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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

If $$y = {e^{{{\tan }^{ - 1}}x}}$$, then

A
$$(1 + {x^2}){y_2} + (2x - 1){y_1} = 0$$
B
$$(1 + {x^2}){y_2} + 2xy = 0$$
C
$$(1 - {x^2}){y_2} - {y_1} = 0$$
D
$$(1 + {x^2}){y_2} + 3x{y_1} + 4y = 0$$

Explanation

$$y = {e^{{{\tan }^{ - 1}}x}}$$

Differentiating both sides with respect to x, we get

$$ \Rightarrow {{dy} \over {dx}} = {e^{{{\tan }^{ - 1}}(x)}} \times {1 \over {1 + {x^2}}}$$

$$ \Rightarrow {y_1}(1 + {x^2}) = {e^{{{\tan }^{ - 1}}x}}$$

$$ \Rightarrow {y_1}(1 + {x^2}) = y$$

Differentiating both side with respect to x, we get

$$ \Rightarrow {y_2}(1 + {x^2}) + {y_1}(2x) = {y_1}$$

$$ \Rightarrow {y_2}(1 + {x^2}) + {y_1}(2x - 1) = 0$$

যখন $$y = {e^{{{\tan }^{ - 1}}x}}$$, তবে

A
$$(1 + {x^2}){y_2} + (2x - 1){y_1} = 0$$
B
$$(1 + {x^2}){y_2} + 2xy = 0$$
C
$$(1 - {x^2}){y_2} - {y_1} = 0$$
D
$$(1 + {x^2}){y_2} + 3x{y_1} + 4y = 0$$
2

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
A bulb is placed at the centre of a circular track of radius 10 m. A vertical wall is erected touching the track at a point P. A man is running along the track with a speed of 10 m/sec. Starting from P the speed with which his shadow is running along the wall when he is at an angular distance of 60$$^\circ$$ from P is
A
30 m/sec
B
40 m/sec
C
60 m/sec
D
80 m/sec

Explanation

Given, radius of circular track is 10 m.


and speed of man = 10 m/sec. i.e. $$v = r{{d\theta } \over {dt}} = 10$$

Now, $$\tan \theta = {y \over r}$$

$$ \Rightarrow y = r\tan \theta $$

$$ \Rightarrow {{dy} \over {dt}} = r{\sec ^2}\theta \,.\,{{d\theta } \over {dt}}$$

$$ = {\sec ^2}\theta \,.\,r{{d\theta } \over {dt}}$$

$$ = {\sec ^2}60^\circ \,.\,(10)$$

$$ = {(2)^2} \times 10$$

$$ = 40$$ m/sec.
10 m ব্যাসার্ধ বিশিষ্ট বৃত্তীয় ট্রাকের কেন্দ্রে একটি বাল্ব রাখা হয়েছে। ট্রাকটিকে P বিন্দুতে স্পর্শ করে এমন একটি উল্লম্ব দেওয়াল আছে। এক ব্যক্তি P বিন্দু থেকে 10 m/sec গতিবেগে ঐ ট্র্যাক বরাবর দৌড়ােচেছন। P থেকে তাঁর অবস্থানের কৌণিক দূরত্ব যখন 60°, তখন ঐ ব্যক্তির ছায়া দেওয়াল বরাবর যাবে যে গতিবেগে তা হল
A
30 m/sec
B
40 m/sec
C
60 m/sec
D
80 m/sec

Explanation

প্রদত্ত, বৃত্তাকার ট্র্যাকের ব্যাসার্ধ 10 মিটার।


এবং মানুষের গতি = 10 m/sec. i.e. $$v = r{{d\theta } \over {dt}} = 10$$

এখন, $$\tan \theta = {y \over r}$$

$$ \Rightarrow y = r\tan \theta $$

$$ \Rightarrow {{dy} \over {dt}} = r{\sec ^2}\theta \,.\,{{d\theta } \over {dt}}$$

$$ = {\sec ^2}\theta \,.\,r{{d\theta } \over {dt}}$$

$$ = {\sec ^2}60^\circ \,.\,(10)$$

$$ = {(2)^2} \times 10$$

$$ = 40$$ m/sec.
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
Let $$g(x) = \int\limits_x^{2x} {{{f(t)} \over t}dt} $$ where x > 0 and f be continuous function and f(2x) = f(x), then
A
g(x) is strictly increasing function
B
g(x) is strictly decreasing function
C
g(x) is constant function
D
g(x) is not derivable function

Explanation

Given,

$$g(x) = \int_x^{2x} {{{f(t)} \over t}dt} $$

and f is a continuous function and $$f(2x) = f(x)$$

As, f is a continuous function so we can apply Leibnitz Rule for differentiation.

$$\therefore$$ $$g'(x) = {d \over {dx}}(2x)\,.\,{{f(2x)} \over {2x}} - {d \over {dx}}(x)\,.\,{{f(x)} \over x}$$

$$ = 2\,.\,{{f(2x)} \over {2x}} - 1\,.\,{{f(x)} \over x}$$

$$ = {{f(x)} \over x} - {{f(x)} \over x}$$ [as f(2x) = f(x)]

$$ = 0$$

$$\therefore$$ $$g'(x) = 0 \Rightarrow g(x) = $$ constant function

মনে কর $$g(x) = \int\limits_x^{2x} {{{f(t)} \over t}dt} $$ যেখানে x > 0 এবং f সন্তত অপেক্ষক এবং f(2x) = f(x), সেক্ষেত্রে
A
g(x) যথার্থ ক্রমবর্ধমান অপেক্ষক
B
g(x) যথার্থ ক্রমহ্রাসমান অপেক্ষক
C
g(x) ধ্রুবক অপেক্ষক
D
g(x) অবকলযোগ্য নয়

Explanation

দেওয়া,

$$g(x) = \int_x^{2x} {{{f(t)} \over t}dt} $$, x > 0

$$ \Rightarrow g'(x) = {{f(2x)} \over {2x}}{d \over {2x}}(2x) - {{f(x)} \over x}{d \over {dx}}(x)$$ (বিভেদের জন্য Leibnitz নিয়ম ব্যবহার করে)

$$ = {{f(2x)} \over {2x}}(2) - {{f(x)} \over x}$$

$$ = {{f(2x) - f(x)} \over x}$$

$$ = {{f(x) - f(x)} \over x}$$ ($$\because$$ $$f(2x) = f(x)$$)

$$ \Rightarrow g(x) = 0$$

$$ \Rightarrow$$ g(x) হল ধ্রুবক অপেক্ষক।
4

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
If the function $$f(x) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$ [a > 0] attains its maximum and minimum at p and q respectively such that p2 = q, then a is equal to
A
2
B
$${1 \over 2}$$
C
$${1 \over 4}$$
D
3

Explanation

Given,

$$f(x) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$

$$f'(x) = 6{x^2} - 18ax + 12{a^2}$$

For maxima or minima f'(x) = 0

$$ \therefore $$ $$6{x^2} - 18ax + 12{a^2}$$ = 0

$${x^2} - 3ax + 2{a^2} = 0$$

$$(x - 2a)(x - a) = 0$$

$$ \Rightarrow $$ x = 2a, a

$$f"(x) = 12x - 18a$$

$$ \therefore $$ $$f"(2a) = 24a - 18a > 0$$

and $$f"(a) = 12a - 18a < 0$$

$$ \therefore $$ f(x) is maximum at x = a

and minimum at x = 2a

$$ \therefore $$ p = a, q = 2a

Given, p2 = q

$$ \Rightarrow $$ a2 = 2a

$$ \Rightarrow $$ a = 2

যদি অপেক্ষক $$f(x) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$ [a > 0], যথাক্রমে p ও q বিন্দুতে সর্বোচ্চ ও সর্বনিম্ন মান পরিগ্রহ করে এবং p2 = q হয়, তবে a-এর মান হবে

A
2
B
$${1 \over 2}$$
C
$${1 \over 4}$$
D
3

Explanation

$$f(x) = 2{x^3} - 9a{x^2} + 12{a^2}x + 1$$

$$\therefore$$ $$f'(x) = 6{x^2} - 18ax + 12{a^2}$$

এখানে, $$f'(p) = 0$$ এবং $$f'(q) = 0$$

$$\therefore$$ $$6{p^2} - 18ap + 12{a^2} = 0$$

এবং $$6{p^2} - 18aq + 12{a^2} = 0$$

$$6({p^2} - {q^2}) - 18a(p - q) = 0$$

বা, $$6(p -q) - 18a = 0$$

বা, $$a = {{p + q} \over 3} = {{{q^2} + q} \over 3}$$

$$\therefore$$ $$6{q^2} - 18aq + 12{a^2} = 0$$

বা, $$6{q^2} - 18q\left( {{{{q^2} + q} \over 3}} \right) + 12{\left( {{{{q^2} + q} \over 3}} \right)^2} = 0$$

বা, $$6{q^2} - {{18{q^3} + 18{q^2}} \over 3} + {{4({q^4} + 2{q^3} + {q^2})} \over 3} = 0$$

বা, $$18{q^2} - 18{q^3} - 18{q^2} + 4{q^4} + 8{q^3} + 4{q^2} = 0$$

বা, $$4{q^4} - 10{q^3} + 4{q^2} = 0$$

বা, $$2{q^2} - 5q + 2 = 0$$

বা, $$2{q^2} - 4q - q + 2 = 0$$

বা, $$2q(q - 2) - 1(q - 2) = 0 \Rightarrow q = 2,\,{1 \over 2}$$

$$\therefore$$ q = 2 হলে $$a = {6 \over 3} = 2$$

$$q = {1 \over 2}$$ হলে $$a = {{{1 \over 4} + {1 \over 2}} \over 3} = {1 \over 4}$$

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