Question

$f\left(x\right)=\{\begin{array}{c}{x}^{2}x\le x_0\\ ax+b,x>x_0\end{array}$. If f(x) is differentiable at $x_0$. Then

• A.
• B.
• C.
• D. None of these

Answer 1

The correct option is B

For f(x) to be diffrentiable at $x_0$ f(x) sould be continuous at $x_0$ and $LHD\&RHD$ of f(x) at $x_0$ should be

equal.

Therefore, $\underset{x\to x_0}{lim}=x_0^2$

$\underset{x\to x_0^{}+}{lim}=a{x}_0+b$

$\Rightarrow x_0^2=a{x}_0+b$.....(1)

LHD $\Rightarrow$ $\underset{x\to x_0^{-}}{lim}=\frac{(x_0+h{)}^2-x_0^2}{h}\underset{x\to 0^{-}}{lim}=\frac{x_0^2+2h{x}_0+h^2-x^2}{h}$

$=\underset{x\to 0^{-}}{lim}=\frac{h}{h}(2x_0+h)=2x_0$

$RHD,\underset{h\to 0^{+}}{lim}=\frac{ah+b-b}{h}=a\underset{h\to 0}{lim}\frac{h}{h}=a$

$\therefore LHD=RHD\Rightarrow a=2x_0$......(2)

Substituting (2) in (1)

$x_0^2=a{x}_0+b$

$x_0^2=2a{x}_0^2+b$

$\Rightarrow b=-x_0^2$.

$\therefore a=2x_0,b=-x_0^2$

Answer option -b