Question

Let $f,g:R\to R$ be two real valued functions defined as $f(x)=\{\begin{array}{ll}-|x+3|& ,x<0\\ e^x& ,x\ge 0\end{array}$ and $g(x)=\{\begin{array}{ll}{x}^2+k_{1}x& ,x<0\\ 4x+k_2& ,x\ge 0\end{array}$, 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. $2\left(2e^4+1\right)$
• C. $4e^4$
• D. $2\left(2e^4-1\right)$

Answer 1

Answer: (D)

$f(x)=\left\{\begin{array}{ccc}x+3& ;& x<-3\\ -(x+3)& ;& -3\le x<0\\ e^x& ;& x\ge 0\end{array}\right\}$
$g(x)=\left\{\begin{array}{ccc}{x}^2+k_{1}x& ;& x<0\\ 4x+k_2& ;& x\ge 0\end{array}\right\}$
$g(f(x))=\left\{\begin{array}{ccc}f(x{)}^2+k_{1}f(x)& ;& f(x)<0\\ 4f(x)+k_2& ;& f(x)\ge 0\end{array}\right\}$
$g(f(x))=\left\{\begin{array}{ccc}(x+3{)}^2+k_1(x+3)& ;& x<-3\\ (x+3{)}^2-k_1(x+3)& ;& -3\le x<0\\ 4e^x+k_2& ;& x>0\end{array}\right\}$
check continuity at $x=0$
$gof(0)=g\left(f\left(0^{-}\right)\right)=g\left(f\left(0^{+}\right)\right)$
$4+k_2=9-3k_1=4+k_2$
$3k_1+k_2=\mathrm{5......}$(a)
differentiate
$(g(f(x)){)}^{\prime }=\left\{\begin{array}{ccc}2(x+3)+k_1& ;& x<-3\\ 2(x+3)-k_1& ;& -3\le x<0\\ 4e^x& ;& x\ge 0\end{array}\right\}$
$6-k_1=4$
$k_1=\mathrm{2.....}$(b)
$\therefore k_1=2,k_2=-1$
$\mathrm{gof}(x)=\left\{\begin{array}{ccc}(x+3{)}^2+2(x+3)& ;& x<-3\\ (x+3{)}^2-2(x+3)& ;& -3\le x<0\\ 4e^x-1& ;& x\ge 0\end{array}\right\}$
$\mathrm{gof}(-4)+\mathrm{gof}(4)=4e^4-2$
$\Rightarrow 2\left(2e^4-1\right)$