Let $\mathrm{f}, \mathrm{g}: \mathrm{R} \rightarrow \mathrm{R}$ be two real valued functions defined as $f(x)=\left\{\begin{array}{cc}-|x+3| & , x<0\\ e^{x} & , \quad x \geq 0\end{array}\right.$ and $g(x)=\left\{\begin{array}{cc}x^{2}+k_{1} x & , x<0 \\ 4 x+k_{2} & , x \geq 0\end{array}\right.$, where $k_{1}$ and $k_{2}$ are real constants. If (gof) is differentiable at $\mathrm{x}=0$, then (gof) $(-4)+$(gof)$(4)$ is equal to :

(D) $2\left(2 \mathrm{e}^{4}-1\right)$

(A) $4\left(\mathrm{e}^{4}+1\right)$

(B) $2\left(2 \mathrm{e}^{4}+1\right)$

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Continuity and Differentiability

Let

f, g : R \to R

be two real valued functions defined as

f (x) = {- ∣ x + 3∣ e^{x}, x < 0, x \geq 0

and

g (x) = {x^{2} + k_{1} x 4 x + k_{2}, x < 0, x \geq 0

, where

k_{1}

and

k_{2}

are real constants. If (gof) is differentiable at

x = 0

, then (gof)

(- 4) +

(gof)

(4)

is equal to :

Answer: (D)

For

x < 0

near the origin,

f (x) = - ∣ x + 3∣ = - (x + 3)

is negative, so the composite uses the lower branch of

g

and equals

(x + 3)^{2} - k_{1} (x + 3)

; for

x \geq 0

f (x) = e^{x} \geq 1

is positive, so the composite equals

4 e^{x} + k_{2}

. Continuity at

x = 0

demands

9 - 3 k_{1} = 4 + k_{2}

. For differentiability the one-sided derivatives must agree: the left derivative

2 (x + 3) - k_{1}

equals

6 - k_{1}

0

, the right derivative

4 e^{x}

equals

4

, so

6 - k_{1} = 4

, giving

k_{1} = 2

and then

k_{2} = - 1

. Now evaluate the composite: at

x = - 4

f (- 4) = - ∣ - 1∣ = - 1

, which is negative, so

(g \circ f) (- 4) = (- 1)^{2} + k_{1} (- 1) = 1 - 2 = - 1

; at

x = 4

f (4) = e^{4}

, so

(g \circ f) (4) = 4 e^{4} + k_{2} = 4 e^{4} - 1

. Their sum is

- 1 + (4 e^{4} - 1) = 4 e^{4} - 2 = 2 (2 e^{4} - 1)

JEE Main · 2022 · 26 Jun 2022 · Shift 1 · Single correct option