JEE Mathematics Functions MCQs Set 18

Answer: (b) 9

Explanation:
1. Reshuffling digits does not change the digit sum, so \( f(n)\equiv n\pmod9 \) and 9 divides \( f(n)-n \).
2. For \( n=12 \): \( f(n)-n=9 \), so no larger number can divide every difference.

Teacher's Note:
1. A number and its digit rearrangement leave the same remainder mod 9.
2. One example with difference 9 caps the answer.

Answer: (c) all odd positive integers n

Explanation:
1. If n is odd, f tends to \( +\infty \) at one end and \( -\infty \) at the other and is continuous, so it takes every real value.
2. If n is even, f is bounded on one side (it has a minimum or a maximum), so it is not onto R.

Teacher's Note:
1. Compare the behaviour of the polynomial at the two ends.
2. Even degree polynomials have a lowest or highest value.

Answer: (c) \( 0 < c < 1 \)

Explanation:
1. Put \( y=f(x) \): \( (y-1)x^2+(4y-2)x+(3cy-c)=0 \). For f to be onto, this must have a real root x for every real y.
2. The discriminant is \( (16-12c)y^2+16(c-1)y+4(1-c)\ge0 \) for all y, which needs \( 16-12c>0 \) and \( 256(c-1)^2-64(4-3c)(1-c)\le0 \).
3. The second condition simplifies to \( -64c(1-c)\le0 \), i.e. \( 0\le c\le1 \).
4. For \( c=0 \) or \( c=1 \) the numerator and denominator share a factor and the value 1 is lost, so we need \( 0 < c < 1 \).

Teacher's Note:
1. Treat y = f(x) as a quadratic in x and demand a real root for every y.
2. Check the end values of c separately.

Answer: (d) 16 + f(0)

Explanation:
1. Let \( h(x)=f(x)-f(x/2) \). The relation reads \( h(x)-h(x/2)=x^2 \). Repeating, \( h(x)=x^2\left(1+\tfrac14+\tfrac1{16}+\cdots\right)+\lim_{k\to\infty}h(x/2^k)=\dfrac43x^2+0 \), because h(0) = 0 and f is continuous.
2. So \( f(x)=\dfrac43x^2+f(x/2) \) and similarly \( f(x)=\dfrac43x^2\cdot\dfrac43+f(0)=\dfrac{16}9x^2+f(0) \).
3. \( f(3)=\dfrac{16}{9}\cdot9+f(0)=16+f(0) \).

Teacher's Note:
1. Define h(x) = f(x) - f(x/2) to turn the relation into a one-step recurrence.
2. Continuity lets you take the limit as the argument tends to 0.

Answer: (a) two solutions

Explanation:
1. \( f(x)=\dfrac{16}9x^2+f(0) \), so the equation becomes \( \dfrac{16}9x^2-x=0 \).
2. \( x\left(\dfrac{16}9x-1\right)=0 \) gives \( x=0 \) and \( x=\dfrac9{16} \): two solutions.

Teacher's Note:
1. Use the explicit form of f.
2. Factor the quadratic.

Answer: (c) \( 4x^{2/3}\left(\dfrac{1+x}{1-x}\right)^{1/3} \)

Explanation:
1. Let \( g(x)=\dfrac{1-x}{1+x} \). Then \( g(g(x))=x \).
2. The given relation: \( f(x)^2f(g)=64x \). Replacing x by g(x): \( f(g)^2f(x)=64g \).
3. Dividing the first by the second: \( \dfrac{f(x)}{f(g)}=\dfrac xg \), so \( f(g)=\dfrac{g\,f(x)}x \).
4. Substituting back: \( f(x)^3\dfrac gx=64x \), so \( f(x)^3=\dfrac{64x^2}{g} \) and \( f(x)=4x^{2/3}\left(\dfrac{1+x}{1-x}\right)^{1/3} \).

Teacher's Note:
1. Use the pairing x and (1 - x)/(1 + x), which is its own inverse.
2. Eliminate f of g(x) by dividing the two equations.

Answer: (a) \( -8\left(\dfrac97\right)^{2/3} \)

Explanation:
1. From \( f(x)=4x^{2/3}\left(\dfrac{1+x}{1-x}\right)^{1/3} \). At \( x=\tfrac97 \): \( \dfrac{1+x}{1-x}=\dfrac{16/7}{-2/7}=-8 \).
2. The real cube root of \( -8 \) is \( -2 \). So \( f\left(\tfrac97\right)=4\left(\tfrac97\right)^{2/3}(-2)=-8\left(\tfrac97\right)^{2/3} \).

Teacher's Note:
1. Derive the formula for f first, then substitute.
2. Take the real cube root of a negative number.

Answer: (c) \( [1,2]\cup(3,5] \)

Explanation:
1. For \( x\in[-1,\sqrt2] \), \( t=g(x)=x^2\in[0,2] \).
2. For \( 0\le t\le1 \): \( f(t)=t+1\in[1,2] \). For \( 1 < t\le2 \): \( f(t)=2t+1\in(3,5] \).
3. The range is \( [1,2]\cup(3,5] \).

Teacher's Note:
1. Find the range of the inner function, then apply f piece by piece.
2. Notice the gap between 2 and 3.

Answer: (c) \( Q_3 \)

Explanation:
1. \( f(x)=0 \) means \( \tan x=x \) (where \( \cos x\ne0 \)). \( f'(x)=x\sin x \), so f increases and is positive on \( (0,\pi] \), and there is no root there.
2. On \( \left(\pi,\tfrac{3\pi}2\right) \), tan x increases from 0 to \( \infty \) and crosses the line y = x once, near \( x\approx4.49 \).
3. Since \( \pi < 4.49 < \tfrac{3\pi}2\approx4.71 \), the root is in the third quadrant.

Teacher's Note:
1. f(x) = 0 is the same as tan x = x.
2. f is positive on (0, pi].

Answer: (a) None of these

Explanation:
1. \( f'(x)=x\sin x>0 \) on \( (0,\pi) \), so f rises from \( f(0)=0 \) and is positive on \( (0,\pi] \).
2. On \( (\pi,2\pi) \), f decreases and becomes zero at \( \alpha\approx4.49 \); beyond it f is negative.
3. So \( f>0 \) exactly on \( (0,\alpha) \), which matches none of the given intervals.

Teacher's Note:
1. Study the sign of the derivative x sin x.
2. f(2 pi) = -2 pi is negative.

Answer: (a) A-q,r; B-r,s; C-p,s; D-p,s

Explanation:
1. (A) \( 2x-x^2-2=-(x-1)^2-1\le-1 \), so \( f(x)\in\left[\tfrac{3\pi}4,\pi\right) \), a proper part of \( \left[\tfrac\pi4,\pi\right) \): into. It is symmetric about \( x=1 \): many-one. So q, r.
2. (B) The factor \( \sin bx \) makes f oscillate with growing amplitude, so it repeats values and takes every real value: many-one and onto. So r, s.
3. (C) \( 2+3x^2 \) is strictly increasing on \( [0,\infty) \) with range \( [2,\infty) \): one-one and onto. So p, s.
4. (D) \( f(f(x))=x \) means f is its own inverse, hence a bijection: one-one and onto. So p, s.

Teacher's Note:
1. Find the range and the repeated values for each function.
2. An involution is invertible, hence a bijection.

Answer: (a) A-q,t; B-s; C-p,q,r; D-p

Explanation:
1. (A) Equality needs \( (x^2-1)\cos x\ge0 \). Of the integers \( -3,\dots,3 \): \( x=\pm1 \) gives \( x^2-1=0 \) (true); \( x=0 \) gives \( (-1)(1) < 0 \); \( |x|=2,3 \) give positive times negative cosine. So the values are 1 and \( -1 \): q, t.
2. (B) Write \( x=n+t \): \( n^2=n+3t \), so \( t=\dfrac{n^2-n}3\in[0,1) \), which needs \( n\in\{-1,0,1,2\} \). Each gives one solution: 4 solutions: s.
3. (C) On \( [-1,1] \), the expression is \( \tfrac\pi2+\tan^{-1}x\in\left[\tfrac\pi4,\tfrac{3\pi}4\right]\approx[0.79,2.36] \), so its integer part is 0, 1 or 2: p, q, r.
4. (D) We need \( \ln(\cos(\sin x))\ge0 \), so \( \cos(\sin x)=1 \), \( \sin x=0 \) and the value of the expression is 0: p.

Teacher's Note:
1. Handle each part separately and write the solution set.
2. In (A), look at the sign of the product (x^2 - 1) cos x.

Answer: (a) A-q; B-r; C-s; D-p

Explanation:
1. (A) For \( -1\le x < 0 \), \( f(x)=x+1\in[0,1) \), so \( f(f(x))=(x+1)^2-1=x^2+2x \): q.
2. (B) The argument is \( \sin2x \). The right side is \( \dfrac{2\cos^2x\left(\sec^2x+2\tan x\right)}2=1+2\sin x\cos x=1+\sin2x \). So \( f(t)=1+t \): r.
3. (C) Let \( h=\sqrt f \). Then \( h(x+y+1)=h(x)+h(y) \), and with continuity \( h(x)=c(x+1) \). \( h(0)=1 \) gives \( c=1 \), so \( f(x)=(x+1)^2 \): s.
4. (D) For \( 4 < x < 5 \), \( \left[\tfrac x4\right]=1 \), so \( f(x)=2x+3 \) and \( f^{-1}(x)=\dfrac{x-3}2 \): p.

Teacher's Note:
1. Simplify each function separately.
2. In (C) the substitution reduces the equation to Cauchy's additive form.

Answer: (a) Both A and R are true, and R is the correct explanation of A.

Explanation:
1. \( f(x)\ge2 \) since \( e^{\cot^2x}\ge1 \). For g: \( \sqrt{2|\sin x|-1}\le1 \) and \( \dfrac{1-\cos2x}{1+\sin^4x}=\dfrac{2\sin^2x}{1+\sin^4x}\le1 \) because \( 1+s^4\ge2s^2 \). So \( g(x)\le2 \).
2. Equality \( f=g \) needs \( f=g=2 \): \( \cot x=0 \), i.e. \( x=(2n+1)\tfrac\pi2 \); there \( g=1+1=2 \). The assertion is true.
3. The reason is true: if \( f\ge k\ge g \) and \( f=g \), then both equal k. It is exactly the argument used in the assertion, so it is the correct explanation.

Teacher's Note:
1. Show f is at least 2 and g is at most 2.
2. Equality forces both values to be 2.

Answer: (c) A is true, but R is false.

Explanation:
1. \( f(1)=f(-1)=1 \), so the assertion is true.
2. The reason is false: \( -2 \) is a pre-image of 2, because \( f(-2)=|-2|=2 \). (It is the negative numbers that are not images of any x.)

Teacher's Note:
1. Find two inputs with the same output.
2. A pre-image of y is any x with f(x) = y.

Answer: (b) 6

Explanation:
1. \( f(x+3)=-f(x) \), so \( f(x+6)=-f(x+3)=f(x) \).
2. The period is 6.

Teacher's Note:
1. Apply the relation twice.
2. The function is anti-periodic with 3 and periodic with 6.

Answer: (b) 6

Explanation:
1. \( \sin(nx) \) has period \( \tfrac{2\pi}n \), and \( \cos(\sin(nx+\pi))=\cos(-\sin nx)=\cos(\sin nx) \), so the numerator has period \( \tfrac\pi n \).
2. \( \tan\left(\tfrac xn\right) \) has period \( n\pi \). Since \( n\pi=n^2\cdot\tfrac\pi n \), the L.C.M. is \( n\pi \).
3. \( n\pi=6\pi \) gives \( n=6 \).

Teacher's Note:
1. Find the period of each factor and take the L.C.M.
2. Cosine of sine has half the period of sine.

Answer: (b) 2

Explanation:
1. \( \sin(\sin\pi x) \) has period 2 (and is not periodic with 1, because \( \sin(\sin\pi(x+1))=-\sin(\sin\pi x) \)).
2. \( e^{\{3x\}} \) has period \( \tfrac13 \). The L.C.M. of 2 and \( \tfrac13 \) is 2 (since \( 2=6\cdot\tfrac13 \)).
3. The period is 2.

Teacher's Note:
1. Find the period of each term separately.
2. Take the L.C.M. of the two periods.

Answer: (c) 1

Explanation:
1. \( x^2-2x+3=(x-1)^2+2>0 \) for all x.
2. So the signum is 1 for every x.

Teacher's Note:
1. Complete the square to see the sign.
2. The signum of a positive number is 1.

Answer: (c) 6

Explanation:
1. \( f(x)=\dfrac{1+x+x^2}x\cdot\dfrac{1+x^4}{x^2}=\left(x+1+\dfrac1x\right)\left(x^2+\dfrac1{x^2}\right) \).
2. For \( x>0 \): \( x+\tfrac1x\ge2 \) so the first factor is at least 3, and \( x^2+\tfrac1{x^2}\ge2 \).
3. Both minima occur at \( x=1 \), so the minimum is \( 3\times2=6 \). So \( K=6 \).

Teacher's Note:
1. Split the expression into two parts of the form t + 1/t.
2. Both parts are smallest at x = 1.

Answer: (c) 5

Explanation:
1. For \( x\in(\sqrt2,\sqrt3) \): \( x^2\in(2,3) \), so \( \{x^2\}=x^2-2 \), and \( \tfrac1x\in(0.57,0.71) \), so \( \left\{\tfrac1x\right\}=\tfrac1x \).
2. \( \tfrac1x=x^2-2 \) gives \( x^3-2x-1=0 \), i.e. \( (x+1)\left(x^2-x-1\right)=0 \), so \( x^2=x+1 \) (with \( x=\tfrac{1+\sqrt5}2\approx1.618 \)).
3. \( x^4=(x+1)^2=3x+2 \) and \( \tfrac1x=x-1 \), so \( \tfrac3x=3x-3 \). Then \( x^4-\tfrac3x=(3x+2)-(3x-3)=5 \).

Teacher's Note:
1. Find which integer parts apply on the given interval.
2. Use x^2 = x + 1 to reduce the powers.

Answer: (b) 3

Explanation:
1. Put \( x=2\cos\theta,\ y=\sin\theta \): \( f=4\cos^2\theta+\sin^2\theta-2\sin\theta\cos\theta=\dfrac52+\dfrac32\cos2\theta-\sin2\theta \).
2. \( \dfrac32\cos2\theta-\sin2\theta \) has amplitude \( \sqrt{\tfrac94+1}=\dfrac{\sqrt{13}}2 \). So \( L=\dfrac{5-\sqrt{13}}2 \) and \( M=\dfrac{5+\sqrt{13}}2 \).
3. \( LM=\dfrac{25-13}4=3 \).

Teacher's Note:
1. Parametrise the ellipse.
2. Use the amplitude of a cos t + b sin t.

Answer: (b) 3

Explanation:
1. The relation is of the form \( f(x+1)+f(x-1)=2\cos\theta\,f(x) \) with \( 2\cos\theta=\sqrt3 \), i.e. \( \theta=\dfrac\pi6 \). The function \( \cos\left(\tfrac{\pi x}6+\varphi\right) \) satisfies it.
2. Its period is \( \dfrac{2\pi}{\pi/6}=12 \). So \( 4\lambda=12 \) and \( \lambda=3 \).

Teacher's Note:
1. Compare with the identity cos(t + a) + cos(t - a) = 2 cos a cos t.
2. The angle is pi/6 per unit step.