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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.

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Solution

## Given that f(x) is differentiable at x = 1. Therefore, f(x) is continuous at x = 1. $\underset{x\to {1}^{-}}{\mathrm{lim}}f\left(x\right)=\underset{x\to {1}^{+}}{\mathrm{lim}}f\left(x\right)=f\left(1\right)\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 1}{\mathrm{lim}}\left({x}^{2}+3x+a\right)=\underset{x\to 1}{\mathrm{lim}}\left(bx+2\right)=1+3+a\phantom{\rule{0ex}{0ex}}⇒1+3+a=b+2\phantom{\rule{0ex}{0ex}}⇒a-b+2=0.....\left(1\right)$ Again, f(x) is differentiable at x = 1. So, (LHD at x = 1) = (RHD at x = 1) $⇒\underset{x\to {1}^{-}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(1\right)}{x-1}=\underset{x\to {1}^{+}}{\mathrm{lim}}\frac{f\left(x\right)-f\left(1\right)}{x-1}$ $\underset{x\to 1}{\mathrm{lim}}\frac{\left({x}^{2}+3x+a\right)-\left(4+a\right)}{x-1}=\underset{x\to 1}{\mathrm{lim}}\frac{\left(bx+2\right)-\left(4+a\right)}{x-1}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 1}{\mathrm{lim}}\frac{{x}^{2}+3x-4}{x-1}=\underset{x\to 1}{\mathrm{lim}}\frac{\left(bx-2-a\right)}{x-1}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 1}{\mathrm{lim}}\frac{\left(x+4\right)\left(x-1\right)}{x-1}=\underset{x\to 1}{\mathrm{lim}}\frac{bx-b}{x-1}\phantom{\rule{0ex}{0ex}}⇒\underset{x\to 1}{\mathrm{lim}}\left(x+4\right)=\underset{x\to 1}{\mathrm{lim}}\frac{b\left(x-1\right)}{x-1}\phantom{\rule{0ex}{0ex}}⇒5=b$ Putting b = 5 in (1), we get a = 3 Hence, a = 3 and b = 5.

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