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Saturday, 31 October

JEE Main 2023 · MathsMultiple choiceSingle correctMediumMulti-step

JEE Main 11 April 2023, Shift 2, Maths Q10

Question 10 of 90 in this shift, Maths question 10 of 30, Section A.

Let ff and gg be two functions defined by f(x)={x+1,x<0∣x−1∣,x≥0f(x) = \begin{cases} x+1, & x<0 \\ |x-1|, & x \ge 0 \end{cases} and g(x)={x+1,x<01,x≥0g(x) = \begin{cases} x+1, & x<0 \\ 1, & x \ge 0 \end{cases} Then (g∘f)(x)(g \circ f)(x) is
  1. (1)differentiable everywhere
  2. (2)not continuous at x=−1x = -1
  3. (3)continuous everywhere but not differentiable at x=1x = 1
  4. (4)continuous everywhere but not differentiable exactly at one pointOfficial answer

Official answer

Option 4

NTA final key.

Chapter
Continuity and Differentiability
Topic
Continuity and Differentiability
Idea tested
Continuity and Differentiability of Composite Functions

Same idea in other shifts

Asked 7× in all
  1. 6 Apr 2026, Shift 2 · Q25Let f(x)={x3+8;x<0,x2−4;x≥0,f(x)=\begin{cases}x^3+8;&x<0,\\x^2-4;&x\ge0,\end{cases} and g(x)={(x−8)1/3;x<0,(x+4)1/2;x≥0.g(x)=\begin{cases}(x-8)^{1/3};&x<0,\\(x+4)^{1/2};&x\ge0.\end{cases}…HardNumerical value
  2. 6 Apr 2024, Shift 2 · Q10Suppose for a differentiable function hh, h(0)=0h(0)=0, h(1)=1h(1)=1 and h′(0)=h′(1)=2h'(0)=h'(1)=2. If g(x)=h(ex)eh(x)g(x)=h(e^x)e^{h(x)}, then g′(0)g'(0) is equal to :MediumSingle correct
  3. 25 Jun 2022, Shift 2 · Q22Let f(x)=[2x2+1]f(x) = \left[2x^2 + 1\right] and g(x)={2x−3,x<02x+3,x≥0g(x) = \begin{cases} 2x - 3, & x < 0 \\ 2x + 3, & x \geq 0 \end{cases}, where [t][t] is the…HardNumerical value
  4. 26 Jun 2022, Shift 1 · Q7Let f,g:R→Rf, g : \mathbf{R} \to \mathbf{R} be two real valued functions defined as…HardSingle correct
  5. 28 Jun 2022, Shift 2 · Q6Let f,g:R→Rf, g : \mathbf{R} \to \mathbf{R} be functions defined by f(x)={[x],x<0∣1−x∣,x≥0f(x) = \begin{cases} [x] , & x < 0 \\ |1-x| , & x \geq 0 \end{cases} and…HardSingle correct
  6. 16 Mar 2021, Shift 2 · Q85Let f:R→Rf:\mathbf R\to\mathbf R and g:R→Rg:\mathbf R\to\mathbf R be defined as f(x)={x+a,x<0∣x−1∣,x≥0f(x)=\begin{cases}x+a, & x<0\\ |x-1|, & x\ge0\end{cases} and…MediumNumerical value

Question text from the official JEE Main paper published by NTA; answer from the final answer key. Chapter, topic and idea tags, difficulty and skill are Lumi’s. Lumi analyses 108 of the 174 JEE Main shifts held from 2017 to 2026.