Question

The value of a for which the function $f\left(x\right)=\{\begin{array}{cc}{\tan }^{-1}a-3x^2,& 0<x<1\\ -6x,& x\ge 1\end{array}$ has a maximum at $x=1$, is?

• A. $0$
• B. $1$
• C. $2$
• D. None of these

Answer 1

The correct option is D None of these

We have,
$f\left(x\right)=\{\begin{array}{cc}{\tan }^{-1}a-3x^2,& 0<x<1\\ -6x,& x\ge 1\end{array}$
If $f\left(x\right)$ attains a maximum at $x=1$, then $f^{\prime }\left(1\right)$ must exist and should be zero. This means that $f\left(x\right)$ must be continuous and differentiable at $x=1$.
We observe that $f\left(x\right)$ will be continuous at $x=1$, if ${\tan }^{-1}a=-3$.
But, (LHD at $x=1$)$=$(RHD at $x=1$)$=-6\ne 0$ for any value of $a$.
Hence, there is no value of $a$ for which $f\left(x\right)$ has a minimum at $x=1$.