If the function $y=f(x)=alog_ex+bx^2+x, x> 0, $ has extreme points at x=1 and x= 2 and y'' at $x=\frac{3}{2}$ is $\lambda $ then $\frac{a\, b}{\lambda } $ is equal to :

-3

The correct answer is Option (1) → -3

$y=a\log_ex+bx^2+x$

$\frac{dy}{dx}=\frac{a}{x}+2bx+1$

Extreme points at $x = 1$ and $x = 2$

At extreme points, $\frac{dy}{dx}=0$

At $x=1$,

$0=a+2b+1$   ...(1)

At $x=2$,

$0=\frac{a}{2}+2b(2)+1$

$0=\frac{a}{2}+4b+1$

$0=a+8b+2$   ...(2)

From (1) and (2)

$a+2b+1=0$

$a+8b+2=0$

$∴a+8b+2-(a+2b+1)=0$

$a+8b+2-a-2b-1=0$

$6b+1=0$

$b=-\frac{1}{6}$

Substitute $b=-\frac{1}{6}$ in (1)

$0=a+2(-\frac{1}{6})+1$

$a+\frac{2}{3}=0$

$a=-\frac{2}{3}$

$∴ab=-\frac{1}{6}.\frac{-2}{3}$

$=\frac{1}{9}$

Taking second derivative of y

$\frac{d^2y}{dx^2}=\frac{-a}{x^2}+2b$

$=\frac{-(-\frac{2}{3})}{x^2}+2(-\frac{1}{6})$

$\frac{d^2y}{dx^2}=\frac{2}{3x^2}-\frac{1}{3}$

Substitute $x=\frac{3}{2}$

$\frac{d^2y}{dx^2}=\frac{2}{3(\frac{3}{2})^2}-\frac{1}{3}$

$=\frac{2}{\frac{27}{4}}-\frac{1}{3}$

$=\frac{8}{27}-\frac{1}{3}$

$=\frac{8}{27}-\frac{9}{27}=-\frac{1}{27}$

Since $\frac{d^2y}{dx^2}=λ$

$∴\frac{ab}{λ}=\frac{\frac{1}{9}}{-\frac{1}{27}}$

$=-3$