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QJEEAdv24Ma8bab: JEE Advanced 2024 Mathematics - Class 12 JEEAdv24Ma8bab Solution: Let and be functions defined by Let . Define the function by…

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

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

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

h(x)=af(x)+b(g(x)+g(12−x))+c(x−g(x))+dg(x),x∈R.h(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 \mathbb{R} .

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

List-I List-II
(P) If a=0a = 0 , b=1b = 1 , c=0c = 0 , and d=0d = 0 , then (1) hh is one-one.
(Q) If a=1a = 1 , b=0b = 0 , c=0c = 0 , and d=0d = 0 , then (2) hh is onto.
(R) If a=0a = 0 , b=0b = 0 , c=1c = 1 , and d=0d = 0 , then (3) hh is differentiable on R\mathbb{R} .
(S) If a=0a = 0 , b=0b = 0 , c=0c = 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

Multiple Choice Options

  • A (P)→(4)(\mathrm{P}) \rightarrow(4) (Q)→(3)(\mathrm{Q}) \rightarrow(3) (R)→(1)(\mathrm{R}) \rightarrow(1) (S) →\rightarrow (2)
  • B (P)→(5)(\mathrm{P}) \rightarrow(5) (Q)→(2)(\mathrm{Q}) \rightarrow(2) (R)→(4)(\mathrm{R}) \rightarrow(4) (S) →(3)\rightarrow(3)
  • C (P)→(5)(\mathrm{P}) \rightarrow(5) (Q)→(3)(\mathrm{Q}) \rightarrow(3) (R)→(2)(\mathrm{R}) \rightarrow(2) (S)→(4)(\mathrm{S}) \rightarrow(4)
  • D (P)→(4)(\mathrm{P}) \rightarrow(4) (Q)→(2)(\mathrm{Q}) \rightarrow(2) (R)→(1)(\mathrm{R}) \rightarrow(1) (S)→(3)(\mathrm{S}) \rightarrow(3)

Step-by-Step Verified Solution

f(x)={x∣x∣sin⁡1xamp;;x≠00amp;;x=0g(x)={1−2xamp;0≤x≤120;amp; otherwise f(x)=\left\{\begin{array}{cc}x|x| \sin \frac{1}{x} & ; x \neq 0 \\0 & ; \quad x=0\end{array} \quad g(x)=\left\{\begin{array}{cc}1-2 x & 0 \leq x \leq \frac{1}{2} \\0 ; & \text { otherwise }\end{array}\right.\right.

g(12−x)={2xamp;;0≤12−x≤120;amp; otherwise ={2x;amp;0≤x≤120;amp; otherwise }g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{cc}2 x & ; \quad 0 \leq \frac{1}{2}-x \leq \frac{1}{2} \\0 ; & \text { otherwise }\end{array}=\left\{\begin{array}{cc}2 x ; & 0 \leq x \leq \frac{1}{2} \\0 ; & \text { otherwise }\end{array}\right\}\right.

g(x)+g(12−x)={1;amp;0≤x≤120amp;; otherwise }g(x)+g\left(\frac{1}{2}-x\right)=\left\{\begin{array}{ll}1 ; & 0 \leq x \leq \frac{1}{2} \\0 & ; \text { otherwise }\end{array}\right\}

(P) \quad Now a=0,b=1,c=0,d=0a=0, b=1, c=0, d=0

∵h(x)=g(x)+g(12−x)={1;amp;0≤x≤120;amp; otherwise \because h(x)=g(x)+g\left(\frac{1}{2}-x\right)= \begin{cases}1 ; & 0 \leq x \leq \frac{1}{2} \\ 0 ; & \text { otherwise }\end{cases}

JEE Advanced 2024 Paper 1 Online Mathematics - Limits, Continuity and Differentiability Question 4 English Explanation 1

Hence Range of h(x) is {0, 1}

(Q) a=1, b=0,c=0, d=0\mathrm{a}=1, \mathrm{~b}=0, \mathrm{c}=0, \mathrm{~d}=0

h(x)=f(x)={x∣x∣sin⁡1xamp;;x≠00amp;;x=0h(x)=f(x)=\left\{\begin{array}{cc}x|x| \sin \frac{1}{x} & ; x \neq 0 \\0 & ; x=0\end{array}\right.

amp;RHD=lim⁡x→0x2sin⁡1x−0x=0amp; LHD =lim⁡x→0−x2sin⁡1x−0x=0\begin{aligned}& R H D=\lim _{x \rightarrow 0} \frac{x^2 \sin \frac{1}{x}-0}{x}=0 \\& \text { LHD }=\lim _{x \rightarrow 0} \frac{-x^2 \sin \frac{1}{x}-0}{x}=0\end{aligned}

Hence h(x)h(x) is differentiable on RR

(R) a=0, b=0,c=1, d=0\mathrm{a}=0, \mathrm{~b}=0, \mathrm{c}=1, \mathrm{~d}=0

h(x)=x−g(x)={3x−1amp;;0≤x≤120;amp; otherwise h(x)=x-g(x)=\left\{\begin{array}{cl}3 x-1 & ; 0 \leq x \leq \frac{1}{2} \\0 ; & \text { otherwise }\end{array}\right.

JEE Advanced 2024 Paper 1 Online Mathematics - Limits, Continuity and Differentiability Question 4 English Explanation 2

∴h(x) is ONTO \therefore \mathrm{h}(\mathrm{x}) \text { is ONTO }

(S) a=0,b=0,c=0,d=1a=0, b=0, c=0, d=1

h(x)=g(x)={1−2xamp;;0≤x≤120;amp; otherwise h(x)=g(x)=\left\{\begin{array}{cl}1-2 x & ; 0 \leq x \leq \frac{1}{2} \\0 ; & \text { otherwise }\end{array}\right.

JEE Advanced 2024 Paper 1 Online Mathematics - Limits, Continuity and Differentiability Question 4 English Explanation 3

Range of h(x) is [0, 1]

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