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

Option: 1

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


Option: 2

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


Option: 3

4 \mathrm{e}^{4}


Option: 4

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


Answers (1)

best_answer

f(x)=\left\{\begin{array}{cl} x+3 ; & x<-3 \\ -(x+3) ; & -3 \leq x<0 \\ e^{x} & ; x \geq 0 \end{array}\right\}

\mathrm{g(x)=\left\{\begin{array}{ll} x^{2}+k_{1} x & ; x<0 \\ 4 x+k_{2} & ; x \geq 0 \end{array}\right\}}

\mathrm{g(f(x))=\left\{\begin{array}{ll} f(x)^{2}+k_{1} f(x) ; & f(x)<0 \\ 4 f(x)+k_{2} ; & f(x) \geq 0 \end{array}\right\} } \\

\mathrm{g(f(x))=\left\{\begin{array}{ll} (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} & ; \quad x>0 \end{array}\right\} }

\mathrm{Check\: Continuity\: at \: x=0}

\mathrm{go f(0)=g\left(f\left(0^{-}\right)\right)=g\left(f\left(0^{+}\right)\right)} \\

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

\mathrm{3 k_{1}+k_{2}=5}              ........(a)

Differentiate

\mathrm{(g(f(x)))=\left\{\begin{array}{cl} 2(x+3)+k_1 ; & x<-3 \\ 2(x+3)-k_1 & ;-3 \leq x<0 \\ 4 e^{x} & ; x \geq 0 \end{array}\right\}}

\mathrm{6-k_{1}=4} \\

\mathrm{k_{1}=2 } \\             ......(b)

\mathrm{\therefore k_{1}=2 \quad, \quad k_{2}=-1}

\mathrm{g o f(x)=\left\{\begin{array}{cc} (x+3)^{2}+2 \cdot(x+3) \cdot ; & x<-3 \\ (x+3)^{2}-2(x+3), & ;-3 \leq x<0 \\ 4 e^{x}-1 & ; x \geq 0 \end{array}\right\}}

\mathrm{g o f(-4)+g o f(4)=4 e^{4}-2 } \\

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

Hence the correct answer is option 4.

Posted by

Ritika Jonwal

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