Question

If $y=a$ $log$ $x+b{x}^2+x$ has its extremum value at $x=1$ and $y=2$, then $(a,b)=$

• A. $(1,\frac{1}{2})$
• B. $(\frac{1}{2},2)$
• C. $(2,\frac{-1}{2})$
• D. $(\frac{-2}{3},\frac{-1}{6})$

Answer 1

The correct option is D $(\frac{-2}{3},\frac{-1}{6})$

Given $y=alogx+b{x}^2+x$

To find maximum or minimum value of function, differentiate function w.r.t. x and equate to zero.

$\therefore y^{\prime }=\frac{d}{dx}[alogx+b{x}^2+x]$

$\therefore y^{\prime }=\frac{a}{x}+b\left(2x\right)+1=0$

$\therefore y^{\prime }=\frac{a}{x}+2bx+1=0$ (1)

Given at $x=1$, function is extreme.

$\therefore$ put $x=1$ in equation (1), we get,

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

$\therefore a+2b=-1$ (2)

Similarly, at $x=2$, function is extreme.

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

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

$\therefore \frac{a+8b+2}{2}=0$

$\therefore a+8b+2=0$

$\therefore a+8b=-2$ (3)

Subtract equation (2) from (3), we get,

$8b-2b=-2-(-1)$

$6b=-1$

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

Put this value in equation (2), we get,

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

$a-\frac{1}{3}=-1$

$a=\frac{1}{3}-1$

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