Question

Let $f\left(x\right)=x^2$ for $x\le \alpha ,f\left(x\right)=ax+b$ for $x>\alpha$.

lf $f\left(x\right)$ derivable at $x=\alpha$ then?

• A. $a=\alpha ,b=-\alpha$
• B. $a={\alpha }^2,b=-\alpha$
• C. $a=2\alpha ,b=-{\alpha }^2$
• D. $a=\alpha ,b=-{\alpha }^2$

Answer 1

The correct option is C $a=2\alpha ,b=-{\alpha }^2$

For $x<\alpha$, $f\left(x\right)=x^2$
$f^{\prime }\left(x\right)=2x$
For, $x>\alpha$, $f\left(x\right)=ax+b$
$f^{\prime }\left(x\right)=a$
For $x$ to be differentiable at $x=\alpha$
It has to be first continuous at $x=\alpha$
${\alpha }^2=a\alpha +b$
And, $a=2\alpha$
Thus $a=2\alpha$ and $b=-{\alpha }^2$