Question

For what values of $‘p’$ the function
$f(n) = \begin{cases} px^2 + 1 & \text{if } x \leq 1\\ x - p & \text{if } x > 1 \end{cases} $ 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
$\Rightarrow \:\: Rf '(1) =\displaystyle \lim_{h \to 0} \frac{f(1+h) - f(1)}{h}$
$ = \displaystyle\lim_{h \to 0} \frac{h}{h} = 1$
$ Lf'(1) =\displaystyle\lim_{h \to 0} \frac{f(1 - h ) - f( 1)}{-h}$
= $ \displaystyle\lim_{h \to 0} \frac{p(1 - h )^2 + 1 - ( p - 1)}{-h}$
= $\displaystyle \lim_{h \to 0} \frac{p(1 - h^2 - 2h) + 1 - ( p + 1)}{-h}$
= $\displaystyle \lim_{h \to 0} \frac{p(h^2 - 2h ) }{-h}$
= $\displaystyle \lim_{h \to 0} (h - 2)p = 2p $
Now, $ Rf ' (1) = Lf ' (1)$
$\Rightarrow \: 1 = 2p$
$\Rightarrow \: p = \frac{1}{2}$