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

Let $\int x^3 \sin x \mathrm{~d} x=g(x)+C$, where $C$ is the constant of integration. If $8\left(g\left(\frac{\pi}{2}\right)+g^{\prime}\left(\frac{\pi}{2}\right)\right)=\alpha \pi^3+\beta \pi^2+\gamma, \alpha, \beta, \gamma \in Z$, then $\alpha+\beta-\gamma$ equals :

A
47
B
55
C
62
D
48
2
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1

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
$256 \pi$
3
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let the shortest distance from $(a, 0), a>0$, to the parabola $y^2=4 x$ be 4 . Then the equation of the circle passing through the point $(a, 0)$ and the focus of the parabola, and having its centre on the axis of the parabola is :

A
$x^2+y^2-8 x+7=0$
B
$x^2+y^2-6 x+5=0$
C
$x^2+y^2-4 x+3=0$
D
$x^2+y^2-10 x+9=0$
4
JEE Main 2025 (Online) 23rd January Evening Shift
MCQ (Single Correct Answer)
+4
-1

Let $x=x(y)$ be the solution of the differential equation $y=\left(x-y \frac{\mathrm{~d} x}{\mathrm{~d} y}\right) \sin \left(\frac{x}{y}\right), y>0$ and $x(1)=\frac{\pi}{2}$. Then $\cos (x(2))$ is equal to :

A
$2\left(\log _e 2\right)-1$
B
$1-2\left(\log _e 2\right)^2$
C
$1-2\left(\log _{\mathrm{e}} 2\right)$
D
$2\left(\log _e 2\right)^2-1$
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