1
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

A spherical chocolate ball has a layer of ice-cream of uniform thickness around it. When the thickness of the ice-cream layer is 1 cm , the ice-cream melts at the rate of $81 \mathrm{~cm}^3 / \mathrm{min}$ and the thickness of the ice-cream layer decreases at the rate of $\frac{1}{4 \pi} \mathrm{~cm} / \mathrm{min}$. The surface area (in $\mathrm{cm}^2$ ) of the chocolate ball (without the ice-cream layer) is :

A
$128 \pi$
B
$196 \pi$
C
$225 \pi$
D
$256 \pi$
2
JEE Main 2025 (Online) 22nd January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $f(x)=\int_0^{x^2} \frac{\mathrm{t}^2-8 \mathrm{t}+15}{\mathrm{e}^{\mathrm{t}}} \mathrm{dt}, x \in \mathbf{R}$. Then the numbers of local maximum and local minimum points of $f$, respectively, are :

A
3 and 2
B
2 and 2
C
2 and 3
D
1 and 3
3
JEE Main 2024 (Online) 8th April Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

If the function $$f(x)=2 x^3-9 \mathrm{ax}^2+12 \mathrm{a}^2 x+1, \mathrm{a}> 0$$ has a local maximum at $$x=\alpha$$ and a local minimum at $$x=\alpha^2$$, then $$\alpha$$ and $$\alpha^2$$ are the roots of the equation :

A
$$x^2-6 x+8=0$$
B
$$8 x^2-6 x+1=0$$
C
$$8 x^2+6 x-1=0$$
D
$$x^2+6 x+8=0$$
4
JEE Main 2024 (Online) 8th April Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $$f(x)=4 \cos ^3 x+3 \sqrt{3} \cos ^2 x-10$$. The number of points of local maxima of $$f$$ in interval $$(0,2 \pi)$$ is

A
1
B
3
C
4
D
2
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