Let the functions f: R → R and g: R → R be defined as : f(x) = \ cases x + 2, & x Then, the number of points in R where (fog)(x) is NOT differentiable is equal to :

Correct answer: B. 1 f(g(x)) = \ cases g(x) + 2, & g(x) = \ cases x ^ 3 + 2, & x (fog(x)) ^ ' = \ cases 3 x ^ 2 , & x At x = 0 L . H . L . ≠ R . H . L . (Discontinuous) At x = 1 L . H . L . = R . H . L . (continuous) L . H . D . = 6 = R . H . D . ⇒ fog (x) is differentiable for all x ∈ R - \ 0\

JEE Main 202116 MarchMorning Shift

Exam: JEE Main
Year: 2021
Date: 16 March
Shift: Morning Shift
Subject: Mathematics
Chapter: Continuity and Differentiability

JEE Main 2021 · 16 March · Morning Shift · Mathematics · Continuity and Differentiability

Let the functions f: R → R and g: R → R be defined as : f(x) = \...

Let the functions $f: R \to R and$ g: R \to R $be defined as :$ f\left(x\right) = \left\{ $\begin{cases}x + 2, & x < 0 \\ {x}^{2}, & x \ge 0\end{cases}$ \right) $and g\left(x\right) = \left\{$ \begin{cases}{x}^{3}, & x < 1 \\ 3x - 2, & x \ge 1\end{cases} $\right)Then, the number of points in$ R $where \left(fog\right)\left(x\right)$ is NOT differentiable is equal to :

Options

A $3$
B $1$
C $0$
D $2$

Correct answer

B. $1$

Step-by-step solution

$f\left(g\left(x\right)\right) = \left\{$ \begin{cases}g\left(x\right) + 2, & g\left(x\right) < 0 \\ {\left(g\left(x\right)\right)}^{2}, & g\left(x\right) \ge 0\end{cases} $\right)$ = \left\{ $\begin{cases}{x}^{3} + 2, & x < 0 \\ {x}^{6}, & x \in [ 0, 1 ) \\ {\left(3x - 2\right)}^{2}, & x \in [ 1, \infty )\end{cases}$ \right)

${\left(fog\left(x\right)\right)}^{'} = \left\{$ \begin{cases}3{x}^{2}, & x < 0 \\ 6{x}^{5}, & x \in [ 0, 1 ) \\ 2\left(3x - 2\right) \times 3, & x \in [ 1, \infty )\end{cases} $\right)At$ x = 0 $$ L . H . L . \ne R . H . L . $(Discontinuous)At$ x = 1 $$ L . H . L . = R . H . L . $(continuous)$ L . H . D . = 6 = R . H . D . $$ ⇒ fog $\left(x\right)$ is differentiable for all $x \in R - \left\{0\right\}$

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