JEE Main 2022 · 26 Jun · Shift 1 · Mathematics Q7

← JEE Main 2022 · 26 Jun · Shift 1

MathematicsSingle correct+4 / -1Question 7 of 90

Let $f,g:\mathbb R\to\mathbb R$ be two real valued functions defined as $f(x)=\begin{cases}-|x+3|, & x<0\\e^x, & x\ge0\end{cases}$ and $g(x)=\begin{cases}x^2+k_1x, & x<0\\4x+k_2, & x\ge0\end{cases}$ , where $k_1$ and $k_2$ are real constants. If $(g\circ f)$ is differentiable at $x=0$ , then $(g\circ f)(-4)+(g\circ f)(4)$ is equal to :

4\left(e^4+1\right)

2\left(2e^4+1\right)

4e^4

2\left(2e^4-1\right)

Answer and worked solution

Official answer: D

Near 0. For $x<0$ (close to 0), $f=-(x+3)\to-3$ ; for $x\ge0$ , $f=e^x$ , $f(0)=1$ .

Continuity. $g(-3)=9-3k_1$ must equal $g(1)=4+k_2$ .

Derivatives. Left: $g'(-3)f'(0^-)=(k_1-6)(-1)=6-k_1$ . Right: $g'(1)\cdot1=4$ . So $k_1=2$ , then $k_2=-1$ .

Values. $f(-4)=-1$ , $g(-1)=1-2=-1$ ; $f(4)=e^4$ , $g(e^4)=4e^4-1$ . Sum $4e^4-2=2\left(2e^4-1\right)$ .

(D)

Topics:Differentiability Chain rule and implicit differentiation Composition and inverse functions

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JEE Main 2022 · 26 Jun · Shift 1, Mathematics question 7