JEE Main 2024 · MathsNumerical answerNumerical valueMediumCalculation

JEE Main 30 January 2024, Shift 1, Maths Q30

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

If the function

f(x)=\begin{cases}\frac{1}{|x|}, & |x|\geq2\\ ax^2+2b, & |x|<2\end{cases}

is differentiable on

\mathbf{R}

, then

48(a+b)

is equal to __________.

Official answer

NTA final key.

Chapter: Continuity and Differentiability
Topic: Continuity and Differentiability
Idea tested: Piecewise Differentiability

Same idea in other shifts

Asked 7× in all

25 Jul 2022, Shift 1 · Q26Let $f(x)=\begin{cases}\left|\,4x^2-8x+5\,\right|, & \text{if } 8x^2-6x+1\ge0\\ [\,4x^2-8x+5\,], & \text{if } 8x^2-6x+1<0,\end{cases}$ …HardNumerical value
27 Jul 2022, Shift 1 · Q20Let a function $f : \mathbb{R} \rightarrow \mathbb{R}$ be defined as :…HardSingle correct
29 Jun 2022, Shift 1 · Q4Let $f: \mathbf{R} \to \mathbf{R}$ be a function defined by :…HardSingle correct
22 Jul 2021, Shift 2 · Q85Let $f:\mathbf{R}\to\mathbf{R}$ be a function defined as…HardNumerical value
25 Jul 2021, Shift 2 · Q72If $f(x)=\begin{cases}\int_0^x(5+|1-t|)\,dt, & x>2\\ 5x+1, & x\le 2\end{cases}$ , thenMediumSingle correct
4 Sep 2020, Shift 2 · Q59The function $f(x) = \begin{cases} \frac{\pi}{4} + \tan^{-1} x, & |x| \leq 1 \\ \frac{1}{2}(|x| - 1), & |x| > 1 \end{cases}$ 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.