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1

### JEE Advanced 2021 Paper 2 Online

Numerical
For any real number x, let [ x ] denote the largest integer less than or equal to x. If $$I = \int\limits_0^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx}$$, then the value of 9I is __________.

## Explanation

Given, $$I = \int_0^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]d} x$$ .... (i)

$${{10x} \over {x + 1}} = 1 \Rightarrow 10x = x + 1 \Rightarrow x = {1 \over 9}$$

$${{10x} \over {x + 1}} = 4 \Rightarrow 10x = 4x + 4 \Rightarrow x = {2 \over 3}$$

$${{10x} \over {x + 1}} = 9 \Rightarrow 10x = 9x + 9 \Rightarrow x = 9$$

From Eq. (i), we get

$$I = \int_0^{1/9} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{1/9}^{2/3} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_{2/3}^9 {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx + \int_9^{10} {\left[ {\sqrt {{{10x} \over {x + 1}}} } \right]dx} } } }$$

$$= \int_0^{1/9} {0.dx + \int_{1/9}^{2/3} {1.dx + \int_{2/3}^9 {2.dx + \int_9^{10} {3.dx} } } }$$

$$= 0 + [x]_{1/9}^{2/3} + [2x]_{2/3}^9 + [3x]_9^{10}$$

$$= {2 \over 3} - {1 \over 9} + 18 - {4 \over 3} + 30 - 27 = 21 - {2 \over 3} - {1 \over 9} = {{182} \over 9}$$

Hence, 9I = 182
2

### JEE Advanced 2021 Paper 2 Online

Numerical
Let $${g_i}:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R,i = 1,2$$, and $$f:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R$$ be functions such that $${g_1}(x) = 1,{g_2}(x) = |4x - \pi |$$ and $$f(x) = {\sin ^2}x$$, for all $$x \in \left[ {{\pi \over 8},{{3\pi } \over 8}} \right]$$. Define $${S_i} = \int\limits_{{\pi \over 8}}^{{{3\pi } \over 8}} {f(x).{g_i}(x)dx}$$, i = 1, 2

The value of $${{48{S_2}} \over {{\pi ^2}}}$$ is ___________.

## Explanation

$${S_2} = \int\limits_{\pi /8}^{3\pi /8} {{{\sin }^2}x\left| {4x - \pi } \right|dx}$$ .... (i)

$${S_2} = \int_{\pi /8}^{3\pi /8} {{{\sin }^2}\left( {{{3\pi } \over 8} + {\pi \over 8} - x} \right)\left| {4\left( {{{3\pi } \over 8} + {\pi \over 8} - x} \right) - \pi } \right|dx}$$ $$[\int_a^b {f(x)dx = \int_a^b {f(a + b - x)dx]} }$$

$$\Rightarrow {S_2} = \int_{\pi /8}^{3\pi /8} {{{\cos }^2}x\left| {\pi - 4x} \right|dx}$$ .... (ii)

Adding Eqs. (i) and (ii), we get

$$2{S_2} = \int_{\pi /8}^{3\pi /8} {\left| {4x - \pi } \right|dx}$$

From figure,

$${A_1} = {1 \over 2} \times {\pi \over 8} \times {\pi \over 2} = {{{\pi ^2}} \over {32}} = {A_2}$$

$$\therefore$$ $$2{S_2} = 2{A_1} = {{{\pi ^2}} \over {16}} \Rightarrow {S_2} = {{{\pi ^2}} \over {32}}$$

Hence, $${{48{S_2}} \over {{\pi ^2}}} = {{48} \over {32}} = {3 \over 2} = 1.5$$
3

### JEE Advanced 2021 Paper 2 Online

Numerical
Let $${g_i}:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R,i = 1,2$$, and $$f:\left[ {{\pi \over 8},{{3\pi } \over 8}} \right] \to R$$ be functions such that $${g_1}(x) = 1,{g_2}(x) = |4x - \pi |$$ and $$f(x) = {\sin ^2}x$$, for all $$x \in \left[ {{\pi \over 8},{{3\pi } \over 8}} \right]$$. Define $${S_i} = \int\limits_{{\pi \over 8}}^{{{3\pi } \over 8}} {f(x).{g_i}(x)dx}$$, i = 1, 2

The value of $${{16{S_1}} \over \pi }$$ is _____________.

## Explanation

$${S_1} = \int_{\pi /8}^{3\pi /8} {{{\sin }^2}x.1\,dx = \int_{\pi /8}^{3\pi /8} {\left( {{{1 - \cos 2x} \over 2}} \right)dx} }$$

$$= \left( {{x \over 2} - {{\sin 2x} \over 4}} \right)_{\pi /8}^{3\pi /8} = {\pi \over 8}$$

$$\therefore$$ $${{16{S_1}} \over \pi } = {{16} \over \pi } \times {\pi \over 8} = 2$$
4

### JEE Advanced 2021 Paper 2 Online

Numerical
Let f1 : (0, $$\infty$$) $$\to$$ R and f2 : (0, $$\infty$$) $$\to$$ R be defined by $${f_1}(x) = \int\limits_0^x {\prod\limits_{j = 1}^{21} {{{(t - j)}^j}dt} }$$, x > 0 and $${f_2}(x) = 98{(x - 1)^{50}} - 600{(x - 1)^{49}} + 2450,x > 0$$, where, for any positive integer n and real numbers a1, a2, ....., an, $$\prod\nolimits_{i = 1}^n {{a_i}}$$ denotes the product of a1, a2, ....., an. Let mi and ni, respectively, denote the number of points of local minima and the number of points of local maxima of function fi, i = 1, 2 in the interval (0, $$\infty$$).

The value of $$6{m_2} + 4{n_2} + 8{m_2}{n_2}$$ is ___________.

## Explanation

$${f_1}(x) = \int_0^x {{{(t - 1)}^1}{{(t - 2)}^2}{{(t - 3)}^3}{{(t - 4)}^4}....{{(t - 21)}^{21}}dt}$$

$$\Rightarrow {f_1}'(x) = (x - 1){(x - 2)^2}{(x - 3)^3}{(x - 4)^4}....{(x - 21)^{21}}$$

Sign Scheme for f1'(x)

From sign scheme of f1'(x), we observe that f(x) has local minima at x = 4k + 1, k$$\in$$W i.e. f1'(x) changes sign from $$-$$ve to + ve which are x = 1, 5, 9, 13, 17, 21 and f(x) has local maxima at x = 4k + 3, k$$\in$$W i.e. f1'(x) changes sign from + ve to $$-$$ ve, which are x = 3, 7, 11, 15, 19.

So, m1 = number of local minima points = 6

and n1 = number of local maxima points = 5

Hence, $$2{m_1} + 3{n_1} + {m_1}{n_1} = 2 \times 6 + 3 \times 5 + 6 \times 5 = 57$$

Also, $${f_2}(x) = 98{(x - 1)^{50}} - 600{(x - 1)^{49}} + 2450$$

$$\Rightarrow {f_2}'(x) = 98 \times 50{(x - 1)^{49}} - 600 \times 49{(x - 1)^{48}}$$

$$\Rightarrow {f_2}'(x) = 98 \times 50{(x - 1)^{48}}(x - 1 - 6)$$

$$\Rightarrow {f_2}'(x) = 98 \times 50{(x - 1)^{48}}(x - 7)$$ Clearly, m2 = 1 and n2 = 0

So, $$6{m_2} + 4{n_2} + 8{m_2}{n_2} = 6 + 0 + 0 = 6$$

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