Composite Functions

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JEE Advanced 2015

Let $f (x) = sin (\frac{π}{6} sin (\frac{π}{2} sin x))$ for all $x \in R$ and $g (x) = \frac{π}{2} sin x$ for all $x \in R$ . Let $(f \circ g) (x)$ denote $f (g (x))$ and $(g \circ f) (x)$ denote $g (f (x))$ . Then which of the following is (are) true?

f

[- 1/2, 1/2]

(B)

Range of

f \circ g

[- 1/2, 1/2]

(C)

lim_{x \to 0} \frac{f ( x )}{g ( x )} = \frac{π}{6}

(D)

There is an

x \in R

such that

(g \circ f) (x) = 1

JEE Advanced 2011

LEVELJEE Main

Let $f (x) = x^{2}$ and $g (x) = sin x$ for all $x \in R$ . Then the set of all $x$ satisfying $(f \circ g \circ g \circ f) (x) = (g \circ g \circ f) (x)$ , where $(f \circ g) (x) = f (g (x))$ , is

(A)

\pm nπ, n \in {0, 1, 2, ...}

(B)

\pm nπ, n \in {1, 2, ...}

(C)

\frac{π}{2} + 2 nπ, n \in {... - 2, - 1, 0, 1, 2, ...}

(D)

2 nπ, n \in {... - 2, - 1, 0, 1, 2, ...}

JEE Advanced 2001

LEVELJEE Main

Let $f (x) = \frac{α x}{x + 1}, x \neq = - 1$ . Then, for what value of $α$ is $f (f (x)) = x$ ?

(A)

2

(B)

- 2

(C)

1

(D)

- 1

JEE Advanced 1998

LEVELBoard

If $g (f (x)) = ∣ sin x ∣$ and $f (g (x)) = (sin x)^{2}$ , then

(A)

f (x) = sin^{2} x, g (x) = x

(B)

f (x) = sin x, g (x) = ∣ x ∣

(C)

f (x) = x^{2}, g (x) = sin x

(D)

f

and

g

cannot be determined.

JEE Advanced 1994

LEVELJEE Main

Let $f (x) = sin x$ and $g (x) = ln ∣ x ∣$ . If the ranges of the composition functions $f \circ g$ and $g \circ f$ are $R_{1}$ and $R_{2}$ respectively, then

(A)

R_{1} = {u : - 1 \leq u < 1}, R_{2} = {v : - \infty < v < 0}

(B)

R_{1} = {u : - \infty < u < 0}, R_{2} = {v : - 1 \leq v \leq 0}

(C)

R_{1} = {u : - 1 < u < 1}, R_{2} = {v : - \infty < v < 0}

(D)

R_{1} = {u : - 1 \leq u \leq 1}, R_{2} = {v : - \infty < v \leq 0}

JEE Main 2019 (9 January)

LEVELBoard

For $x \in R - {0, 1}$ , let $f_{1} (x) = \frac{1}{x}$ , $f_{2} (x) = 1 - x$ and $f_{3} (x) = \frac{1}{1 - x}$ 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 :-

(A)

f_{3} (x)

(B)

f_{1} (x)

(C)

f_{2} (x)

(D)

\frac{1}{x}

JEE Main 2019 (8 April Shift 1)

LEVELBoard

If $f (x) = lo g_{e} (\frac{1 - x}{1 + x}), ∣ x ∣ < 1$ , then $f (\frac{2 x}{1 + x ^{2}})$ is equal to :

(A)

2 f (x)

(B)

2 f (x^{2})

(C)

(f (x))^{2}

(D)

- 2 f (x)

JEE Main 2019 (12 April Shift 1)

LEVELJEE Main

For $x \in (0, 3/2)$ , let $f (x) = x$ , $g (x) = tan x$ and $h (x) = \frac{1 - x ^{2}}{1 + x ^{2}}$ . If $ϕ (x) = ((h o f) o g) (x)$ , then $ϕ (π /3)$ is equal to :

(A)

tan (π /12)

(B)

tan (7 π /12)

(C)

tan (11 π /12)

(D)

tan (5 π /12)

JEE Main 2020 - 7 Jan (Morning)

LEVELJEE Main

If $g (x) = x^{2} + x - 1$ and $(g o f) (x) = 4 x^{2} - 10 x + 5$ , then $f (\frac{5}{4})$ is equal to.

(A)

\frac{1}{2}

(B)

- \frac{1}{2}

(C)

- \frac{1}{3}

(D)

\frac{1}{3}

JEE Main 2020 - 6 Sep (Evening)

LEVELJEE Main

For a suitably chosen real constant $a$ , let a function, $f : R - {- a} \to R$ be defined by $f (x) = \frac{a - x}{a + x}$ . Further suppose that for any real number $x \neq = - a$ and $f (x) \neq = - a$ , $(f o f) (x) = x$ . Then $f (- \frac{1}{2})$ is equal to

(A)

- 3

(B)

\frac{1}{3}

(C)

- \frac{1}{3}

(D)

3

JEE Main 2021 (February)

LEVELJEE Main

Let $f : R \to R$ be defined as $f (x) = 2 x - 1$ and $g : R - {1} \to R$ be defined as $g (x) = \frac{x - 1}{x}$ . Then the composition function $f (g (x))$ is :

(A)

both one-one and onto

(B)

onto but not one-one

(C)

neither one-one nor onto

(D)

one-one but not onto

JEE Main 2021 (February)

LEVELJEE Main

Let $A = {1, 2, 3, \dots, 10}$ and $f : A \to A$ be defined as $f (k) = {k + 1 k - 1 if k is odd if k is even$ . Then the number of possible functions $g : A \to A$ such that $g \circ f = f$ is:

(A)

1 0^{5}

(B)

10 C_{5}

(C)

5^{5}

(D)

5!

JEE Main 2021 (February)

LEVELJEE Main

Let $f (x) = sin^{- 1} x$ and $g (x) = \frac{x ^{2} - x - 2}{2 x ^{2} - x - 6}$ . If $g (2) = lim_{x \to 2} g (x)$ , then the domain of the function $f \circ g$ is :

(A)

(- \infty, - 2] \cup [- \frac{4}{3}, \infty)

(B)

(- \infty, - 1] \cup [2, \infty)

(C)

(- \infty, - 2] \cup [- 1, \infty)

(D)

(- \infty, - 2] \cup [- \frac{3}{2}, \infty)

JEE Main 2022 (25 June Shift 1)

LEVELJEE Main

Let $f : R \to R$ be a function defined by $f (x) = [2 (1 - \frac{x ^{25}}{2}) (2 + x^{25})]^{\frac{1}{50}}$ . If the function $g (x) = f (f (f (x))) + f (f (x))$ , then the greatest integer less than or equal to $g (1)$ is ______.

JEE Main 2022 (26 June Shift 1)

LEVELJEE Main

Let $f (x) = \frac{x - 1}{x + 1}, x \in R - {0, - 1, 1}$ . If $f^{n + 1} (x) = f (f^{n} (x))$ for all $n \in N$ , then $f^{6} (6) + f^{7} (7)$ is equal to:

(A)

\frac{7}{6}

(B)

- \frac{3}{2}

(C)

\frac{7}{12}

(D)

- \frac{11}{12}

JEE Main 2022 (27 June Shift 2)

LEVELJEE Main

Let $S = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10}$ . Define $f : S \to S$ as $f (n) = {2 n, 2 n - 11, if n = 1, 2, 3, 4, 5 if n = 6, 7, 8, 9, 10$ . Let $g : S \to S$ be a function such that $f o g (n) = {n + 1, n - 1, if n is odd if n is even$ , then $g (10) ((g (1) + g (2) + g (3) + g (4) + g (5))$ is equal to:

JEE Main 2022 (29 June Shift 2)

LEVELJEE Main

Let $f (x)$ and $g (x)$ be two real polynomials of degree 2 and 1 respectively. If $f (g (x)) = 8 x^{2} - 2 x$ , and $g (f (x)) = 4 x^{2} + 6 x + 1$ , then the value of $f (2) + g (2)$ is ______.

JEE Main 2023 (11 Apr Shift 2)

LEVELJEE Main

Let $f$ and $g$ be two functions defined by $f (x) = {x + 1, ∣ x - 1∣, x < 0 x \geq 0$ and $g (x) = {x + 1, 1, x < 0 x \geq 0$ . Then $(g \circ f) (x)$ is

(A)

Continuous everywhere but not differentiable exactly at one point

(B)

Continuous everywhere but not differentiable at

x = 1

(C)

Differentiable everywhere

(D)

Not continuous at

x = 1

JEE Main 2023 (13 April Shift 1)

LEVELBoard

For $x \in R$ , two real valued functions $f (x)$ and $g (x)$ are such that, $g (x) = x + 1$ and $f (g (x)) = x + 3 - x$ . Then $f (0)$ is equal to

- 3

JEE Main 2023 (25 January Shift 1)

LEVELJEE Main

For some $a, b, c \in N$ , let $f (x) = a x^{b} + 3$ and $g (x) = x^{b} + c, x \in R$ . If $(f \circ g)^{- 1} (x) = (\frac{x - 7}{2})^{1/3}$ , then $(f \circ g) (a c) + (g \circ f) (b)$ is equal to

JEE Main 2023 (10 Apr Shift 1)

LEVELJEE Main

If $f (x) = \frac{( t a n 1 ^{\circ} ) ^{x} + l o g _{e} ( 123 )}{x l o g _{e} ( 1234 ) - ( t a n 1 ^{\circ} )}$ , $x > 0$ , then the least value of $f (f (x)) + f (f (\frac{4}{x}))$ is

JEE Main 2024 (04 Apr Shift 2)

LEVELJEE Main

Consider the function $f : R \to R$ defined by $f (x) = \frac{2 x}{1 + 9 x ^{2}}$ . If the composition of $f, 10 times (f \circ f \circ f \circ \dots \circ f) (x) = \frac{2 ^{10} x}{1 + 9 α x ^{2}}$ , then the value of $3 α + 1$ is equal to ______

JEE Main 2024 (01 Feb Shift 1)

LEVELJEE Main

Let $f : R \to R$ and $g : R \to R$ be defined as $f (x) = {lo g_{e} x e^{- x}, x > 0, x \leq 0$ and $g (x) = {x e^{x}, x \geq 0, x < 0$ . Then, $g \circ f : R \to R$ is :

(A)

one-one but not onto

(B)

neither one-one nor onto

(C)

onto but not one-one

(D)

both one-one and onto

JEE Main 2024 (27 Jan Shift 2)

LEVELJEE Main

Let $f : R - {- \frac{1}{2}} \to R$ and $g : R - {- \frac{5}{2}} \to R$ be defined as $f (x) = \frac{2 x + 3}{2 x + 1}$ and $g (x) = \frac{∣ x ∣ + 1}{2 x + 5}$ . Then the domain of the function $f \circ g$ is :

(A)

R - {- \frac{5}{2}}

(B)

R

(C)

R - {- \frac{7}{4}}

(D)

R - {- \frac{5}{2}, - \frac{7}{4}}

JEE Main 2024 (29 Jan Shift 1)

LEVELJEE Main

If $f (x) = {2 + 2 x, 1 - \frac{x}{3}, - 1 \leq x < 0 0 \leq x \leq 3; g (x) = {- x, x, - 3 \leq x \leq 0 0 < x \leq 1$ , then range of $(f \circ g (x))$ is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x2 and g(x)=sinx for all x∈R. Then the set of all x satisfying (f∘g∘g∘f)(x)=(g∘g∘f)(x), where (f∘g)(x)=f(g(x)), is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x+1αx​,x=−1. Then, for what value of α is f(f(x))=x?

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If g(f(x))=∣sinx∣ and f(g(x))=(sinx​)2, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=sinx and g(x)=ln∣x∣. If the ranges of the composition functions f∘g and g∘f are R1​ and R2​ respectively, then

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=loge​(1+x1−x​),∣x∣<1, then f(1+x22x​) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈(0,3/2), let f(x)=x​, g(x)=tanx and h(x)=1+x21−x2​. If ϕ(x)=((hof)og)(x), then ϕ(π/3) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If g(x)=x2+x−1 and (gof)(x)=4x2−10x+5, then f(45​) is equal to.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R be defined as f(x)=2x−1 and g:R−{1}→R be defined as g(x)=xx−1​. Then the composition function f(g(x)) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={1,2,3,…,10} and f:A→A be defined as f(k)={k+1k−1​if k is oddif k is even​. Then the number of possible functions g:A→A such that g∘f=f is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=sin−1x and g(x)=2x2−x−6x2−x−2​. If g(2)=limx→2​g(x), then the domain of the function f∘g is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=x+1x−1​,x∈R−{0,−1,1}. If fn+1(x)=f(fn(x)) for all n∈N, then f6(6)+f7(7) is equal to:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x) and g(x) be two real polynomials of degree 2 and 1 respectively. If f(g(x))=8x2−2x, and g(f(x))=4x2+6x+1, then the value of f(2)+g(2) is ______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f and g be two functions defined by f(x)={x+1,∣x−1∣,​x<0x≥0​ and g(x)={x+1,1,​x<0x≥0​. Then (g∘f)(x) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For x∈R, two real valued functions f(x) and g(x) are such that, g(x)=x​+1 and f(g(x))=x+3−x​. Then f(0) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } For some a,b,c∈N, let f(x)=axb+3 and g(x)=xb+c,x∈R. If (f∘g)−1(x)=(2x−7​)1/3, then (f∘g)(ac)+(g∘f)(b) is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)=xloge​(1234)−(tan1∘)(tan1∘)x+loge​(123)​, x>0, then the least value of f(f(x))+f(f(x4​)) is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R→R and g:R→R be defined as f(x)={loge​xe−x​,x>0,x≤0​ and g(x)={xex​,x≥0,x<0​. Then, g∘f:R→R is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f:R−{−21​}→R and g:R−{−25​}→R be defined as f(x)=2x+12x+3​ and g(x)=2x+5∣x∣+1​. Then the domain of the function f∘g is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If f(x)={2+2x,1−3x​,​−1≤x<00≤x≤3​;g(x)={−x,x,​−3≤x≤00<x≤1​, then range of (f∘g(x)) is

Let $f (x) = x^{2}$ and $g (x) = sin x$ for all $x \in R$ . Then the set of all $x$ satisfying $(f \circ g \circ g \circ f) (x) = (g \circ g \circ f) (x)$ , where $(f \circ g) (x) = f (g (x))$ , is

Let $f (x) = \frac{α x}{x + 1}, x \neq = - 1$ . Then, for what value of $α$ is $f (f (x)) = x$ ?

If $g (f (x)) = ∣ sin x ∣$ and $f (g (x)) = (sin x)^{2}$ , then

Let $f (x) = sin x$ and $g (x) = ln ∣ x ∣$ . If the ranges of the composition functions $f \circ g$ and $g \circ f$ are $R_{1}$ and $R_{2}$ respectively, then

If $f (x) = lo g_{e} (\frac{1 - x}{1 + x}), ∣ x ∣ < 1$ , then $f (\frac{2 x}{1 + x ^{2}})$ is equal to :

For $x \in (0, 3/2)$ , let $f (x) = x$ , $g (x) = tan x$ and $h (x) = \frac{1 - x ^{2}}{1 + x ^{2}}$ . If $ϕ (x) = ((h o f) o g) (x)$ , then $ϕ (π /3)$ is equal to :

If $g (x) = x^{2} + x - 1$ and $(g o f) (x) = 4 x^{2} - 10 x + 5$ , then $f (\frac{5}{4})$ is equal to.

Let $f : R \to R$ be defined as $f (x) = 2 x - 1$ and $g : R - {1} \to R$ be defined as $g (x) = \frac{x - 1}{x}$ . Then the composition function $f (g (x))$ is :

Let $A = {1, 2, 3, \dots, 10}$ and $f : A \to A$ be defined as $f (k) = {k + 1 k - 1 if k is odd if k is even$ . Then the number of possible functions $g : A \to A$ such that $g \circ f = f$ is:

Let $f (x) = sin^{- 1} x$ and $g (x) = \frac{x ^{2} - x - 2}{2 x ^{2} - x - 6}$ . If $g (2) = lim_{x \to 2} g (x)$ , then the domain of the function $f \circ g$ is :

Let $f (x) = \frac{x - 1}{x + 1}, x \in R - {0, - 1, 1}$ . If $f^{n + 1} (x) = f (f^{n} (x))$ for all $n \in N$ , then $f^{6} (6) + f^{7} (7)$ is equal to:

Let $f (x)$ and $g (x)$ be two real polynomials of degree 2 and 1 respectively. If $f (g (x)) = 8 x^{2} - 2 x$ , and $g (f (x)) = 4 x^{2} + 6 x + 1$ , then the value of $f (2) + g (2)$ is ______.

Let $f$ and $g$ be two functions defined by $f (x) = {x + 1, ∣ x - 1∣, x < 0 x \geq 0$ and $g (x) = {x + 1, 1, x < 0 x \geq 0$ . Then $(g \circ f) (x)$ is

For $x \in R$ , two real valued functions $f (x)$ and $g (x)$ are such that, $g (x) = x + 1$ and $f (g (x)) = x + 3 - x$ . Then $f (0)$ is equal to

For some $a, b, c \in N$ , let $f (x) = a x^{b} + 3$ and $g (x) = x^{b} + c, x \in R$ . If $(f \circ g)^{- 1} (x) = (\frac{x - 7}{2})^{1/3}$ , then $(f \circ g) (a c) + (g \circ f) (b)$ is equal to

If $f (x) = \frac{( t a n 1 ^{\circ} ) ^{x} + l o g _{e} ( 123 )}{x l o g _{e} ( 1234 ) - ( t a n 1 ^{\circ} )}$ , $x > 0$ , then the least value of $f (f (x)) + f (f (\frac{4}{x}))$ is

Let $f : R \to R$ and $g : R \to R$ be defined as $f (x) = {lo g_{e} x e^{- x}, x > 0, x \leq 0$ and $g (x) = {x e^{x}, x \geq 0, x < 0$ . Then, $g \circ f : R \to R$ is :

Let $f : R - {- \frac{1}{2}} \to R$ and $g : R - {- \frac{5}{2}} \to R$ be defined as $f (x) = \frac{2 x + 3}{2 x + 1}$ and $g (x) = \frac{∣ x ∣ + 1}{2 x + 5}$ . Then the domain of the function $f \circ g$ is :

If $f (x) = {2 + 2 x, 1 - \frac{x}{3}, - 1 \leq x < 0 0 \leq x \leq 3; g (x) = {- x, x, - 3 \leq x \leq 0 0 < x \leq 1$ , then range of $(f \circ g (x))$ is