Question

$f\left(x\right)=\{\begin{array}{cc}ax;& x<2\\ a{x}^2-bx+3;& x\ge 2\end{array}$ If $f$ is differentiable for all $x$ then

• A. $a=\frac{3}{4},b=\frac{9}{4}$
• B. $a=\frac{3}{2},b=\frac{9}{2}$
• C. $a=1,b=2$
• D. $a=\frac{3}{4},b=\frac{9}{2}$

Answer 1

The correct option is A $a=\frac{3}{4},b=\frac{9}{4}$

For a function to be differentiable it has to be continuous.

$\Rightarrow f\left(2^{-}\right)=f\left(2^{+}\right)=f\left(2\right)\Rightarrow 2a=4a-2b+3$

$\Rightarrow 2a-2b+3=0........\left(i\right)$

Now,

$f^{\prime }\left(x\right)=\{\begin{array}{cc}a;& x<2\\ 2ax-b;& x\ge 2\end{array}$

$\because$ L.H.D $=$ R.H.D$\Rightarrow a=4a-b\Rightarrow b=3a...\left(ii\right)$

Solving (i) and (ii) we get, $a=\frac{3}{4},b=\frac{9}{4}$