Question

If $f\left(x\right)=\{\begin{array}{ccc}ax& a<1& \\ a{x}^2+bx+2& a\ge 1& .\end{array}$
Then the values of $a$, $b$ for which $f\left(x\right)$ is differentiable, are

• A. $a=\frac{3}{4}$, $b=\frac{1}{4}$
• B. $a=2$, $b=-2$
• C. $a=\frac{3}{2}$, $b=\frac{1}{4}$
• D. $a=\frac{3}{4}$, $b=-2$

Answer 1

Answer: (B)

$a=2$, $b=-2$

Given the function $f\left(x\right)$ is differentiable, then it is differentiable at $x=1$ and it will be continuous there at.

Then using continuity condition at $x=1$,

$\underset{x\to 1^{-}}{lim}f\left(x\right)=\underset{x\to 1+}{lim}f\left(x\right)$

or, $a=a+b+2$

or, $b=-2$.

Now using differentiability at $x=1$ we get,

$\underset{x\to 1^{-}}{lim}{f}^{\prime }\left(x\right)=\underset{x\to 1^{+}}{lim}{f}^{\prime }\left(x\right)$

or, $a=2a+b$

or, $a=-b$

or, $a=2$.

$\therefore a=2,b=-2$.

so, option (B) is true.