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1

### JEE Advanced 2019 Paper 2 Offline

MCQ (Single Correct Answer)
Let f(x) = sin($$\pi$$ cos x) and g(x) = cos(2$$\pi$$ sin x) be two functions defined for x > 0. Define the following sets whose elements are written in the increasing order :

X = {x : f(x) = 0}, Y = {x : f'(x) = 0}

Z = {x : g(x) = 0}, W = {x : g'(x) = 0}

List - I contains the sets X, Y, Z and W. List - II contains some information regarding these sets.

A
(II), (Q), (T)
B
(II), (R), (S)
C
(I), (P), (R)
D
(I), (Q), (U)

## Explanation

For, X = {x : f(x) = 0}, x > 0

Now, f(x) = 0

$$\Rightarrow$$ sin($$\pi$$ cos x) = 0, x > 0

$$\Rightarrow$$ $$\pi$$ cos x = n$$\pi$$, n $$\in$$ Integer.

$$\Rightarrow$$ cos x = n

$$\Rightarrow$$ cos x = $$-$$1, 0, 1

{$$\because$$ cos x $$\in$$[$$-$$1, 1]}

When cos x = ± 1$$\Rightarrow$$ x = n$$\pi$$

When cos x = 0 $$\Rightarrow$$ x = (2n + 1)$${{\pi \over 2}}$$

Hence, (i) $$\to$$ (P), (Q)

For, Y = {x : f'(x) = 0}, x > 0

Now, f'(x) = 0

$$\Rightarrow$$ $$-$$$$\pi$$ sin x cos($$\pi$$ cos x) = 0

$$\Rightarrow$$ either sin x = 0 $$\Rightarrow$$ x = n$$\pi$$, n is an integer,

or cos($$\pi$$ cos x) = 0

$$\Rightarrow$$ $$\pi$$ cos x = (2n + 1)$${{\pi \over 2}}$$, n is an integer

$$\Rightarrow$$ cos x = $${{{2n + 1} \over 2}}$$

$$\Rightarrow$$ $$\cos x = \pm {1 \over 2}$$ {$$\because$$ cos x $$\in$$[$$-$$1, 1]}

$$\Rightarrow$$ x = $$2n\pi \pm {\pi \over 3}$$ or $$2n\pi \pm {{2\pi } \over 3}$$, n is an integer.

So, (ii) $$\to$$ (Q), (T)

Hence, option (a) is correct.
2

### JEE Advanced 2019 Paper 2 Offline

MCQ (Single Correct Answer)
Let f(x) = sin($$\pi$$ cos x) and g(x) = cos(2$$\pi$$ sin x) be two functions defined for x > 0. Define the following sets whose elements are written in the increasing order:

X = {x : f(x) = 0}, Y = {x : f'(x) = 0}

Z = {x : g(x) = 0}, W = {x : g'(x) = 0}

List - I contains the sets X, Y, Z and W. List - II contains some information regarding these sets.

Which of the following is the only CORRECT combination?
A
(IV), (P), (R), (S)
B
(III), (P), (Q), (U)
C
(III), (R), (U)
D
(IV), (Q), (T)

## Explanation

For Z = {x : g(x) = 0}, x > 0

$$\because$$ g(x) = cos(2$$\pi$$ sin x) = 0

$$\Rightarrow$$ $$2\pi \sin x = (2n + 1){\pi \over 2},\,n \in$$ Integer

$$\Rightarrow$$ $$\sin x = - {3 \over 4}, - {1 \over 4},{1 \over 4},{3 \over 4}$$ [$$\because$$ sin x $$\in$$ [$$-$$1, 1]]

here values of sin x, $$- {3 \over 4}, - {1 \over 4},{1 \over 4},{3 \over 4}$$ are in an A.P. but corresponding values of x are not in an AP so, (iii) $$\to$$ R.

For W = {x : g'(x) = 0}, x > 0

So, g'(x) = $$-$$2 $$\pi$$ cos x sin(2$$\pi$$ sin x) = 0

$$\Rightarrow$$ either cos x = 0 or sin(2$$\pi$$ sin x) = 0

$$\Rightarrow$$ either $$x = (2n + 1){\pi \over 2}$$ or 2$$\pi$$ sin x = n$$\pi$$, n$$\in$$ Integers.

$$\because$$ $$2\pi \sin x = nx$$

$$\Rightarrow$$ $$\sin x = {n \over 2} = - 1, - {1 \over 2},0,{1 \over 2},1$$ {$$\because$$ sin x $$\in$$[$$-$$1, 1)}

$$\because$$ $$x = n\pi ,\,(2n + 1){\pi \over 2}$$ or $$x = n\pi + {( - 1)^n}\left( { \pm {\pi \over 6}} \right)$$

$$\Rightarrow$$ (iv) $$\to$$ P, R, S

Hence, option (a) is correct.
3

### JEE Advanced 2016 Paper 2 Offline

MCQ (Single Correct Answer)
The value of

$$\sum\limits_{k = 1}^{13} {{1 \over {\sin \left( {{\pi \over 4} + {{\left( {k - 1} \right)\pi } \over 6}} \right)\sin \left( {{\pi \over 4} + {{k\pi } \over 6}} \right)}}}$$ is equal to
A
$$3 - \sqrt 3$$
B
$$2\left( {3 - \sqrt 3 } \right)$$
C
$$2\left( {\sqrt 3 - 1} \right)\,\,\,$$
D
$$2\left( {2 - \sqrt 3 } \right)$$

## Explanation

It is given that,

$$\sum\limits_{k = 1}^{13} {{1 \over {\sin \left( {{\pi \over 4} + {{(k - 1)\pi } \over 6}} \right)\sin \left( {{\pi \over 4} + {{k\pi } \over 6}} \right)}}}$$

Let $$\alpha = {\pi \over 4}$$ and $$\beta = {\pi \over 6}$$. Therefore,

$$\sum\limits_{k = 1}^{13} {{1 \over {\sin (\alpha + k\beta )sin(\alpha + (k - 1)\beta )}}}$$

$$= {1 \over {\sin \beta }}\sum\limits_{k = 1}^{13} {{{\sin ((\alpha + k\beta ) - (\alpha + (k - 1)\beta ))} \over {\sin (\alpha + k\beta )\sin (\alpha + (k - 1)\beta )}}}$$

$$= {1 \over {\sin \beta }}\sum\limits_{k = 1}^{13} {(\cot (\alpha + (k - 1)\beta ) - \cot (\alpha + k\beta ))}$$

$$= {1 \over {\sin \beta }}\{ [\cot (\alpha ) - \cot (\alpha + \beta )] + [\cot (\alpha + \beta ) - \cot (\alpha + 2\beta )] + ...... + [\cot (\alpha + 12\beta ) - \cot (\alpha + 13\beta )]\}$$

$$= {1 \over {\sin \beta }}(\cot \alpha - \cot (\alpha + 13\beta ))$$

$$= {1 \over {\sin (\pi /6)}}\left( {\cot {\pi \over 4} - \cot \left( {{\pi \over 4} + {{13\pi } \over 6}} \right)} \right)$$

$$= 2(1 - 2 + \sqrt 3 ) = 2(\sqrt 3 - 1)$$

4

### JEE Advanced 2016 Paper 1 Offline

MCQ (Single Correct Answer)
Let $$S = \left\{ {x \in \left( { - \pi ,\pi } \right):x \ne 0, \pm {\pi \over 2}} \right\}.$$ The sum of all distinct solutions of the equation $$\sqrt 3 \,\sec x + \cos ec\,x + 2\left( {\tan x - \cot x} \right) = 0$$ in the set S is equal to
A
$$- {{7\pi } \over 9}$$
B
$$- {{2\pi } \over 9}$$
C
0
D
$${{5\pi } \over 9}$$

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