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1

JEE Main 2021 (Online) 27th August Evening Shift

The area of the region bounded by the parabola (y $$-$$ 2)2 = (x $$-$$ 1), the tangent to it at the point whose ordinate is 3 and the x-axis is :
A
9
B
10
C
4
D
6

Explanation

y = 3 $$\Rightarrow$$ x = 2

Point is (2, 3)

Diff. w.r.t x

2 (y $$-$$ 2) y' = 1

$$\Rightarrow$$ $$y' = {1 \over {2(y - 2)}}$$

$$\Rightarrow y{'_{(2,3)}} = {1 \over 2}$$

$$\Rightarrow {{y - 3} \over {x - 2}} = {1 \over 2} \Rightarrow x - 2y + 4 = 0$$

Area $$= \int\limits_0^3 {\left( {{{(y - 2)}^2} + 1 - (2y - 4)} \right)} \,dy$$

= 9 sq. units

2

JEE Main 2021 (Online) 27th August Morning Shift

$$\int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx}$$ is equal to :
A
6
B
8
C
5
D
10

Explanation

Let $$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 44x + 484)}}dx}$$

$$I = \int\limits_6^{16} {{{{{\log }_e}{x^2}} \over {{{\log }_e}{x^2} + {{\log }_e}({x^2} - 22)}}dx}$$ .... (1)

We know,

$$\int\limits_a^b {f(x)dx} = \int\limits_a^b {f(a + b - x)} \,dx$$ (king)

So, $$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{{(22 - x)}^2} + {{\log }_e}{{(21 - (22 - x))}^2}}}}$$

$$I = \int\limits_6^{16} {{{{{\log }_e}{{(22 - x)}^2}} \over {{{\log }_e}{x^2} + {{\log }_e}{{(22 - x)}^2}}}dx}$$ .... (2)

(1) + (2)

$$2I = \int\limits_6^{16} {1.\,dx} = 10$$

I = 5
3

JEE Main 2021 (Online) 27th August Morning Shift

If $${U_n} = \left( {1 + {1 \over {{n^2}}}} \right)\left( {1 + {{{2^2}} \over {{n^2}}}} \right)^2.....\left( {1 + {{{n^2}} \over {{n^2}}}} \right)^n$$, then $$\mathop {\lim }\limits_{n \to \infty } {({U_n})^{{{ - 4} \over {{n^2}}}}}$$ is equal to :
A
$${{{e^2}} \over {16}}$$
B
$${4 \over e}$$
C
$${{16} \over {{e^2}}}$$
D
$${4 \over {{e^2}}}$$

Explanation

$${U_n} = \prod\limits_{r = 1}^n {{{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}}$$

$$L = \mathop {\lim }\limits_{n \to \infty } {({U_n})^{ - 4/{n^2}}}$$

$$\log L = \mathop {\lim }\limits_{n \to \infty } {{ - 4} \over {{n^2}}}\sum\limits_{r = 1}^n {\log {{\left( {1 + {{{r^2}} \over {{n^2}}}} \right)}^r}}$$

$$\Rightarrow \log L = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 1}^n { - {{4r} \over n}.{1 \over n}\log \left( {1 + {{{r^2}} \over {{n^2}}}} \right)}$$

$$\Rightarrow \log L = - 4\int\limits_0^1 {x\log (1 + {x^2})\,dx}$$

put 1 + x2 = t

Now, 2xdx = dt

$$= - 2\int\limits_1^2 {\log (t)dt = - 2[t\log t - t]_1^2}$$

$$\Rightarrow \log L = - 2(2\log 2 - 1)$$

$$\therefore$$ $$L = {e^{ - 2(2\log 2 - 1)}}$$

$$= {e^{ - 2\left( {\log \left( {{4 \over e}} \right)} \right)}}$$

$$= {e^{\log {{\left( {{4 \over e}} \right)}^2}}}$$

$$= {\left( {{e \over 4}} \right)^2} = {{{e^2}} \over {16}}$$
4

JEE Main 2021 (Online) 26th August Evening Shift

The value of $$\int\limits_{ - {\pi \over 2}}^{{\pi \over 2}} {\left( {{{1 + {{\sin }^2}x} \over {1 + {\pi ^{\sin x}}}}} \right)} \,dx$$ is
A
$${\pi \over 2}$$
B
$${{5\pi } \over 4}$$
C
$${{3\pi } \over 4}$$
D
$${{3\pi } \over 2}$$

Explanation

$$I = \int\limits_0^{{\pi \over 2}} {{{(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}} + {{{\pi ^{\sin x}}(1 + {{\sin }^2}x)} \over {(1 + {\pi ^{\sin x}})}}} dx$$

$$I = \int_0^{{\pi \over 2}} {(1 + {{\sin }^2}x)\,dx}$$

$$I = {\pi \over 2} + {\pi \over 2}.{1 \over 2} = {{3\pi } \over 4}$$

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