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  •  <img alt="\begin{aligned}\text{if } \mathrm{ f(x)}&=\mathrm{e^x, x<2}\\ &\mathrm{a+b x, x \geq 2} \end{aligned}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cbegin%7Bali

 \begin{aligned}\text{if } \mathrm{ f(x)}&=\mathrm{e^x, x<2}\\ &\mathrm{a+b x, x \geq 2} \end{aligned}
is differentiable for all \mathrm{x \in R} then
 

Option: 1

\mathrm{ a+b=10}


Option: 2

\mathrm{ a+2 b=e^{20}}


Option: 3

\mathrm{ b=e^2 }


Option: 4

none of these


Answers (1)

best_answer

\mathrm{f(x)} is differentiable at \mathrm{x=2} also.

\begin{aligned} &\mathrm{So,~ \lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h}.}\\ &\mathrm{Now,~ \lim _{h \rightarrow 0} \frac{f(2+h)-f(2)}{h}=\lim _{h \rightarrow 0} \frac{a+b(2+h)-(a+2 b)}{h}=b . }\\ & \begin{aligned} \mathrm{\lim _{h \rightarrow 0} \frac{f(2-h)-f(2)}{-h} }& =\mathrm{\lim _{h \rightarrow 0} \frac{e^{2-h}-(a+2 b)}{-h}} \\ & \mathrm{=\lim _{h \rightarrow 0} \frac{e^{-h}-(a+2 b) e^{-2}}{-h \cdot e^{-2}} }\\ & \mathrm{=\lim _{h \rightarrow 0} \frac{\left\{1-h+\frac{h^2}{2 !}-\ldots\right\}-(a+2 b) e^{-2}}{-h e^{-2}}} . \end{aligned} \end{aligned}

This can be equal to b if \mathrm{1-(a+2 b) e^{-2}=0}  and \mathrm{e^{-2}=\frac{1}{b}}.

\begin{aligned} \therefore \quad \mathrm{1-(a+2 b) \frac{1}{b}=0} & \Rightarrow \mathrm{b-(a+2 b)=0 }\\ & \Rightarrow \mathrm{a+b=0 .} \end{aligned}

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