Question

Let If f(x) is differentiable at x = 1, then a = ____________.

Solution

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. $\therefore Lf\text{'}\left(1\right)=Rf\text{'}\left(1\right)$ $⇒\underset{h\to 0}{\mathrm{lim}}\frac{f\left(1-h\right)-f\left(1\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{f\left(1+h\right)-f\left(1\right)}{h}$ $⇒\underset{h\to 0}{\mathrm{lim}}\frac{\left(1-h+\frac{5}{2}\right)-\left(1+\frac{5}{2}\right)}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{\left[a{\left(1+h\right)}^{2}+3\right]-\left(1+\frac{5}{2}\right)}{h}$ $⇒\underset{h\to 0}{\mathrm{lim}}\frac{-h}{-h}=\underset{h\to 0}{\mathrm{lim}}\frac{a{\left(1+h\right)}^{2}-\frac{1}{2}}{h}$ $⇒1=\underset{h\to 0}{\mathrm{lim}}\frac{\left(a-\frac{1}{2}\right)+a\left(2h+{h}^{2}\right)}{h}$ $⇒1=\underset{h\to 0}{\mathrm{lim}}\frac{\left(a-\frac{1}{2}\right)+ah\left(2+h\right)}{h}$ $⇒1=\underset{h\to 0}{\mathrm{lim}}\frac{\left(a-\frac{1}{2}\right)}{h}+\underset{h\to 0}{\mathrm{lim}}a\left(2+h\right)$ Here, LHS is finite. So, for RHS to be finite, we must have $a-\frac{1}{2}=0$ $⇒a=\frac{1}{2}$ Thus, the value of a is $\frac{1}{2}$. Let If f(x) is differentiable at x = 1, then a = .MathematicsRD Sharma XII Vol 1 (2019)All

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