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 If \mathrm{f(x)=x, x \leq 1}, and \mathrm{f(x)=x^2+b x+c, x>1}, and \mathrm{f^{\prime}(x)} exists finitely for all \mathrm{x \in R} then
 

Option: 1

\mathrm{ b=-1, c \in R}


Option: 2

\mathrm{ c=1, b \in R}


Option: 3

\mathrm{ b=1, c=-1}


Option: 4

\mathrm{ b=-1, c=1}


Answers (1)

best_answer

\mathrm{f(x)}  is differentiable at \mathrm{x=1}  also.

\begin{aligned} &\mathrm{ \therefore \lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h} . }\\ & \text { Now, } \mathrm{\lim _{h \rightarrow 0} \frac{f(1+h)-f(1)}{h}=\lim _{h \rightarrow 0} \frac{(1+h)^2+b(1+h)+c-1}{h} }\\ & \quad=\mathrm{\lim _{h \rightarrow 0} \frac{h^2+(2+b) h+b+c}{h} .} \\ &\mathrm{ \lim _{h \rightarrow 0} \frac{f(1-h)-f(1)}{-h}=\lim _{h \rightarrow 0} \frac{1-h-1}{-h}=1 .} \end{aligned}

The two limits can be equal if  \mathrm{2+b=1, b+c=0}.

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Rishabh

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