auxiliary angles like \(\sin(A - B)\) simplifies linear combinations of sine and cosine.

(viii) \(\lim_{x \to 0} \frac{\cos 3x - \cos x}{x^2}\)
Answer: Use the sum-to-product formula for cosines: \(\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right)\sin\left(\frac{C-D}{2}\right)\)
\(= \lim_{x \to 0} \frac{-2 \sin\left(\frac{3x+x}{2}\right)\sin\left(\frac{3x-x}{2}\right)}{x^2}\)
\(= \lim_{x \to 0} \frac{-2 \sin(2x)\sin(x)}{x^2}\)
\(= -2 \lim_{x \to 0} \left( \frac{\sin 2x}{2x} \times 2 \times \frac{\sin x}{x} \right)\)
\(= -2 \times 1 \times 2 \times 1 = -4\)
In simple words: Convert the difference of cosines into a product using trigonometric identities.

Exam Tip: Always convert difference of trigonometric functions into products when evaluating limits at zero.

(ix) \(\lim_{x \to 0} \frac{\tan 2x - \sin 2x}{x^3}\)
Answer: Rewrite \(\tan 2x = \frac{\sin 2x}{\cos 2x}\):
\(= \lim_{x \to 0} \frac{\frac{\sin 2x}{\cos 2x} - \sin 2x}{x^3}\)
\(= \lim_{x \to 0} \frac{\sin 2x (1 - \cos 2x)}{x^3 \cos 2x}\)
Since \(1 - \cos 2x = 2 \sin^2 x\):
\(= \lim_{x \to 0} \frac{\sin 2x \times 2 \sin^2 x}{x^3 \cos 2x}\)
\(= \lim_{x \to 0} \left( \frac{\sin 2x}{2x} \times 2 \times 2 \times \left(\frac{\sin x}{x}\right)^2 \times \frac{1}{\cos 2x} \right)\)
\(= 1 \times 4 \times (1)^2 \times \frac{1}{\cos 0} = 4 \times 1 = 4\)
In simple words: Express tangent in terms of sine and cosine, then factor out standard limit components.

Exam Tip: Group terms into \(\frac{\sin x}{x}\) forms to easily evaluate the limit.

(x) \(\lim_{x \to 0} \frac{1 - \cos mx}{1 - \cos nx}\)
Answer: Use \(1 - \cos \theta = 2 \sin^2\left(\frac{\theta}{2}\right)\):
\(= \lim_{x \to 0} \frac{2 \sin^2\left(\frac{mx}{2}\right)}{2 \sin^2\left(\frac{nx}{2}\right)}\)
\(= \lim_{x \to 0} \left( \frac{\sin\left(\frac{mx}{2}\right)}{\frac{mx}{2}} \times \frac{\frac{mx}{2}}{\frac{nx}{2}} \times \frac{\frac{nx}{2}}{\sin\left(\frac{nx}{2}\right)} \right)^2\)
\(= \left( 1 \times \frac{m}{n} \times 1 \right)^2 = \frac{m^2}{n^2}\)
In simple words: Adjust the angles in numerator and denominator to match the standard limit \(\lim \frac{\sin \theta}{\theta} = 1\).

Exam Tip: Multiply and divide by the required angle arguments to apply standard formulas.

(xi) \(\lim_{x \to 0} \frac{\tan x - \sin x}{x^3}\)
Answer: Rewrite \(\tan x = \frac{\sin x}{\cos x}\):
\(= \lim_{x \to 0} \frac{\frac{\sin x}{\cos x} - \sin x}{x^3} = \lim_{x \to 0} \frac{\sin x(1 - \cos x)}{x^3 \cos x}\)
\(= \lim_{x \to 0} \left( \frac{\sin x}{x} \times \frac{1 - \cos x}{x^2} \times \frac{1}{\cos x} \right)\)
\(= 1 \times \frac{1}{2} \times \frac{1}{1} = \frac{1}{2}\)
In simple words: Factor out \(\sin x\) and group with standard limits for \(\frac{\sin x}{x}\) and \(\frac{1-\cos x}{x^2}\).

Exam Tip: Breaking complex trigonometric expressions into products of known standard limits is extremely efficient.

(xii) \(\lim_{x \to 0} (1+x)^{\frac{2}{x}}\)
Answer: Use the standard limit \(\lim_{x \to 0} (1+x)^{\frac{1}{x}} = e\):
\(= \left( \lim_{x \to 0} (1+x)^{\frac{1}{x}} \right)^2 = e^2\)
In simple words: Express the exponent in terms of \(\frac{1}{x}\) to match the exponential limit definition of \(e\).

Exam Tip: Remember the fundamental limit \(\lim_{x \to 0} (1+x)^{\frac{1}{x}} = e\).

(xiii) \(\lim_{x \to 0} \frac{e^{2x} - 1}{x}\)
Answer: Use the standard limit \(\lim_{t \to 0} \frac{e^t - 1}{t} = 1\). Put \(t = 2x\), so as \(x \to 0\), \(t \to 0\):
\(= \lim_{2x \to 0} \frac{e^{2x} - 1}{2x} \times 2 = 1 \times 2 = 2\)
In simple words: Adjust the denominator to match the exponent of \(e\) using substitution.

Exam Tip: For exponential limits, ensure the denominator matches the power of \(e\).

(xiv) \(\lim_{x \to 0} \frac{5^x - 1}{x}\)
Answer: Use the standard limit formula \(\lim_{x \to 0} \frac{a^x - 1}{x} = \log a\):
Here \(a = 5\), so the limit is \(\log 5\) (or \(3 \log 5\) if coefficient was 3, but here formula gives \(\log 5\)).
In simple words: Directly apply the standard formula for exponential limits with base \(a\).

Exam Tip: \(\lim_{x \to 0} \frac{a^x - 1}{x} = \log_e a\) is a direct textbook result.

(xv) \(\lim_{x \to 0} \frac{\log(1+7x)}{x}\)
Answer: Use the standard logarithmic limit \(\lim_{t \to 0} \frac{\log(1+t)}{t} = 1\). Put \(t = 7x\):
\(= \lim_{7x \to 0} \frac{\log(1+7x)}{7x} \times 7 = 1 \times 7 = 7\)
In simple words: Multiply and divide by 7 to make the denominator match the argument inside the logarithm.

Q2. Let f(x) be a function defined by \(f(x) = \begin{cases} 6 - 3x & \text{if } x \le 3 \\ 2x - k & \text{if } x > 3 \end{cases}\). Find k, if \(\lim_{x \to 3} f(x)\) exists.
Answer: For the limit to exist, left-hand limit (LHL) must equal right-hand limit (RHL) at \(x = 3\):
\(\text{LHL} = \lim_{x \to 3^-} (6 - 3x) = 6 - 3(3) = 6 - 9 = -3\)
\(\text{RHL} = \lim_{x \to 3^+} (2x - k) = 2(3) - k = 6 - k\)
Equating LHL and RHL:
\(6 - k = -3\)
\(k = 6 + 3 = 9\) (Wait, source answer is -6. Let us re-verify: if \(f(x) = 2x - k\) for \(x > 3\), then \(6 - k = -3 \implies k = 9\). Let us check the function definition in text: \(6x - 3 \dots\) or \(6 - 3x\). OCR says \(6x - 3\) or \(6 - 3x\). If LHL is evaluated as -3 and RHL as \(6 - k\), then \(6 - k = -3 \implies k = 9\). If the question had \(f(x) = \begin{cases} 6x - 3 & x \le 3 \\ 2x - k & x > 3 \end{cases}\), then at 3, \(6(3)-3 = 15 \implies 6-k=15 \implies k=-9\). Regardless, follow the condition \(\text{LHL} = \text{RHL}\)).
In simple words: Set the limit from the left equal to the limit from the right at the given point.
Exam Tip: A limit exists if and only if Left-Hand Limit = Right-Hand Limit.

Q3. Let f(x) be a function defined by \(f(x) = \begin{cases} -2x + 3 & \text{if } x < 1 \\ 1 & \text{if } 1 \le x \le 2 \\ 2x - 3 & \text{if } x > 2 \end{cases}\). Find \(\lim_{x \to 1} f(x)\) and \(\lim_{x \to 2} f(x)\).
Answer: For \(\lim_{x \to 1} f(x)\):
LHL at \(x = 1\): \(\lim_{x \to 1^-} (-2x + 3) = -2(1) + 3 = 1\)
RHL at \(x = 1\): \(\lim_{x \to 1^+} (1) = 1\)
Since LHL = RHL = 1, \(\lim_{x \to 1} f(x) = 1\).
For \(\lim_{x \to 2} f(x)\):
LHL at \(x = 2\): \(\lim_{x \to 2^-} (1) = 1\)
RHL at \(x = 2\): \(\lim_{x \to 2^+} (2x - 3) = 2(2) - 3 = 1\)
Since LHL = RHL = 1, \(\lim_{x \to 2} f(x) = 1\).
In simple words: Evaluate the limits from both sides for each point to check existence and value.

Exam Tip: Check the appropriate sub-function definition depending on whether you approach from values smaller or greater than the target point.

Q4. (i) If \(\lim_{x \to a} \frac{x^7 + a^7}{x + a} = 7\), find the value of a. (ii) If \(\lim_{\phi \to 0} (k\phi)\csc \phi = \lim_{\phi \to 0} \cos(\phi)\csc(k\phi)\), prove that \(k = \pm 1\).
Answer: (i) Using standard formula \(\lim_{x \to a} \frac{x^n + a^n}{x + a}\) is not standard, but using standard limit \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\). Here we have \(\frac{x^7 + a^7}{x + a}\). By direct substitution or algebraic expansion, for \(x \to -a\), \(\frac{(-a)^7 + a^7}{-a + a}\) gives 0/0. Using L'Hopital's rule:
\(\lim_{x \to a} \frac{7x^6}{1} = 7a^6\)
Given \(7a^6 = 7 \implies a^6 = 1 \implies a = \pm 1\).
(ii) Evaluate both limits:
LHS: \(\lim_{\phi \to 0} \frac{k\phi}{\sin \phi} = k \times 1 = k\)
RHS: \(\lim_{\phi \to 0} \frac{\cos \phi}{\sin(k\phi)} = \frac{1}{k}\)
Equating LHS and RHS: \(k = \frac{1}{k} \implies k^2 = 1 \implies k = \pm 1\).
In simple words: Evaluate limits on both sides using standard trigonometric limits like \(\lim \frac{\sin \theta}{\theta} = 1\).

Exam Tip: Convert cosecant to sine in the denominator before evaluating limits involving trigonometric functions.

Q5. Find the positive integer n, so that \(\lim_{x \to 3} \frac{x^n - 3^n}{x - 3} = 108\), find the value of a. (Note: standard problem asks for n).
Answer: Use the standard limit formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\):
Here \(a = 3\), so:
\(n(3)^{n-1} = 108\)
Test integer values for \(n\):
If \(n = 4\): \(4(3)^{4-1} = 4(3^3) = 4(27) = 108\).
Thus, \(n = 4\).
In simple words: Use the algebraic limit formula for powers and test small integers to find n.

Exam Tip: \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\) is one of the most important algebraic limit formulas.

Q6. Let f(x) be a function defined by \(f(x) = \begin{cases} \frac{x^2 - 6x}{|x - 2|} & \text{if } x \neq 0 \\ 0 & \text{if } x = 0 \end{cases}\). Find \(\lim_{x \to 0} f(x)\).
Answer: Note: The question likely checks limit as \(x \to 2\) or \(x \to 0\). Evaluating at \(x \to 0\):
\(\lim_{x \to 0} \frac{x^2 - 6x}{|x - 2|} = \frac{0^2 - 6(0)}{|0 - 2|} = \frac{0}{2} = 0\). (If approaching \(x \to 2\), the denominator becomes 0, leading to non-existence. As stated by standard text for this problem, limit does not exist at the critical point where denominator modulus changes sign or becomes zero).
In simple words: Check the left-hand and right-hand limits around the critical point of the modulus function.

Exam Tip: Always expand modulus functions into piecewise linear functions based on the sign of their argument.

Q7. Find all the possible values of a, if \(\lim_{x \to a} \frac{x^9 - a^9}{x - a} = \lim_{x \to 5} (4 + x)\).
Answer: Evaluate LHS using formula \(\lim_{x \to a} \frac{x^n - a^n}{x - a} = n a^{n-1}\):
\(\text{LHS} = 9 a^{n-1} = 9 a^{8}\)
Evaluate RHS:
\(\text{RHS} = 4 + 5 = 9\)
Equate LHS and RHS:
\(9 a^8 = 9\)
\(a^8 = 1\)
\(a = \pm 1\)
In simple words: Apply the standard power limit formula on the left and basic substitution on the right, then solve for a.

Exam Tip: Equate the evaluated results of both sides to form a simple algebraic equation in a.

Q8. If \(\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2}\), find all the possible values of k.
Answer: Evaluate LHS:
\(\lim_{x \to 1} \frac{x^4 - 1}{x - 1} = 4(1)^{4-1} = 4\)
Evaluate RHS using standard limits or L'Hopital's Rule:
\(\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2} = \lim_{x \to k} \frac{(x-x)(...) }{}\) or via L'Hopital's: \(\lim_{x \to k} \frac{3x^2}{2x} = \frac{3k}{2}\)
Equate LHS and RHS:
\(\frac{3k}{2} = 4\)
\(3k = 8 \implies k = \frac{8}{3}\)
In simple words: Evaluate both limits independently using standard formulas and equate them to find k.

Exam Tip: L'Hopital's rule quickly resolves limits of algebraic fractions containing parameters like k.

Q9. Let f(x) be a function defined by \(f(x) = \begin{cases} x < 1 & \text{if } x < 1 \\ 2 & \text{if } x = 1 \\ x + 2 & \text{if } x > 1 \end{cases}\). Find \(\lim_{x \to 1} f(x)\).
Answer: Calculate Left-Hand Limit (LHL) at \(x = 1\):
\(\text{LHL} = \lim_{x \to 1^-} x = 1\)
Calculate Right-Hand Limit (RHL) at \(x = 1\):
\(\text{RHL} = \lim_{x \to 1^+} (x + 2) = 1 + 2 = 3\)
Since LHL \(\neq\) RHL (\(1 \neq 3\)), the limit does not exist.
In simple words: Find limits from left and right; if they are unequal, the limit does not exist.

Exam Tip: The value of the function at the point (\(f(1) = 2\)) has no bearing on whether the limit exists.

Q10. Differentiate each of the following functions with respect to x, from first principle:
(i) \(f(x) = 15x\)
Answer: Using first principle \(f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}\):
\(f'(x) = \lim_{h \to 0} \frac{15(x+h) - 15x}{h} = \lim_{h \to 0} \frac{15h}{h} = 15\)
In simple words: Apply the definition of derivative as a limit of the difference quotient.

Exam Tip: The derivative of a linear function \(mx\) is always the slope \(m\).

(ii) \(f(x) = x^3 - 1\)
Answer: Using first principle:
\(f'(x) = \lim_{h \to 0} \frac{((x+h)^3 - 1) - (x^3 - 1)}{h}\)
\(= \lim_{h \to 0} \frac{x^3 + 3x^2h + 3xh^2 + h^3 - 1 - x^3 + 1}{h}\)
\(= \lim_{h \to 0} \frac{h(3x^2 + 3xh + h^2)}{h} = \lim_{h \to 0} (3x^2 + 3xh + h^2) = 3x^2\)
In simple words: Expand binomial cube, cancel terms, factor out h, and take the limit as h approaches zero.

Exam Tip: Always expand powers using binomial theorem for first principle polynomial derivations.

(iii) \(f(x) = x - \frac{1}{x}\)
Answer: Using first principle:
\(f'(x) = \lim_{h \to 0} \frac{\left( (x+h) - \frac{1}{x+h} \right) - \left( x - \frac{1}{x} \right)}{h}\)
\(= \lim_{h \to 0} \frac{h + \frac{1}{x} - \frac{1}{x+h}}{h} = \lim_{h \to 0} \left( 1 + \frac{\frac{x+h - x}{x(x+h)}}{h} \right)\)
\(= \lim_{h \to 0} \left( 1 + \frac{h}{h x(x+h)} \right) = \lim_{h \to 0} \left( 1 + \frac{1}{x(x+h)} \right) = 1 + \frac{1}{x^2}\)
In simple words: Group terms, find common denominator, cancel h, and evaluate the limit.

Exam Tip: Combine fractional terms carefully before dividing by h in the denominator.

(iv) \(f(x) = \tan 2x\)
Answer: Using first principle:
\(f'(x) = \lim_{h \to 0} \frac{\tan(2(x+h)) - \tan 2x}{h} = \lim_{h \to 0} \frac{\sin(2x+2h) \cos 2x - \cos(2x+2h)\sin 2x}{h \cos(2x+2h)\cos 2x}\)
\(= \lim_{h \to 0} \frac{\sin(2x+2h - 2x)}{h \cos(2x+2h)\cos 2x} = \lim_{h \to 0} \frac{\sin 2h}{h} \times \frac{1}{\cos(2x+2h)\cos 2x}\)
\(= 2 \times \frac{1}{\cos^2 2x} = 2 \sec^2 2x\)
In simple words: Convert tangents to sines and cosines, apply subtraction formula, and use standard limit \(\frac{\sin 2h}{2h}\).

(v) \(f(x) = \cos(2x-3)\)
Answer: Using first principle and cosine difference formula \(\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)\):
\(f'(x) = \lim_{h \to 0} \frac{\cos(2(x+h)-3) - \cos(2x-3)}{h}\)
\(= \lim_{h \to 0} \frac{-2 \sin\left(2x+h-3\right)\sin(h)}{h}\)
\(= -2 \sin(2x-3) \lim_{h \to 0} \frac{\sin h}{h} = -2 \sin(2x-3)\)
In simple words: Apply cosine sum-to-product identity and evaluate the standard limit of sine over h.

Exam Tip: Product-to-sum identities transform trigonometric difference quotients into manageable limits.

(vi) \(f(x) = \sec 3x\)
Answer: Using first principle: \(f(x) = \frac{1}{\cos 3x}\)
\(f'(x) = \lim_{h \to 0} \frac{\frac{1}{\cos(3x+3h)} - \frac{1}{\cos 3x}}{h} = \lim_{h \to 0} \frac{\cos 3x - \cos(3x+3h)}{h \cos 3x \cos(3x+3h)}\)
Using cosine difference formula: \(\cos 3x - \cos(3x+3h) = 2 \sin\left(3x + \frac{3h}{2}\right)\sin\left(\frac{3h}{2}\right)\)
\(= \lim_{h \to 0} \frac{2 \sin\left(3x+\frac{3h}{2}\right)\sin\left(\frac{3h}{2}\right)}{h \cos 3x \cos(3x+3h)} = 3 \sec 3x \tan 3x\)
In simple words: Express secant as cosine reciprocal, subtract, and apply trigonometric identities.

Exam Tip: Reciprocal trigonometric functions are best converted to sine and cosine before applying limits.

(vii) \(f(x) = 120\)
Answer: Using first principle:
\(f'(x) = \lim_{h \to 0} \frac{120 - 120}{h} = \lim_{h \to 0} 0 = 0\)
In simple words: The derivative of any constant number is zero because constant functions never change.

Exam Tip: Constant rule: \(\frac{d}{dx}(c) = 0\).

(viii) \(f(x) = \sqrt{5x - 7}\)
Answer: Using first principle and rationalizing the numerator:
\(f'(x) = \lim_{h \to 0} \frac{\sqrt{5(x+h)-7} - \sqrt{5x-7}}{h}\)
Multiply and divide by conjugate \((\sqrt{5x+5h-7} + \sqrt{5x-7})\):
\(= \lim_{h \to 0} \frac{(5x+5h-7) - (5x-7)}{h(\sqrt{5x+5h-7} + \sqrt{5x-7})}\)
\(= \lim_{h \to 0} \frac{5h}{h(\sqrt{5x+5h-7} + \sqrt{5x-7})} = \frac{5}{2\sqrt{5x-7}}\)
In simple words: Rationalize the numerator by multiplying with the conjugate sum of square roots.

Exam Tip: Radical functions in first principle are always rationalized to eliminate square roots from the numerator.

Q11. Evaluate each of the following, from first principle:
(i) \(f'(4)\) if \(f(x) = x^2\)
Answer: First find general derivative \(f'(x) = \lim_{h \to 0} \frac{(x+h)^2 - x^2}{h} = \lim_{h \to 0} \frac{2xh + h^2}{h} = 2x\).
Substitute \(x = 4\): \(f'(4) = 2(4) = 8\).
In simple words: Find the derivative function first, then substitute the given point value.

Exam Tip: Always find the derivative formula for general x before substituting specific point values.

(ii) \(f'(16)\) if \(f(x) = 16\) (Constant function)
Answer: Since \(f(x)\) is a constant function, \(f'(x) = 0\) for all x, so \(f'(16) = 0\).
In simple words: Derivative of a constant is always zero.

(iii) \(f'(5)\) if \(f(x) = 2x\)
Answer: Derivative of \(2x\) is constant 2 for all x, so \(f'(5) = 2\).
In simple words: The slope of a straight line \(2x\) is 2 everywhere.

Exam Tip: The derivative of a linear term \(ax\) is simply coefficient a.

(iv) \(f'(9)\) if \(f(x) = \sqrt{x}\)
Answer: Find general derivative using first principle:
\(f'(x) = \lim_{h \to 0} \frac{\sqrt{x+h} - \sqrt{x}}{h} = \lim_{h \to 0} \frac{h}{h(\sqrt{x+h} + \sqrt{x})} = \frac{1}{2\sqrt{x}}\)
Substitute \(x = 9\): \(f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{2(3)} = \frac{1}{6}\).
In simple words: Rationalize numerator to find derivative formula, then substitute 9.

Exam Tip: Standard result: \(\frac{d}{dx}(\sqrt{x}) = \frac{1}{2\sqrt{x}}\).

Q12. Find the derivative of each of the following functions, with respect to x:
(i) \(f(x) = (3x+7)^5\)
Answer: Use Chain Rule: \(\frac{d}{dx}(u^n) = n u^{n-1} \frac{du}{dx}\)
\(f'(x) = 5(3x+7)^4 \times \frac{d}{dx}(3x+7) = 5(3x+7)^4 \times 3 = 15(3x+7)^4\).
In simple words: Bring power to front, decrease power by 1, and multiply by derivative of inside function.

Exam Tip: Chain rule is essential for composite functions raised to a power.

(ii) \(f(x) = x^2 \sin x\)
Answer: Use Product Rule \((uv)' = u'v + uv'\):
\(f'(x) = \frac{d}{dx}(x^2) \cdot \sin x + x^2 \cdot \frac{d}{dx}(\sin x)\)
\(= 2x \sin x + x^2 \cos x\).
In simple words: Derivative of first times second plus first times derivative of second.

Exam Tip: Always use product rule when two distinct functions of x are multiplied together.

(iii) \(f(x) = \frac{x}{1 + \tan x}\)
Answer: Use Quotient Rule \(\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}\):
\(f'(x) = \frac{(1+\tan x) \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(1+\tan x)}{(1+\tan x)^2}\)
\(= \frac{(1+\tan x)(1) - x(\sec^2 x)}{(1+\tan x)^2} = \frac{1 + \tan x - x \sec^2 x}{(1+\tan x)^2}\).
In simple words: Low d-high minus high d-low over low squared.

Exam Tip: Be careful with signs and derivative of tangent during quotient rule expansion.

(iv) \(f(x) = x \sec^3 x\)
Answer: Use Product Rule and Chain Rule:
\(f'(x) = 1 \cdot \sec^3 x + x \cdot 3 \sec^2 x \cdot (\sec x \tan x)\)
\(= \sec^3 x + 3x \sec^3 x \tan x = \sec^3 x(1 + 3x \tan x)\).
In simple words: Apply product rule between x and \(\sec^3 x\), using chain rule for the secant power.

Exam Tip: Factor out common terms like \(\sec^3 x\) to simplify the final derivative expression.

(v) \(f(x) = \frac{\cos x - 5}{\sin x}\) (or variant)
Answer: Using quotient rule:
\(f'(x) = \frac{\sin x \cdot (-\sin x) - (\cos x - 5)(\cos x)}{\sin^2 x}\)
\(= \frac{-\sin^2 x - \cos^2 x + 5 \cos x}{\sin^2 x} = \frac{-(1) + 5 \cos x}{\sin^2 x} \dots\) (matches standard derivative form).
In simple words: Apply quotient rule and use Pythagorean trigonometric identity \(\sin^2 x + \cos^2 x = 1\).

Exam Tip: Simplify trigonometric derivative numerators using fundamental identities.

(vi) \(f(x) = (1 - x^2)(3x + 4)\)
Answer: Using product rule or expanding first:
Method 1 (Product Rule):
\(f'(x) = (-2x)(3x + 4) + (1 - x^2)(3)\)
\(= -6x^2 - 8x + 3 - 3x^2 = -9x^2 - 8x + 3\).
In simple words: Differentiate using product rule or multiply out polynomials first to differentiate term by term.

(vii) \(f(x) = \frac{\sin x + \cos x}{\sin x - \cos x}\)
Answer: Using quotient rule:
\(f'(x) = \frac{(\sin x - \cos x)(\cos x - \sin x) - (\sin x + \cos x)(\cos x + \sin x)}{(\sin x - \cos x)^2}\)
\(= \frac{-(\sin x - \cos x)^2 - (\sin x + \cos x)^2}{(\sin x - \cos x)^2}\)
Expand and simplify numerators: \(-(\sin^2 x + \cos^2 x - 2\sin x\cos x) - (\sin^2 x + \cos^2 x + 2\sin x\cos x) = -1 - 1 = -2\).
\(f'(x) = \frac{-2}{(\sin x - \cos x)^2}\).
In simple words: Apply quotient rule, expand squares, and use \(\sin^2 x + \cos^2 x = 1\).

Exam Tip: Recognizing algebraic expansions like \((a-b)^2 + (a+b)^2 = 2(a^2+b^2)\) speeds up simplification.

(viii) \(f(x) = \frac{x^3 - 4}{x + 1}\)
Answer: Using quotient rule:
\(f'(x) = \frac{(x + 1)(3x^2) - (x^3 - 4)(1)}{(x + 1)^2}\)
\(= \frac{3x^3 + 3x^2 - x^3 + 4}{(x + 1)^2} = \frac{2x^3 + 3x^2 + 4}{(x + 1)^2}\).
In simple words: Apply quotient rule formula, expand numerator terms, and combine like powers of x.

Exam Tip: Always expand numerator products fully in quotient rule problems.

Q13. (i) If \(y = \csc x + \cot x\), show that \(\sin x \frac{dy}{dx} + y = 0\). (ii) If \(y = a \sin x + b \cos x\), show that \(y^2 + \left(\frac{dy}{dx}\right)^2 = a^2 + b^2\).
Answer: (i) Given \(y = \csc x + \cot x\):
\(\frac{dy}{dx} = -\csc x \cot x - \csc^2 x = -\csc x (\cot x + \csc x) = -\csc x \cdot y\)
Multiply both sides by \(\sin x\):
\(\sin x \frac{dy}{dx} = \sin x (-\csc x \cdot y) = -1 \cdot y \implies \sin x \frac{dy}{dx} + y = 0\).
(ii) Given \(y = a \sin x + b \cos x\):
\(\frac{dy}{dx} = a \cos x - b \sin x\)
Calculate \(y^2 + \left(\frac{dy}{dx}\right)^2\):
\((a \sin x + b \cos x)^2 + (a \cos x - b \sin x)^2\)
\(= a^2 \sin^2 x + b^2 \cos^2 x + 2ab \sin x \cos x + a^2 \cos^2 x + b^2 \sin^2 x - 2ab \sin x \cos x\)
\(= a^2(\sin^2 x + \cos^2 x) + b^2(\cos^2 x + \sin^2 x) = a^2(1) + b^2(1) = a^2 + b^2\).
In simple words: Differentiate the given expressions and substitute into the target identity using trigonometric formulas.

Exam Tip: Algebraic expansion of derivative squares often results in cross-term cancellation via trigonometric identities.

Q14. The letters of the word 'SOCIETY' are placed at random in a row. What is the probability that the three vowels come together?
Answer: Total letters in 'SOCIETY' = 7 letters (S, O, C, I, E, T, Y).
Vowels = O, I, E (3 vowels). Consonants = S, C, T, Y (4 consonants).
Total ways to arrange 7 letters = \(7! = 5040\).
Treat the 3 vowels (O, I, E) as a single block. Then we have 4 consonants + 1 vowel block = 5 units.
Number of ways to arrange these 5 units = \(5! = 120\).
The 3 vowels within their block can be arranged among themselves in \(3! = 6\) ways.
Favorable arrangements = \(5! \times 3! = 120 \times 6 = 720\).
Probability = \(\frac{720}{5040} = \frac{1}{7}\).
In simple words: Group vowels together as one single item, arrange all items, then multiply by internal vowel permutations.

Exam Tip: "Come together" problems are always solved by bundling the required items into a single block.

Q15. Four digit numbers are formed by using the digits 1, 2, 3, 4, and 5 without repeating any digit. Find the probability that a number, chosen at random, is an odd number.
Answer: Total digits available = 5 (1, 2, 3, 4, 5).
Total 4-digit numbers without repetition = \(P(5,4) = \frac{5!}{(5-4)!} = 5 \times 4 \times 3 \times 2 = 120\).
For a number to be odd, its units place must be filled with an odd digit (1, 3, or 5). Thus, 3 choices for the units place.
Remaining 3 places can be filled from the remaining 4 digits in \(P(4,3) = 4 \times 3 \times 2 = 24\) ways.
Favorable odd numbers = \(3 \times 24 = 72\).
Probability = \(\frac{72}{120} = \frac{3}{5}\).
In simple words: Fix the last digit to be odd, then calculate permutations for the remaining positions.

Exam Tip: Always fill restricted positions (like odd/even units place) first in permutation problems.

Q16. Two dice are thrown simultaneously. Find the probability of getting a doublet.
Answer: Total possible outcomes when throwing two dice = \(6 \times 6 = 36\).
A doublet means both dice show the same number: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
Number of favorable outcomes = 6.
Probability = \(\frac{6}{36} = \frac{1}{6}\).
In simple words: Count matching pairs out of 36 total outcomes.

Exam Tip: Doublets are pairs where both dice coordinates are identical: (x, x).

Q17. What is the chance that a leap year, selected at random, will contain 53 Sundays?
Answer: A leap year has 366 days = 52 weeks and 2 extra days.
The 52 weeks guarantee 53 Sundays? No, 52 weeks give 52 Sundays. The remaining 2 extra days can be any consecutive pair of days of the week:
(Monday, Tuesday), (Tuesday, Wednesday), (Wednesday, Thursday), (Thursday, Friday), (Friday, Saturday), (Saturday, Sunday), (Sunday, Monday).
Total possible pairs = 7.
Pairs containing Sunday = (Saturday, Sunday) and (Sunday, Monday) — exactly 2 pairs.
Probability = \(\frac{2}{7}\).
In simple words: A leap year has 52 full weeks plus 2 leftover days. Check how many day pairs include Sunday.

Exam Tip: For leap years, the probability of 53 Sundays or 53 Mondays is always 2/7.

Q18. Find the probability of an event of getting Prime Number or Even number in a single throw of dice.
Answer: Sample space for a single die = \(\{1, 2, 3, 4, 5, 6\}\), total outcomes = 6.
Prime numbers = \(\{2, 3, 5\}\).
Even numbers = \(\{2, 4, 6\}\).
Union of Prime or Even numbers = \(\{2, 3, 4, 5, 6\}\) (total 5 numbers).
Probability = \(\frac{5}{6}\).
In simple words: List all prime numbers and even numbers on a die, combine them without duplicates, and divide by 6.

Exam Tip: For 'or' conditions in probability, find the union of the two sets of favorable outcomes.

Q19. Find the probability of an event of getting two heads in a tossing of three coins.
Answer: Total outcomes when tossing 3 coins = \(2^3 = 8\):
{HHH, HHT, HTH, THH, HTT, THT, TTH, TTT}.
Outcomes with exactly two heads = {HHT, HTH, THH} (3 outcomes).
Probability = \(\frac{3}{8}\).
In simple words: Count all outcomes with exactly two heads out of 8 total possibilities.

Q20. Two unbiased dice are thrown. Find the probability that neither a doublet nor a total of 10 will appear.
Answer: Total outcomes = 36.
Doublets = 6 outcomes: (1,1), (2,2), (3,3), (4,4), (5,5), (6,6).
Outcomes with a sum of 10 = 3 outcomes: (4,6), (5,5), (6,4).
Notice (5,5) is counted in both doublets and sum of 10.
Total favorable outcomes for (doublet OR sum of 10) = \(6 + 3 - 1 = 8\).
Outcomes where neither occurs = \(36 - 8 = 28\).
Probability = \(\frac{28}{36} = \frac{7}{9}\).
In simple words: Find total outcomes with doublets or sums of 10, subtract overlaps, and subtract from 36.
Exam Tip: Use the addition rule of probability \(P(A \cup B) = P(A) + P(B) - P(A \cap B)\) for overlapping conditions.

Q21. A card is drawn from a deck of 52 cards. Find the probability of getting a king or a heart or a red card.
Answer: Total cards = 52.
Let K = King cards (4 cards), H = Heart cards (13 cards), R = Red cards (26 cards).
Notice that all 4 Kings include 2 red kings (King of Hearts, King of Diamonds), which are already inside the 26 red cards.
Also, all 13 Heart cards are already inside the 26 red cards, except the remaining 2 black kings (King of Spades, King of Clubs).
Thus, the set of cards that are "a king or a heart or a red card" is simply the union of Red cards (26) plus the 2 black kings = 28 cards.
Probability = \(\frac{28}{52} = \frac{7}{13}\).
In simple words: Since all hearts and red kings are already red cards, just add the 2 remaining black kings to the 26 red cards.

Exam Tip: Use set inclusion-exclusion principle to avoid double-counting overlapping card categories.

Q22. The probabilities that a student will receive A, B, C, or D grade are 0.40, 0.35, 0.15, and 0.10 respectively. Find the probability that a student will receive (i) not an A grade (ii) B or C grade (iii) at most C grade.
Answer: Given: \(P(A) = 0.40, P(B) = 0.35, P(C) = 0.15, P(D) = 0.10\).
(i) Probability of not an A grade: \(1 - P(A) = 1 - 0.40 = 0.60\).
(ii) Probability of B or C grade: \(P(B) + P(C) = 0.35 + 0.15 = 0.50\).
(iii) Probability of at most C grade (grades C or D): \(P(C) + P(D) = 0.15 + 0.10 = 0.25\).
In simple words: Sum the relevant individual probabilities or subtract from 1 for complement events.

Exam Tip: "At most C" means grade C or lower (i.e., C or D).