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If a function f(x) defined by f(x)=\left\{\begin{matrix} ae^{2}+be^{-x} ,& -1\leq x<1\\ cx^{2} ,&1\leq x\leq 3 \\ ax^{2}+2cx ,& 3< x\leq 4 \end{matrix}\right. be continuous for some a, b, c  \epsilon R and f'(0)+f'(2)=e, then the value of a is:    
Option: 1 \frac{1}{e^{2}-3e+13}
 
Option: 2 \frac{e}{e^{2}-3e-13}
Option: 3 \frac{e}{e^{2}+3e+13}  
Option: 4 \frac{e}{e^{2}-3e+13}

Answers (1)

best_answer

f(x)=\left\{\begin{array}{cc} a e^{x}+b e^{-x} & ,-1 \leq x<1 \\ c x^{2} & , 1 \leq x \leq 3 \\ a x^{2}+2 c x, & 3<x \leq 4 \end{array}\right.

For continuity at x = 1

\\\lim _{x \rightarrow 1^{-}} f(x)=\operatorname{\lim}_{x \rightarrow 1^{+}} f(x) \\ \Rightarrow a e+b e^{-1}=c \\ \Rightarrow b=c e-a e^{2}

For continuity at x = 3

\begin{array}{l} \operatorname{lim}_{x \rightarrow 3^{-}} f(x)=\operatorname{lim}_{x \rightarrow 3^{+}} f(x) \\ \Rightarrow 9 c=9 a+6 c \\ \Rightarrow c=3 a \end{array}

\\f(0)+f^{\prime}(2)=e \Rightarrow\left(a e^{x}-b e^{x}\right)_{x=0}+(2 c x)_{x=2}=e \\ \Rightarrow a-b+4 c=e \\

\\a-3 a e+a e^{2}+12 a=e \\ \Rightarrow a\left(e^{2}+13-3 e\right)=e \\ \Rightarrow a=\frac{e}{e^{2}-3 e+13} \\

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himanshu.meshram

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