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Saturday, 31 October

JEE Main 2022 · MathsMultiple choiceSingle correctHardMulti-step

JEE Main 26 June 2022, Shift 1, Maths Q7

Question 7 of 90 in this shift, Maths question 7 of 30, Section A.

Let f,g:R→Rf, g : \mathbf{R} \to \mathbf{R} be two real valued functions defined as f(x)={−∣x+3∣,x<0ex,x≥0f(x) = \begin{cases} -|x+3|, & x < 0 \\ e^{x}, & x \geq 0 \end{cases} and g(x)={x2+k1x,x<04x+k2,x≥0g(x) = \begin{cases} x^2 + k_1 x, & x < 0 \\ 4x + k_2, & x \geq 0 \end{cases}, where k1k_1 and k2k_2 are real constants. If (g∘f)(g \circ f) is differentiable at x=0x = 0, then (g∘f) (−4)+(g∘f) (4)(g \circ f)\,(-4) + (g \circ f)\,(4) is equal to :
  1. (1)4(e4+1)4(e^4 + 1)
  2. (2)2(2e4+1)2(2e^4 + 1)
  3. (3)4e44e^4
  4. (4)2(2e4−1)2(2e^4 - 1)Official answer

Official answer

Option 4

NTA final key.

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. 11 Apr 2023, Shift 2 · Q10Let 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…MediumSingle correct
  4. 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
  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.