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1

### JEE Advanced 2021 Paper 2 Online

MCQ (More than One Correct Answer)
For any real numbers $$\alpha$$ and $$\beta$$, let $${y_{\alpha ,\beta }}(x)$$, x$$\in$$R, be the solution of the differential equation $${{dy} \over {dx}} + \alpha y = x{e^{\beta x}},y(1) = 1$$. Let $$S = \{ {y_{\alpha ,\beta }}(x):\alpha ,\beta \in R\}$$. Then which of the following functions belong(s) to the set S?
A
$$f(x) = {{{x^2}} \over 2}{e^{ - x}} + \left( {e - {1 \over 2}} \right){e^{ - x}}$$
B
$$f(x) = - {{{x^2}} \over 2}{e^{ - x}} + \left( {e + {1 \over 2}} \right){e^{ - x}}$$
C
$$f(x) = {{{e^x}} \over 2}\left( {x - {1 \over 2}} \right) + \left( {e - {{{e^2}} \over 4}} \right){e^{ - x}}$$
D
$$f(x) = {{{e^x}} \over 2}\left( {{1 \over 2} - x} \right) + \left( {e + {{{e^2}} \over 4}} \right){e^{ - x}}$$

## Explanation

Given, $${{dy} \over {dx}} + \alpha y = x\,.\,{e^{\beta x}}$$ which is a linear differential equation.

Integrating factor $$(IF) = {e^{\int {\alpha dx} }} = {e^{\alpha x}}$$

So, the solution is $$y \times {e^{\alpha x}} = \int {x{e^{\beta x}}\,.\,{e^{\alpha x}}dx}$$

$$\Rightarrow y \times {e^{\alpha x}} = \int {x{e^{(\alpha + \beta )x}}dx}$$ .... (i)

Case (I) If $$\alpha$$ + $$\beta$$ = 0

From Eq. (i), we get

$$\Rightarrow y{e^{\alpha x}} = \int {x{e^{0.x}}dx = \int {xdx = {{{x^2}} \over 2} + C} }$$ .... (ii)

Given, y(1) = 1 i.e. when x = 1, then y = 1

From Eq. (ii), we get

$$1.{e^\alpha } = {1 \over 2} + C \Rightarrow C = {e^\alpha } - {1 \over 2}$$

From Eq. (ii), we get

$$y{e^{\alpha x}} = {{{x^2} - 1} \over 2} + {e^\alpha }$$

For $$\alpha$$ = 1

$$y{e^x} = {{{x^2} - 1} \over 2} + e \Rightarrow y = {{{x^2}} \over 2}{e^{ - x}} + \left( {e - {1 \over 2}} \right){e^{ - x}}$$

Option (a) is correct.

Case (II) If $$\alpha$$ + $$\beta$$ $$\ne$$ 0

$$\Rightarrow y{e^{\alpha x}} = x.{{{e^{(\alpha + \beta )x}}} \over {(\alpha + \beta )}} - \int {1 \times {{{e^{(\alpha + \beta )x}}} \over {(\alpha + \beta )}}dx}$$

$$\Rightarrow y{e^{\alpha x}} = x.{{{e^{(\alpha + \beta )x}}} \over {(\alpha + \beta )}} - {{{e^{(\alpha + \beta )x}}} \over {{{(\alpha + \beta )}^2}}} + {c_1}$$

$$\Rightarrow y = {{x\,.\,{e^{\beta x}}} \over {(\alpha + \beta )}} - {{{e^{\beta x}}} \over {{{(\alpha + \beta )}^2}}} + {c_1}{e^{ - \alpha x}}$$ (Cancelling e$$\alpha$$x from both sides)

$$\Rightarrow y = {{{e^{\beta x}}} \over {\alpha + \beta }}\left( {x - {1 \over {\alpha + \beta }}} \right) + {c_1}{e^{ - \alpha x}}$$ .... (iii)

Putting $$\alpha$$ = $$\beta$$ = 1 in Eq. (iii), we get

$$y = {{{e^x}} \over 2}\left( {x - {1 \over 2}} \right) + {c_1}{e^{ - x}}$$

Given, y(1) = 1

$$\therefore$$ $$1 = {e \over 2} \times {1 \over 2} + {{{c_1}} \over e} \Rightarrow {c_1} = e - {{{e^2}} \over 4}$$

So, $$y = {{{e^x}} \over 2}\left( {x - {1 \over 2}} \right) + \left( {e - {{{e^2}} \over 4}} \right){e^{ - x}}$$ $$\to$$ option (c) is correct.
2

### JEE Advanced 2021 Paper 2 Online

MCQ (More than One Correct Answer)
Let $$f:\left[ { - {\pi \over 2},{\pi \over 2}} \right] \to R$$ be a continuous function such that $$f(0) = 1$$ and $$\int_0^{{\pi \over 3}} {f(t)dt = 0}$$. Then which of the following statements is(are) TRUE?
A
The equation $$f(x) - 3\cos 3x = 0$$ has at least one solution in $$\left( {0,{\pi \over 3}} \right)$$
B
The equation $$f(x) - 3\sin 3x = - {6 \over \pi }$$ has at least one solution in $$\left( {0,{\pi \over 3}} \right)$$
C
$$\mathop {\lim }\limits_{x \to 0} {{x\int_0^x {f(t)dt} } \over {1 - {e^{{x^2}}}}} = - 1$$
D
$$\mathop {\lim }\limits_{x \to 0} {{\sin x\int_0^x {f(t)dt} } \over {{x^2}}} = - 1$$

## Explanation

Given, f(0) = 1 and $$\int_0^{\pi /3} {f(t)\,dt = 0}$$

For option (a)

Consider a function

$$g(x) = \int_0^\pi {f(t)\,dt - \sin 3x}$$

g(x) is continuous and differentiable function and g(0) = g($$\pi$$/3) = 0

$$\therefore$$ By Rolle's theorem, g'(x) = 0 has at least one solution in (0, $$\pi$$/3).

i.e. g'(x) = f(x) $$\times$$ 1 $$-$$ 3cos 3x = 0 for some $$x \in \left( {0,{\pi \over 3}} \right)$$

For option (b)

Consider the function

$$\phi (x) = \int_0^x {f(t)dt + \cos 3x + {6 \over \pi }x}$$

$$\phi$$(x) is continuous and differentiable function as well as $$\phi$$(0) = $$\phi$$($$\pi$$/3) = 1

Hence, by Rolle's theorem, $$\phi$$'(x) = 0 has at least one solution in (0, $$\pi$$ / 3).

i.e. $$\phi$$'(x) = f(x) $$\times$$ 1 $$-$$ 3sin 3x + $${6 \over \pi } = 0$$ for some $$x \in \left( {0,{\pi \over 3}} \right)$$.

For option (c)

Let $$L = \mathop {\lim }\limits_{x \to 0} {{x\int_0^x {f(t)dt} } \over {1 - {e^{{x^2}}}}}$$ (form $${0 \over 0}$$)

$$\Rightarrow L = \mathop {\lim }\limits_{x \to 0} {{xf(x) + \int_0^x {f(t)dt} } \over { - 2x{e^{{x^2}}}}}$$ (Using L-Hospital Rule)

Again, using L'-Hospital Rule ($$\therefore$$ form $${0 \over 0}$$)

$$L = \mathop {\lim }\limits_{x \to 0} {{xf'(x) + f(x) + f(x)} \over { - 4{x^2}{e^{{x^2}}} - 2{e^{{x^2}}}}} = {{0 + 2f(0)} \over { - 0 - 2}} = - 1$$ ($$\because$$ f(0) = 1)

For option (d)

Let $$P = \mathop {\lim }\limits_{x \to 0} {{\sin x.\int_0^x {f(t)dt} } \over {{x^2}}}$$ (form $${0 \over 0}$$)

Applying L-Hospital Rule,

$$P = \mathop {\lim }\limits_{x \to 0} {{\sin x.f(x) + \cos x.\int_0^x {f(t)dt} } \over {2x}}$$ (form $${0 \over 0}$$)

Again using L-Hospital Rule,

$$P = \mathop {\lim }\limits_{x \to 0} {{[\cos x.f(x) + \sin x.f'(x) + \cos x.f(x) - \sin x\int_0^x {f(t)dt]} } \over 2}$$

$$\Rightarrow P = {{1 \times f(0) + 0 \times f'(0) + 1 \times f(0) - 0 \times 0} \over 2}$$

$$\Rightarrow P = {{1 + 0 + 1 - 0} \over 2} = 1$$
3

### JEE Advanced 2020 Paper 2 Offline

MCQ (More than One Correct Answer)
Let b be a nonzero real number. Suppose f : R $$\to$$ R is a differentiable function such that f(0) = 1. If the derivative f' of f satisfies the equation $$f'(x) = {{f(x)} \over {{b^2} + {x^2}}}$$

for all x$$\in$$R, then which of the following statements is/are TRUE?
A
If b > 0, then f is an increasing function
B
If b < 0, then f is a decreasing function
C
f(x) f($$-$$x) = 1 for all x$$\in$$R
D
f(x) $$-$$f($$-$$x) = 0 for all x$$\in$$R

## Explanation

Given differential equation

$$f'(x) = {{f(x)} \over {{b^2} + {x^2}}}$$

$$\Rightarrow \int {{{f'(x)} \over {f(x)}}dx = \int {{{dx} \over {{b^2} + {x^2}}}} }$$

$$\Rightarrow {\log _e}|f(x)| = {1 \over b}{\tan ^{ - 1}}\left( {{x \over b}} \right) + c$$

$$\because$$ f(0) = 1, so c = 0

$$\therefore$$ $$|f(x)| = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$$

$$\Rightarrow f(x) = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$$ or $$- {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$$

$$\because$$ $$f(0) = 1 \Rightarrow f(x) = {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}$$

and $$f'(x) = {1 \over {{b^2}}}{{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {1 + {{\left( {{x \over b}} \right)}^2}}} = {{{e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}} \over {{b^2} + {x^2}}}$$

$$\therefore$$ $$f'(x) > 0\forall x \in R$$ and $$b \in {R_0}$$.

Therefore, f(x) is an increasing function $$\forall b \in {R_0}$$.

and f(x) f($$-$$x)

$$= {e^{{1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}}\,.\,{e^{ - {1 \over b}{{\tan }^{ - 1}}\left( {{x \over b}} \right)}} = 1$$
4

### JEE Advanced 2020 Paper 1 Offline

MCQ (More than One Correct Answer)
Which of the following inequalities is/are TRUE?
A
$$\int_0^1 {x\cos xdx\, \ge \,{3 \over 8}}$$
B
$$\int_0^1 {x\sin xdx\, \ge \,{3 \over {10}}}$$
C
$$\int_0^1 {{x^2}\cos xdx\, \ge \,{1 \over 2}}$$
D
$$\int_0^1 {{x^2}\sin xdx\, \ge \,{2 \over 9}}$$

## Explanation

$$\because$$ $$\cos x = 1 - {{{x^2}} \over {2!}} + {{{x^4}} \over {4!}} - {{{x^6}} \over {6!}} + ...$$

and $$\sin x = x - {{{x^3}} \over {3!}} + {{{x^5}} \over {5!}} - {{{x^7}} \over {7!}} + ....$$

$$\therefore$$ $$\int_0^1 {x\cos xdx} \ge \int_0^1 {\left( {x - {{{x^3}} \over {2!}}} \right)} \,dx$$

$$= \left[ {{{{x^2}} \over 2} - {{{x^4}} \over 8}} \right]_0^1 = {1 \over 2} - {1 \over 8} = {3 \over 8}$$

$$\Rightarrow \int_0^1 {x\cos xdx\, \ge \,{3 \over 8}}$$

and, $$\int_0^1 {x\sin dx \ge \int_0^1 {\left( {{x^2} - {{{x^4}} \over 6}} \right)} \,dx}$$

$$\left[ {{{{x^3}} \over 3} - {{{x^5}} \over {30}}} \right]_0^1 = {1 \over 3} - {1 \over {30}} = {9 \over {30}} = {3 \over {10}}$$

$$\Rightarrow \int_0^1 {x\sin xdx\, \ge \,{3 \over {10}}}$$

and, $$\int_0^1 {{x^2}\cos xdx\, \ge \,\int_0^1 {\left( {{x^3} - {{{x^5}} \over 2}} \right)\,dx} }$$

$$= \left[ {{{{x^4}} \over 4} - {{{x^6}} \over {12}}} \right]_0^1 = {1 \over 4} - {1 \over {12}} = {2 \over {12}} = {1 \over 6}$$

$$\therefore$$ $$\int_0^1 {{x^2}\cos xdx\, \ge \,{1 \over 6}}$$

and, $$\int_0^1 {{x^2}\sin xdx\, \ge \int_0^1 {\left( {{x^3} - {{{x^5}} \over 6}} \right)\,dx} }$$

$$= \left[ {{{{x^4}} \over 4} - {{{x^6}} \over {36}}} \right]_0^1 = {1 \over 4} - {1 \over {36}} = {8 \over {36}} = {2 \over 9}$$

$$\therefore$$ $$\int_0^1 {{x^2}\sin xdx\, \ge \,{2 \over 9}}$$

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