CBSE Class 11 Mathematics Relations and Functions Assignment Set 01

Relations and Functions MCQ Questions with Answers Class 11 Mathematics

Question 1. Find the domain and range of following functions:
(i) \(f(x) = \sqrt{x - 3}\)
(ii) \(f(x) = \sqrt{25 - x^2}\)
(iii) \(f(x) = 5 - |x + 1|\)
(iv) \(f(x) = \frac{1}{\sqrt{16 - x^2}}\)
Answer: For (i): For \(f(x) = \sqrt{x - 3}\) to be real, \(x - 3 \geq 0 \implies x \geq 3\).
Domain = \([3, \infty)\), Range = \([0, \infty)\).
For (ii): For \(f(x) = \sqrt{25 - x^2}\) to be real, \(25 - x^2 \geq 0 \implies x^2 \leq 25 \implies -5 \leq x \leq 5\).
Domain = \([-5, 5]\), Range = \([0, 5]\).
For (iii): \(f(x) = 5 - |x + 1|\). Defined for all real numbers \(x\).
Domain = \((-\infty, \infty)\), Range = \((-\infty, 5]\) since \(|x + 1| \geq 0\).
For (iv): For \(f(x) = \frac{1}{\sqrt{16 - x^2}}\) to be real and non-zero denominator, \(16 - x^2 > 0 \implies x^2 < 16 \implies -4 < x < 4\).
Domain = \((-4, 4)\), Range = \(\left[\frac{1}{4}, \infty\right)\).
In simple words: The domain is all possible input values of x where the function is defined, and the range is the set of all resulting output values.

Exam Tip: Always check for square roots (expression inside must be greater than or equal to zero) and denominators (must not equal zero).

Question 2. Draw the graph of function \(f(x) = \begin{cases} 1 + 2x, & x < 0 \\ 3 + 5x, & x \ge 0 \end{cases}\). Hence find its range.
Answer: For \(x < 0\), the line is \(y = 1 + 2x\). As \(x \to -\infty\), \(y \to -\infty\), and as \(x \to 0^{-}\), \(y \to 1\).
For \(x \ge 0\), the line is \(y = 3 + 5x\). At \(x = 0\), \(y = 3\), and as \(x \to \infty\), \(y \to \infty\).
Since the function values cover all reals from \(-\infty\) up to 1 (exclusive) and from 3 to \(\infty\) (inclusive), the range is \((-\infty, 1) \cup [3, \infty)\).
In simple words: Graph the two linear pieces separately for their respective domains and observe the vertical span to find the range.

Exam Tip: Pay close attention to strict and non-strict inequalities at junction points like \(x = 0\) to determine whether endpoints are included.

Question 3. If \(f(x) = 2x - 3\) and \(g(x) = x^2 - 1\) then find:
(i) \(3g + 2f\)
(ii) \(g/f\)
(iii) \(2g + 5f\)
Answer: (i) \((3g + 2f)(x) = 3g(x) + 2f(x) = 3(x^2 - 1) + 2(2x - 3)\)
\(= 3x^2 - 3 + 4x - 6 = 3x^2 + 4x - 9\)
(ii) \(\left(\frac{g}{f}\right)(x) = \frac{g(x)}{f(x)} = \frac{x^2 - 1}{2x - 3}\), where \(2x - 3 \neq 0 \implies x \neq \frac{3}{2}\)
(iii) \((2g + 5f)(x) = 2g(x) + 5f(x) = 2(x^2 - 1) + 5(2x - 3)\)
\(= 2x^2 - 2 + 10x - 15 = 2x^2 + 10x - 17\)
In simple words: Combine the functions algebraically by substituting their expressions directly.

Exam Tip: Always state the domain restriction for quotient functions where the denominator can be zero.

Question 4. Determine a quadratic function \(f(x) = ax^2 + bx + c\) if \(f(0) = 6\), \(f(2) = 11\) and \(f(-3) = 6\).
Answer: Given \(f(x) = ax^2 + bx + c\).
Using \(f(0) = 6\): \(a(0)^2 + b(0) + c = 6 \implies c = 6\).
Using \(f(2) = 11\): \(a(2)^2 + b(2) + 6 = 11 \implies 4a + 2b = 5\).--(1)
Using \(f(-3) = 6\): \(a(-3)^2 + b(-3) + 6 = 6 \implies 9a - 3b = 0 \implies 3a = b\).--(2)
Substitute (2) into (1): \(4a + 2(3a) = 5 \implies 10a = 5 \implies a = \frac{1}{2}\).
From (2), \(b = 3\left(\frac{1}{2}\right) = \frac{3}{2}\).
Thus, the quadratic function is \(f(x) = \frac{1}{2}x^2 + \frac{3}{2}x + 6\).
In simple words: Substitute the given coordinate conditions into the general quadratic equation to form a system of linear equations in a, b, and c.

Exam Tip: Solving for c first using \(f(0)\) simplifies the remaining equations into a two-variable system.

Question 5. If \(R = \{(x, y) : y = x + 1\) and \(y \in \{0, 1, 2, 3, 4, 5\}\}\):
(i) List the elements of R
(ii) Represent R by an arrow diagram
(iii) Find \(R^{-1}\).
Answer: Assuming \(x\) takes values from the same set such that \(y = x + 1 \in \{0, 1, 2, 3, 4, 5\}\):
Possible values for \(y\) are \(0, 1, 2, 3, 4, 5\).
If \(y = 0 \implies x = -1\) (not in set, assuming domain is the same set). Let us take \(x \in \{0, 1, 2, 3, 4, 5\}\).
Then \(y = x + 1\) gives pairs: \((0, 1), (1, 2), (2, 3), (3, 4), (4, 5)\).
(i) \(R = \{(0, 1), (1, 2), (2, 3), (3, 4), (4, 5)\}\)
(ii) Arrow diagram maps elements \(\{0, 1, 2, 3, 4\}\) to \(\{1, 2, 3, 4, 5\}\) with arrows from \(x\) to \(x + 1\).
(iii) \(R^{-1} = \{(1, 0), (2, 1), (3, 2), (4, 3), (5, 4)\}\)
In simple words: The inverse relation \(R^{-1}\) is obtained by simply swapping the x and y coordinates of each pair in R.

Exam Tip: Always double-check that every generated pair satisfies the given condition and set constraints.

Question 6. Let \(f\) be a subset of \(Q \times Z\) defined by \(f = \{(m, n) : m, n \in Z\}\). Is \(f\) a function from \(Q\) to \(Z\)? Justify your answer.
Answer: No, \(f\) is not a function from \(Q\) to \(Z\).
Justification: Let \(m = \frac{1}{2} \in Q\). Since \(\frac{1}{2}\) is not an integer, there is no pair in \(f\) with first element \(\frac{1}{2}\), so elements of \(Q\) do not all have images. Furthermore, a rational number like \(\frac{1}{3}\) cannot be the first coordinate if \(m \in Z\). Thus, not every element of the domain \(Q\) has a unique image in \(Z\).
In simple words: A relation is a function only if every single element in the domain is paired with exactly one element in the codomain.

Question 7. Let \(A = \{6, 7, 8, 10\}\) and \(B = \{2, 4, 5\}\) and \(R\) is a relation from \(A\) to \(B\) defined as \(aRb\) if and only if \(a\) is divisible by \(b\). Write \(R\) in roster form.
Answer: Checking divisibility \(a \div b\) for each \(a \in A\) and \(b \in B\):
For \(a = 6\): divisible by 2 (since \(6/2 = 3\)). Pair: \((6, 2)\).
For \(a = 7\): not divisible by 2, 4, or 5.
For \(a = 8\): divisible by 2 and 4 (since \(8/2 = 4, 8/4 = 2\)). Pairs: \((8, 2), (8, 4)\).
For \(a = 10\): divisible by 2 and 5 (since \(10/2 = 5, 10/5 = 2\)). Pairs: \((10, 2), (10, 5)\).
Roster form: \(R = \{(6, 2), (8, 2), (8, 4), (10, 2), (10, 5)\}\)
In simple words: Test each element of set A against each element of set B to see if the first is evenly divisible by the second.

Exam Tip: "a is divisible by b" means \(a\) is a multiple of \(b\), or \(a/b\) leaves no remainder.

Question 8. If \(A = \{1, 2, 3\}\) and \(B = \{4, 5\}\) and \(C = \{5, 6\}\) verify that:
(i) \(A \times (B \cup C) = (A \times B) \cup (A \times C)\)
(ii) \(A \times (B - C) = (A \times B) - (A \times C)\)
Answer: Given \(A = \{1, 2, 3\}\), \(B = \{4, 5\}\), \(C = \{5, 6\}\).
(i) \(B \cup C = \{4, 5, 6\}\)
LHS: \(A \times (B \cup C) = \{1, 2, 3\} \times \{4, 5, 6\}\)
\(= \{(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)\}\)
RHS: \(A \times B = \{(1,4), (1,5), (2,4), (2,5), (3,4), (3,5)\}\)
\(A \times C = \{(1,5), (1,6), (2,5), (2,6), (3,5), (3,6)\}\)
\((A \times B) \cup (A \times C) = \{(1,4), (1,5), (1,6), (2,4), (2,5), (2,6), (3,4), (3,5), (3,6)\}\)
Hence, LHS = RHS.
(ii) \(B - C = \{4\}\) (elements in B that are not in C)
LHS: \(A \times (B - C) = \{1, 2, 3\} \times \{4\} = \{(1,4), (2,4), (3,4)\}\)
RHS: \((A \times B) - (A \times C)\)
\(= \{(1,4), (1,5), (2,4), (2,5), (3,4), (3,5)\} - \{(1,5), (1,6), (2,5), (2,6), (3,5), (3,6)\}\)
\(= \{(1,4), (2,4), (3,4)\}\)
Hence, LHS = RHS.
In simple words: Cartesian product distributes over union and set difference.

Exam Tip: Always calculate set operations inside parentheses first before performing the Cartesian product.

Question 9. If \(f(x) = \frac{x - 1}{x + 1}\) then show that \(f\left(\frac{1}{x}\right) = -f(x)\).
Answer: Given \(f(x) = \frac{x - 1}{x + 1}\).
Substitute \(x\) with \(\frac{1}{x}\):
\(f\left(\frac{1}{x}\right) = \frac{\frac{1}{x} - 1}{\frac{1}{x} + 1}\)
\(= \frac{\frac{1 - x}{x}}{\frac{1 + x}{x}}\)
\(= \frac{1 - x}{1 + x}\)
\(= -\left(\frac{x - 1}{x + 1}\right) = -f(x)\).
Hence proved.
In simple words: Replace every instance of x with 1/x and simplify the complex fraction to arrive at the negative of the original function.

Exam Tip: Multiplying numerator and denominator by x simplifies complex fractions quickly.

Question 10. Find the domain of following functions:
(i) \(f(x) = \frac{1}{\sqrt{|x| - x}}\)
(ii) \(f(x) = \frac{1}{x^2 + 3x + 2}\)
(iii) \(f(x) = \frac{|x - x|}{2x}\)
Answer: (i) For \(f(x) = \frac{1}{\sqrt{|x| - x}}\), we need \(|x| - x > 0 \implies |x| > x\).
This inequality holds true only when \(x < 0\).
Domain = \((-\infty, 0)\).
(ii) For \(f(x) = \frac{1}{x^2 + 3x + 2}\), the denominator must not be zero:
\(x^2 + 3x + 2 = 0 \implies (x + 1)(x + 2) = 0 \implies x = -1\) or \(x = -2\).
Domain = \(\mathbb{R} \setminus \{-1, -2\}\) or \((-\infty, -2) \cup (-2, -1) \cup (-1, \infty)\).
(iii) For \(f(x) = \frac{|x| - x}{2x}\) (assuming OCR typo for numerator \(|x|-x\)): denominator \(2x \neq 0 \implies x \neq 0\). Also numerator is defined for all reals. Since \(|x|-x = 0\) for \(x > 0\), the function is valid for all \(x \neq 0\).
Domain = \(\mathbb{R} \setminus \{0\}\).
In simple words: Exclude values where denominators become zero or square roots contain negative numbers.

Exam Tip: Factor quadratic denominators completely to easily identify excluded points.

Assignment on Limits and Derivatives

I) Evaluate the limits:

Question 1. \(\lim_{x \to 2} \frac{x^3 - 3x^2 + 4}{x^4 - 8x^2 + 16}\)
Answer: Substitute \(x = 2\): Numerator = \(2^3 - 3(2^2) + 4 = 8 - 12 + 4 = 0\).
Denominator = \(2^4 - 8(2^2) + 16 = 16 - 32 + 16 = 0\) (0/0 indeterminate form).
Factorize numerator: \(x^3 - 3x^2 + 4 = (x - 2)(x^2 - x - 2) = (x - 2)^2(x + 1)\).
Factorize denominator: \(x^4 - 8x^2 + 16 = (x^2 - 4)^2 = (x - 2)^2(x + 2)^2\).
\(\lim_{x \to 2} \frac{(x - 2)^2(x + 1)}{(x - 2)^2(x + 2)^2} = \lim_{x \to 2} \frac{x + 1}{(x + 2)^2}\)
\(= \frac{2 + 1}{(2 + 2)^2} = \frac{3}{16}\).
In simple words: Factor both polynomials completely to cancel out the zero-making factor (x - 2) squared.

Exam Tip: When direct substitution yields 0/0, use synthetic division or polynomial factoring to eliminate the indeterminate factor.

Question 2. \(\lim_{x \to 3} \left( \frac{1}{x - 3} - \frac{2}{x^2 - 4x + 3} \right)\)
Answer: Factor the second denominator: \(x^2 - 4x + 3 = (x - 3)(x - 1)\).
Combine terms under a common denominator:
\(\frac{1}{x - 3} - \frac{2}{(x - 3)(x - 1)} = \frac{(x - 1) - 2}{(x - 3)(x - 1)}\)
\(= \frac{x - 3}{(x - 3)(x - 1)}\)
Cancel \((x - 3)\):
\(\lim_{x \to 3} \frac{1}{x - 1} = \frac{1}{3 - 1} = \frac{1}{2}\).
In simple words: Find a common denominator to combine the rational expressions into a single fraction before evaluating.

Exam Tip: Never evaluate limits of separate terms if it results in infinity minus infinity (\(\infty - \infty\)); always combine them first.

Question 3. \(\lim_{x \to \sqrt{3}} \frac{x^4 - 9}{x^4 + 4\sqrt{3}x - 15}\)
Answer: Substitute \(x = \sqrt{3}\): Numerator = \((\sqrt{3})^4 - 9 = 9 - 9 = 0\).
Denominator = \((\sqrt{3})^4 + 4\sqrt{3}(\sqrt{3}) - 15 = 9 + 12 - 15 = 6 \neq 0\).
Since the denominator is non-zero, direct substitution is valid:
\(= \frac{0}{6} = 0\).
In simple words: Always test direct substitution first; if the denominator is not zero, that value is the limit.

Exam Tip: Do not unnecessarily factor expressions if direct substitution yields a well-defined non-zero denominator.

Question 4. \(\lim_{x \to 3} \frac{x - 3}{\sqrt{x - 2} - \sqrt{4 - x}}\)
Answer: Substitute \(x = 3\): Numerator = \(3 - 3 = 0\).
Denominator = \(\sqrt{3 - 2} - \sqrt{4 - 3} = \sqrt{1} - \sqrt{1} = 0\) (0/0 form).
Rationalize the denominator by multiplying by its conjugate \((\sqrt{x - 2} + \sqrt{4 - x})\):
\(= \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{(\sqrt{x - 2} - \sqrt{4 - x})(\sqrt{x - 2} + \sqrt{4 - x})}\)
\(= \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{(x - 2) - (4 - x)}\)
\(= \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{x - 2 - 4 + x}\)
\(= \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{2x - 6}\)
\(= \lim_{x \to 3} \frac{(x - 3)(\sqrt{x - 2} + \sqrt{4 - x})}{2(x - 3)}\)
Cancel \((x - 3)\):
\(= \lim_{x \to 3} \frac{\sqrt{x - 2} + \sqrt{4 - x}}{2}\)
\(= \frac{\sqrt{3 - 2} + \sqrt{4 - 3}}{2} = \frac{1 + 1}{2} = 1\).
In simple words: Multiply numerator and denominator by the conjugate of the radical expression to eliminate square roots.

Exam Tip: Conjugate multiplication is the standard technique when limits involve square root binomials.

Question 5. \(\lim_{x \to 4} \frac{3 - \sqrt{5 + x}}{1 - \sqrt{5 - x}}\)
Answer: Substitute \(x = 4\): Numerator = \(3 - \sqrt{9} = 0\), Denominator = \(1 - \sqrt{1} = 0\).
Rationalize both numerator and denominator by multiplying by their respective conjugates:
Conjugate of numerator is \((3 + \sqrt{5 + x})\), conjugate of denominator is \((1 + \sqrt{5 - x})\).
\(= \lim_{x \to 4} \frac{(3 - \sqrt{5 + x})(3 + \sqrt{5 + x})(1 + \sqrt{5 - x})}{(1 - \sqrt{5 - x})(1 + \sqrt{5 - x})(3 + \sqrt{5 + x})}\)
\(= \lim_{x \to 4} \frac{(9 - (5 + x))(1 + \sqrt{5 - x})}{(1 - (5 - x))(3 + \sqrt{5 + x})}\)
\(= \lim_{x \to 4} \frac{(4 - x)(1 + \sqrt{5 - x})}{(x - 4)(3 + \sqrt{5 + x})}\)
Note that \((4 - x) = -(x - 4)\):
\(= \lim_{x \to 4} \frac{-(x - 4)(1 + \sqrt{5 - x})}{(x - 4)(3 + \sqrt{5 + x})} = \lim_{x \to 4} \frac{-(1 + \sqrt{5 - x})}{3 + \sqrt{5 + x}}\)
\(= \frac{-(1 + \sqrt{5 - 4})}{3 + \sqrt{5 + 4}} = \frac{-(1 + 1)}{3 + 3} = \frac{-2}{6} = -\frac{1}{3}\).
In simple words: Rationalize both top and bottom terms simultaneously to cancel out the common factor of (x - 4).

Exam Tip: Be careful with negative signs when flipping binomials like (4 - x) to -(x - 4).

Question 6. \(\lim_{x \to 0} \frac{\cos Ax - \cos Bx}{x^2}\)
Answer: Use the trigonometric transformation formula: \(\cos C - \cos D = -2 \sin\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)\).
\(= \lim_{x \to 0} \frac{-2 \sin\left(\frac{Ax + Bx}{2}\right) \sin\left(\frac{Ax - Bx}{2}\right)}{x^2}\)
\(= \lim_{x \to 0} \frac{-2 \sin\left(\frac{A + B}{2}x\right) \sin\left(\frac{A - B}{2}x\right)}{x^2}\)
Rewrite by adjusting denominators using standard limit \(\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1\):
\(= -2 \cdot \lim_{x \to 0} \left[ \frac{\sin\left(\frac{A + B}{2}x\right)}{\frac{A + B}{2}x} \cdot \frac{A + B}{2} \cdot \frac{\sin\left(\frac{A - B}{2}x\right)}{\frac{A - B}{2}x} \cdot \frac{A - B}{2} \right]\)
\(= -2 \cdot 1 \cdot \left(\frac{A + B}{2}\right) \cdot 1 \cdot \left(\frac{A - B}{2}\right)\)
\(= -2 \cdot \frac{A^2 - B^2}{4} = \frac{B^2 - A^2}{2}\).
In simple words: Transform the difference of cosines into a product of sines, then apply the standard limit formula.

Question 7. \(\lim_{x \to 0} \frac{\cot 2x - \csc 2x}{x}\)
Answer: Express cotangent and cosecant in terms of sine and cosine:
\(\cot 2x = \frac{\cos 2x}{\sin 2x}, \quad \csc 2x = \frac{1}{\sin 2x}\)
\(= \lim_{x \to 0} \frac{\frac{\cos 2x - 1}{\sin 2x}}{x} = \lim_{x \to 0} \frac{\cos 2x - 1}{x \sin 2x}\)
Using \(\cos 2x - 1 = -2 \sin^2 x\):
\(= \lim_{x \to 0} \frac{-2 \sin^2 x}{x \sin 2x}\)
Since \(\sin 2x = 2 \sin x \cos x\):
\(= \lim_{x \to 0} \frac{-2 \sin^2 x}{x(2 \sin x \cos x)} = \lim_{x \to 0} \frac{-\sin x}{x \cos x}\)
\(= -\left(\lim_{x \to 0} \frac{\sin x}{x}\right) \cdot \left(\lim_{x \to 0} \frac{1}{\cos x}\right)\)
\(= -(1) \cdot \left(\frac{1}{1}\right) = -1\).
In simple words: Convert trigonometric functions into sine and cosine, apply double angle formulas, and simplify.

Exam Tip: \(\lim_{x \to 0} \frac{\sin x}{x} = 1\) is the fundamental trigonometric limit; try to structure your expression to use it.

Question 8. \(\lim_{x \to \pi} \frac{1 + \cos^3 x}{\sin^2 x}\)
Answer: Substitute \(x = \pi\): Numerator = \(1 + (-1)^3 = 0\), Denominator = \(0^2 = 0\) (0/0 form).
Factor numerator using sum of cubes \(a^3 + b^3 = (a + b)(a^2 - ab + b^2)\):
\(1 + \cos^3 x = (1 + \cos x)(1 - \cos x + \cos^2 x)\)
Denominator \(\sin^2 x = 1 - \cos^2 x = (1 - \cos x)(1 + \cos x)\)
\(= \lim_{x \to \pi} \frac{(1 + \cos x)(1 - \cos x + \cos^2 x)}{(1 - \cos x)(1 + \cos x)}\)
Cancel \((1 + \cos x)\):
\(= \lim_{x \to \pi} \frac{1 - \cos x + \cos^2 x}{1 - \cos x}\)
Substitute \(x = \pi\):
\(= \frac{1 - \cos(\pi) + \cos^2(\pi)}{1 - \cos(\pi)} = \frac{1 - (-1) + (-1)^2}{1 - (-1)} = \frac{1 + 1 + 1}{1 + 1} = \frac{3}{2}\).
In simple words: Factor the sum of cubes in the numerator and use the Pythagorean identity for the denominator.

Exam Tip: Convert all trigonometric functions to cosine when dealing with polynomial-like trigonometric expressions at \(x = \pi\).

Question 9. \(\lim_{x \to \pi} \frac{1 - \sin\left(\frac{x}{2}\right)}{\cos\left(\frac{x}{2}\right)\left[\cos\left(\frac{x}{4}\right) - \sin\left(\frac{x}{4}\right)\right]}\)
Answer: Let \(\frac{x}{4} = \theta\), so \(x = 4\theta\). As \(x \to \pi\), \(\theta \to \frac{\pi}{4}\).
Numerator: \(1 - \sin(2\theta) = (\cos \theta - \sin \theta)^2\) (since \(1 = \sin^2\theta + \cos^2\theta\) and \(\sin 2\theta = 2\sin\theta\cos\theta\)).
Denominator: \(\cos(2\theta)[\cos \theta - \sin \theta] = (\cos^2\theta - \sin^2\theta)(\cos \theta - \sin \theta) = (\cos \theta - \sin \theta)(\cos \theta + \sin \theta)(\cos \theta - \sin \theta)\)
\(= (\cos \theta - \sin \theta)^2(\cos \theta + \sin \theta)\)
Substitute back into the limit expression:
\(= \lim_{\theta \to \frac{\pi}{4}} \frac{(\cos \theta - \sin \theta)^2}{(\cos \theta - \sin \theta)^2(\cos \theta + \sin \theta)}\)
Cancel \((\cos \theta - \sin \theta)^2\):
\(= \lim_{\theta \to \frac{\pi}{4}} \frac{1}{\cos \theta + \sin \theta} = \frac{1}{\cos\left(\frac{\pi}{4}\right) + \sin\left(\frac{\pi}{4}\right)} = \frac{1}{\frac{1}{\sqrt{2}} + \frac{1}{\sqrt{2}}} = \frac{1}{\frac{2}{\sqrt{2}}} = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}\).
In simple words: Substitute a dummy variable to simplify half-angle terms and expand using standard trigonometric identities.

Question 10. \(\lim_{x \to \frac{\pi}{6}} \frac{\sqrt{3} \sin x - \cos x}{x - \frac{\pi}{6}}\)
Answer: Rewrite numerator using auxiliary angle method by dividing and multiplying by 2:
\(\sqrt{3} \sin x - \cos x = 2 \left( \frac{\sqrt{3}}{2} \sin x - \frac{1}{2} \cos x \right) = 2 \left( \sin x \cos\left(\frac{\pi}{6}\right) - \cos x \sin\left(\frac{\pi}{6}\right) \right)\)
Using sine subtraction formula \(\sin(A - B) = \sin A \cos B - \cos A \sin B\):
\(= 2 \sin\left(x - \frac{\pi}{6}\right)\)
Substitute into the limit:
\(= \lim_{x \to \frac{\pi}{6}} \frac{2 \sin\left(x - \frac{\pi}{6}\right)}{x - \frac{\pi}{6}}\)
Let \(x - \frac{\pi}{6} = h\). As \(x \to \frac{\pi}{6}\), \(h \to 0\):
\(= 2 \cdot \lim_{h \to 0} \frac{\sin h}{h} = 2 \cdot 1 = 2\).
In simple words: Transform the numerator into a single sine function using angle subtraction formulas and substitute \(x - \pi/6 = h\).

Exam Tip: Limits of the form \(\frac{f(x) - f(a)}{x - a}\) represent the derivative at a point, or can be solved using simple variable substitution.

Question II. If \(\lim_{x \to 2} \frac{x^4 - 1}{x - 1} = \lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2}\), then find the value of K.
Answer: Evaluate LHS limit \(\lim_{x \to 2} \frac{x^4 - 1}{x - 1}\) (Note: assuming typo in source limit point, evaluating at \(x \to 1\) or substituting directly. If \(x \to 2\), direct substitution gives \(\frac{2^4 - 1}{2 - 1} = 15\)):
Let us evaluate RHS limit \(\lim_{x \to k} \frac{x^3 - k^3}{x^2 - k^2}\):
Factor numerator and denominator:
\(\frac{(x - k)(x^2 + kx + k^2)}{(x - k)(x + k)} = \frac{x^2 + kx + k^2}{x + k}\)
Substitute \(x = k\):
\(= \frac{k^2 + k(k) + k^2}{k + k} = \frac{3k^2}{2k} = \frac{3k}{2}\)
Equating LHS and RHS (assuming LHS limit evaluated at \(x \to 2\) as given):
\(15 = \frac{3k}{2} \implies 3k = 30 \implies k = 10\).
In simple words: Evaluate both sides independently by factoring and substituting, then solve the resulting equation for k.

Exam Tip: Algebraic factoring formulas for cubes and squares are essential for simplifying limits with unknown constants.

III) Differentiate the following by the method of First Principle:

Question 1. \(\sin x\)
Answer: Let \(f(x) = \sin x\). By First Principle, \(f'(x) = \lim_{h \to 0} \frac{f(x + h) - f(x)}{h}\)
\(f'(x) = \lim_{h \to 0} \frac{\sin(x + h) - \sin x}{h}\)
Using transformation formula \(\sin C - \sin D = 2 \cos\left(\frac{C+D}{2}\right) \sin\left(\frac{C-D}{2}\right)\):
\(= \lim_{h \to 0} \frac{2 \cos\left(\frac{2x + h}{2}\right) \sin\left(\frac{h}{2}\right)}{h}\)
\(= \lim_{h \to 0} \cos\left(x + \frac{h}{2}\right) \cdot \frac{\sin\left(\frac{h}{2}\right)}{\frac{h}{2}}\)
\(= \cos(x + 0) \cdot 1 = \cos x\).
In simple words: Apply the definition of derivative using limits and use trigonometric sum-to-product formulas.

Exam Tip: Memorize the standard first principle derivations as they frequently appear in board exams.

Question 2. \(\tan x\)
Answer: Let \(f(x) = \tan x\).
\(f'(x) = \lim_{h \to 0} \frac{\tan(x + h) - \tan x}{h}\)
Convert to sine and cosine:
\(= \lim_{h \to 0} \frac{\frac{\sin(x + h)}{\cos(x + h)} - \frac{\sin x}{\cos x}}{h} = \lim_{h \to 0} \frac{\sin(x + h)\cos x - \cos(x + h)\sin x}{h \cos(x + h)\cos x}\)
Using sine subtraction formula \(\sin(A - B)\):
\(= \lim_{h \to 0} \frac{\sin(x + h - x)}{h \cos(x + h)\cos x} = \lim_{h \to 0} \frac{\sin h}{h} \cdot \frac{1}{\cos(x + h)\cos x}\)
\(= 1 \cdot \frac{1}{\cos x \cos x} = \sec^2 x\).
In simple words: Express tangent as sine over cosine, find a common denominator, and use the sine subtraction identity.

Question 3. \(x^2 \cos x\)
Answer: Let \(f(x) = x^2 \cos x\).
\(f'(x) = \lim_{h \to 0} \frac{(x + h)^2 \cos(x + h) - x^2 \cos x}{h}\)
Add and subtract \(x^2 \cos(x + h)\) in the numerator:
\(= \lim_{h \to 0} \frac{(x + h)^2 \cos(x + h) - x^2 \cos(x + h) + x^2 \cos(x + h) - x^2 \cos x}{h}\)
\(= \lim_{h \to 0} \left[ \frac{[(x + h)^2 - x^2]\cos(x + h)}{h} + \frac{x^2[\cos(x + h) - \cos x]}{h} \right]\)
\(= \lim_{h \to 0} (2x + h)\cos(x + h) + x^2 \lim_{h \to 0} \frac{-2 \sin\left(x + \frac{h}{2}\right)\sin\left(\frac{h}{2}\right)}{h}\)
\(= 2x \cos x + x^2(-2 \sin x \cdot \frac{1}{2}) = 2x \cos x - x^2 \sin x\).
In simple words: Use the addition-subtraction trick to split the limit into parts matching the product rule components.

Exam Tip: The algebraic splitting technique is very helpful for products of functions in first-principle problems.

Question 4. \(\tan x^2\)
Answer: Let \(f(x) = \tan(x^2)\).
\(f'(x) = \lim_{h \to 0} \frac{\tan((x + h)^2) - \tan(x^2)}{h}\)
Convert to sine and cosine:
\(= \lim_{h \to 0} \frac{\frac{\sin((x + h)^2)}{\cos((x + h)^2)} - \frac{\sin(x^2)}{\cos(x^2)}}{h} = \lim_{h \to 0} \frac{\sin((x + h)^2)\cos(x^2) - \cos((x + h)^2)\sin(x^2)}{h \cos((x + h)^2)\cos(x^2)}\)
\(= \lim_{h \to 0} \frac{\sin((x + h)^2 - x^2)}{h \cos((x + h)^2)\cos(x^2)}\)
Since \((x + h)^2 - x^2 = 2xh + h^2\):
\(= \lim_{h \to 0} \frac{\sin(2xh + h^2)}{h} \cdot \frac{1}{\cos((x + h)^2)\cos(x^2)}\)
Multiply and divide by \((2x + h)\):
\(= \lim_{h \to 0} \frac{\sin(2xh + h^2)}{2xh + h^2} \cdot (2x + h) \cdot \frac{1}{\cos((x + h)^2)\cos(x^2)}\)
\(= 1 \cdot (2x + 0) \cdot \frac{1}{\cos^2(x^2)} = 2x \sec^2(x^2)\).
In simple words: Apply trigonometric conversion and adjust the denominator to use the standard sine limit alongside the chain rule result.

Exam Tip: Recognizing composite function structures helps in adjusting terms to match standard limits.

Question 5. \(\frac{2x + 3}{x - 2}\)
Answer: Let \(f(x) = \frac{2x + 3}{x - 2}\).
\(f'(x) = \lim_{h \to 0} \frac{\frac{2(x + h) + 3}{(x + h) - 2} - \frac{2x + 3}{x - 2}}{h}\)
Find common denominator for the numerator:
\(= \lim_{h \to 0} \frac{[2(x + h) + 3](x - 2) - [2x + 3](x + h - 2)}{h(x + h - 2)(x - 2)}\)
Expand numerators:
\([2x + 2h + 3][x - 2] = 2x^2 - 4x + 2xh - 4h + 3x - 6 = 2x^2 + 2xh - x - 4h - 6\)
\([2x + 3][x + h - 2] = 2x^2 + 2xh - 4x + 3x + 3h - 6 = 2x^2 + 2xh - x + 3h - 6\)
Subtract the two expressions:
\((2x^2 + 2xh - x - 4h - 6) - (2x^2 + 2xh - x + 3h - 6) = -7h\)
Substitute back into limit:
\(= \lim_{h \to 0} \frac{-7h}{h(x + h - 2)(x - 2)} = \lim_{h \to 0} \frac{-7}{(x + h - 2)(x - 2)}\)
\(= \frac{-7}{(x - 2)(x - 2)} = \frac{-7}{(x - 2)^2}\).
In simple words: Combine fractions in the numerator, expand and simplify algebraic terms, and cancel h.

Question i) If \(y = \frac{\sec x - \tan x}{\sec x + \tan x}\), show that \(\frac{dy}{dx} = \sec x(\tan x + \sec x)\).
Answer: Simplify \(y\) using sine and cosine:
\(y = \frac{\frac{1}{\cos x} - \frac{\sin x}{\cos x}}{\frac{1}{\cos x} + \frac{\sin x}{\cos x}} = \frac{1 - \sin x}{1 + \sin x}\)
Differentiate using quotient rule or rationalize first:
\(y = \frac{(1 - \sin x)(1 - \sin x)}{(1 + \sin x)(1 - \sin x)} = \frac{(1 - \sin x)^2}{\cos^2 x} = \left(\frac{1 - \sin x}{\cos x}\right)^2 = (\sec x - \tan x)^2\)
Now differentiate with respect to \(x\) using chain rule:
\(\frac{dy}{dx} = 2(\sec x - \tan x) \cdot (\sec x \tan x - \sec^2 x)\)
\(= 2(\sec x - \tan x) \cdot \sec x(\tan x - \sec x)\)
Note that \((\sec x - \tan x) = -(\tan x - \sec x)\):
\(= -2(\sec x - \tan x) \cdot \sec x (\sec x - \tan x) = 2\sec x(\sec x - \tan x)^2\) (Wait, let us check direct quotient rule).
Using quotient rule on \(y = \frac{1 - \sin x}{1 + \sin x}\):
\(\frac{dy}{dx} = \frac{(-\cos x)(1 + \sin x) - (1 - \sin x)(\cos x)}{(1 + \sin x)^2}\)
\(= \frac{-\cos x - \sin x \cos x - \cos x + \sin x \cos x}{(1 + \sin x)^2} = \frac{-2\cos x}{(1 + \sin x)^2}\)
Multiply numerator and denominator by appropriate terms to match target \(\sec x(\tan x + \sec x)\):
\(= \frac{-2\cos x}{(1 + \sin x)^2} \cdot \frac{(1 - \sin x)^2}{(1 - \sin x)^2} = \frac{-2\cos x(1 - \sin x)^2}{\cos^4 x} = \frac{-2(1 - \sin x)^2}{\cos^3 x}\)...
Alternatively, from \(y = (\sec x - \tan x)^2\):
\(\frac{dy}{dx} = 2(\sec x - \tan x)(\sec x \tan x - \sec^2 x) = 2(\sec x - \tan x)\sec x(\tan x - \sec x) = -2\sec x(\sec x - \tan x)^2\).
In simple words: Simplify the trigonometric expression before differentiating to make calculations much easier.

Exam Tip: Simplifying rational trigonometric expressions using basic identities often avoids cumbersome quotient rule expansions.

Question ii) If \(y = \frac{1 - \tan^2\left(\frac{x}{2}\right)}{1 + \tan^2\left(\frac{x}{2}\right)}\), then find \(\frac{dy}{dx}\).
Answer: Recall the double angle identity for cosine: \(\cos 2\theta = \frac{1 - \tan^2\theta}{1 + \tan^2\theta}\).
Here, \(\theta = \frac{x}{2}\), so \(y = \cos\left(2 \cdot \frac{x}{2}\right) = \cos x\).
Differentiate with respect to \(x\):
\(\frac{dy}{dx} = -\sin x\).
In simple words: Recognize that the given fraction is the standard trigonometric expansion for \(\cos x\), reducing the derivative to a single step.

Exam Tip: Memorize standard half-angle and double-angle substitutions in terms of tangent.

Question iii) If \(y = \sqrt{\frac{1 - x}{1 + x}}\), prove that \((1 - x^2)\frac{dy}{dx} + y = 0\).
Answer: Given \(y = \sqrt{\frac{1 - x}{1 + x}} \implies y^2 = \frac{1 - x}{1 + x}\).
Differentiate implicitly with respect to \(x\):
\(2y \frac{dy}{dx} = \frac{(-1)(1 + x) - (1 - x)(1)}{(1 + x)^2}\)
\(2y \frac{dy}{dx} = \frac{-1 - x - 1 + x}{(1 + x)^2} = \frac{-2}{(1 + x)^2}\)
Divide by 2: \(y \frac{dy}{dx} = \frac{-1}{(1 + x)^2}\)
Multiply both sides by \((1 - x^2) = (1 - x)(1 + x)\):
\((1 - x^2) \frac{dy}{dx} = (1 - x)(1 + x) \frac{dy}{dx} = \frac{1 - x}{1 + x} \cdot \frac{-1}{y} = y^2 \cdot \left(\frac{-1}{y}\right) = -y\)
Therefore, \((1 - x^2)\frac{dy}{dx} + y = 0\).
Hence proved.
In simple words: Squaring both sides before differentiating eliminates the square root and simplifies the quotient rule.

Exam Tip: Squaring radical equations prior to differentiation is a powerful technique for proving differential identities.

Question iv) If \(y = \frac{\cos x - \sin x}{\cos x + \sin x}\), prove that \(\frac{dy}{dx} + y^2 + 1 = 0\).
Answer: Divide numerator and denominator by \(\cos x\):
\(y = \frac{1 - \tan x}{1 + \tan x} = \tan\left(\frac{\pi}{4} - x\right)\)
Differentiate with respect to \(x\):
\(\frac{dy}{dx} = \sec^2\left(\frac{\pi}{4} - x\right) \cdot (-1) = -\sec^2\left(\frac{\pi}{4} - x\right)\)
We know that \(\sec^2 \theta = 1 + \tan^2 \theta\), so:
\(\frac{dy}{dx} = -\left(1 + \tan^2\left(\frac{\pi}{4} - x\right)\right) = -(1 + y^2) = -1 - y^2\)
Rearranging terms:
\(\frac{dy}{dx} + y^2 + 1 = 0\).
Hence proved.
In simple words: Convert the expression into a tangent of difference formula to differentiate easily using standard secant squared identities.

Exam Tip: \(\frac{\cos x - \sin x}{\cos x + \sin x} = \tan\left(\frac{\pi}{4} - x\right)\) is a standard trigonometric substitution worth memorizing.

Question v) If \(y = \frac{x}{x + 5}\), prove that \(x \frac{dy}{dx} = y(y - 1)\).
Answer: Find \(\frac{dy}{dx}\) using quotient rule:
\(\frac{dy}{dx} = \frac{(1)(x + 5) - x(1)}{(x + 5)^2} = \frac{x + 5 - x}{(x + 5)^2} = \frac{5}{(x + 5)^2}\)
Now consider RHS: \(y(y - 1) = \left(\frac{x}{x + 5}\right) \left(\frac{x}{x + 5} - 1\right)\)
\(= \left(\frac{x}{x + 5}\right) \left(\frac{x - (x + 5)}{x + 5}\right) = \left(\frac{x}{x + 5}\right) \left(\frac{-5}{x + 5}\right) = \frac{-5x}{(x + 5)^2}\)
Multiply LHS by \(x\):
\(x \frac{dy}{dx} = x \left(\frac{5}{(x + 5)^2}\right) = \frac{5x}{(x + 5)^2}\)
Wait, let us check signs: \(y - 1 = \frac{x - x - 5}{x + 5} = \frac{-5}{x + 5}\), so \(y(y-1) = \frac{-5x}{(x + 5)^2}\).
Thus \(x \frac{dy}{dx} = -y(y-1)\) or depending on exact problem formulation, verify steps. Re-verifying RHS: \(y(y-1) = \frac{x}{x+5}(\frac{-5}{x+5}) = \frac{-5x}{(x+5)^2}\), and \(x \frac{dy}{dx} = \frac{5x}{(x+5)^2}\).
In simple words: Compute the derivative using the quotient rule and substitute both sides independently to verify the identity.

Exam Tip: When proving algebraic differential identities, working on LHS and RHS separately often makes matching terms easier.

Question vi) If \(y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\), prove that \(2xy \frac{dy}{dx} = \frac{x}{a} - \frac{a}{x}\).
Answer: Rewrite \(y = \frac{1}{\sqrt{a}}x^{1/2} + \sqrt{a}x^{-1/2}\).
Differentiate with respect to \(x\):
\(\frac{dy}{dx} = \frac{1}{\sqrt{a}} \left(\frac{1}{2}x^{-1/2}\right) + \sqrt{a} \left(-\frac{1}{2}x^{-3/2}\right)\)
\(= \frac{1}{2\sqrt{a}\sqrt{x}} - \frac{\sqrt{a}}{2x\sqrt{x}} = \frac{1}{2x} \left( \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}} \right)\)
Multiply both sides by \(2xy\):
\(2xy \frac{dy}{dx} = 2xy \cdot \frac{1}{2x} \left( \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}} \right) = y \left( \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}} \right)\)
Substitute \(y = \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}}\):
\(= \left( \sqrt{\frac{x}{a}} + \sqrt{\frac{a}{x}} \right) \left( \sqrt{\frac{x}{a}} - \sqrt{\frac{a}{x}} \right)\)
Using difference of squares \((A + B)(A - B) = A^2 - B^2\):
\(= \left(\sqrt{\frac{x}{a}}\right)^2 - \left(\sqrt{\frac{a}{x}}\right)^2 = \frac{x}{a} - \frac{a}{x}\).
Hence proved.
In simple words: Rewrite radical terms with fractional exponents, differentiate using the power rule, and use algebraic identities.

Exam Tip: Converting roots to fractional powers makes applying the power rule straightforward.