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If the derivative of the function
$f(x)=\left\{\begin{array}{cc}a x^{2}+b & x < -1 \\ b x^{2}+a x+4 & x \geq-1\end{array}\right.$
is every where continuous, then what are the values of a and $\mathrm{b}$ ?
Options:
$f(x)=\left\{\begin{array}{cc}a x^{2}+b & x < -1 \\ b x^{2}+a x+4 & x \geq-1\end{array}\right.$
is every where continuous, then what are the values of a and $\mathrm{b}$ ?
Solution:
2876 Upvotes
Verified Answer
The correct answer is:
$\quad a=2, b=3$
Derivative of $f(x)=\left\{\begin{array}{ll}a x^{2}+b & x < -1 \\ b x^{2}+a x+a & x \geq-1\end{array}\right.$ is
$f^{\prime}(x)=\left\{\begin{array}{ll}2 a x & x < -1 \\ 2 b x+a, & x \geq-1\end{array}\right.$
If $\mathrm{f}^{\prime}(\mathrm{x})$ is continuous everywhere then it is also continuous at $\mathrm{x}=-1$
$\left.\mathrm{f}^{\prime}(\mathrm{x})\right|_{\mathrm{x}=-1}=-2 \mathrm{a}=-2 \mathrm{~b}+\mathrm{a}$
or, $3 \mathrm{a}=2 \mathrm{~b}$
From the given choice $\mathrm{a}=2, \mathrm{~b}=3$ satisfied this equation.
$f^{\prime}(x)=\left\{\begin{array}{ll}2 a x & x < -1 \\ 2 b x+a, & x \geq-1\end{array}\right.$
If $\mathrm{f}^{\prime}(\mathrm{x})$ is continuous everywhere then it is also continuous at $\mathrm{x}=-1$
$\left.\mathrm{f}^{\prime}(\mathrm{x})\right|_{\mathrm{x}=-1}=-2 \mathrm{a}=-2 \mathrm{~b}+\mathrm{a}$
or, $3 \mathrm{a}=2 \mathrm{~b}$
From the given choice $\mathrm{a}=2, \mathrm{~b}=3$ satisfied this equation.
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