Question

Assertion :$f\left(x\right)=\{\begin{array}{cc}{x}^2& \text{for}x\le c\\ ax+b& \text{for}x>c\end{array}$
If $f\left(x\right)$ is differentiable at $x=c$, then $a=2c$ and $b=-c^2$ Reason: A continuous function is differentiable everywhere.

• A. Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
• B. Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
• C. Assertion is correct but Reason is incorrect
• D. Assertion is incorrect but Reason is correct

Answer 1

The correct option is C Assertion is correct but Reason is incorrect

It is given that $f\left(x\right)$ is differentiable at $x=c$ and every differentiable function is countinuous. So, $f\left(x\right)$ is continuous at $x=c$

$\therefore \underset{x\to c^{-}}{lim}f\left(x\right)=\underset{x\to c^{+}}{lim}f\left(x\right)=f\left(c\right)$

$\Rightarrow \underset{x\to c}{lim}{x}^2=\underset{x\to c}{lim}(ax+b)$ [Using def. of f(x)]

$\Rightarrow c^2=ac+b\Rightarrow b=c^2-ac$ .....$\left(i\right)$

Now, $f\left(x\right)$ is differentiable at $x=c$

$\Rightarrow$ LHD${}_{x=c}$ $=$ RHD${}_{x=c}$

$\Rightarrow \underset{x\to c^{-}}{lim}\frac{f\left(x\right)-f\left(c\right)}{x-c}=\underset{x\to c^{+}}{lim}\frac{f\left(x\right)-f\left(c\right)}{x-c}$

$\Rightarrow \underset{x\to c}{lim}\frac{x^2-c^2}{x-c}=\underset{x\to c}{lim}\frac{(ax+b)-c^2}{x-c}$ [Using def. of f(x)]

$\Rightarrow \underset{x\to c}{lim}\frac{(x+c)(x-c)}{x-c}=\underset{x\to c}{lim}\frac{ax+c^2-ac-c^2}{x-c}$

$\Rightarrow \underset{x\to c}{lim}(x+c)=\underset{x\to c}{lim}a$

$\therefore a=2c$ ......$\left(ii\right)$

From $\left(i\right)$ and $\left(ii\right)$, we get

$c^2=2c^2+b\Rightarrow b=-c^2$

Hence, $a=2c$ and $b=-c^2$

A differentiable function is continuous everywhere, however, the converse is not always true. Hence, assertion is correct but reason is incorrect.