Question

Find the values of a and b so that the function $f\left(x\right)\left\{\begin{array}{ll}{x}^2+3x+a,& \mathrm{if}x\le 1\\ bx+2,& \mathrm{if}x>1\end{array}\right.$ is differentiable at each x ∈ R.

Answer 1

Given:
$f\left(x\right)=\left\{\begin{array}{l}{x}^2+3x+a,x\le 1\\ bx+2,x>1\end{array}\right.$

It is given that the function is differentiable at each $x\in R$ and every differentiable function is continuous.
So, $f\left(x\right)$ is continuous at $x=1$.

Therefore,

$\underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right)$

$$\begin{gathered} \Rightarrow \underset{x\to 1}{\lim }{x}^2+3x+a=\underset{x\to 1}{\lim }bx+2=a+4\left[\mathrm{Using}\mathrm{def}.\mathrm{of}\mathit{}f\left(x\right)\right] \\ \Rightarrow a+4=b+2=a+4...\left(\mathrm{i}\right) \end{gathered}$$


Since, $f\left(x\right)$ is differentiable at $x=1$. So,

(LHD at x = 1) = (RHD at x = 1)

$$\begin{gathered} \underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1} \\ \Rightarrow \underset{x\to 1}{\lim }\frac{x^2+3x+a-a-4}{x-1}=\underset{x\to 1}{\lim }\frac{bx+2-4-a}{x-1}\left[\mathrm{Using}\mathrm{def}.\mathrm{of}f\left(x\right)\right] \\ \Rightarrow \underset{x\to 1}{\lim }\frac{(x+4)(x-1)}{x-1}=\underset{x\to 1}{\lim }\frac{bx-2-a}{x-1} \\ \Rightarrow \underset{x\to 1}{\lim }\frac{(x+4)(x-1)}{x-1}=\underset{x\to 1}{\lim }\frac{bx-b}{x-1}\left[\mathrm{Using}\left(\mathrm{i}\right)\right] \\ \Rightarrow \underset{x\to 1}{\lim }\frac{(x+4)(x-1)}{x-1}=\underset{x\to 1}{\lim }\frac{b(x-1)}{x-1} \\ \Rightarrow 5=b \end{gathered}$$

From $\left(i\right)$, we have

$$\begin{gathered} a+4=b+2 \\ \Rightarrow a+4=5+2 \\ \Rightarrow a=7-4 \\ \Rightarrow a=3 \end{gathered}$$
Hence, $a=3,b=5$.