Let f(x)=ax,x<2ax2+bx+3,x≥2 If f is differentiable for all x then the value of (a, b) is equal to

# Let $f\left(x\right)=\left\{\begin{array}{cc}ax,& x<2\\ a{x}^{2}+bx+3,& x\ge 2\end{array}\right\$ If $f$ is differentiable for all $x$ then the value of  is equal to

1. A

(1, 2)

2. B

$\left(\frac{3}{2},\frac{9}{2}\right)$

3. C

$\left(\frac{3}{4},-\frac{9}{2}\right)$

4. D

$\left(\frac{3}{4},-\frac{9}{4}\right)$

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### Solution:

$\underset{x\to 2-}{lim} f\left(x\right)=\underset{x\to 2-}{lim} ax=2a$ and $\underset{x\to 2+}{lim} f\left(x\right)=\underset{x\to 2+}{lim}$

$\left(a{x}^{2}+bx+3\right)=4a+2b+3$

Since is differentiable so continuous, hence

$2a+2b+3=0$

Also $\underset{h\to 0+}{lim} \frac{f\left(2+h\right)-f\left(2\right)}{h}$

$=\underset{h\to 0+}{lim} \frac{a\left(2+h{\right)}^{2}+b\left(2+h\right)+3-\left(4a+2b+3\right)}{h}$

$=4a+b$

$\underset{h\to 0-}{lim} \frac{f\left(2+h\right)-f\left(2\right)}{h}$

$=\underset{h\to 0-}{lim} \frac{a\left(2+h\right)-\left(4a+2b+3\right)}{h}$

So $a=4a+b⇒-3a=b$. Putting this in (i),

we get $a=\frac{3}{4},b=-\frac{9}{4}$