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1

### 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 $$c \in (0,1)$$
2

### 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

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

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

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

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

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

### 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

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

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

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

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

$$\Rightarrow I = - b{e^{ - a}}$$
4

### WB JEE 2021

MCQ (Single Correct Answer) English
Bengali
Let f(x) be continuous periodic function with period T. Let $$I = \int\limits_a^{a + T} {f(x)\,dx}$$. Then
A
I is linear function in 'a'
B
I does not depend on 'a'
C
0 < I < a2 + 1 where I depends on 'a'
D
I is quadratic function in 'a'

## Explanation

f(x) be a continuous periodic function with period T.

Given,

$$I = \int\limits_a^{a + T} {f(x)\,dx}$$

$$I = \int\limits_0^T {f(x)\,dx}$$

Hence, I does not depend on 'a'.
মনে কর f(x) একটি সন্তত, পর্যাবৃত্ত অপেক্ষক ও পর্যায়কাল T। মনে কর $$I = \int\limits_a^{a + T} {f(x)\,dx}$$. সেক্ষেত্রে
A
I, 'a' -এর একরৈখিক অপেক্ষক
B
I, 'a' -এর উপর নির্ভরশীল নয়
C
0 < I < a2 + 1 যেখানে I, a-এর উপর নির্ভরশীল
D
I, 'a' -এর দ্বিঘাত অপেক্ষক

## Explanation

f(x) একটি সন্তত, পর্যাবৃত্ত অপেক্ষক ও পর্যায়কাল T

প্রদত্ত,

$$I = \int\limits_a^{a + T} {f(x)\,dx}$$

$$I = \int\limits_0^T {f(x)\,dx}$$

তাই I, 'a' -এর উপর নির্ভরশীল নয়।

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