Question

Assertion :If $f$ is differentiable on an open interval $(a,b)$ such that $|f^{\prime }\left(x\right)|\le M$ for all $xϵ(a,b)$, then $|f\left(x\right)-f\left(y\right)|\le M|x-y|$ for all $x,yϵ(a,b)$. Reason: If $f\left(x\right)$ is a continuous function defined on $[a,b]$ such that it is differentiable on $(a,b)$, then there exists $cϵ(a,b)$ such that $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$

• 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 A Both Assertion and Reason are correct and Reason is the correct explanation for Assertion

Lagrange's mean value theorem states "If f(x) is a continuous function defined on [a, b] such that it is differentiable on (a, b), then there exists $cϵ(a,b)$ such that $f^{\prime }\left(c\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$"
Hence, statement 2 is true.
$f^{\prime }\left(x\right)=\frac{f\left(b\right)-f\left(a\right)}{b-a}$ for all $x\in$(a,b)
Since, $|f^{\prime }\left(x\right)|\le M$

$⟹\left|\frac{f\left(b\right)-f\left(a\right)}{b-a}\right|=|f^{\prime }\left(x\right)|$

$⟹\left|\frac{f\left(b\right)-f\left(a\right)}{a-b}\right|=|f^{\prime }\left(x\right)|$
$⟹|f\left(a\right)-f\left(b\right)|\le M|a-b|$

$⟹|f\left(x\right)-f\left(y\right)|\le M|x-y|$

Thus, the assertion is correct.