Question

Assertion :If $f\left(x\right)=\tan x,x\in [0,\frac{\pi }{7}]$ then $\frac{\pi }{7}<f\left(\frac{\pi }{7}\right)<\frac{2\pi }{7}$ Reason: ${\sec }^{2}x$ is strictly increasing in $[0,\frac{\pi }{7}]$

• 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

We have,

$f\left(x\right)=\tan x,x\in [0,\frac{\pi }{7}]$

and $f^{\prime }\left(x\right)={\sec }^{2}x,x\in [0,\frac{\pi }{7}]$

Applying Lagrange's theorem on $f\left(x\right)$ in the interval $[0,\frac{\pi }{7}]$, we have

$f^{\prime }\left(c\right)=\frac{f\left(\frac{\pi }{7}\right)-f\left(0\right)}{\left(\frac{\pi }{7}\right)-0}$ for some $c$ in $[0,\frac{\pi }{7}]$

Since, ${\sec }^{2}x$ is strictly increasing in $[0,\frac{\pi }{7}]$, therefore

we have, $f^{\prime }\left(0\right)<f^{\prime }\left(c\right)<f^{\prime }\left(\frac{\pi }{7}\right)$

$\Rightarrow 1<\frac{f\left(\frac{\pi }{7}\right)}{\left(\frac{\pi }{7}\right)}<{\sec }^2\left(\frac{\pi }{7}\right)<{\sec }^2\left(\frac{\pi }{4}\right)=2$

$\Rightarrow \frac{\pi }{7}<f\left(\frac{\pi }{7}\right)<\frac{2\pi }{7}$

Thus both the Assertion and Reason are correct and the reason is the correct explanation for the assertion.