Let f: R → R be a function defined by : f(x) = \ cases max _ t ≤ x \ t ^ 3 - 3t\ ; & x ≤ 2 x ^ 2 + 2x - 6; & 2 5 cases ) Where [t] is the greatest integer less than or equal to t . Let m be the number of points where f

Correct answer: C. (4, (27)/(4)) f(x) = \ array ll x ^ 3 - 3x, x ≤ - 1 2, - 1 5 array Clearly f(x) is not differentiable at x = 2, 3, 4, 5 ⇒ m = 4 I = ∫ _ - 2 ^ 2 f (x)dx = ∫ _ - 2 ^ - 1 ( x ^ 3 - 3x)dx + ∫ _ - 1 ^ 2 2 · dx = (27)/(4) Hence the ordered pair is(4, (27)/(4))

JEE Main 202229 JuneMorning Shift

Exam: JEE Main
Year: 2022
Date: 29 June
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2022 · 29 June · Morning Shift · Mathematics · Continuity and Differentiability

Let f: R → R be a function defined by : f(x) = \ cases max _ t ≤ x...

Let

f: R \to R

be a function defined by :

f\left(x\right) = \left\{

\begin{cases}\mathop{max}_{t \le x}\left\{{t}^{3} - 3t\right\}; & x \le 2 \\ {x}^{2} + 2x - 6; & 2 5\end{cases}

\right)<br />Where

\left[t\right]

is the greatest integer less than or equal to

. Let

be the number of points where

is not differentiable and

I = {∫}_{- 2}^{2}f\left(x\right)dx

. Then the ordered pair \left(m, I\right)

is equal to

Options

A $\left(3, \dfrac{27}{4}\right)$
B $\left(3, \dfrac{23}{4}\right)$
C $\left(4, \dfrac{27}{4}\right)$
D $\left(4, \dfrac{23}{4}\right)$

Correct answer

C. $\left(4, \dfrac{27}{4}\right)$

Step-by-step solution

$f\left(x\right) =$ \left\{ $\begin{array}{ll}{x}^{3} - 3x, x \le - 1 \\ 2, - 1 5\end{array}$ \right

Clearly $f\left(x\right)$ is not differentiable at $x = 2, 3, 4, 5 ⇒ m = 4$

$I = {∫}_{- 2}^{2}f \left(x\right)dx = {∫}_{- 2}^{- 1}\left({x}^{3} - 3x\right)dx + {∫}_{- 1}^{2}2 \cdot dx = \dfrac{27}{4}$

Hence the $ordered pair is\left(4, \dfrac{27}{4}\right)$

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