Question

Assertion :Let $f\left(x\right)=x^2+7x+4$ be a polynomial function, then $f^{\prime }\left(2\right)=11$. Reason: A polynomial 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 B Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion

The derivative of $f$ at $x$ is given by,
$f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{f(x+h)-f\left(x\right)}{h}$
$\Rightarrow f^{\prime }\left(x\right)=\underset{h\to 0}{lim}\frac{\left\{\right(x+h{)}^2+7(x+h)+4\}-\{x^2+7x+4\}}{h}$
$=\underset{h\to 0}{lim}\frac{x^2+h^2+2hx+7x+7h+4-x^2-7x-4}{h}$

$=\underset{h\to 0}{lim}\frac{2hx+7h}{h}$

$=2x+7$

$\Rightarrow f^{\prime }\left(2\right)=2\times 2+7=11$

We know that a polynomial function is everywhere differentiable. Therefore, $f\left(x\right)$ is everywhere differentiable. However, this reason is not the correct explanation of the assertion.