Question

The function $f\left(x\right)=(\begin{array}{c}{x}^2+3x+a,\mathrm{if}x\le 1\\ \mathrm{bx}+2,\mathrm{if}x>1\end{array}$
is differentiable at each

$x\in R$. Then, the value of a is and b is .

Answer 1

We have,
$f\left(x\right)=\left(\begin{array}{c}{x}^2+3x+a,\mathrm{if}x\le 1\\ \mathrm{bx}+2,\mathrm{if}x>1\end{array}\mathrm{It}\mathrm{is}\mathrm{given}\mathrm{that}f\right(x)\mathrm{is}\mathrm{differentiable}\mathrm{at}x=1\mathrm{and}\mathrm{every}\mathrm{differentiable}\mathrm{function}\mathrm{is}\mathrm{continuous}.\mathrm{So},f(x)\mathrm{is}\mathrm{continuous}\mathrm{at}x=1\therefore \underset{x\to 1^{-}}{\lim }f(x)=\underset{x\to 1^{+}}{\lim }f(x)=f(1\left)\underset{x\to 1^{-}}{\lim }\right(x^2+3x+a)=\underset{x\to 1^{+}}{\lim }\mathrm{bx}+2=4+a\Rightarrow 4+a=b+2=4+a\Rightarrow a=b-2\mathrm{Now},f(x)\mathrm{is}\mathrm{differentiable}\mathrm{at}x=1\Rightarrow (\mathrm{LHD}\mathrm{at}x=1)=(\mathrm{RHD}\mathrm{at}x=1)\Rightarrow \underset{x\to 1^{-}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{f\left(x\right)-f\left(1\right)}{x-1}\Rightarrow \underset{x\to 1^{-}}{\lim }\frac{\left(x^2+3x+a\right)-4-a}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{\mathrm{bx}+2-4-a}{x-1}\Rightarrow \underset{x\to 1^{-}}{\lim }\frac{x^2+3x-4}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{\mathrm{bx}+2-4-a}{x-1}\Rightarrow \underset{x\to 1^{-}}{\lim }\frac{x^2+4x-x-4}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{\mathrm{bx}-2-\left(b-2\right)}{x-1}\Rightarrow \underset{x\to 1^{-}}{\lim }\frac{\left(x+4\right)\left(x-1\right)}{x-1}=\underset{x\to 1^{+}}{\lim }\frac{\mathrm{bx}-b}{x-1}\Rightarrow \underset{x\to 1^{-}}{\lim }x+4=\underset{x\to 1^{+}}{\lim }\frac{b\left(x-1\right)}{x-1}\Rightarrow 5=b\therefore a=3$