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

The total number of real solutions of the equation

$ \theta = \tan^{-1}(2 \tan \theta) - \frac{1}{2} \sin^{-1}\left(\frac{6 \tan \theta}{9 + \tan^2 \theta}\right) $

is

(Here, the inverse trigonometric functions $\sin^{-1} x$ and $\tan^{-1} x$ assume values in $[ -\frac{\pi}{2}, \frac{\pi}{2}]$ and $( -\frac{\pi}{2}, \frac{\pi}{2})$, respectively.)

A

1

B

2

C

3

D

5

2
JEE Advanced 2024 Paper 2 Online
MCQ (Single Correct Answer)
+3
-1

Considering only the principal values of the inverse trigonometric functions, the value of

$$ \tan \left(\sin ^{-1}\left(\frac{3}{5}\right)-2 \cos ^{-1}\left(\frac{2}{\sqrt{5}}\right)\right) $$

is

A
$\frac{7}{24}$
B
$\frac{-7}{24}$
C
$\frac{-5}{24}$
D
$\frac{5}{24}$
3
JEE Advanced 2023 Paper 2 Online
MCQ (Single Correct Answer)
+3
-1
For any $y \in \mathbb{R}$, let $\cot ^{-1}(y) \in(0, \pi)$ and $\tan ^{-1}(y) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$. Then the sum of all the solutions of the equation

$\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3}$ for $0<|y|<3$, is equal to :
A
$2 \sqrt{3}-3$
B
$3-2 \sqrt{3}$
C
$4 \sqrt{3}-6$
D
$6-4 \sqrt{3}$
4
JEE Advanced 2014 Paper 2 Offline
MCQ (Single Correct Answer)
+3
-1
Match List $$I$$ with List $$II$$ and select the correct answer using the code given below the lists:

$$\,\,\,\,$$ $$\,\,\,\,$$ $$\,\,\,\,$$ List-$$I$$
(P.)$$\,\,\,\,$$ Let $$y\left( x \right) = \cos \left( {3{{\cos }^{ - 1}}x} \right),x \in \left[ { - 1,1} \right],x \ne \pm {{\sqrt 3 } \over 2}.$$ Then $${1 \over {y\left( x \right)}}\left\{ {\left( {{x^2} - 1} \right){{{d^2}y\left( x \right)} \over {d{x^2}}} + x{{dy\left( x \right)} \over {dx}}} \right\}$$ equals
(Q.)$$\,\,\,\,$$ Let $${A_1},{A_2},....,{A_n}\left( {n > 2} \right)$$ be the vertices of a regular polygon of $$n$$ sides with its centre at the origin. Let $${\overrightarrow {{a_k}} }$$ be the position vector of the point $${A_k},k = 1,2,......,n.$$ $$$f\left| {\sum\nolimits_{k = 1}^{n - 1} {\left( {\overrightarrow {{a_k}} \times \overrightarrow {{a_{k + 1}}} } \right)} } \right| = \left| {\sum\limits_{k = 1}^{n - 1} {\left( {\overrightarrow {{a_k}} .\,\overrightarrow {{a_{k + 1}}} } \right)} } \right|,$$$ then the minimum value of $$n$$ is
(R.)$$\,\,\,\,$$ If the normal from the point $$P(h, 1)$$ on the ellipse $${{{x^2}} \over 6} + {{{y^2}} \over 3} = 1$$ is perpendicular to the line $$x+y=8,$$ then the value of $$h$$ is
(S.)$$\,\,\,\,$$ Number of positive solutions satisfying the equation $${\tan ^{ - 1}}\left( {{1 \over {2x + 1}}} \right) + {\tan ^{ - 1}}\left( {{1 \over {4x + 1}}} \right) = {\tan ^{ - 1}}\left( {{2 \over {{x^2}}}} \right)$$ is

$$\,\,\,\,$$ $$\,\,\,\,$$ $$\,\,\,\,$$List-$$II$$
(1.)$$\,\,\,\,$$ $$1$$
(2.)$$\,\,\,\,$$ $$2$$
(3.)$$\,\,\,\,$$ $$8$$
(4.)$$\,\,\,\,$$ $$9$$

A
$$P = 4,Q = 3,R = 2,S = 1$$
B
$$P = 2,Q = 4,R = 3,S = 1$$
C
$$P = 4,Q = 3,R = 1,S = 2$$
D
$$P = 2,Q = 4,R = 1,S = 3$$
JEE Advanced Subjects