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Mathematics · Functions

JEE Advanced 2024 — Paper 1 — Question 17

Let f:R→Rf: R \rightarrow R and g:R→Rg: R \rightarrow R be functions defined by

f(x)={x∣x∣sin⁡(1x),x≠0,0,x=0, and g(x)={1−2x,0≤x≤12,0, otherwise f(x)=\left\{\begin{array}{cl} x|x| \sin \left(\frac{1}{x}\right), & x \neq 0, \\ 0, & x=0, \end{array} \text { and } g(x)=\left\{\begin{array}{cc} 1-2 x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text { otherwise } \end{array}\right.\right.

Let a,b,c,d∈Ra, b, c, d \in R. Define the function h:R→Rh: R \rightarrow R by

h(x)=af(x)+b(g(x)+g(12−x))+c(x−g(x))+dg(x),x∈Rh(x)=a f(x)+b\left(g(x)+g\left(\frac{1}{2}-x\right)\right)+c(x-g(x))+d g(x), x \in R

Match each entry in List-I to the correct entries in List-II.

List - IList - II
(P) If a=0,b=1,c=0a=0, b=1, c=0 and d=0d=0, then(1) h is one-one.
(Q) If a=1,b=0,c=0a=1, b=0, c=0 and d=0d=0, then(2) h is onto.
(R) If a=0,b=0,c=1a=0, b=0, c=1 and d=0d=0, then(3) hh is differentiable on RR.
(S) If a=0,b=0,c=0a=0, b=0, c=0 and d=1d=1, then(4) the range of hh is [0,1][0,1].
(5) the range of hh is {0,1}\{0,1\}.

The correct option is:

  1. Option A:

    (P) →\rightarrow (4)(Q) \rightarrow(3)$$(R) \rightarrow(1)$$(S) \rightarrow(2)

  2. Option B:

    (P) \rightarrow(5)$$(Q) \rightarrow($$(R) \rightarrow(4) \quad(S) \rightarrow(3)

  3. Option C:

    (P) →\rightarrow (5)(Q) →\rightarrow (3)(R)→(2)(S)→(4)(R) \rightarrow(2) \quad(S) \rightarrow(4)

    Correct
  4. Option D:

    (P) →(4)\rightarrow(4)(Q) →\rightarrow (2)(R) \rightarrow(1)$$(S) \rightarrow(3)

Answer: C

Step-by-step solution

Compute g(x)+g(12−x)g(x)+g(\frac12-x). For x∈[0,12]x\in[0,\frac12], g(x)=1−2xg(x)=1-2x, g(12−x)=1−2(12−x)=2xg(\frac12-x)=1-2(\frac12-x)=2x, so sum=1.

For x∉[0,12]x \notin[0,\frac12], both are 0, so sum=0. For (P) a=0,b=1,c=0,d=0a=0,b=1,c=0,d=0: h(x)=g(x)+g(12−x)h(x)=g(x)+g(\frac12-x) gives 1 on [0,12][0,\frac12] and 0 elsewhere → range {0,1}\{0,1\} → matches (5). For (Q) a=1,b=0,c=0,d=0a=1,b=0,c=0,d=0: h(x)=f(x)h(x)=f(x).

ff is differentiable on R\mathbb{R} (check x=0x=0: lim⁡x→0f(x)−0x=0\lim_{x\to0}\frac{f(x)-0}{x}=0, and elsewhere product of differentiable functions) → (3). For (R) a=0,b=0,c=1,d=0a=0,b=0,c=1,d=0: h(x)=x−g(x)h(x)=x-g(x).

For x<0x<0, h(x)=xh(x)=x; for 0≤x≤120\le x\le\frac12, h(x)=3x−1h(x)=3x-1; for x>12x>\frac12, h(x)=xh(x)=x. This is continuous and piecewise linear with slopes 1,3,1. As x→±∞x\to\pm\infty, h(x)→±∞h(x)\to\pm\infty → hh is onto → (2). For (S) a=0,b=0,c=0,d=1a=0,b=0,c=0,d=1: h(x)=g(x)h(x)=g(x).

On [0,12][0,\frac12], g(x)=1−2xg(x)=1-2x gives values from 1 to 0; elsewhere 0 → range [0,1][0,1] → (4). Match: P→5,  Q→3,  R→2,  S→4P\to5,\; Q\to3,\; R\to2,\; S\to4 → option C.

Answer key and solution verified before publishing.

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Exam
JEE Advanced 2024
Paper
Paper 1
Subject
Mathematics
Chapter
Functions
Topic
Domain & range of functions