JEE Mathematics Relations and Functions MCQs Set 12

Answer: (d) many-one into

Explanation:
1. \( f(0)=f(3)=2 \), so f is many-one.
2. \( f(x)\ge0 \) so the negative reals are not in the range; f is into.

Teacher's Note:
1. Find a repeated value and note the range is non-negative.

Answer: (a) one-one onto

Explanation:
1. For \( x>e \), \( \log x>1 \), \( \log\log x>0 \) and \( \log\log\log x \) is strictly increasing in x, so f is one-one.
2. As x runs over \( (e,\infty) \), \( \log\log x \) takes all values in \( (0,\infty) \) and its log takes every real value, so f is onto R.

Teacher's Note:
1. Each logarithm is increasing, so the composition is increasing.

Answer: (a) \( f:Z\to Z \) defined by \( y=x+2 \)

Explanation:
1. \( y=x+2 \) is one-one and every integer \( y \) comes from \( x=y-2 \).
2. \( y=2x \) misses odd integers, \( y=x+|x| \) is 0 for all \( x\le0 \), and \( y=x^2 \) is not one-one.

Teacher's Note:
1. Check both injectivity and surjectivity on the stated sets.

Answer: (b) f is one-one onto

Explanation:
1. \( f'(x)=3x^2+6x+12-2\cos x=3(x+1)^2+9-2\cos x>0 \), so f is strictly increasing and one-one.
2. As \( x\to\pm\infty \), \( f\to\pm\infty \), so f is onto.

Teacher's Note:
1. A cubic with a positive derivative everywhere is a bijection of R.

Answer: (a) \( \begin{cases}0,&x>0\\-4x,&x<0\end{cases} \) and 0

Explanation:
1. \( g(x)=0 \) for \( x\ge0 \) and \( -2x \) for \( x<0 \); \( f(t)=2t \) for \( t\ge0 \). So \( f(g(x))=0 \) for \( x>0 \) and \( -4x \) for \( x<0 \).
2. \( f(x)\ge0 \), so \( g(f(x))=|f(x)|-f(x)=0 \).

Teacher's Note:
1. Evaluate the inner function first on each sign of x.

Answer: (b) \( \frac x{\sqrt{1+3x^2}} \)

Explanation:
1. \( f(f(x))=\frac{x}{\sqrt{1+2x^2}} \).
2. \( f(f(f(x)))=\frac{x}{\sqrt{1+3x^2}} \) (each application adds \( x^2 \) to the term under the root).

Teacher's Note:
1. Compute \( f\circ f \) first and watch the pattern.

Answer: (a) \( 7+(4-x)^{1/3} \)

Explanation:
1. Put \( y=4-(x-7)^3 \); then \( (x-7)^3=4-y \).
2. \( x=7+(4-y)^{1/3} \), so \( f^{-1}(x)=7+(4-x)^{1/3} \).

Teacher's Note:
1. Solve \( y=f(x) \) for x and rename the variable.

Answer: (b) \( \{(1,3),(3,1),(4,3)\} \)

Explanation:
1. \( 1\to2\to3 \), so \( (1,3) \).
2. \( 3\to5\to1 \), so \( (3,1) \); \( 4\to1\to3 \), so \( (4,3) \).

Teacher's Note:
1. Follow each element through f and then g.

Answer: (c) \( x \)

Explanation:
1. \( f(f(x))=\frac{4\cdot\frac{4x+3}{6x-4}+3}{6\cdot\frac{4x+3}{6x-4}-4}=\frac{16x+12+18x-12}{24x+18-24x+16}=\frac{34x}{34}=x \).

Teacher's Note:
1. A Mobius function with \( a=-d \) is its own inverse.

Answer: (b) \( g(y)=\frac{4y}{4-3y} \)

Explanation:
1. \( y(3x+4)=4x \) gives \( 3xy+4y=4x \), so \( x(4-3y)=4y \).
2. \( x=\frac{4y}{4-3y} \).

Teacher's Note:
1. Collect the x terms on one side.

Answer: (c) are equal

Explanation:
1. \( f(-1)=2,f(0)=0,f(1)=0,f(2)=2 \).
2. \( g(-1)=2,g(0)=0,g(1)=0,g(2)=2 \).
3. The values agree for every element of A, so \( f=g \).

Teacher's Note:
1. Two functions are equal when they have the same domain, codomain and values.

Answer: (d) \( (0,1) \)

Explanation:
1. For \( 0<x<1 \): \( f(g(x))=\operatorname{sgn}(0)=0 \) but \( g(f(x))=[1]=1 \).
2. At \( x=0 \) both are 0 and at \( x=1 \) both are 1. So they differ exactly on \( (0,1) \).

Teacher's Note:
1. Check the endpoints separately from the interior.

Answer: (d) none of these

Explanation:
1. \( f_1 \) sends \( x=0 \) to \( 1\notin Y \), and \( x=4 \) to \( 5\in Y \), so it is not a map into Y.
2. \( f_2 \) and \( f_3 \) assign several y to one x, so they are not functions.

Teacher's Note:
1. A function gives exactly one image in Y for every x in X.

Answer: (a) 2

Explanation:
1. \( f(x+y)+f(x-y)=\frac{a^{x+y}+a^{-x-y}+a^{x-y}+a^{-x+y}}2=\frac{(a^x+a^{-x})(a^y+a^{-y})}2 \).
2. This is \( 2f(x)f(y) \), so \( k=2 \).

Teacher's Note:
1. Factor the numerator as a product of two sums.

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

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

Teacher's Note:
1. Use the domain to pick the correct root.

Answer: (d) \( f:R\to R^+ \) defined by \( y=e^{|x|} \)

Explanation:
1. The first three are bijections.
2. \( e^{|x|} \) is an even function, so \( f(1)=f(-1) \) and it is not one-one; no inverse exists.

Teacher's Note:
1. A function has an inverse only if it is both one-one and onto.

Answer: (b) \( \frac12\left(1+\sqrt{1+4\log_2x}\right) \)

Explanation:
1. \( y=2^{x^2-x} \) gives \( x^2-x=\log_2y \), so \( x=\frac{1\pm\sqrt{1+4\log_2y}}2 \).
2. Since \( x\ge1 \), take the plus sign.

Teacher's Note:
1. Take logarithms to base 2 and solve the quadratic.

Answer: (c) \( \pi-\sin^{-1}x \)

Explanation:
1. On \( \left[\frac\pi2,\frac{3\pi}2\right] \), \( \sin x=y \) gives \( \pi-x\in\left[-\frac\pi2,\frac\pi2\right] \).
2. So \( \pi-x=\sin^{-1}y \) and \( x=\pi-\sin^{-1}y \).

Teacher's Note:
1. Reduce the angle into the principal range of \( \sin^{-1} \).

Answer: (c) \( \frac{x^2-y^2}4 \)

Explanation:
1. Let \( a=x+y \) and \( b=x-y \), so \( x=\frac{a+b}2 \) and \( y=\frac{a-b}2 \).
2. \( f(a,b)=\frac{(a+b)(a-b)}4=\frac{a^2-b^2}4 \).

Teacher's Note:
1. Change variables to the sum and difference.

Answer: (d) none of these

Explanation:
1. At \( x=2 \): \( f(4)=3\ne1=(f(2))^2 \).
2. At \( x=y=1 \): \( f(2)=1\ne0 \). At \( x=-2 \): \( f(|{-2}|)=1\ne3=|f(-2)| \). So all three fail.

Teacher's Note:
1. Test each identity with a specific number.

Answer: (c) \( \frac1{a^2-b^2}\left[x(a-5b)+\frac1x(5a-b)\right] \)

Explanation:
1. Replace x by \( \frac1x \): \( af\left(\frac1x\right)+bf(x)=\frac1x+5x \).
2. Multiply the first equation by a and the second by b and subtract: \( (a^2-b^2)f(x)=a\left(x+\frac5x\right)-b\left(\frac1x+5x\right)=x(a-5b)+\frac1x(5a-b) \).

Teacher's Note:
1. Substituting \( \frac1x \) gives a second equation; solve the pair for \( f(x) \).

Answer: (d) none of these

Explanation:
1. Let \( A=2\ln x \) and \( B=2\ln y \). The expression is \( \cos A\cos B-\frac12[\cos(A-B)+\cos(A+B)] \).
2. Since \( \cos(A-B)+\cos(A+B)=2\cos A\cos B \), the value is 0.

Teacher's Note:
1. Use the product-to-sum identity.

Answer: (b) \( \frac{3f(x)+1}{f(x)+3} \)

Explanation:
1. Let \( t=f(x) \); then \( x=\frac{1+t}{1-t} \).
2. \( f(2x)=\frac{2x-1}{2x+1}=\frac{2(1+t)-(1-t)}{2(1+t)+(1-t)}=\frac{1+3t}{3+t} \).

Teacher's Note:
1. Express x in terms of \( f(x) \) first.

Answer: (c) \( f(x)=x^m \), \( g(x)=x^n \); \( m\ne n \); \( m,n\in I \)

Explanation:
1. \( f(g(x))=x^{mn}=g(f(x)) \) for the power functions in option (c).
2. The other pairs fail, e.g. \( (x+1)^2\ne x^2+1 \).

Teacher's Note:
1. Powers of x compose by multiplying exponents, which is commutative.

Answer: (b) 1

Explanation:
1. \( g(x)=1+\{x\}\ge1>0 \) for every x.
2. So \( f[g(x)]=1 \).

Teacher's Note:
1. The fractional part is never negative.

Answer: (a) \( f(x)=\sin^2x \), \( g(x)=\sqrt x \)

Explanation:
1. With \( f=\sin^2x \) and \( g=\sqrt x \): \( g(f(x))=\sqrt{\sin^2x}=|\sin x| \).
2. \( f(g(x))=\sin^2\sqrt x \), as required.

Teacher's Note:
1. Try the natural candidates and verify both compositions.

Answer: (a) 28

Explanation:
1. The polynomial solutions are \( f(x)=1\pm x^n \); here \( f(x)=x^n+1 \).
2. \( 5^n+1=126 \) gives \( n=3 \), so \( f(3)=27+1=28 \).

Teacher's Note:
1. The relation is satisfied by \( f(x)=x^n+1 \).

Answer: (c) \( f(x)=\sqrt x \), \( g(x)=\sin^2x \)

Explanation:
1. With \( f=\sqrt x \) and \( g=\sin^2x \): \( f(g(x))=\sqrt{\sin^2x}=|\sin x| \).
2. \( g(f(x))=\sin^2\sqrt x \), as required.

Teacher's Note:
1. Check both compositions for each candidate pair.

Answer: (b) 13

Explanation:
1. \( f(2)=2f(0)-f(1)=1 \).
2. \( f(3)=2f(1)-f(2)=5 \), \( f(4)=2f(2)-f(3)=-3 \).
3. \( f(5)=2f(3)-f(4)=10+3=13 \).

Teacher's Note:
1. Use the recurrence repeatedly.

Answer: (b) \( \sin^{-1}\left(\frac{x-2}2\right)+\frac\pi6 \)

Explanation:
1. \( f(x)=2\sin\left(x-\frac\pi6\right)+2 \), and \( x-\frac\pi6\in\left[\frac\pi6,\frac\pi2\right] \) lies in the principal range of \( \sin^{-1} \).
2. \( x-\frac\pi6=\sin^{-1}\frac{y-2}2 \), so \( f^{-1}(x)=\sin^{-1}\frac{x-2}2+\frac\pi6 \).

Teacher's Note:
1. Write \( \sqrt3\sin x-\cos x \) as a single sine.