f′(x)=x1+2bx+a f has extremevalues and differentiable ⇒f′(−1)=0⇒a−2b=1 f′(2)=0⇒a+4b=−21⇒a=21,b=−41 f′′(−1),f′′(2) are negative. f has local maxima at −1,2
Let a, b ∈ R be such that the function f given by f(x)=ln |x|+b x^2+a x, x ≠ 0 has extreme values at x=-1 and x=2. Statement 1: f has local maximum…
JEE Main 2012 — Mathematics Calculus
Let a,b∈R be such that the function f given by f(x)=ln∣x∣+bx2+ax,x=0 has extreme values at x=−1 and x=2. Statement 1: f has local maximum at x=−1 and at x=2. Statement 2: a=21 and b=4−1
Held on 30 Apr 2012 · Verified 6 Jul 2026.
Options
- A
Statement 1 is false, statement 2 is true
- B
Statement 1 is true, statement 2 is true; statement 2 is a correct explanation for statement 1
- C
Statement 1 is true, statement 2 is true; statement 2 is not a correct explanation for statement 1
- D
Statement 1 is true, statement 2 is false
Did you get this right?
Sign in to track your attempts and accuracy.
Your note
Sign in to keep a private note on this question. Nothing you write is ever public.
JEE Main Calculus in other years
More JEE Main Calculus questions
If $\int\left(\frac{1-5 \cos ^{2} x}{\sin ^{5} x \cos ^{2} x}\right) d x=f(x)+\mathrm{C}$, where C is the constant of integration, then $f\left(\frac{\pi}{6}\right)-f\left(\frac{\pi}{4}\right)$ is equal to
If the solution curve $y=f(x)$ of the differential equation $\left(x^{2}-4\right) y^{\prime}-2 x y+2 x\left(4-x^{2}\right)^{2}=0, x>2$, passes through the point $(3,15)$, then the local maximum value of $f$ is $\_\_\_\_$.
Let $[\cdot]$ be the greatest integer function. If $\alpha=\int_{0}^{64}\left(x^{1 / 3}-\left[x^{1 / 3}\right]\right) \mathrm{d} x$, then $\frac{1}{\pi} \int_{0}^{\alpha \pi}\left(\frac{\sin ^{2} \theta}{\sin ^{6} \theta+\cos ^{6} \theta}\right) \mathrm{d} \theta$ is equal to $\_\_\_\_$ .
Let $P_{1}: y=4 x^{2}$ and $P_{2}: y=x^{2}+27$ be two parabolas. If the area of the bounded region enclosed between $P_{1}$ and $P_{2}$ is six times the area of the bounded region enclosed between the line $y=\alpha x, \alpha>0$ and $P_{1}$, then $\alpha$ is equal to :
If $y=y(x)$ satisfies the differential equation $16(\sqrt{x+9 \sqrt{x}})(4+\sqrt{9+\sqrt{x}}) \cos y \mathrm{~d} y=(1+2 \sin y) \mathrm{d} x, x>0$ and $y(256)=\frac{\pi}{2}, y(49)=\alpha$, then $2 \sin \alpha$ is equal to :
Other Mathematics topics for JEE Main
Keep practicing Calculus
Work through every JEE Main Calculus PYQ, year by year.