JEE Mathematics Functions MCQs Set 21

Answer: (b) \( \frac{x+\sqrt{x^2-4}}{2} \)

Explanation:
1. Put \( y=x+\frac1x \): \( x^2-yx+1=0 \), so \( x=\frac{y\pm\sqrt{y^2-4}}{2} \).
2. Since \( x\ge1 \), take the plus sign: \( f^{-1}(x)=\frac{x+\sqrt{x^2-4}}2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) \( (-3,\infty)-\{-1,-2\} \)

Explanation:
1. The logarithm needs \( x+3>0 \), i.e. \( x>-3 \).
2. The denominator \( (x+1)(x+2) \) is zero at \( -1 \) and \( -2 \), which must be removed.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (b) 2

Explanation:
1. \( f(0)=1 \), so \( g(1)=0 \).
2. \( f'(x)=3x^2+\frac12e^{x/2} \), \( f'(0)=\frac12 \), so \( g'(1)=\frac1{f'(0)}=2 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( \left[-\frac\pi4,\frac\pi4\right] \)

Explanation:
1. Here \( g(f(x))=(\sin x+\cos x)^2-1=\sin2x \).
2. \( \sin2x \) is one-one (increasing) when \( 2x\in\left[-\frac\pi2,\frac\pi2\right] \), i.e. \( x\in\left[-\frac\pi4,\frac\pi4\right] \).

Teacher's Note:
1. Reduce the expression to a standard form before reading off the answer.
2. Verify with a simple check of the final result.

Answer: (d) \( g \) is continuous if \( f \) is continuous

Explanation:
1. The modulus of a continuous function is continuous (composition of continuous functions).
2. Counterexamples for the others: \( f(x)=x \) is onto but \( |x| \) is not; \( f(x)=x \) is one-one but \( |x| \) is not; \( f(x)=x \) is differentiable but \( |x| \) is not differentiable at 0.

Teacher's Note:
1. Test each option with f(x) = x.
2. Only continuity survives taking the absolute value.

Answer: (c) one-to-one and onto

Explanation:
1. \( f'(x)=2+\cos x\ge1>0 \), so \( f \) is strictly increasing, hence one-to-one.
2. It is continuous with \( f(x)\to\pm\infty \) as \( x\to\pm\infty \), so it takes every real value: onto.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) (i) and (iii) only

Explanation:
1. \( f(-x)=\log\frac{1-x}{1+x}=-f(x) \), so \( f \) is odd.
2. \( f(x_1)+f(x_2)=\log\frac{(1+x_1)(1+x_2)}{(1-x_1)(1-x_2)} \); and \( \frac{1+\frac{x_1+x_2}{1+x_1x_2}}{1-\frac{x_1+x_2}{1+x_1x_2}}=\frac{(1+x_1)(1+x_2)}{(1-x_1)(1-x_2)} \), so (iii) holds.
3. Statement (iv) fails; for example it gives \( f(x)^2 \) instead of a logarithm.

Teacher's Note:
1. Inverse hyperbolic tangent has the addition formula in (iii).
2. Test statement (iv) with small values of x.

Answer: (d) \( \sqrt x-1\ \text{for}\ x\ge0 \)

Explanation:
1. The reflection in \( y=x \) is the inverse function. From \( y=(x+1)^2 \) with \( x>-1 \) we get \( x+1=\sqrt y \), so \( g(x)=\sqrt x-1 \) for \( x\ge0 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) odd and strictly increasing in \( (-\infty,\infty) \)

Explanation:
1. \( g(-u)=2\tan^{-1}(e^{-u})-\frac\pi2=2\left(\frac\pi2-\tan^{-1}e^u\right)-\frac\pi2=-g(u) \): odd.
2. \( g'(u)=\frac{2e^u}{1+e^{2u}}>0 \): strictly increasing.

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (c) \( (-\infty,-1]\cup[1,\infty) \)

Explanation:
1. \( f'(x)=3x^2+3(a^2-1) \).
2. For invertibility \( f \) must be monotonic, so \( f'(x)\ge0 \) for all \( x \), which needs \( a^2-1\ge0 \).
3. At \( a=\pm1 \), \( f(x)=x^3+1 \) is still one-one. So \( a\le-1 \) or \( a\ge1 \).

Teacher's Note:
1. A cubic is invertible when it never turns back, i.e. its derivative does not change sign.
2. Check the boundary values a = 1 and a = -1 separately.

Answer: (c) \( (-\infty,1) \)

Explanation:
1. \( 2^y=2-2^x \) must be positive, so \( 2^x<2 \) and \( x<1 \).
2. Every \( x<1 \) gives a real y, so the domain is \( (-\infty,1) \).

Teacher's Note:
1. The exponential 2^y is always positive.
2. This restricts 2 - 2^x to be positive.

Answer: (c) 7

Explanation:
1. For additive \( f \) on rationals, \( f(x)=kx \) with \( k=f(1) \).
2. Then \( \sum_{n=1}^5\sum_{r=1}^nkr^2=k(1+5+14+30+55)=105k=315 \), so \( k=3 \).
3. Hence \( f\left(\frac73\right)=3\cdot\frac73=7 \).

Teacher's Note:
1. Reduce the problem to a standard result and then substitute.
2. Verify the final value with a simple case.

Answer: (d) \( -1 \)

Explanation:
1. \( f(f(x))=\frac{\alpha\cdot\frac{\alpha x}{x+1}}{\frac{\alpha x}{x+1}+1}=\frac{\alpha^2x}{(\alpha+1)x+1} \).
2. Setting equal to \( x \) for all \( x \): \( \alpha^2=(\alpha+1)x+1 \) needs \( \alpha+1=0 \) and \( \alpha^2=1 \). So \( \alpha=-1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) 1

Explanation:
1. \( g(x)=1+\{x\}\ge1>0 \) for all \( x \) (fractional part is non-negative).
2. So \( f(g(x))=1 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.

Answer: (d) \( \frac{7n(n+1)}{2} \)

Explanation:
1. Additivity gives \( f(r)=r\,f(1)=7r \) for natural numbers r.
2. Sum \( =7\cdot\frac{n(n+1)}2 \).

Teacher's Note:
1. Induction on r: f(r + 1) = f(r) + f(1).
2. Then use the sum of the first n natural numbers.

Answer: (c) \( \frac13 \)

Explanation:
1. At \( x=0 \): \( f(0)=2 \). Differentiating: \( e^{-x}(f'-f)=\sqrt{x^4+1} \), so \( f'(0)-2=1 \) and \( f'(0)=3 \).
2. Then \( (f^{-1})'(2)=\frac1{f'(0)}=\frac13 \).

Teacher's Note:
1. Identify the standard result or condition that applies and use it directly.
2. Substitute carefully and verify the answer with a quick check.