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=3/4, b=9/4
• B. a=1, b=2
• C. a=3/2, b=9/2
• D. a=3/4, b=9/2

Answer 1

The correct option is A a=3/4, b=9/4

Since $f$ is differentiable for all x, in particular it is continuous at 2.
But $f\left(2\right)=4a-2b+3$ and ${lim}_{x\to 2-}f\left(x\right)=2a$
so $4a-2b+3=2a$. i.e. $2a-2b+3=0$ (i)
Also $f^{\prime }(2-)=a$ and
$f^{\prime }(2+)=\underset{h\to 0+}{lim}\frac{a{(2+h)}^2-b(2+h)+3-(4a-2b+3)}{h}$
$=\underset{h\to 0+}{lim}\frac{a({(2+h)}^2-4)-b(h-2)}{h}=4a-b$
Thus $4a-b=a\Rightarrow 3a=b$ (ii)
Solving (i) and (ii) $a=3/4$ and $b=9/4.$