Question

For what values of $‘p’$ the function
$f(n)=\{\begin{array}{ll}p{x}^2+1& \text{if }x\le 1\\ x-p& \text{if }x>1\end{array}$ is derivable at x = 1?

• A. $\frac{1}{2}$
• B. 2
• C. $-\frac{1}{2}$
• D. -2

Answer 1

Answer: (A)

For f (x) to be derivable at x = 1
RHD of f (x) = LHD of f(x) at x = 1
$\Rightarrow R{f}^{\prime }(1)=\underset{h\to 0}{\lim }\frac{f(1+h)-f(1)}{h}$
$=\underset{h\to 0}{\lim }\frac{h}{h}=1$
$L{f}^{\prime }(1)=\underset{h\to 0}{\lim }\frac{f(1-h)-f(1)}{-h}$
= $\underset{h\to 0}{\lim }\frac{p(1-h{)}^2+1-(p-1)}{-h}$
= $\underset{h\to 0}{\lim }\frac{p(1-h^2-2h)+1-(p+1)}{-h}$
= $\underset{h\to 0}{\lim }\frac{p(h^2-2h)}{-h}$
= $\underset{h\to 0}{\lim }(h-2)p=2p$
Now, $R{f}^{\prime }(1)=L{f}^{\prime }(1)$
$\Rightarrow 1=2p$
$\Rightarrow p=\frac{1}{2}$