Question

If y = a ln |x| + bx² + x has its extreme values at x = –1 and

x = 2 then P ≡ (a, b) is

• A. (2, –1)
• B. $\left( 2, - \frac{1}{2} \right)$
• C. $\left( - 2,\frac{1}{2} \right)$
• D. None of these

Answer 1

Answer: (B)

$\left( 2, - \frac{1}{2} \right)$

Answer 2

Since$\frac{dy}{dx} = \frac{a}{x} + 2bx + 1$, $\left. \ \frac{dy}{dx} \right|_{x = - 1} = 0and\left. \ \frac{dy}{dx} \right|_{x = 2} = 0$

⇒ a + 2b –1 = 0, a + 8 b + 2 = 0 ⇒ a = 2, b = –$\frac{1}{2}$