Question

The values of $a$ and $b$ for which the function $y=a{\log }_{e}x+b{x}^2+x$, has extremum at the points $x_1=1$ and $x_2=2$ are

• A. $a=\frac{2}{3},b=-\frac{1}{6}$
• B. $a=-\frac{2}{3},b=-\frac{1}{6}$
• C. $a=-\frac{2}{3},b=\frac{1}{6}$
• D. $a=-\frac{1}{3},b=-\frac{1}{6}$

Answer 1

Answer: (B)

$a=-\frac{2}{3},b=-\frac{1}{6}$

Given $y=a{\log }_{e}x+b{x}^2+x....\left(i\right)$
$\therefore \frac{dy}{dx}=\frac{a}{x}+2bx+\mathrm{1....}\left(ii\right)$
At the point of maxima $\frac{dy}{dx}=0$
$\therefore \frac{a}{x}+2bx+1=\mathrm{0....}\left(iii\right)$ (By Eq.(ii))
According to question for $x=1$ and $x=2$ satisfy
Eq(iii)
$\therefore a+2b+1=\mathrm{0....}\left(iv\right)$
And $\frac{a}{2}+4b+1=\mathrm{0....}\left(v\right)$
$a=-\frac{2}{3},b=-\frac{1}{6}$By solving Eqs. (iv) and (v)