1
JEE Advanced 2025 Paper 2 Online
Numerical
+4
-0
Change Language
The sum of the spin only magnetic moment values (in B.M.) of $\left[\mathrm{Mn}(\mathrm{Br})_6\right]^{3-}$ and $\left[\mathrm{Mn}(\mathrm{CN})_6\right]^{3-}$ is _________.
Your input ____
2
JEE Advanced 2025 Paper 2 Online
Numerical
+4
-0
Change Language

A linear octasaccharide (molar mass $=1024 \mathrm{~g} \mathrm{~mol}^{-1}$ ) on complete hydrolysis produces three monosaccharides: ribose, 2-deoxyribose and glucose. The amount of 2-deoxyribose formed is $58.26 \%(\mathrm{w} / \mathrm{w})$ of the total amount of the monosaccharides produced in the hydrolyzed products. The number of ribose unit(s) present in one molecule of octasaccharide is $\qquad$ .

Use: Molar mass $\left(\right.$ in g $\left.\mathrm{mol}^{-1}\right)$ : ribose $=150,2$-deoxyribose $=134$, glucose $=180$;

Atomic mass (in amu): $\mathrm{H}=1, \mathrm{O}=16$

Your input ____
3
JEE Advanced 2025 Paper 2 Online
MCQ (Single Correct Answer)
+3
-1
Change Language

Let $x_0$ be the real number such that $e^{x_0} + x_0 = 0$. For a given real number $\alpha$, define

$$g(x) = \frac{3x e^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)}$$

for all real numbers $x$.

Then which one of the following statements is TRUE?

A

For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$

B

For $\alpha = 2$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 1$

C

For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = 0$

D

For $\alpha = 3$, $\displaystyle \lim_{x \to x_0} \left| \frac{g(x) + e^{x_0}}{x - x_0} \right| = \frac{2}{3}$

4
JEE Advanced 2025 Paper 2 Online
MCQ (Single Correct Answer)
+3
-1
Change Language

Let ℝ denote the set of all real numbers. Then the area of the region

$ \left\{ (x, y) \in \mathbb{R} \times \mathbb{R} : x > 0, y > \frac{1}{x}, 5x - 4y - 1 > 0, 4x + 4y - 17 < 0 \right\} $

is

A

$\frac{17}{16} - \log_e{4}$

B

$\frac{33}{8} - \log_e{4}$

C

$\frac{57}{8} - \log_e{4}$

D

$\frac{17}{2} - \log_e{4}$

JEE Advanced Papers
EXAM MAP