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  •  If the derivative of the function <img alt="\mathrm{f(x)=\left\{\begin{array}{cc} a x^2+b, & x<-1 \\ b x^2+a x+4, & x \geq-1 \end{array}\right. }" src="https://entrancecorne

 If the derivative of the function
\mathrm{f(x)=\left\{\begin{array}{cc} a x^2+b, & x<-1 \\ b x^2+a x+4, & x \geq-1 \end{array}\right. }
is everywhere continuous, then

Option: 1

\mathrm{a=2, b=3}


Option: 2

\mathrm{a=3, b=2}


Option: 3

\mathrm{a=-2, b=-3}


Option: 4

\mathrm{a=-3, b=-2}


Answers (1)

best_answer

\text { We have, } f(x)= \begin{cases}a x^2+b, & x<-1 \\ b x^2+a x+4, & x \geq-1\end{cases}

\mathrm{\Rightarrow f^{\prime}(x)= \begin{cases}2 a x, & x<-1 \\ 2 b x+a, & x \geq-1\end{cases}}
Since f(x) is differentiable at x=-1, therefore it is co at x=-1 and hence

\mathrm{ \lim _{x \rightarrow 1^{-}} f(x)=\lim _{x \rightarrow 1^{+}} f(x) \Rightarrow a+b=b-a+4= }
As  \mathrm{ f^{\prime}(x) }   is everywhere continuous, therefore it is co at x=-1. Consequently

\mathrm{ \lim _{x \rightarrow-1^{-}} f^{\prime}(x)=\lim _{x \rightarrow-1^{+}} f^{\prime}(x) \Rightarrow-2 a=-2 b+a \\ }
\mathrm{ \Rightarrow 3 a=2 b \Rightarrow b=3 }
 Hence, a=2, b=3.

Posted by

shivangi.bhatnagar

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