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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

A particle moving in a straight line starts from rest and the acceleration at any time t is $$a - k{t^2}$$ where a and k are positive constants. The maximum velocity attained by the particle is

A
$${2 \over 3}\sqrt {{{{a^3}} \over k}} $$
B
$${1 \over 3}\sqrt {{{{a^3}} \over k}} $$
C
$$\sqrt {{{{a^3}} \over k}} $$
D
$$2\sqrt {{{{a^3}} \over k}} $$

স্থিতাবস্থা থেকে যাত্রা শুরু করে সরলরেখায় গতিশীল কোনও কণার t সময়ে ত্বরণ $$a - k{t^2}$$, a এবং k ধনাত্মক ধ্রুবক হলে, উহার সর্বোচ্চ গতিবেগ হবে

A
$${2 \over 3}\sqrt {{{{a^3}} \over k}} $$
B
$${1 \over 3}\sqrt {{{{a^3}} \over k}} $$
C
$$\sqrt {{{{a^3}} \over k}} $$
D
$$2\sqrt {{{{a^3}} \over k}} $$
2

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
If the tangent at the point P with co-ordinates (h, k) on the curve y2 = 2x3 is perpendicular to the straight line 4x = 3y, then
A
(h, k) = (0, 0) only
B
(h, k) = $$\left( {{1 \over 8}, - {1 \over {16}}} \right)$$ only
C
(h, k) = (0, 0) or $$\left( {{1 \over 8},{1 \over {16}}} \right)$$
D
no such point P exists

Explanation

Given equation

$${y^2} = 2{x^3}$$

$$ \Rightarrow 2y{{dy} \over {dx}} = 6{x^2}$$

$$ \Rightarrow {{dy} \over {dx}} = {{3{x^2}} \over 4}$$

$$ \Rightarrow {\left( {{{dy} \over {dx}}} \right)_{(h,k)}} = {{3{h^2}} \over k}$$

Slope of tangent is perpendicular to line 4x = 3y

$$\therefore$$ $$\left( {{{3{h^2}} \over k}} \right)\left( {{4 \over 3}} \right) = - 1$$

$$ \Rightarrow 4{h^2} = - k$$ .... (i)

and $${k^2} = 2{h^3}$$ ..... (ii)

From Eqs. (i) and (ii)

$${(4{h^2})^2} = 2{h^3}$$

$$ \Rightarrow 16{h^4} = 2{h^3}$$

$$ \Rightarrow h = {1 \over 8}$$

$$\therefore$$ $$k = - 4{\left( {{1 \over 8}} \right)^2}$$

$$ \Rightarrow k = - 4 \times {1 \over {64}} = {{ - 1} \over {64}}$$

$$\therefore$$ $$(h,k) = \left( {{1 \over 8},{{ - 1} \over {64}}} \right)$$ only
বক্ররেখা y2 = 2x3 -এর উপরিস্থ P(h, k) বিন্দুতে অঙ্কিত স্পর্শক 4x = 3y রেখার উপর লম্ব হলে
A
(h, k) = (0, 0) শুধুমাত্র
B
(h, k) = $$\left( {{1 \over 8}, - {1 \over {16}}} \right)$$ শুধুমাত্র
C
(h, k) = (0, 0) or $$\left( {{1 \over 8},{1 \over {16}}} \right)$$
D
এ ধরণের P বিন্দুর অস্তিত্ব নেই

Explanation

দেওয়া সমীকরণ

$${y^2} = 2{x^3}$$

$$ \Rightarrow 2y{{dy} \over {dx}} = 6{x^2}$$

$$ \Rightarrow {{dy} \over {dx}} = {{3{x^2}} \over 4}$$

$$ \Rightarrow {\left( {{{dy} \over {dx}}} \right)_{(h,k)}} = {{3{h^2}} \over k}$$

স্পর্শকের ঢাল 4x = 3y রেখার লম্ব

$$\therefore$$ $$\left( {{{3{h^2}} \over k}} \right)\left( {{4 \over 3}} \right) = - 1$$

$$ \Rightarrow 4{h^2} = - k$$ .... (i)

এবং $${k^2} = 2{h^3}$$ ..... (ii)

সমীকরণ (i) এবং (ii) থেকে

$${(4{h^2})^2} = 2{h^3}$$

$$ \Rightarrow 16{h^4} = 2{h^3}$$

$$ \Rightarrow h = {1 \over 8}$$

$$\therefore$$ $$k = - 4{\left( {{1 \over 8}} \right)^2}$$

$$ \Rightarrow k = - 4 \times {1 \over {64}} = {{ - 1} \over {64}}$$

$$\therefore$$ $$(h,k) = \left( {{1 \over 8},{{ - 1} \over {64}}} \right)$$ শুধুমাত্র
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
Two particles A and B move from rest along a straight line with constant accelerations f and f' respectively. If A takes m sec. more than that of B and describes n units more than that of B in acquiring the same velocity, then
A
$$(f + f'){m^2} = ff'n$$
B
$$(f - ff'){m^2} = ff'n$$
C
$$(f' - f)n = {1 \over 2}ff'{m^2}$$
D
$${1 \over 2}(f + f')m = ff'{n^2}$$

Explanation

For particle A,

Initial velocity = o

Final velocity = v

Acceleration = f

Time taken by A = t + m sec

distance travel by A = x + n

For particle B,

Initial velocity = o

Final velocity = v

Acceleration = f'

Time taken by B = t sec

Distance travel by B = x

For particle A,

v = f(t + m)

For particle B,

v = f' t

$$\therefore$$ f'(t) = f(t + m)

$$\Rightarrow$$ f'(t) $$-$$ ft = fm

$$ \Rightarrow {t^2} = {{{f^2}{m^2}} \over {{{(f' - f)}^2}}}$$

Also, for particle A,

$$n + x = {1 \over 2}f{(t + m)^2}$$

For particle B,

$$x = {1 \over 2}f'{t^2}$$

$$\therefore$$ $$x + {1 \over 2}f'{t^2} = {1 \over 2}f{(t + m)^2}$$

$$ \Rightarrow n + {1 \over 2}f'{t^2} = {1 \over 2}f{t^2} + ftm + {1 \over 2}f{m^2}$$

$$ \Rightarrow n + {1 \over 2}(f' - f){t^2} = fm \times {{fm} \over {f' - f}} + {1 \over 2}f{m^2}$$

$$ \Rightarrow n + {1 \over 2}(f' - f) \times {{{f^2}{m^2}} \over {{{(f' - f)}^2}}} = {{{f^2}{m^2}} \over {(f' - f)}} + {1 \over 2}f{m^2}$$

$$ \Rightarrow n + {{{f^2}{m^2}} \over {2(f' - f)}} = {{{f^2}{m^2}} \over {f' - f}} + {1 \over 2}f{m^2}$$

$$ \Rightarrow n - {{{f^2}{m^2}} \over {2(f' - f)}} = {1 \over 2}f{m^2}$$

$$ \Rightarrow n(f' - f) - {{{f^2}{m^2}} \over 2} = {1 \over 2}f(f' - f){m^2}$$

$$ \Rightarrow n(f' - f) - {{{f^2}{m^2}} \over 2} = {1 \over 2}ff'{m^2} - {1 \over 2}{f^2}{m^2}$$

$$ \Rightarrow n(f' - f) = {1 \over 2}ff'{m^2}$$

স্থিতাবস্থা থেকে দুটি বস্তু কণা A ও B যথাক্রমে সম ত্বরণ f ও f' সহ একটি সরলরেখা বরাবর যাত্রা করে। A-র সময় লাগে B-র চেয়ে m sec. বেশি। A, B এর চেয়ে n একক অধিকপথ অতিক্রম করে ঐ গতিবেগ অর্জনের জন্য। সেক্ষেত্রে
A
$$(f + f'){m^2} = ff'n$$
B
$$(f - ff'){m^2} = ff'n$$
C
$$(f' - f)n = {1 \over 2}ff'{m^2}$$
D
$${1 \over 2}(f + f')m = ff'{n^2}$$

Explanation

For particle A,

Initial velocity = o

Final velocity = v

Acceleration = f

Time taken by A = t + m sec

distance travel by A = x + n

For particle B,

Initial velocity = o

Final velocity = v

Acceleration = f'

Time taken by B = t sec

Distance travel by B = x

For particle A,

v = f(t + m)

For particle B,

v = f' t

$$\therefore$$ f'(t) = f(t + m)

$$\Rightarrow$$ f'(t) $$-$$ ft = fm

$$ \Rightarrow {t^2} = {{{f^2}{m^2}} \over {{{(f' - f)}^2}}}$$

Also, for particle A,

$$n + x = {1 \over 2}f{(t + m)^2}$$

For particle B,

$$x = {1 \over 2}f'{t^2}$$

$$\therefore$$ $$x + {1 \over 2}f'{t^2} = {1 \over 2}f{(t + m)^2}$$

$$ \Rightarrow n + {1 \over 2}f'{t^2} = {1 \over 2}f{t^2} + ftm + {1 \over 2}f{m^2}$$

$$ \Rightarrow n + {1 \over 2}(f' - f){t^2} = fm \times {{fm} \over {f' - f}} + {1 \over 2}f{m^2}$$

$$ \Rightarrow n + {1 \over 2}(f' - f) \times {{{f^2}{m^2}} \over {{{(f' - f)}^2}}} = {{{f^2}{m^2}} \over {(f' - f)}} + {1 \over 2}f{m^2}$$

$$ \Rightarrow n + {{{f^2}{m^2}} \over {2(f' - f)}} = {{{f^2}{m^2}} \over {f' - f}} + {1 \over 2}f{m^2}$$

$$ \Rightarrow n - {{{f^2}{m^2}} \over {2(f' - f)}} = {1 \over 2}f{m^2}$$

$$ \Rightarrow n(f' - f) - {{{f^2}{m^2}} \over 2} = {1 \over 2}f(f' - f){m^2}$$

$$ \Rightarrow n(f' - f) - {{{f^2}{m^2}} \over 2} = {1 \over 2}ff'{m^2} - {1 \over 2}{f^2}{m^2}$$

$$ \Rightarrow n(f' - f) = {1 \over 2}ff'{m^2}$$

4

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
Consider the curve $$y = b{e^{ - x/a}}$$, where a and b are non-zero real numbers. Then
A
$${x \over a} + {y \over b}$$ = 1 is tangent to the curve at (0, 0)
B
$${x \over a} + {y \over b}$$ = 1 is tangent to the curve, where the curve crosses the axis of y
C
$${x \over a} + {y \over b}$$ = 1 is tangent to the curve at (a, 0)
D
$${x \over a} + {y \over b}$$ = 1 is tangent to the curve at (2a, 0)

Explanation

Given, curve $$y = b{e^{ - {x \over a}}}$$

$$ \Rightarrow {{dy} \over {dx}} = - {b \over a}{e^{ - {x \over a}}}$$

$$ \therefore $$ $$\left( {{{dy} \over {dx}}} \right)(0,\,b) = - {b \over a}$$

$$ \therefore $$ Equation of tangent

$$y - b = - {b \over a}(x - 0)$$

$$ \Rightarrow y - b = - {{bx} \over a}$$

$$ \Rightarrow {y \over b} - 1 = {{ - x} \over a}$$

$$ \Rightarrow {x \over a} + {y \over b} = 1$$

বক্ররেখা $$y = b{e^{{{ - x} \over a}}}$$ বিবেচনা করো, যেখানে a ও b অ-শূন্য বাস্তব সংখ্যা। সেক্ষেত্রে

A
$${x \over a} + {y \over b} = 1$$, মূলবিন্দুতে বক্ররেখার স্পর্শক
B
$${x \over a} + {y \over b} = 1$$, বক্ররেখা যে বিন্দুতে y-অক্ষকে ছেদ করে সেই বিন্দুতে স্পর্শক
C
$${x \over a} + {y \over b} = 1,\,(a,0)$$ বিন্দুতে বক্ররেখার স্পর্শক
D
$${x \over a} + {y \over b} = 1,\,(2a,0)$$ বিন্দুতে বক্ররেখার স্পর্শক

Explanation

$$y = b{e^{ - {x \over a}}}$$

বা, $${e^{{x \over a}}}y = b$$

বা, $${x \over a} + \log y = \log b$$

বা, $${1 \over a} + {1 \over y}.{{dy} \over {dx}} = 0$$

বা, $${{dy} \over {dx}} = - {y \over a}$$

(0, 0), (a, 0) ও (2a, 0) বিন্দুতে $${{dy} \over {dx}} = 0$$

$$\therefore$$ (A), (C) ও (D) সিদ্ধ হয় না।

(B) এর ক্ষেত্রে $${{dy} \over {dx}} = - {b \over a}$$

$$\therefore$$ $$y = b$$

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