Question

The number of integral value(s) of $a$ for which function $f\left(x\right)=\{\begin{array}{cc}\frac{x^2-5x+6}{x-2},& x<2\\ a,& x=2\\ x^2+6,& x>2\end{array}$
is strictly increasing at $x=2$, is equal to

Answer 1

Given : $f\left(x\right)=\{\begin{array}{cc}x-3,& x<2\\ a,& x=2\\ x^2+6,& x>2\end{array}$
is strictly increasing at $x=2$
$\Rightarrow f(2-h)<f\left(2\right)<f(2+h),$ where $h\to 0^{+}\cdots \left(i\right)$
Now, $f(2-h)=2-h-3=-1-h,$ where $(h\to 0^{+})\cdots \left(ii\right)$
$f(2+h)=10+h^2+4h,$ where $(h\to 0^{+})\cdots \left(iii\right)$
From $\left(i\right)$ ,$\left(ii\right)$ and $\left(iii\right)$
$\Rightarrow -1\le a\le 10$
$\therefore$ number of integral values of $a$ is equal to $12.$