Question

Let \[f\left( x \right)\begin{cases}a x^2 + 1, & x > 1 \\ x + 1/2, & x \leq 1\end{cases}\] . Then, f (x) is derivable at x = 1, if 

पर्याय

• a = 2

• a = 1

• a = 0

• a = 1/2

Answer 1

(d) a = 1/2 

Given:  

`f(x) = {(ax^2 +1 , x>1),(x +1/2, xle 1):}`

The function is derivable at x = 1, iff left hand derivative and right hand derivative of the function are equal at x = 1.

\[\left( \text { LHD at x } = 1 \right) = \lim_{x \to 1^-} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[\left( \text { LHD at x } = 1 \right) = \lim_{h \to 0} \frac{f\left( 1 - h \right) - f\left( 1 \right)}{1 - h - 1}\]
\[\left( \text { LHD at x } = 1 \right) = \lim_{h \to 0} \frac{f\left( 1 - h \right) - f\left( 1 \right)}{- h}\]
\[\left( \text { LHD at x = 1 } \right) = \lim_{h \to 0} \frac{\left( 1 - h + \frac{1}{2} \right) - \frac{3}{2}}{- h} = 1\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{x \to 1^+} \frac{f\left( x \right) - f\left( 1 \right)}{x - 1}\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{h \to 0} \frac{f\left( 1 + h \right) - f\left( 1 \right)}{1 + h - 1}\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{h \to 0} \frac{f\left( 1 + h \right) - f\left( 1 \right)}{h}\]
\[\left( \text { RHD at x = 1 } \right) = \lim_{h \to 0} \frac{a \left( 1 + h \right)^2 + 1 - \frac{3}{2}}{h}\]
\[\left( \text { RHD at x } = 1 \right) = \lim_{h \to 0} \frac{a\left( 1 + h^2 + 2h \right) - \frac{1}{2}}{h}\]
\[ \because\text {  LHD = RHD }\]
\[ \Rightarrow a - \frac{1}{2} = 0\]
\[ \Rightarrow a = \frac{1}{2}\]