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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

If I is the greatest of $${I_1} = \int\limits_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} $$, $${I_2} = \int\limits_0^1 {{e^{ - {x^2}}}{{\cos }^2}x\,dx} $$, $${I_3} = \int\limits_0^1 {{e^{ - {x^2}}}dx} $$, $${I_4} = \int\limits_0^1 {{e^{ - {x^2}/2}}dx} $$, then

A
I = I1
B
I = I2
C
I = I3
D
I = I4

$${I_1} = \int\limits_0^1 {{e^{ - x}}{{\cos }^2}x\,dx} $$, $${I_2} = \int\limits_0^1 {{e^{ - {x^2}}}{{\cos }^2}x\,dx} $$, $${I_3} = \int\limits_0^1 {{e^{ - {x^2}}}dx} $$, $${I_4} = \int\limits_0^1 {{e^{ - {x^2}/2}}dx} $$ দেওয়া আছে। এদের মধ্যে বৃহত্তম I হলে

A
I = I1
B
I = I2
C
I = I3
D
I = I4
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

Let $$f(x) = \int\limits_{\sin x}^{\cos x} {{e^{ - {t^2}}}dt} $$. Then $$f'\left( {{\pi \over 4}} \right)$$ equals

A
$$\sqrt {{1 \over e}} $$
B
$$ - \sqrt {{2 \over e}} $$
C
$$\sqrt {{2 \over e}} $$
D
$$ - \sqrt {{1 \over e}} $$

Explanation

$$f(x) = \int\limits_{\sin x}^{\cos x} {{e^{ - {t^2}}}dt} $$

Differentiating both sides using Newton Leibnitz formula,

$$f'(x) = {e^{ - {{\cos }^2}x}}\,.\,( - \sin x) - {e^{ - {{\sin }^2}x}}\,.\,(\cos x)$$

$$\therefore$$ $$f'\left( {{\pi \over 4}} \right) = {e^{ - {1 \over 2}}}\,.\,\left( { - {1 \over {\sqrt 2 }}} \right) - {e^{ - {1 \over 2}}}\,.\,\left( {{1 \over {\sqrt 2 }}} \right)$$

$$ = - {2 \over {\sqrt 2 }}{e^{ - {1 \over 2}}}$$

$$ = - \sqrt 2 {e^{ - {1 \over 2}}}$$

$$ = - \sqrt {{2 \over e}} $$

মনে কর $$f(x) = \int\limits_{\sin x}^{\cos x} {{e^{ - {t^2}}}dt} $$ । তবে $$f'\left( {{\pi \over 4}} \right)$$ এর মান হবে

A
$$\sqrt {{1 \over e}} $$
B
$$ - \sqrt {{2 \over e}} $$
C
$$\sqrt {{2 \over e}} $$
D
$$ - \sqrt {{1 \over e}} $$
3

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

Let $$\mathop {\lim }\limits_{ \in \to 0 + } \int\limits_ \in ^x {{{bt\cos 4t - a\sin 4t} \over {{t^2}}}dt = {{a\sin 4x} \over x} - 1,\left( {0 < x < {\pi \over 4}} \right)} $$. Then a and b are given by

A
$$a = 2,b = 2$$
B
$$a = {1 \over 4},b = 1$$
C
$$a = - 1,b = 4$$
D
$$a = 2,b = 4$$

Explanation

$$\mathop {\lim }\limits_{ \in \to {0^ + }} \int_ \in ^x {{{bt\cos 4t - a\sin 4t} \over {{t^2}}}dt = {{a\sin 4x} \over x} - 1} $$

$$ \Rightarrow \int_{{0^ + }}^x {{{bt\cos 4t - a\sin 4t} \over {{t^2}}}dt = {{a\sin 4x} \over x} - 1} $$

Differentiating both sides we get,

$${{bx\cos 4x - a\sin 4x} \over {{x^2}}} = {{xa\cos 4x \times 4 - a\sin 4x} \over {{x^2}}}$$

$$ \Rightarrow bx\cos 4x - a\sin 4x = 4ax\cos 4x - a\sin 4x$$

Comparing coefficient of cos 4x and sin 4x both side we get,

b = 4a an $$-$$a = $$-$$a

$$\therefore$$ $${b \over a} = 4$$

By checking options, when $$a = {1 \over 4}$$ and b = 1, then,

$${b \over a} = 4$$

মনে কর $$\mathop {\lim }\limits_{ \in \to 0 + } \int\limits_ \in ^x {{{bt\cos 4t - a\sin 4t} \over {{t^2}}}dt = {{a\sin 4x} \over x} - 1,\left( {0 < x < {\pi \over 4}} \right)} $$ । সেক্ষেত্রে a ও b এর মান হল

A
$$a = 2,b = 2$$
B
$$a = {1 \over 4},b = 1$$
C
$$a = - 1,b = 4$$
D
$$a = 2,b = 4$$
4

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
$${1 \over 2}$$

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