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1

### JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
The option(s) with the values of a and $$L$$ that satisfy the following equation is (are) $${{\int\limits_0^{4\pi } {{e^t}\left( {{{\sin }^6}at + {{\cos }^4}at} \right)dt} } \over {\int\limits_0^\pi {{e^t}\left( {{{\sin }^6}at + {{\cos }^4}at} \right)dt} }} = L?$$\$
A
$$a = 2,L = {{{e^{4\pi }} - 1} \over {{e^\pi } - 1}}$$
B
$$a = 2,L = {{{e^{4\pi }} + 1} \over {{e^\pi } + 1}}$$
C
$$a = 4,L = {{{e^{4\pi }} - 1} \over {{e^\pi } - 1}}$$
D
$$a = 4,L = {{{e^{4\pi }} + 1} \over {{e^\pi } + 1}}$$

## Explanation

Let $${I_1} = \int_0^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt}$$

$$= \int_0^\pi {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_\pi ^{2\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_{2\pi }^{3\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt + \int_{3\pi }^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} } } }$$

$$\therefore$$ $${I_1} = {I_2} + {I_3} + {I_4} + {I_5}$$ ...... (i)

Now, $${I_3} = \int_\pi ^{2\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt}$$

Put $$t = \pi + t \Rightarrow dt = dt$$

$$\therefore$$ $${I_3} = \int_0^\pi {{e^{\pi + t}}.\,({{\sin }^6}at + {{\cos }^6}at)dt}$$

$$= {e^t}\,.\,{I_2}$$ ..... (ii)

Now, $${I_4} = \int_{2\pi }^{3\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt}$$

Put $$t = 2\pi + t \Rightarrow dt = dt$$

$$\therefore$$ $${I_4} = \int_0^\pi {{e^{t + 2\pi }}({{\sin }^6}at + {{\cos }^6}at)dt}$$

$$= {e^{2\pi }}.{I_2}$$ ...... (iii)

and $${I_5} = \int_{3\pi }^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt}$$

Put $$t = 3\pi + t$$

$$\therefore$$ $${I_5} = \int_0^\pi {{e^{3\pi + t}}({{\sin }^6}at + {{\cos }^6}at)dt}$$

$$= {e^{3\pi }}.\,{I_2}$$ ...... (iv)

From Eqs. (i), (ii), (iii) and (iv), we get

$${I_1} = {I_2} + {e^\pi }.\,{I_2} + {e^{2\pi }}.\,{I_2} + {e^{3\pi }}.\,{I_2}$$

$$= (1 + {e^\pi } + {e^{2\pi }} + {e^{3\pi }}){I_2}$$

$$\therefore$$ $$L = {{\int_0^{4\pi } {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} } \over {\int_0^\pi {{e^t}({{\sin }^6}at + {{\cos }^6}at)dt} }}$$

$$= (1 + {e^\pi } + {e^{2\pi }} + {e^{3\pi }})$$

$$= {{1.\,({e^{4\pi }} - 1)} \over {{e^\pi } - 1}}$$ for $$a \in R$$

2

### JEE Advanced 2015 Paper 2 Offline

MCQ (More than One Correct Answer)
Let $$f\left( x \right) = 7{\tan ^8}x + 7{\tan ^6}x - 3{\tan ^4}x - 3{\tan ^2}x$$ for all $$x \in \left( { - {\pi \over 2},{\pi \over 2}} \right).$$
Then the correct expression(s) is (are)
A
$$\int\limits_0^{\pi /4} {xf\left( x \right)dx = {1 \over {12}}}$$
B
$$\int\limits_0^{\pi /4} {f\left( x \right)dx = 0}$$
C
$$\int\limits_0^{\pi /4} {xf\left( x \right)dx = {1 \over {6}}}$$
D
$$\int\limits_0^{\pi /4} {f\left( x \right)dx = 1}$$

## Explanation

$$f(x) = 7{\tan ^8}x + 7{\tan ^6}x - 3{\tan ^4}x - 3{\tan ^2}x\,\forall x \in \left( {{{ - \pi } \over 2},{\pi \over 2}} \right)$$

$$= 7{\tan ^6}x\,.\,{\sec ^2}x - 3{\tan ^2}x\,.\,{\sec ^2}x$$

$$= (7{\tan ^6}x - 3{\tan ^2}x)\,.\,{\sec ^2}x$$

$$\Rightarrow \int\limits_0^{\pi /4} {f(x)dx = \int\limits_0^{\pi /4} {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}dx = \int\limits_0^1 {(7 + 603{t^2})dt = [{t^7} - {t^3}]_0^1 = 0} } }$$

Also, $$I = \int\limits_0^{\pi /4} {xf(x)dx}$$

$$= \left| {x\,.\,\int {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}xdx} } \right|_0^{\pi /4} - \int\limits_0^{\pi /4} {1\,.\,\int {(7{{\tan }^6}x - 3{{\tan }^2}x){{\sec }^2}x\,dx} }$$

$$= \left| {x\,.\,({{\tan }^7}x - {{\tan }^3}x} \right|_0^{\pi /4} - \int\limits_0^{\pi /4} {({{\tan }^7}x - {{\tan }^3}x)dx}$$

$$= 0 - \int\limits_0^{\pi /4} {{{\tan }^3}x({{\tan }^4}x - 1)dx}$$

$$= - \int\limits_0^{\pi /4} {{{\tan }^3}x({{\tan }^2}x - 1)({{\sec }^2}x)dx}$$

$$= - \int\limits_0^1 {({t^5} - {t^3})dt}$$

$$= - \left[ {{{{t^6}} \over 6} - {{{t^4}} \over 4}} \right]_0^1 = \left[ {{1 \over 4} - {1 \over 6}} \right] = {1 \over {12}}$$

3

### JEE Advanced 2014 Paper 1 Offline

MCQ (More than One Correct Answer)
Let a $$\in$$ R and f : R $$\to$$ R be given by f(x) = x5 $$-$$ 5x + a. Then,
A
f(x) has three real roots , if a > 4
B
f(x) has only one real root, if a > 4
C
f(x) has three real roots, if a < $$-$$4
D
f(x) has three real roots, if $$-$$4 < a < 4

## Explanation

In the equation $$f(x) = {x^5} - 5x + a$$, there are different polynomials depending on the parameter a. Now, for the roots of each of these, in general, f(x) = 0.

That is, $$a = 5x - {x^5} = x(5 - {x^5})$$

Hence, the parameter a is a function of x. That is,

$$a(x) = x(5 - {x^5})$$

Now, $$a'(x) = 5 - 5{x^4}$$

Therefore, extrema occurs at a'(x) = 0

That is, when x4 = 1 or x = 1 and x = $$-$$1 (only real roots considered)

Here, $$a''(x) = - 20{x^3}$$

$$a''(1) < 0$$ (maximum)

$$a''( - 1) > 0$$ (minimum)

Hence, the maximum value is

a(1) = 4

The minimum value is

a($$-$$1) = $$-$$4

Hence, when $$-$$4 < a < 4, there are three points that is, x values where f(x) = 0, that is, three roots of f(x) for any value of a lying in ($$-$$4, 4)}. .... (1)

When | a | > 4, there is only one x for which f(x) = 0 ...... (2)

Hence, from statements (1) and (2), we can conclude that options (B) and (D) are correct.
4

### JEE Advanced 2014 Paper 1 Offline

MCQ (More than One Correct Answer)
Let $$f:\left( {0,\infty } \right) \to R$$ be given by $$f\left( x \right) = \int\limits_{{1 \over x}}^x {{e^{ - \left( {t + {1 \over t}} \right){{dt} \over t}}}} .$$ Then
A
$$f(x)$$ is monotonically increasing on $$\left[ {1,\infty } \right)$$
B
$$f(x)$$ is monotonically decreasing on $$(0,1)$$
C
$$f(x)$$ $$+ f\left( {{1 \over x}} \right) = 0$$, for all $$x \in \left( {0,\infty } \right)$$
D
$$f\left( {{2^x}} \right)$$ is an odd function of $$x$$ on $$R$$

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