Question

For $nϵN$, let $f\left(x\right)=min\{1-{\tan }^{n}x,1-{\sin }^{n}x,1-x^n\}$, $xϵ(-\frac{\pi }{2},\frac{\pi }{2})$. The left hand derivative of $f$ at $x=\frac{\pi }{4}$ is

• A. $-2n$
• B. $-2(n+1)$
• C. $-n\frac{\pi }{4}$
• D. $-n{\left(\frac{\pi }{4}\right)}^{n-1}$

Answer 1

The correct option is A $-2n$

Given, $f\left(x\right)=min\{1-{\tan }^{n}x,1-{\sin }^{n}x,1-x^n\}$
For $x\in (0,\frac{\pi }{4})$, we have

$\tan x>x>\sin x$

$\Rightarrow 1-{\tan }^{n}x<1-x^n<1-{\sin }^{n}x$

$\Rightarrow f\left(x\right)=1-{\tan }^{n}x$

So, $f\left(\frac{\pi }{4}\right)=min\{0,1-{\left(\frac{1}{\sqrt{2}}\right)}^n,1-{\left(\frac{\pi }{4}\right)}^n\}$

$\Rightarrow f\left(\frac{\pi }{4}\right)=0$

Now, $f^{\prime }\left({\frac{\pi }{4}}^{-}\right)=\underset{h\to 0^{-}}{lim}\frac{f(\frac{\pi }{4}+h)-f\left(\frac{\pi }{4}\right)}{h}$

$=\underset{h\to 0^{-}}{lim}\frac{1-{\tan }^n(\frac{\pi }{4}+h)}{h}$

$=\underset{h\to 0^{-}}{lim}\frac{1-{\left(\frac{1+\tan h}{1-\tan h}\right)}^n}{h}$

$=\underset{h\to 0^{-}}{lim}\frac{{(1-\tan h)}^n-{(1+\tan h)}^n}{h{(1-\tan h)}^n}$

$=\underset{h\to 0^{-}}{lim}\frac{-2[n\tan h+\frac{n(n-1)}{2}{\tan }^{3}h+.....]}{h(1-n\tan h+\frac{n(n-1)}{2}{\tan }^{2}h+....)}$

As, $\underset{h\to 0}{lim}\tan h=0$

So, using this result , we get
$=\underset{h\to 0^{-}}{lim}\frac{-2n\tan h}{h}$

$=-2n$ $(\underset{h\to 0}{lim}\frac{\tan h}{h}=1)$