Question

Let $f\left(x\right)=\{\begin{array}{cc}{\tan }^{-1}\alpha -5x^2,& 0<x<1\\ -6x,& x\ge 1\end{array}.$
If $f\left(x\right)$ has a maximum at $x=1,$ then

• A. $\alpha \in (\tan 1,\infty )$
• B. $\alpha \in (\frac{\pi }{4},\infty )$
• C. $\alpha \in (-\infty ,-\frac{\pi }{4})$
• D. $\alpha \in (-\infty ,-\tan 1)$

Answer 1

The correct option is D $\alpha \in (-\infty ,-\tan 1)$

Given :
$f\left(x\right)=\{\begin{array}{cc}{\tan }^{-1}\alpha -5x^2,& 0<x<1\\ -6x,& x\ge 1\end{array}$
Since $f\left(x\right)$ has local maximum at $x=1.$
Then, $f\left(1\right)>f(1-h)$ as $h\to 0^{+}$
$\Rightarrow -6>{\tan }^{-1}\alpha -5(1-h{)}^2$ as $h\to 0^{+}$
$\Rightarrow -6>{\tan }^{-1}\alpha -5$
$\Rightarrow {\tan }^{-1}\alpha <-1$
$\Rightarrow \alpha <-\tan 1$
$\Rightarrow \alpha \in (-\infty ,-\tan 1)$