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1

### JEE Advanced 2014 Paper 2 Offline

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
$$P = 2,Q = 4,R = 1,S = 3$$
2

### JEE Advanced 2013 Paper 2 Offline

Match List $$I$$ with List $$II$$ and select the correct answer using the code given below the lists:

List $$I$$
$$P.$$$$\,\,\,\,\,$$ $${\left( {{1 \over {{y^2}}}{{\left( {{{\cos \left( {{{\tan }^{ - 1}}y} \right) + y\sin \left( {{{\tan }^{ - 1}}y} \right)} \over {\cot \left( {{{\sin }^{ - 1}}y} \right) + \tan \left( {{{\sin }^{ - 1}}y} \right)}}} \right)}^2} + {y^4}} \right)^{1/2}}$$ takes value

$$Q.$$ $$\,\,\,\,$$ If $$\cos x + \cos y + \cos z = 0 = \sin x + \sin y + \sin z$$ then
possible value of $$\cos {{x - y} \over 2}$$ is

$$R.$$ $$\,\,\,\,\,$$ If $$\cos \left( {{\pi \over 4} - x} \right)\cos 2x + \sin x\sin 2\sec x = \cos x\sin 2x\sec x +$$
$$\cos \left( {{\pi \over 4} + x} \right)\cos 2x$$ then possible value of $$\sec x$$ is

$$S.$$ $$\,\,\,\,\,$$ If $$\cot \left( {{{\sin }^{ - 1}}\sqrt {1 - {x^2}} } \right) = \sin \left( {{{\tan }^{ - 1}}\left( {x\sqrt 6 } \right)} \right),\,\,x \ne 0,$$
Then possible value of $$x$$ is

List $$II$$
$$1.$$ $$\,\,\,\,\,$$ $${1 \over 2}\sqrt {{5 \over 3}}$$

$$2.$$ $$\,\,\,\,\,$$ $$\sqrt 2$$

$$3.$$ $$\,\,\,\,\,$$ $${1 \over 2}$$

$$1.$$ $$\,\,\,\,$$ $$1$$

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

### JEE Advanced 2013 Paper 1 Offline

The value of $$\cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}} \left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} \right)$$ is
A
$${{23} \over {25}}$$
B
$${{25} \over {23}}$$
C
$${{23} \over {24}}$$
D
$${{24} \over {23}}$$

## Explanation

$$\cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + \sum\limits_{k = 1}^n {2k} } \right)} } \right)$$

$$= \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {1 + 2 \times {{n(n + 1)} \over 2}} \right)} } \right)$$

$$= \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}(1 + n(n + 1))} } \right)$$

$$= \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}\left( {{{n(n + 1) + 1} \over {(n + 1) - n}}} \right)} } \right)$$

$$= \cot \left( {\sum\limits_{n = 1}^{23} {{{\cot }^{ - 1}}n - {{\cot }^{ - 1}}(n + 1)} } \right)$$

$$= \cot (({\cot ^{ - 1}}1 + {\cot ^{ - 1}}2 + {\cot ^{ - 1}}3 + .... + {\cot ^{ - 1}}23) - ({\cot ^{ - 1}}2 + {\cot ^{ - 1}}3 + .... + {\cot ^{ - 1}}23 + {\cot ^{ - 1}}24))$$

$$= \cot ({\cot ^{ - 1}}1 - {\cot ^{ - 1}}24)$$

$$= \cot \left( {{{\cot }^{ - 1}}{{24 \times 1 + 1} \over {24 - 1}}} \right) = {{25} \over {23}}$$

4

### IIT-JEE 2008

If $$0 < x < 1$$, then
$$\sqrt {1 + {x^2}} {\left[ {{{\left\{ {x\cos \left( {{{\cot }^{ - 1}}x} \right) + \sin \left( {{{\cot }^{ - 1}}x} \right)} \right\}}^2} - 1} \right]^{1/2}} =$$
A
$${x \over {\sqrt {1 + {x^2}} }}$$
B
$$x$$
C
$$x\sqrt {1 + {x^2}}$$
D
$$\sqrt {1 + {x^2}}$$

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