Q: For x \in (0, \pi/2), let f(x)=\sqrt{x}, g(x)=\tan x and h(x)=\dfrac{1-x^2}{1+x^2}. If \varphi(x)=((h\circ f)\circ g)(x), then \varphi\left(\dfrac{\pi}{3}\right) is equal to

\tan\dfrac{11\pi}{12}. \varphi(x)=h(f(g(x)))=h(\sqrt{\tan x})=\dfrac{1-\tan x}{1+\tan x}=\tan\left(\dfrac{\pi}{4}-x\right). So \varphi\left(\dfrac{\pi}{3}\right)=\tan\left(\dfrac{\pi}{4}-\dfrac{\pi}{3}\right)=\tan\left(-\dfrac{\pi}{12}\right)=-\tan\dfrac{\pi}{12}=\tan\dfrac{11\pi}{12}.

Q: If f(x)=\log_e\left(\dfrac{1-x}{1+x}\right), |x|<1, then f\left(\dfrac{2x}{1+x^2}\right) is equal to

2f(x). f\left(\dfrac{2x}{1+x^2}\right)=\log_e\left(\dfrac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right)=\log_e\left(\dfrac{(1-x)^2}{(1+x)^2}\right)=2\log_e\left(\dfrac{1-x}{1+x}\right)=2f(x).

Q: For x\in R-\{0,1\}, let f_1(x)=\dfrac{1}{x}, f_2(x)=1-x and f_3(x)=\dfrac{1}{1-x} be three given functions. If a function J(x) satisfies (f_2\circ J\circ f_1)(x)=f_3(x), then J(x) is equal to

f_3(x). f_2(J(f_1(x)))=f_3(x)\Rightarrow 1-J\left(\dfrac1x\right)=\dfrac{1}{1-x}\Rightarrow J\left(\dfrac1x\right)=1-\dfrac{1}{1-x}=\dfrac{-x}{1-x}. Putting \dfrac1x=X gives J(X)=\dfrac{1}{1-X}=f_3(X), so J(x)=f_3(x).

Q: Let a,b,c\in R. If f(x)=ax^2+bx+c is such that a+b+c=3 and f(x+y)=f(x)+f(y)+xy, \forall\, x,y\in R, then \displaystyle\sum_{n=1}^{10} f(n) is equal to

330. Comparing f(x+y)=f(x)+f(y)+xy with f(x)=ax^2+bx+c gives c=0 and 2a=1\Rightarrow a=\dfrac12; with a+b+c=3, b=\dfrac52. So f(x)=\dfrac{x^2}{2}+\dfrac{5x}{2}. \sum_{n=1}^{10}f(n)=\dfrac12(385)+\dfrac52(55)=192.5+137.5=330.

Q: Let f(x)=x^2 and g(x)=\sin x for all x\in R. Then, the set of all x satisfying (fogogof)(x)=(gogof)(x), where (fog)(x)=f(g(x)), is

\pm\sqrt{n\pi}, n\in\{0,1,2,\ldots\}. (gof)(x)=\sin x^2, so (gogof)(x)=\sin(\sin x^2) and (fogogof)(x)=(\sin(\sin x^2))^2. Setting them equal: \sin(\sin x^2)\{\sin(\sin x^2)-1\}=0\Rightarrow \sin(\sin x^2)=0 (since \sin(\sin x^2)=1 is impossible as |\sin\theta|\le1 strictly less near required value) \Rightarrow x^2=n\pi\Rightarrow x=\pm\sqrt{n\pi}, n\in\{0,1,2,\ldots\}.

Q: Let f(x)=\dfrac{\alpha x}{x+1}, x\neq -1. Then, for what value of \alpha is f[f(x)]=x?

-1. f[f(x)]=\dfrac{\alpha^2x}{(\alpha+1)x+1}=x\Rightarrow x[\alpha^2-(\alpha+1)x-1]=0 must hold for all x \Rightarrow \alpha+1=0 and \alpha^2-1=0\Rightarrow \alpha=-1 (checking, \alpha=1 does not satisfy the identity).

Question 1

For x \in (0, \pi/2), let f(x)=\sqrt{x}, g(x)=	an x and h(x)=\dfrac{1-x^2}{1+x^2}. If \varphi(x)=((h\circ f)\circ g)(x), then \varphi\left(\dfrac{\pi}{3}ight) is equal to

Accepted Answer

an\dfrac{11\pi}{12}. \varphi(x)=h(f(g(x)))=h(\sqrt{	an x})=\dfrac{1-	an x}{1+	an x}=	an\left(\dfrac{\pi}{4}-xight). So \varphi\left(\dfrac{\pi}{3}ight)=	an\left(\dfrac{\pi}{4}-\dfrac{\pi}{3}ight)=	an\left(-\dfrac{\pi}{12}ight)=-	an\dfrac{\pi}{12}=	an\dfrac{11\pi}{12}.

Question 2

Let f(x)=x^2, x\in R. For any A\subseteq R, define g(A)=\{x\in R: f(x)\in A\}. If S=[0,4], then which one of the following statements is not true?

Accepted Answer

g(f(S))=g(S). g(S)=\{x: x^2\in[0,4]\}=[-2,2], so f(g(S))=[0,4]=S. Also f(S)=[0,16], so g(f(S))=\{x: x^2\in[0,16]\}=[-4,4]
eq g(S)=[-2,2]. Hence g(f(S))=g(S) is not true.

Question 3

Let \displaystyle\sum_{k=1}^{10} f(a+k)=16(2^{10}-1), where the function f satisfies f(x+y)=f(x)f(y) for all natural numbers x,y and f(1)=2. Then, the natural number 'a' is

Accepted Answer

3. Since f(x+y)=f(x)f(y) and f(1)=2, we get f(x)=2^x. So \sum_{k=1}^{10}f(a+k)=2^a\sum_{k=1}^{10}2^k=2^a\cdot\dfrac{2(2^{10}-1)}{2-1}=2^{a+1}(2^{10}-1). Equating to 16(2^{10}-1) gives 2^{a+1}=16=2^4\Rightarrow a+1=4\Rightarrow a=3.

Question 4

If f(x)=\log_e\left(\dfrac{1-x}{1+x}\right), |x|<1, then f\left(\dfrac{2x}{1+x^2}\right) is equal to

Accepted Answer

2f(x). f\left(\dfrac{2x}{1+x^2}\right)=\log_e\left(\dfrac{1-\frac{2x}{1+x^2}}{1+\frac{2x}{1+x^2}}\right)=\log_e\left(\dfrac{(1-x)^2}{(1+x)^2}\right)=2\log_e\left(\dfrac{1-x}{1+x}\right)=2f(x).

Question 5

For x\in R-\{0,1\}, let f_1(x)=\dfrac{1}{x}, f_2(x)=1-x and f_3(x)=\dfrac{1}{1-x} be three given functions. If a function J(x) satisfies (f_2\circ J\circ f_1)(x)=f_3(x), then J(x) is equal to

Accepted Answer

f_3(x). f_2(J(f_1(x)))=f_3(x)\Rightarrow 1-J\left(\dfrac1x\right)=\dfrac{1}{1-x}\Rightarrow J\left(\dfrac1x\right)=1-\dfrac{1}{1-x}=\dfrac{-x}{1-x}. Putting \dfrac1x=X gives J(X)=\dfrac{1}{1-X}=f_3(X), so J(x)=f_3(x).

Question 6

Let a,b,c\in R. If f(x)=ax^2+bx+c is such that a+b+c=3 and f(x+y)=f(x)+f(y)+xy, \forall\, x,y\in R, then \displaystyle\sum_{n=1}^{10} f(n) is equal to

Accepted Answer

330. Comparing f(x+y)=f(x)+f(y)+xy with f(x)=ax^2+bx+c gives c=0 and 2a=1\Rightarrow a=\dfrac12; with a+b+c=3, b=\dfrac52. So f(x)=\dfrac{x^2}{2}+\dfrac{5x}{2}. \sum_{n=1}^{10}f(n)=\dfrac12(385)+\dfrac52(55)=192.5+137.5=330.

Question 7

Let f(x)=x^2 and g(x)=\sin x for all x\in R. Then, the set of all x satisfying (fogogof)(x)=(gogof)(x), where (fog)(x)=f(g(x)), is

Accepted Answer

\pm\sqrt{n\pi}, n\in\{0,1,2,\ldots\}. (gof)(x)=\sin x^2, so (gogof)(x)=\sin(\sin x^2) and (fogogof)(x)=(\sin(\sin x^2))^2. Setting them equal: \sin(\sin x^2)\{\sin(\sin x^2)-1\}=0\Rightarrow \sin(\sin x^2)=0 (since \sin(\sin x^2)=1 is impossible as |\sin	heta|\le1 strictly less near required value) \Rightarrow x^2=n\pi\Rightarrow x=\pm\sqrt{n\pi}, n\in\{0,1,2,\ldots\}.

Question 8

Let f(x)=\dfrac{\alpha x}{x+1}, x
eq -1. Then, for what value of \alpha is f[f(x)]=x?

Accepted Answer

-1. f[f(x)]=\dfrac{\alpha^2x}{(\alpha+1)x+1}=x\Rightarrow x[\alpha^2-(\alpha+1)x-1]=0 must hold for all x \Rightarrow \alpha+1=0 and \alpha^2-1=0\Rightarrow \alpha=-1 (checking, \alpha=1 does not satisfy the identity).

Question 9

Let g(x)=1+x-[x] and f(x)=\begin{cases}-1, & x<0\0,& x=0\1,& x>0\end{cases}, then for all x, f[g(x)] is equal to

Accepted Answer

1. g(x)=1+(x-[x]) is always greater than 1 since the fractional part x-[x]\in[0,1), so g(x)>0 for all x. Hence f[g(x)]=1 for all x.

Question 10

If g\{f(x)\}=|\sin x| and f\{g(x)\}=(\sin\sqrt{x})^2, then

Accepted Answer

f and g cannot be determined. Each of the pairs f(x)=\sin^2x,\,g(x)=\sqrt x; f(x)=\sin x,\,g(x)=|x|; and f(x)=x^2,\,g(x)=\sin\sqrt x satisfies both given relations, so f and g are not uniquely determined.

Question 11

If f(x)=\cos(\log x), then f(x)f(y)-\dfrac12\left[f\left(\dfrac{x}{y}\right)+f(xy)\right] has the value

Accepted Answer

None of these. f(x)f(y)-\dfrac12\left[f(x/y)+f(xy)\right]=\cos(\log x)\cos(\log y)-\dfrac12\left[2\cos(\log x)\cos(\log y)\right]=0, which is none of the given values.

Question 12

Let f(x)=|x-1|. Then,

Accepted Answer

None of the above. f(x^2)=|x^2-1|
eq (x-1)^2=\{f(x)\}^2 in general, so (a) is false; f(x+y)=|x+y-1|
eq f(x)+f(y) in general, so (b) is false; f(|x|)=||x|-1|
eq |x-1|=|f(x)| in general, so (c) is false. Hence, the answer is none of the above.

Question 13

Let f(x)=\sin\left(\dfrac{\pi}{6}\sin\left(\dfrac{\pi}{2}\sin x\right)\right) for all x\in R and g(x)=\dfrac{\pi}{2}\sin x for all x\in R. Let (fog)(x) denote f\{g(x)\} and (gof)(x) denote g\{f(x)\}. Then, which of the following is/are true?

Accepted Answer

Range of f is \left[-\dfrac12,\dfrac12ight]. Since \dfrac{\pi}{2}\sin x\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}ight], f(x)\in\left[-\dfrac12,\dfrac12ight], so (a) is true. fog(x)=f(t) with t=g(x)\in\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}ight], which also gives range \left[-\dfrac12,\dfrac12ight], so (b) is true. \lim_{x	o0}\dfrac{f(x)}{g(x)}=\dfrac{\pi}{6} using \lim_{	heta	o0}\sin	heta/	heta=1, so (c) is true. (gof)(x)=1\Rightarrow \sin\{f(x)\}=\dfrac2\pi, but f(x)\in\left[-\dfrac12,\dfrac12ight] makes \sin\{f(x)\}\in\left[-\dfrac12,\dfrac12ight] approximately, which never equals \dfrac2\pi\approx0.637, so (d) is false.

Question 14

If f(x)=\cos[\pi^2]x+\cos[-\pi^2]x, where [x] stands for the greatest integer function, then

Accepted Answer

f\left(\dfrac{\pi}{2}ight)+1=0. Since [\pi^2]=9 and [-\pi^2]=-10, f(x)=\cos9x+\cos10x. Then f(\pi/2)=\cos(9\pi/2)+\cos5\pi=0+(-1)=-1\Rightarrow f(\pi/2)+1=0; f(\pi)=\cos9\pi+\cos10\pi=-1+1=0\Rightarrow f(\pi)-1=-1
eq0; f(\pi/4)=\cos(9\pi/4)+\cos(10\pi/4)=\dfrac{1}{\sqrt2}+0=\dfrac{1}{\sqrt2}. Hence, (a) and (c) are correct.

Question 15

Let g(x) be a function defined on [-1,1]. If the area of the equilateral triangle with two of its vertices at (0,0) and [x, g(x)] is \dfrac{\sqrt3}{4}, then the function g(x) is

Accepted Answer

g(x)=\sqrt{1-x^2}. Area =\dfrac{\sqrt3}{4}(BC)^2=\dfrac{\sqrt3}{4}\Rightarrow BC=1\Rightarrow x^2+g^2(x)=1\Rightarrow g(x)=\pm\sqrt{1-x^2}. Hence, (b) and (c) are correct.

Question 16

Given, y=f(x)=\dfrac{x+2}{x-1}, then

Accepted Answer

x=f(y). Solving y=\dfrac{x+2}{x-1} for x gives x(y-1)=y+2\Rightarrow x=\dfrac{y+2}{y-1}=f(y), so (a) is true. f(1) is not defined since the domain is R-\{1\}, so (b) is false. \dfrac{dy}{dx}=\dfrac{-3}{(x-1)^2}<0, so y decreases with x, making (c) false. f is clearly a rational function of x, so (d) is true.

Composite of Functions — JEE Main Maths PYQ MCQs with Solutions

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