If y = a ln |x| + bx2 + x has its extreme values at x = -1 and x = 2 then P ≡ (a , b) is
Answer & explanation
Correct answer: option 2
Since $\frac{dy}{dx}=\frac{a}{x}+2bx+1$, $\frac{dy}{dx}|_{x=-1}=0$ and $\frac{dy}{dx}|_{x=2}=0$
⇒ a + 2b –1 = 0 , a + 8b + 2 = 0 ⇒ a = 2 , b = $-\frac{1}{2}$.
Hence (B) is the correct answer.