Verified Solution • 2025–26 Session moderate

QJEEM23Ma3805: JEE Main 2023 Mathematics - Class 12 JEEM23Ma3805 Solution: Let and be two functions defined by and Then is :

Let ff and gg be two functions defined by

f(x)={x+1,amp;xlt;0∣x−1∣,amp;x≥0f(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ |x-1|, & x \geq 0\end{array}\right. and g(x)={x+1,amp;xlt;01,amp;x≥0\mathrm{g}(x)=\left\{\begin{array}{cc}x+1, & x < 0 \\ 1, & x \geq 0\end{array}\right.

Then (g∘f)(x)(g \circ f)(x) is :

Multiple Choice Options

  • A continuous everywhere but not differentiable at x=1x=1
  • B differentiable everywhere
  • C not continuous at x=−1x=-1
  • D continuous everywhere but not differentiable exactly at one point

Step-by-Step Verified Solution

amp; Sol. f(x)={x+1,xlt;0\1−x,0≤xlt;1x−1,1≤xamp;g(x)={x+1,xlt;01,x≥0amp;g(f(x))={x+2,xlt;−11,x≥−1\begin{aligned}& \text { Sol. } f(x)=\left\{\begin{array}{c}x+1, x<0 \\\1-x, 0 \leq x<1 \\x-1,1 \leq x\end{array}\right. \\\\& g(x)=\left\{\begin{array}{c}x+1, x<0 \\1, x \geq 0\end{array}\right. \\\\& g(f(x))=\left\{\begin{array}{c}x+2, x<-1 \\1, x \geq-1\end{array}\right.\end{aligned}

JEE Main 2023 (Online) 11th April Evening Shift Mathematics - Limits, Continuity and Differentiability Question 49 English Explanation
∴g(f(x))\therefore \mathrm{g}(\mathrm{f}(\mathrm{x})) is continuous everywhere

g(f(x))\mathrm{g}(\mathrm{f}(\mathrm{x})) is not differentiable at x=−1\mathrm{x}=-1 Differentiable everywhere else.

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