Question

Let $f\left(x\right)=\left\{\begin{array}{cc}a{x}^2+3,& x>1\\ x+\frac{5}{2},& x\le 1\end{array}\right..$ If f(x) is differentiable at x = 1, then a = ____________.

Answer 1

The given function $f\left(x\right)=\left\{\begin{array}{cc}a{x}^2+3,& x>1\\ x+\frac{5}{2},& x\le 1\end{array}\right.$ is differentiable at x = 1.

$\therefore Lf\text{'}\left(1\right)=Rf\text{'}\left(1\right)$

$\Rightarrow \underset{h\to 0}{\lim }\frac{f\left(1-h\right)-f\left(1\right)}{-h}=\underset{h\to 0}{\lim }\frac{f\left(1+h\right)-f\left(1\right)}{h}$

$\Rightarrow \underset{h\to 0}{\lim }\frac{\left(1-h+{\displaystyle \frac{5}{2}}\right)-\left(1+{\displaystyle \frac{5}{2}}\right)}{-h}=\underset{h\to 0}{\lim }\frac{\left[a{\left(1+h\right)}^2+3\right]-\left(1+{\displaystyle \frac{5}{2}}\right)}{h}$

$\Rightarrow \underset{h\to 0}{\lim }\frac{-h}{-h}=\underset{h\to 0}{\lim }\frac{a{\left(1+h\right)}^2-{\displaystyle \frac{1}{2}}}{h}$

$\Rightarrow 1=\underset{h\to 0}{\lim }\frac{\left(a-{\displaystyle \frac{1}{2}}\right)+a\left(2h+h^2\right)}{h}$

$\Rightarrow 1=\underset{h\to 0}{\lim }\frac{\left(a-{\displaystyle \frac{1}{2}}\right)+ah\left(2+h\right)}{h}$

$\Rightarrow 1=\underset{h\to 0}{\lim }\frac{\left(a-{\displaystyle \frac{1}{2}}\right)}{h}+\underset{h\to 0}{\lim }a\left(2+h\right)$

Here, LHS is finite.

So, for RHS to be finite, we must have

$a-\frac{1}{2}=0$

$\Rightarrow a=\frac{1}{2}$

Thus, the value of a is $\frac{1}{2}$.


Let $f\left(x\right)=\left\{\begin{array}{cc}a{x}^2+3,& x>1\\ x+\frac{5}{2},& x\le 1\end{array}\right..$ If f(x) is differentiable at x = 1, then a = $\underline{\frac{1}{2}}$.