1
JEE Main 2025 (Online) 28th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

The sum of all local minimum values of the function

$$\mathrm{f}(x)=\left\{\begin{array}{lr} 1-2 x, & x<-1 \\ \frac{1}{3}(7+2|x|), & -1 \leq x \leq 2 \\ \frac{11}{18}(x-4)(x-5), & x>2 \end{array}\right.$$

is

A
$\frac{167}{72}$
B
$\frac{157}{72}$
C
$\frac{171}{72}$
D
$\frac{131}{72}$
2
JEE Main 2025 (Online) 24th January Evening Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Let $(2,3)$ be the largest open interval in which the function $f(x)=2 \log _{\mathrm{e}}(x-2)-x^2+a x+1$ is strictly increasing and (b, c) be the largest open interval, in which the function $\mathrm{g}(x)=(x-1)^3(x+2-\mathrm{a})^2$ is strictly decreasing. Then $100(\mathrm{a}+\mathrm{b}-\mathrm{c})$ is equal to :

A
360
B
420
C
160
D
280
3
JEE Main 2025 (Online) 24th January Morning Shift
MCQ (Single Correct Answer)
+4
-1
Change Language

Consider the region $R=\left\{(x, y): x \leq y \leq 9-\frac{11}{3} x^2, x \geq 0\right\}$. The area, of the largest rectangle of sides parallel to the coordinate axes and inscribed in R , is:

A
$\frac{821}{123}$
B
$\frac{567}{121}$
C
$\frac{730}{119}$
D
$\frac{625}{111}$
4
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
$256 \pi$
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