Question

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

• A. $4\left(e^{4}+1\right)$
• B. $2\left(2 e^{4}+1\right)$
• C. $4 e ^{4}$
• D. $2\left(2 e^{4}-1\right)$

Answer 1

Answer: (D)

$f(x)=\begin{Bmatrix} x+3 & ; & x<-3 \\ -(x+3) & ; & -3 \leq x<0 \\ e^{x} & ; & x \geq 0\end{Bmatrix}$
$g(x)=\begin{Bmatrix}x^{2}+k_{1} x & ; & x < 0 \\ 4 x+k_{2} & ; & x \geq 0\end{Bmatrix}$
$g(f(x))=\begin{Bmatrix}f(x)^{2}+k_{1} f(x) & ; & f(x) < 0 \\ 4 f(x)+k_{2} & ; & f(x) \geq 0\end{Bmatrix}$
$g(f(x))= \begin{Bmatrix}(x+3)^{2}+k_{1}(x+3) & ; & x < -3 \\ (x+3)^{2}-k_{1}(x+3) & ; & -3 \leq x<0 \\ 4 e^{x}+k_{2} & ; & x>0\end{Bmatrix}$
check continuity at $x=0$
$g o f(0)=g\left(f\left(0^{-}\right)\right)=g\left(f\left(0^{+}\right)\right) $
$4+k_{2}=9-3 k_{1}=4+k_{2} $
$3 k_{1}+k_{2}=5......$(a)
differentiate
$(g(f(x)))^{\prime}= \begin{Bmatrix}2(x+3)+k_{1} & ; & x<-3 \\ 2(x+3)-k_{1} & ; & -3 \leq x<0 \\ 4 e^{x} & ; & x \geq 0\end{Bmatrix}$
$6-k_{1}=4$
$k _{1}=2.....$(b)
$\therefore k _{1}=2, k _{2}=-1$
$\operatorname{gof}( x )=\begin{Bmatrix}( x +3)^{2}+2( x +3) & ; & x <-3 \\ ( x +3)^{2}-2( x +3) & ; & -3 \leq x <0 \\ 4 e ^{ x }-1 & ; & x \geq 0\end{Bmatrix}$
$\operatorname{gof}(-4)+\operatorname{gof}(4)=4 e ^{4}-2$
$\Rightarrow 2\left(2 e ^{4}-1\right)$