JEE Mathematics Relations and Functions MCQs Set 13

Answer: (a) \( 2f(x) \)

Explanation:
1. Substitute \( \frac{2x}{1+x^2} \) in place of x: \( \frac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}} = \frac{(1-x)^2}{(1+x)^2} \).
2. For \( |x| \lt 1 \) this equals \( \left(\frac{1-x}{1+x}\right)^2 \), so \( f\left(\frac{2x}{1+x^2}\right) = \log_e\left(\frac{1-x}{1+x}\right)^2 \).
3. Using \( \log A^2 = 2\log A \) (the base is positive for \( |x| \lt 1 \)), the value is \( 2f(x) \). Answer: (a)

Teacher's Note:
1. Whenever the argument of a function is \( \frac{2x}{1+x^2} \), expect a perfect square to appear after simplifying.
2. Always check that the quantity inside the logarithm is positive before bringing out the power.

Answer: (d) \( 2f_1(x) \cdot f_1(y) \)

Explanation:
1. The even part is \( f_1(x) = \frac{a^x + a^{-x}}{2} \) and the odd part is \( f_2(x) = \frac{a^x - a^{-x}}{2} \).
2. \( f_1(x+y) + f_1(x-y) = \frac{1}{2}\left[a^{x+y} + a^{-x-y} + a^{x-y} + a^{y-x}\right] \).
3. Factorise: \( = \frac{1}{2}\left(a^x + a^{-x}\right)\left(a^y + a^{-y}\right) = 2\cdot\frac{a^x+a^{-x}}{2}\cdot\frac{a^y+a^{-y}}{2} \).
4. This is \( 2f_1(x)f_1(y) \). Answer: (d)

Teacher's Note:
1. Every function splits uniquely as \( \frac{f(x)+f(-x)}{2} + \frac{f(x)-f(-x)}{2} \) (even part plus odd part).
2. Here \( f_1 \) behaves like cosh, so the result is the identity \( \cosh(x+y)+\cosh(x-y) = 2\cosh x\cosh y \).

Answer: (d) 13.9

Explanation:
1. Take the population as 100. Then \( n(A)=25 \), \( n(B)=20 \), \( n(A \cap B)=8 \).
2. Readers of A only: \( 25-8 = 17 \). Readers of B only: \( 20-8 = 12 \).
3. Those who look into advertisements: \( 0.30 \times 17 + 0.40 \times 12 + 0.50 \times 8 = 5.1 + 4.8 + 4 = 13.9 \).
4. So 13.9% of the population look into advertisements. Answer: (d)

Teacher's Note:
1. Split the readers into three disjoint groups (A only, B only, both) before applying the different percentages.
2. Assuming a population of 100 turns percentages directly into counts.

Answer: (c) \( (-1,0) \cup (1,2) \cup (2,\infty) \)

Explanation:
1. The fraction needs \( 4-x^2 \ne 0 \), so \( x \ne \pm 2 \).
2. The logarithm needs \( x^3 - x \gt 0 \), i.e. \( x(x-1)(x+1) \gt 0 \). By the wavy-curve method this holds for \( x \in (-1,0) \cup (1,\infty) \).
3. Remove \( x = 2 \) from \( (1,\infty) \) (and \( x=-2 \) is already outside): \( (-1,0) \cup (1,2) \cup (2,\infty) \). Answer: (c)

Teacher's Note:
1. The domain of a sum of functions is the intersection of the individual domains.
2. Mark the roots on a number line and test the sign in each interval for polynomial inequalities.

Answer: (c) \( g(f(S)) = g(S) \)

Explanation:
1. \( g(S) = \{x : x^2 \in [0,4]\} = [-2, 2] \), so \( f(g(S)) = [0,4] = S \). Statement (a) is true.
2. \( f(S) = [0,16] \) and \( g(f(S)) = \{x : x^2 \in [0,16]\} = [-4,4] \ne S \). Statement (b) is true.
3. \( g(S) = [-2,2] \) but \( g(f(S)) = [-4,4] \), so statement (c) is false.
4. \( f(g(S)) = [0,4] \) while \( f(S) = [0,16] \), so (d) is true. The statement that is not true is (c). Answer: (c)

Teacher's Note:
1. Here g is the pre-image (inverse image) under f, not an inverse function.
2. Compute each set explicitly on the number line and compare.

Answer: (d) If \( (A - C) \subseteq B \), then \( A \subseteq B \)

Explanation:
1. Since \( A \cap B \subseteq C \), we get \( A \cap B \subseteq B \cap C \), so \( B \cap C \ne \varnothing \). Statement (a) is true.
2. \( A = (A \cap B) \cup (A - B) \), both parts lie in C, so \( A \subseteq C \). Statement (b) is true.
3. \( (C \cup A) \cap (C \cup B) = C \cup (A \cap B) = C \). Statement (c) is true.
4. Statement (d) fails: \( A - C \subseteq B \) only says that the part of A outside C lies in B. Take \( A = \{1,2\} \), \( B = \{2\} \), \( C = \{1,2\} \): then \( A \cap B = \{2\} \subseteq C \) and \( A - C = \varnothing \subseteq B \), but \( A \not\subseteq B \). Answer: (d)

Teacher's Note:
1. Venn diagrams are a quick way to test set statements, and a small counter-example settles a false one.
2. Splitting a set as \( A=(A\cap B)\cup(A-B) \) is a useful trick.

Answer: (a) injective but not surjective

Explanation:
1. Injective: if \( \frac{2x_1}{x_1-1} = \frac{2x_2}{x_2-1} \), cross-multiplying gives \( -2x_1 = -2x_2 \), so \( x_1 = x_2 \). Hence f is one-one.
2. Surjective: \( y = \frac{2x}{x-1} \Rightarrow x = \frac{y}{y-2} \). For \( y = 2 \) there is no x, so 2 is not attained (note also \( f(x)=2 \) would need \( x = x-1 \), impossible).
3. Hence f is injective but not surjective. Answer: (a)

Teacher's Note:
1. To test one-one, assume \( f(x_1)=f(x_2) \) and prove \( x_1=x_2 \).
2. For onto, solve for x in terms of y and look for values of y that make the solution impossible.

Answer: (c) -2

Explanation:
1. Differentiate: \( f'(x) = 3x^2 + 2xf'(1) + f''(2) \), \( f''(x) = 6x + 2f'(1) \), \( f'''(x) = 6 \). So \( f'''(3) = 6 \).
2. Put \( x=1 \) in \( f' \): \( f'(1) = 3 + 2f'(1) + f''(2) \). Put \( x=2 \) in \( f'' \): \( f''(2) = 12 + 2f'(1) \).
3. Then \( f'(1) = 3 + 2f'(1) + 12 + 2f'(1) \), so \( 3f'(1) = -15 \), giving \( f'(1) = -5 \) and \( f''(2) = 2 \).
4. \( f(x) = x^3 - 5x^2 + 2x + 6 \), so \( f(2) = 8 - 20 + 4 + 6 = -2 \). Answer: (c)

Teacher's Note:
1. Constants such as f'(1), f''(2) and f'''(3) are just numbers, so differentiate the expression and substitute the right values.
2. Solve the resulting simple linear equations in f'(1).

Answer: (c) 38

Explanation:
1. \( n(M) = \lfloor 140/2 \rfloor = 70 \), \( n(P) = \lfloor 140/3 \rfloor = 46 \), \( n(C) = \lfloor 140/5 \rfloor = 28 \).
2. \( n(M \cap P) = \lfloor 140/6 \rfloor = 23 \), \( n(P \cap C) = \lfloor 140/15 \rfloor = 9 \), \( n(M \cap C) = \lfloor 140/10 \rfloor = 14 \), \( n(M \cap P \cap C) = \lfloor 140/30 \rfloor = 4 \).
3. \( n(M \cup P \cup C) = 70 + 46 + 28 - 23 - 9 - 14 + 4 = 102 \).
4. Students in none of the courses: \( 140 - 102 = 38 \). Answer: (c)

Teacher's Note:
1. Use the inclusion-exclusion formula for three sets.
2. The number of multiples of k up to N is the integer part of N/k.

Answer: (a) \( \left[-\frac{1}{2}, \frac{1}{2}\right] \)

Explanation:
1. Put \( y = \frac{x}{1+x^2} \), i.e. \( yx^2 - x + y = 0 \).
2. For real x the discriminant must be non-negative: \( 1 - 4y^2 \ge 0 \) (for \( y = 0 \) we get \( x = 0 \), which is fine).
3. So \( -\frac{1}{2} \le y \le \frac{1}{2} \). Answer: (a)

Teacher's Note:
1. Writing y in terms of x as a quadratic in x and applying \( D \ge 0 \) is the standard way to find a range.
2. Check the boundary values (here \( y=\pm\frac12 \) at \( x=\pm1 \)).

Answer: (a) \( \frac{1}{12} \)

Explanation:
1. \( \sin^4 x + \cos^4 x = 1 - 2\sin^2 x\cos^2 x \), so \( f_4(x) = \frac{1}{4} - \frac{1}{2}\sin^2 x\cos^2 x \).
2. \( \sin^6 x + \cos^6 x = 1 - 3\sin^2 x\cos^2 x \), so \( f_6(x) = \frac{1}{6} - \frac{1}{2}\sin^2 x\cos^2 x \).
3. Subtracting: \( f_4(x) - f_6(x) = \frac{1}{4} - \frac{1}{6} = \frac{1}{12} \). Answer: (a)

Teacher's Note:
1. Learn the two identities \( \sin^4+\cos^4=1-2s^2c^2 \) and \( \sin^6+\cos^6=1-3s^2c^2 \).
2. The x-dependent terms cancel, which tells you the answer is a constant.

Answer: (a) \( 2^{50}(2^{50} - 1) \)

Explanation:
1. Total non-empty subsets of S: \( 2^{100} - 1 \).
2. The product is odd only if every element is odd. S has 50 odd numbers, so non-empty subsets of odd numbers number \( 2^{50} - 1 \).
3. Required count \( = (2^{100} - 1) - (2^{50} - 1) = 2^{100} - 2^{50} = 2^{50}(2^{50} - 1) \). Answer: (a)

Teacher's Note:
1. When counting an event like 'product is even', subtract the complement (all odd elements) from the total.
2. A set with n elements has \( 2^n - 1 \) non-empty subsets.

Answer: (c) 7

Explanation:
1. Since f changes from increasing to decreasing at \( x = 1 \), \( f'(1) = 0 \).
2. \( f'(x) = 3x^2 - 6(a-2)x + 3a \), so \( 3 - 6(a-2) + 3a = 0 \), i.e. \( 15 - 3a = 0 \) and \( a = 5 \).
3. \( f(x) = x^3 - 9x^2 + 15x + 7 \), so \( f(x) - 14 = x^3 - 9x^2 + 15x - 7 = (x-1)^2(x-7) \).
4. Hence \( \frac{f(x)-14}{(x-1)^2} = x - 7 \), which is zero at \( x = 7 \). Answer: (c)

Teacher's Note:
1. A turning point of a differentiable function has zero derivative.
2. Factorise the cubic using the known double root \( x=1 \).

Answer: (b) \( A \subset B \)

Explanation:
1. A is the open square \( 4 \lt a \lt 6 \), \( 4 \lt b \lt 6 \).
2. B is the region \( \frac{(a-6)^2}{9} + \frac{(b-5)^2}{4} \le 1 \), an ellipse with centre \( (6,5) \), semi-axes 3 along the a-axis and 2 along the b-axis.
3. The corners of the square are \( (4,4), (4,6), (6,4), (6,6) \). At \( (4,4) \): \( \frac{4}{9} + \frac{1}{4} = \frac{25}{36} \lt 1 \); at \( (6,6) \): \( 0 + \frac{1}{4} \lt 1 \). All corners lie inside the ellipse, so the whole square does.
4. B extends beyond the square (for example \( (9,5) \in B \) but not in A), so \( A \subset B \). Answer: (b)

Teacher's Note:
1. Rewrite each condition geometrically: a square and an ellipse.
2. For convex regions it is enough to test the corners of the smaller region.

Answer: (c) surjective but not injective

Explanation:
1. \( f\left(\frac{1}{x}\right) = \frac{1/x}{1 + 1/x^2} = \frac{x}{1+x^2} = f(x) \), so \( f(2) = f\left(\frac{1}{2}\right) \) and f is many-one (not injective).
2. Solving \( y = \frac{x}{1+x^2} \) gives \( yx^2 - x + y = 0 \), which has real roots when \( 1 - 4y^2 \ge 0 \), i.e. \( y \in \left[-\frac{1}{2}, \frac{1}{2}\right] \). So the range equals the codomain.
3. Hence f is surjective but not injective. Answer: (c)

Teacher's Note:
1. If \( f(x)=f(1/x) \) for all x, the function cannot be one-one.
2. Compare the range with the stated codomain to decide on surjectivity.

Answer: (a) 330

Explanation:
1. Put \( y = 0 \): \( f(0) = 0 \), so \( c = 0 \). Put \( y = -x \): \( 0 = f(x) + f(-x) - x^2 \), so \( 2ax^2 = x^2 \) and \( a = \frac{1}{2} \).
2. From \( a + b + c = 3 \): \( b = \frac{5}{2} \). So \( f(n) = \frac{n^2 + 5n}{2} \).
3. \( \sum_{n=1}^{10} f(n) = \frac{1}{2}\cdot\frac{10 \cdot 11 \cdot 21}{6} + \frac{5}{2}\cdot\frac{10 \cdot 11}{2} = \frac{385}{2} + \frac{275}{2} = 330 \). Answer: (a)

Teacher's Note:
1. Special substitutions such as \( y=0 \) and \( y=-x \) quickly fix the coefficients in a functional equation.
2. Use the sum formulas for \( \sum n \) and \( \sum n^2 \) once f(n) is known.

Answer: (c) contains exactly two elements

Explanation:
1. Replace x by \( \frac{1}{x} \): \( f\left(\frac{1}{x}\right) + 2f(x) = \frac{3}{x} \).
2. Eliminate \( f(1/x) \) between the two equations: \( 3f(x) = \frac{6}{x} - 3x \), so \( f(x) = \frac{2}{x} - x \).
3. \( f(x) = f(-x) \) gives \( \frac{2}{x} - x = -\frac{2}{x} + x \), i.e. \( \frac{4}{x} = 2x \) and \( x^2 = 2 \).
4. So \( x = \pm\sqrt{2} \) and S has exactly two elements. Answer: (c)

Teacher's Note:
1. When f(x) and f(1/x) both appear, swap x and 1/x to get a second equation and eliminate.
2. Find f(x) explicitly before solving the condition \( f(x)=f(-x) \).

Answer: (d) Y

Explanation:
1. \( X = \{0, 9, 54, 243, \dots\} \) (for \( n = 1, 2, 3, 4, \dots \)) and \( Y = \{0, 9, 18, 27, \dots\} \).
2. Using the binomial theorem, \( 4^n - 3n - 1 = (1+3)^n - 3n - 1 = 9\left[{}^nC_2 + 3\,{}^nC_3 + \dots\right] \), which is a multiple of 9.
3. So every element of X is a multiple of 9 and hence belongs to Y, i.e. \( X \subset Y \). Therefore \( X \cup Y = Y \). Answer: (d)

Teacher's Note:
1. Expanding \( (1+3)^n \) by the binomial theorem shows that \( 4^n-3n-1 \) is divisible by 9.
2. If \( X\subset Y \) then \( X\cup Y=Y \).

Answer: (d) \( 1 + \{g(x)\}^5 \)

Explanation:
1. Since g is the inverse of f, \( f(g(x)) = x \).
2. Differentiate: \( f'(g(x)) \cdot g'(x) = 1 \), so \( g'(x) = \frac{1}{f'(g(x))} \).
3. \( f'(g(x)) = \frac{1}{1 + \{g(x)\}^5} \), hence \( g'(x) = 1 + \{g(x)\}^5 \). Answer: (d)

Teacher's Note:
1. The derivative of an inverse function is the reciprocal of the derivative of f evaluated at \( g(x) \).
2. Do not replace \( g(x) \) by x in \( f' \); the argument is \( g(x) \).

Answer: (b) Statement I is true, Statement II is false

Explanation:
1. Statement I: reflexive since \( x - x = 0 \) is an integer; symmetric since \( y - x \in Z \Rightarrow x - y \in Z \); transitive since the sum of two integers \( (y-x) + (z-y) = z - x \) is an integer. So A is an equivalence relation and Statement I is true.
2. Statement II: B is reflexive (take \( \alpha = 1 \)) but not symmetric. For example \( (0, 1) \in B \) because \( 0 = 0 \cdot 1 \), but \( (1, 0) \notin B \) because \( 1 = \alpha \cdot 0 \) is impossible.
3. So B is not an equivalence relation and Statement II is false. Answer: (b)

Teacher's Note:
1. To disprove an equivalence relation, one failed property with a concrete counter-example is enough.
2. The relation B fails symmetry because zero can be written as a multiple of anything but not the other way round.

Answer: (b) \( (-\infty, 0) \)

Explanation:
1. The expression under the root must be strictly positive: \( |x| - x \gt 0 \), i.e. \( |x| \gt x \).
2. This is true only when \( x \lt 0 \) (for \( x \ge 0 \), \( |x| = x \)).
3. Domain \( = (-\infty, 0) \). Answer: (b)

Teacher's Note:
1. A square root in the denominator needs a strictly positive radicand.
2. \( |x|>x \) holds exactly when x is negative.

Answer: (c) Statement I is true, Statement II is true; Statement II is not a correct explanation of Statement I

Explanation:
1. Reflexive: take \( P = I \), then \( A = I^{-1}AI \).
2. Symmetric: \( A = P^{-1}BP \Rightarrow B = PAP^{-1} = (P^{-1})^{-1}A(P^{-1}) \), so \( (B, A) \in R \) with the invertible matrix \( P^{-1} \).
3. Transitive: \( A = P^{-1}BP \) and \( B = Q^{-1}CQ \) give \( A = (QP)^{-1}C(QP) \), so \( (A, C) \in R \). Hence Statement I is true.
4. Statement II is the standard reversal law and is true. It is used in the transitivity step, but it does not explain why R is reflexive and symmetric, so it is not a complete explanation. Answer: (c)

Teacher's Note:
1. Matrices related by \( A=P^{-1}BP \) are called similar, and similarity is an equivalence relation.
2. In assertion-type questions, check each statement separately and only then decide whether one explains the other.