Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Functions question

2024 · 29 Jan · Shift 1 · Q40
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Functions
  5. /2024 · 29 Jan · Shift 1 · Q40

Functions question

2024 · 29 Jan · Shift 1 · Q40

JEE MainMathematicsFunctionsMCQ+4 / −1
If f(x)={2+2x,−1≤x<01−x3,0≤x≤3;g(x)={−x,−3≤x≤0x,0<x≤1f(x)=\left\{\begin{array}{cc}2+2 x, & -1 \leq x \lt 0 \\ 1-\frac{x}{3}, & 0 \leq x \leq 3\end{array} ; g(x)=\left\{\begin{array}{cc}-x, & -3 \leq x \leq 0 \\ x, & 0 \lt x \leq 1\end{array}\right.\right.f(x)={2+2x,1−3x​,​−1≤x<00≤x≤3​;g(x)={−x,x,​−3≤x≤00<x≤1​, then range of (fog)(x)(f o g)(x)(fog)(x) is
  1. A
    [0,1)[0,1)[0,1)
  2. B
    [0,3)[0,3)[0,3)
  3. C
    (0,1](0,1](0,1]
  4. D
    [0,1][0,1][0,1]
View written solutionFree

Correct answer: D

  1. We need the range of (f∘g)(x)=f(g(x))(f\circ g)(x)=f(g(x))(f∘g)(x)=f(g(x)).

Given

f(x)={2+2x,−1≤x<01−x3,0≤x≤3f(x)=\begin{cases} 2+2x, & -1\le x<0\\ 1-\dfrac{x}{3}, & 0\le x\le 3 \end{cases}f(x)={2+2x,1−3x​,​−1≤x<00≤x≤3​

and

g(x)={−x,−3≤x≤0x,0<x≤1g(x)=\begin{cases} -x, & -3\le x\le 0\\ x, & 0<x\le 1 \end{cases}g(x)={−x,x,​−3≤x≤00<x≤1​
  1. First find the range of g(x)g(x)g(x), because that becomes the input to fff.
  • For −3≤x≤0-3\le x\le 0−3≤x≤0, g(x)=−xg(x)=-xg(x)=−x, so g(x)∈[0,3]g(x)\in [0,3]g(x)∈[0,3].
  • For 0<x≤10<x\le 10<x≤1, g(x)=xg(x)=xg(x)=x, so g(x)∈(0,1]g(x)\in (0,1]g(x)∈(0,1].

Combining both,

Range(g)=[0,3].\text{Range}(g)=[0,3].Range(g)=[0,3].
  1. Therefore, in f(g(x))f(g(x))f(g(x)), the input to fff is always in [0,3][0,3][0,3]. So only the second branch of fff is used:
f(t)=1−t3for t∈[0,3].f(t)=1-\frac{t}{3}\qquad \text{for } t\in[0,3].f(t)=1−3t​for t∈[0,3].

Hence

(f∘g)(x)=1−g(x)3.(f\circ g)(x)=1-\frac{g(x)}{3}.(f∘g)(x)=1−3g(x)​.
  1. Since g(x)g(x)g(x) takes all values from 000 to 333, let t=g(x)t=g(x)t=g(x) with t∈[0,3]t\in[0,3]t∈[0,3]. Then
y=1−t3,t∈[0,3].y=1-\frac{t}{3}, \qquad t\in[0,3].y=1−3t​,t∈[0,3].

Now evaluate endpoints:

  • At t=0t=0t=0, y=1y=1y=1.
  • At t=3t=3t=3, y=0y=0y=0.

Since this is a linear decreasing function, the full range is

[0,1].[0,1].[0,1].
  1. Checking options:
  • A: [0,1)[0,1)[0,1) — wrong, because 111 is included.
  • B: [0,3)[0,3)[0,3) — wrong.
  • C: (0,1](0,1](0,1] — wrong, because 000 is included.
  • D: [0,1][0,1][0,1] — correct.

Therefore, the range of (f∘g)(x)(f\circ g)(x)(f∘g)(x) is

[0,1].[0,1].[0,1].
PreviousNext

More from Functions

  • If the domain of the function f(x)=cos−1(42−∣x∣​)+{loge​(3−x)}−1 is [−α,β)−{γ}, then α+β+γ is equal to :2024 · MCQ
  • Let A={1,2,3,…,7} and let P(A) denote the power set of A. If the number of functions f:A→P(A) such that a∈f(a),∀a∈A…2024 · Numerical
  • If the domain of the function f(x)=loge​(4x2+x−32x+3​)+cos−1(x+22x−1​) is (α,β], then the value of 5β−4α is equal to2024 · MCQ
  • If f(x)=6x−44x+3​,xeq32​ and (f∘f)(x)=g(x), where g:R−{32​}→R−{32​}, then (gogog)(4) is equal to2024 · MCQ
  • Let f(x)=​1+sin2xsin2xsin2x​cos2x1+cos2xcos2x​sin2xsin2x1+sin2x​​,x∈[6π​,3π​]…2023 · MCQ
  • Let f:R−0,1→R be a function such that f(x)+f(1−x1​)=1+x. Then f(2) is equal to2023 · MCQ
  • Let the sets A and B denote the domain and range respectively of the function f(x)=⌈x⌉−x​1​, where ⌈x⌉ denotes the smallest integer greater than or equal to x. Then among the statements (S1) : A∩B=(1,∞)−N…2023 · MCQ
  • Let R={a,b,c,d,e} and S={1,2,3,4}. Total number of onto functions f:R→S such that f(a)eq1, is equal to ​…2023 · Numerical