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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

The value of $$\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{\sin x}}} \over {{{(\cos x)}^{\sin x}} + {{(\sin x)}^{\cos x}}}}dx} $$ is

A
$${\pi \over 4}$$
B
0
C
$${\pi \over 2}$$
D
$${1 \over 2}$$

Explanation

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

Replace x with $$\left( {{\pi \over 2} - x} \right)$$,

$$\therefore$$ $$I = \int\limits_0^{{\pi \over 2}} {{{{{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{\sin \left( {{\pi \over 2} - x} \right)}}} \over {{{\left[ {\cos \left( {{\pi \over 2} - x} \right)} \right]}^{\sin \left( {{\pi \over 2} - x} \right)}} + {{\left[ {\sin \left( {{\pi \over 2} - x} \right)} \right]}^{\cos \left( {{\pi \over 2} - x} \right)}}}}dx} $$

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

Adding 1 and 2 we get,

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

$$ = \int\limits_0^{{\pi \over 2}} {1\,dx} $$

$$ = {\pi \over 2}$$

$$ \Rightarrow I = {\pi \over 4}$$

$$\int\limits_0^{{\pi \over 2}} {{{{{(\cos x)}^{\sin x}}} \over {{{(\cos x)}^{\sin x}} + {{(\sin x)}^{\cos x}}}}dx} $$ -এর মান হল

A
$${\pi \over 4}$$
B
0
C
$${\pi \over 2}$$
D
$${1 \over 2}$$
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

Let f be derivable in [0, 1], then

A
there exists $$c \in (0,1)$$ such that $$\int\limits_0^c {f(x)dx = (1 - c)f(c)} $$
B
there does not exist any point $$d \in (0,1)$$ for which $$\int\limits_0^d {f(x)dx = (1 - d)f(d)} $$
C
$$\int\limits_0^c {f(x)dx} $$ does not exist, for any $$c \in (0,1)$$
D
$$\int\limits_0^c {f(x)dx} $$ is independent of $$c,c \in (0,1)$$

Explanation

Let $$f(x) = x$$ which is derivable in [0, 1].

Option A :

$$\int_0^c {f(x)dx = (1 - c)f(c)} $$

$$ \Rightarrow \int_0^c {x\,dx = (1 - c)\,.\,c} $$

$$ \Rightarrow \left[ {{{{x^2}} \over 2}} \right]_0^c = (1 - c)\,c$$

$$ \Rightarrow {{{c^2}} \over 2} = (1 - c)\,c$$

$$\therefore$$ $$c = 0$$

or $${c \over 2} = 1 - c$$

$$ \Rightarrow c = 2 - 2c$$

$$ \Rightarrow 3c = 2$$

$$ \Rightarrow c = {2 \over 3}$$

$$\therefore$$ c = 0 does not belongs to (0, 1) but $$c = {2 \over 3}$$ belongs to (0, 1)

$$\therefore$$ Option A is correct.

Option B :

$$\int_0^d {f(x)dx = (1 - d)f(d)} $$

$$ \Rightarrow \int_0^d {x\,dx = (1 - d)\,.\,d} $$

$$ \Rightarrow {{{d^2}} \over 2} = (1 - d)d$$

$$\therefore$$ $$d = 0$$

or

$${d \over 2} = 1 - d$$

$$ \Rightarrow d = {2 \over 3}$$ (which belongs to in between (0, 1))

$$\therefore$$ Option B is incorrect.

Option C :

$$\int_0^c {f(x)dx} $$

$$ = \int_0^c {x\,dx} $$

$$ = \left[ {{{{x^2}} \over 2}} \right]_0^c$$

$$ = {{{c^2}} \over 2}$$

$${{{c^2}} \over 2}$$ exist all values of c between 0 and 1.

$$\therefore$$ Option C is incorrect.

Option D :

$$\int_0^c {f(x) = dx} $$

$$ = \int_0^c {x\,dx} $$

$$ = \left[ {{{{x^2}} \over 2}} \right]_0^c$$

$$ = {{{c^2}} \over 2}$$

$$\therefore$$ $$\int_0^c {f(x)\,dx} $$ is not independent of c.

$$\therefore$$ Option D is incorrect.

মনে কর f, [0, 1] -এ অন্তরকলনযোগ্য। সেক্ষেত্রে

A
(0, 1)-এ এমন c বিন্দুর অস্তিত্ব আছে যে $$\int\limits_0^c {f(x)dx = (1 - c)f(c)} $$ হয়
B
এমন কোন $$d \in (0,1)$$ এর অস্তিত্ব নেই যার জন্য $$\int\limits_0^d {f(x)dx = (1 - d)f(d)} $$ হবে
C
$$\int\limits_0^c {f(x)dx} $$ এর অস্তিত্ব নেই যেখানে $$c \in (0,1)$$
D
$$\int\limits_0^c {f(x)dx} ,c$$ এর উপর নির্ভরশীল নয় যেখানে $$c \in (0,1)$$
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
Let $$I = \int_{\pi /4}^{\pi /3} {{{\sin x} \over x}dx} $$. Then
A
$${{\sqrt 3 } \over 8} \le I \le {{\sqrt 2 } \over 6}$$
B
$${{\sqrt 3 } \over {2\pi }} \le I \le {{2\sqrt 3 } \over \pi }$$
C
$${{\sqrt 3 } \over 9} \le I \le {{\sqrt 2 } \over {16}}$$
D
$$\pi \le I \le {{4\pi } \over 3}$$

Explanation

We have, $$I = \int_{\pi /4}^{\pi /3} {{{\sin x} \over x}dx} $$

Let $$f(x) = {{\sin x} \over x}$$

f(x) is decreasing functions

$$ \Rightarrow f(\pi /3) < f(x) < f(\pi /4)$$

$$ \Rightarrow {{3\sqrt 3 } \over {2\pi }} < f(x) < {{2\sqrt 2 } \over \pi }$$

$$ \Rightarrow {{3\sqrt 3 } \over {2\pi }}\int_{\pi /4}^{\pi /3} {dx} < I < {{2\sqrt 2 } \over \pi }\int_{\pi /4}^{\pi /3} {dx} $$

$$ \Rightarrow {{3\sqrt 3 } \over {2\pi }}\left( {{\pi \over 3} - {\pi \over 4}} \right) < I < {{2\sqrt 2 } \over \pi }(\pi /3 - \pi /4)$$

$$ \Rightarrow {{\sqrt 3 } \over 8} < I < {{\sqrt 2 } \over 6}$$
মনে কর $$I = \int_{\pi /4}^{\pi /3} {{{\sin x} \over x}dx} $$। সেক্ষেত্রে
A
$${{\sqrt 3 } \over 8} \le I \le {{\sqrt 2 } \over 6}$$
B
$${{\sqrt 3 } \over {2\pi }} \le I \le {{2\sqrt 3 } \over \pi }$$
C
$${{\sqrt 3 } \over 9} \le I \le {{\sqrt 2 } \over {16}}$$
D
$$\pi \le I \le {{4\pi } \over 3}$$

Explanation

আমাদের কাছে, $$I = \int_{\pi /4}^{\pi /3} {{{\sin x} \over x}dx} $$

Let $$f(x) = {{\sin x} \over x}$$

f(x) হ্রাস করা অপেক্ষক

$$ \Rightarrow f(\pi /3) < f(x) < f(\pi /4)$$

$$ \Rightarrow {{3\sqrt 3 } \over {2\pi }} < f(x) < {{2\sqrt 2 } \over \pi }$$

$$ \Rightarrow {{3\sqrt 3 } \over {2\pi }}\int_{\pi /4}^{\pi /3} {dx} < I < {{2\sqrt 2 } \over \pi }\int_{\pi /4}^{\pi /3} {dx} $$

$$ \Rightarrow {{3\sqrt 3 } \over {2\pi }}\left( {{\pi \over 3} - {\pi \over 4}} \right) < I < {{2\sqrt 2 } \over \pi }(\pi /3 - \pi /4)$$

$$ \Rightarrow {{\sqrt 3 } \over 8} < I < {{\sqrt 2 } \over 6}$$
4

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
If $$b = \int\limits_0^1 {{{{e^t}} \over {t + 1}}dt} $$, then $$\int\limits_{a - 1}^a {{{{e^{ - t}}} \over {t - a - 1}}} $$ is
A
bea
B
be$$-$$a
C
$$-$$ be$$-$$a
D
$$-$$ bea

Explanation

We have, $$b = \int\limits_0^1 {{{{e^t}} \over {t + 1}}dt} $$

Let $$I = \int\limits_{a - 1}^a {{{{e^{ - t}}} \over {t - a - 1}}} dt$$

$$I = \int\limits_{a - 1}^a {{{{e^{ - (a + a - 1 - t)}}} \over {a + a - 1 - t - a - 1}}} dt$$

$$\because$$ $$\left[ {\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } } \right]$$

$$I = \int\limits_{a - 1}^a {{{{e^{t - 2a + 1}}} \over {a - 2 - t}}dt} $$

put $$t - (a - 1) = x \Rightarrow dt = dx$$

when t = a, x = 1 and t = a $$-$$ 1 $$\Rightarrow$$ x = 0

$$\therefore$$ $$I = \int\limits_0^1 {{{{e^{x + a - 1 - 2a + 1}}} \over {a - 2 - x - a + 1}}dx} $$

$$ \Rightarrow I = \int\limits_0^1 {{{{e^x}\,.\,{e^{ - a}}} \over { - (x + 1)}}dx} $$

$$ \Rightarrow I = - {e^{ - a}}\int\limits_0^1 {{{{e^x}} \over {x + 1}}dx} $$

$$ \Rightarrow I = - {e^{ - a}}(b)$$ $$\because$$ $$\left[ {\int\limits_0^1 {{{{e^t}} \over {t + 1}}dt = b} } \right]$$

$$ \Rightarrow I = - b{e^{ - a}}$$
যদি $$b = \int\limits_0^1 {{{{e^t}} \over {t + 1}}dt} $$ হয়, সেক্ষেত্রে $$\int\limits_{a - 1}^a {{{{e^{ - t}}} \over {t - a - 1}}} $$ হবে
A
bea
B
be$$-$$a
C
$$-$$ be$$-$$a
D
$$-$$ bea

Explanation

আমাদের কাছে, $$b = \int\limits_0^1 {{{{e^t}} \over {t + 1}}dt} $$

মনে কর $$I = \int\limits_{a - 1}^a {{{{e^{ - t}}} \over {t - a - 1}}} dt$$

$$I = \int\limits_{a - 1}^a {{{{e^{ - (a + a - 1 - t)}}} \over {a + a - 1 - t - a - 1}}} dt$$

$$\because$$ $$\left[ {\int\limits_a^b {f(x)dx = \int\limits_a^b {f(a + b - x)dx} } } \right]$$

$$I = \int\limits_{a - 1}^a {{{{e^{t - 2a + 1}}} \over {a - 2 - t}}dt} $$

বসিয়ে $$t - (a - 1) = x \Rightarrow dt = dx$$

যখন t = a, x = 1 এবং t = a $$-$$ 1 $$\Rightarrow$$ x = 0

$$\therefore$$ $$I = \int\limits_0^1 {{{{e^{x + a - 1 - 2a + 1}}} \over {a - 2 - x - a + 1}}dx} $$

$$ \Rightarrow I = \int\limits_0^1 {{{{e^x}\,.\,{e^{ - a}}} \over { - (x + 1)}}dx} $$

$$ \Rightarrow I = - {e^{ - a}}\int\limits_0^1 {{{{e^x}} \over {x + 1}}dx} $$

$$ \Rightarrow I = - {e^{ - a}}(b)$$ $$\because$$ $$\left[ {\int\limits_0^1 {{{{e^t}} \over {t + 1}}dt = b} } \right]$$

$$ \Rightarrow I = - b{e^{ - a}}$$

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