Question

The number of integral values of $a$ for which the function $f\left(x\right)=\{\begin{array}{cc}{x}^3+1,& x\le 0\\ a^2-5a+7+x& x>0\end{array}$
has maxima at $x=0$ are

• A. 0.0
• B. 1
• C. 2
• D. infinite

Answer 1

The correct option is A 0.0

If $f\left(x\right)$ has a maxima at $x=0$, then

$\begin{array}{cc}f(0+h)\le f\left(0\right)& \\ f(0-h)\le f\left(0\right)& \end{array}\}$

$\underset{h\to 0}{lim}-h^3+1\le 1\Rightarrow 1\le 1$
$\underset{h\to 0}{lim}(a^2-5a+7+h)\le 1$
$\Rightarrow a^2-5a+7\le 1$
$\Rightarrow a^2-5a+6\le 0$
$\Rightarrow (a-2)(a-3)\le 0$
$\Rightarrow 2\le a\le 3$

For $a=2,\underset{h\to 0^{+}}{lim}(1+h)$ which is greater than 1.
$\therefore a\ne 2$
For $a=3,\underset{h\to 0^{+}}{lim}(1+h)$ which is greater than 1.
$\therefore a\ne 3$

Hence, there is no integral values of $x$ for which the given function can have a maxima at $x=0$.