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JEE Main 22 July 2021, Shift 2, Maths Q85

Question 85 of 90 in this shift, Maths question 25 of 30, Section B.

Let f:R→Rf:\mathbf{R}\to\mathbf{R} be a function defined as f(x)={3(1−∣x∣2)if ∣x∣≤20if ∣x∣>2f(x)=\begin{cases}3\left(1-\frac{|x|}{2}\right) & \text{if } |x|\leq2\\ 0 & \text{if } |x|>2\end{cases} Let g:R→Rg:\mathbf{R}\to\mathbf{R} be given by g(x)=f(x+2)−f(x−2)g(x)=f(x+2)-f(x-2). If n and m denote the number of points in R\mathbf{R} where gg is not continuous and not differentiable, respectively, then n+m\mathrm{n}+\mathrm{m} is equal to ________.

Official answer

4

NTA final key.

Same idea in other shifts

Asked 7× in all
  1. 30 Jan 2024, Shift 1 · Q30If the function f(x)={1∣x∣,∣x∣≥2ax2+2b,∣x∣<2f(x)=\begin{cases}\frac{1}{|x|}, & |x|\geq2\\ ax^2+2b, & |x|<2\end{cases} is differentiable on R\mathbf{R}, then…MediumNumerical value
  2. 25 Jul 2022, Shift 1 · Q26Let f(x)={∣ 4x2−8x+5 ∣,if 8x2−6x+1≥0[ 4x2−8x+5 ],if 8x2−6x+1<0,f(x)=\begin{cases}\left|\,4x^2-8x+5\,\right|, & \text{if } 8x^2-6x+1\ge0\\ [\,4x^2-8x+5\,], & \text{if } 8x^2-6x+1<0,\end{cases}…HardNumerical value
  3. 27 Jul 2022, Shift 1 · Q20Let a function f:R→Rf : \mathbb{R} \rightarrow \mathbb{R} be defined as :…HardSingle correct
  4. 29 Jun 2022, Shift 1 · Q4Let f:R→Rf: \mathbf{R} \to \mathbf{R} be a function defined by :…HardSingle correct
  5. 25 Jul 2021, Shift 2 · Q72If f(x)={∫0x(5+∣1−t∣) dt,x>25x+1,x≤2f(x)=\begin{cases}\int_0^x(5+|1-t|)\,dt, & x>2\\ 5x+1, & x\le 2\end{cases}, thenMediumSingle correct
  6. 4 Sep 2020, Shift 2 · Q59The function f(x)={π4+tan⁡−1x,∣x∣≤112(∣x∣−1),∣x∣>1f(x) = \begin{cases} \frac{\pi}{4} + \tan^{-1} x, & |x| \leq 1 \\ \frac{1}{2}(|x| - 1), & |x| > 1 \end{cases} is :MediumSingle correct

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.