Question

If y = a log x + bx² + x has its extreme values at x = 1 and x = 2, then (a, b) = ____________________.

Answer 1

It is given that, $y=a\log x+b{x}^2+x$ has its extreme values at x = 1 and x = 2.

$\therefore \frac{dy}{dx}=0$ at x = 1 and x = 2

$y=a\log x+b{x}^2+x$

Differentiating both sides with respect to x, we get

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

Now,

${\left(\frac{dy}{dx}\right)}_{x=1}=0$

$\Rightarrow a+2b+1=0$

$\Rightarrow a+2b=-1.....\left(1\right)$

Also,

${\left(\frac{dy}{dx}\right)}_{x=2}=0$

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

$\Rightarrow a+8b=-2.....\left(2\right)$

Subtracting (1) from (2), we get

6b = −1

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

Putting $b=-\frac{1}{6}$ in (1), we get

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

$\Rightarrow a=-1+\frac{1}{3}=-\frac{2}{3}$

Thus, the values of a and b are $-\frac{2}{3}$ and $-\frac{1}{6}$, respectively.

Hence, the ordered pair (a, b) is $\left(-\frac{2}{3},-\frac{1}{6}\right)$.


If y = alogx + bx² + x has its extreme values at x = 1 and x = 2, then (a, b) = $\underline{\left(-\frac{2}{3},-\frac{1}{6}\right)}$.