Question

Find the values of a and b, if the function f defined by $f\left(x\right)=\left\{\begin{array}{ccc}{x}^2+3x+a& ,& x⩽1\\ bx+2& ,& x>1\end{array}\right.$ is differentiable at x = 1.

Answer 1

Given that f(x) is differentiable at x = 1. Therefore, f(x) is continuous at x = 1.

$$\begin{gathered} \underset{x\to 1^{-}}{\lim }f\left(x\right)=\underset{x\to 1^{+}}{\lim }f\left(x\right)=f\left(1\right) \\ \Rightarrow \underset{x\to 1}{\lim }\left(x^2+3x+a\right)=\underset{x\to 1}{\lim }\left(bx+2\right)=1+3+a \\ \Rightarrow 1+3+a=b+2 \\ \Rightarrow a-b+2=0.....\left(1\right) \end{gathered}$$

Again, f(x) is differentiable at x = 1. So,

(LHD at x = 1) = (RHD at x = 1)
$\Rightarrow \underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}$
$$\begin{gathered} \underset{x\to 1}{\lim }\frac{\left(x^2+3x+a\right)-\left(4+a\right)}{x-1}=\underset{x\to 1}{\lim }\frac{\left(bx+2\right)-\left(4+a\right)}{x-1} \\ \Rightarrow \underset{x\to 1}{\lim }\frac{x^2+3x-4}{x-1}=\underset{x\to 1}{\lim }\frac{\left(bx-2-a\right)}{x-1} \\ \Rightarrow \underset{x\to 1}{\lim }\frac{\left(x+4\right)\left(x-1\right)}{x-1}=\underset{x\to 1}{\lim }\frac{bx-b}{x-1} \\ \Rightarrow \underset{x\to 1}{\lim }\left(x+4\right)=\underset{x\to 1}{\lim }\frac{b\left(x-1\right)}{x-1} \\ \Rightarrow 5=b \end{gathered}$$

Putting b = 5 in (1), we get

a = 3

Hence, a = 3 and b = 5.