NEW
New Website Launch
Experience the best way to solve previous year questions with mock tests (very detailed analysis), bookmark your favourite questions, practice etc...
1

WB JEE 2022

MCQ (More than One Correct Answer)
English
Bengali

From a balloon rising vertically with uniform velocity v ft/sec a piece of stone is let go. The height of the balloon above the ground when the stone reaches the ground after 4 sec is [g = 32 ft/sec2]

A
220 ft
B
240 ft
C
256 ft
D
260 ft

v ft/sec সমবেগে উল্লম্বভাবে ঊর্ধ্বমুখী একটি বেলুন থেকে একটি প্রস্তরখণ্ড ফেলে দেওয়া হল। 4 sec পরে যখন প্রস্তরখণ্ডটি ভূমি স্পর্শ করে তখন বেলুনের উচ্চতা হবে [g = 32 ft/sec2]

A
220 ft
B
240 ft
C
256 ft
D
260 ft
2

WB JEE 2021

MCQ (More than One Correct Answer)
English
Bengali
The greatest and least value of $$f(x) = {\tan ^{ - 1}} - {1 \over 2}\,ln \,x\,on\,\left[ {{1 \over {\sqrt 3 }},\sqrt 3 } \right]$$ are
A
$${f_{\min }} = \sqrt 3 - 1$$
B
$${f_{\max }} = {\pi \over 6} + {1 \over 4}\ln 3$$
C
$${f_{\min }} = {\pi \over 3} - {1 \over 4}\ln 3$$
D
$${f_{\max }} = {\pi \over {12}} + \ln 5$$

Explanation

We have,

$$f(x) = {\tan ^{ - 1}} - {1 \over 2}\log x\,on\,\left[ {{1 \over {\sqrt 3 }},\sqrt 3 } \right]$$

$$f'(x) = {1 \over {1 + {x^2}}} - {1 \over {2x}}$$

$$f'(x) = {{2x - 1 - {x^2}} \over {2x(1 + {x^2})}}$$

$$f'(x) = {{ - ({x^2} - 2x + 1)} \over {2x(1 + {x^2})}}$$

$$f'(x) = {{ - {{(x - 1)}^2}} \over {2x(1 + {x^2})}}$$

f'(x) is decreasing function $$\forall$$ x > 0

fmax at $$x = {1 \over {\sqrt 3 }}$$ and fmin at x = $${\sqrt 3 }$$

$$\therefore$$ $$f\left( {{1 \over {\sqrt 3 }}} \right) = {\tan ^{ - 1}}{1 \over {\sqrt 3 }} - {1 \over 2}\log {1 \over {\sqrt 3 }}$$

$$ = {\pi \over 6} + {1 \over 4}\log 3$$

$$f\left( {\sqrt 3 } \right) = {\tan ^{ - 1}}\sqrt 3 - {1 \over 2}\log \sqrt 3 $$

$$ = {\pi \over 3} - {1 \over 4}\log 3$$

$$\therefore$$ $${f_{\max }} = {\pi \over 6} + {1 \over 4}\log 3$$

$${f_{\min }} = {\pi \over 3} - {1 \over 4}\log 3$$
অন্তরাল $$\left[ {{1 \over {\sqrt 3 }},\sqrt 3 } \right]$$ -তে $$f(x) = {\tan ^{ - 1}} - {1 \over 2}\,ln \,x$$ -এর সর্বোচচ (max) ও সর্বনিম্ন (min) মান হবে
A
$${f_{\min }} = \sqrt 3 - 1$$
B
$${f_{\max }} = {\pi \over 6} + {1 \over 4}\ln 3$$
C
$${f_{\min }} = {\pi \over 3} - {1 \over 4}\ln 3$$
D
$${f_{\max }} = {\pi \over {12}} + \ln 5$$

Explanation

আমাদের কাছে,

$$ \left[ {{1 \over {\sqrt 3 }},\sqrt 3 } \right]$$ - এ $$f(x) = {\tan ^{ - 1}} - {1 \over 2}\,ln\,x$$

$$f'(x) = {1 \over {1 + {x^2}}} - {1 \over {2x}}$$

$$f'(x) = {{2x - 1 - {x^2}} \over {2x(1 + {x^2})}}$$

$$f'(x) = {{ - ({x^2} - 2x + 1)} \over {2x(1 + {x^2})}}$$

$$f'(x) = {{ - {{(x - 1)}^2}} \over {2x(1 + {x^2})}}$$

f'(x) অপেক্ষক হ্রাস পাচ্ছে $$\forall$$ x > 0

fmax এ $$x = {1 \over {\sqrt 3 }}$$ এবং fmin এ x = $${\sqrt 3 }$$

$$\therefore$$ $$f\left( {{1 \over {\sqrt 3 }}} \right) = {\tan ^{ - 1}}{1 \over {\sqrt 3 }} - {1 \over 2}\,ln {1 \over {\sqrt 3 }}$$

$$ = {\pi \over 6} + {1 \over 4}\,ln \,3$$

$$f\left( {\sqrt 3 } \right) = {\tan ^{ - 1}}\sqrt 3 - {1 \over 2}\, ln\,\sqrt 3 $$

$$ = {\pi \over 3} - {1 \over 4}\,ln\, 3$$

$$\therefore$$ $${f_{\max }} = {\pi \over 6} + {1 \over 4}\,ln\, 3$$

$${f_{\min }} = {\pi \over 3} - {1 \over 4}\,ln\, 3$$
3

WB JEE 2020

MCQ (More than One Correct Answer)
English
Bengali
Tangent is drawn at any point P(x, y) on a curve, which passes through (1, 1). The tangent cuts X-axis and Y-axis at A and B respectively. If AP : BP = 3 : 1, then
A
The differential equation of the curve is $$3x{{dy} \over {dx}} + y = 0$$
B
the differential equation of the curve is $$3x{{dy} \over {dx}} - y = 0$$
C
the curve passes through $$\left( {{1 \over 8},2} \right)$$
D
the normal at (1, 1) is x + 3y = 4

(1, 1) বিন্দুগামী একটি বক্ররেখার উপরিস্থ P(x, y) বিন্দুতে স্পর্শক আঁকা হল। ওই স্পর্শক দুই অক্ষকে যথাক্রমে A ও B বিন্দুতে ছেদ করে। যদি $$AP:BP = 3:1$$ হয়, তবে

A
ওই বক্ররেখার অবকল সমীকরণ হবে $$3x{{dy} \over {dx}} + y = 0$$
B
ওই বক্ররেখার অবকল সমীকরণ হবে $$3x{{dy} \over {dx}} - y = 0$$
C
বক্ররেখাটি $$\left( {{1 \over 8},2} \right)$$ বিন্দুগামী
D
(1, 1) বিন্দুতে অভিলম্বের সমীকরণ $$x + 3y = 4$$

Explanation

$$Y - y = y'(X - x)$$, যেখানে $$y' = {{dy} \over {dx}}$$

x-অক্ষের উপর A বিন্দুর স্থানাঙ্ক $$\left( {x - {y \over {y'}},0} \right)$$ [y = 0 বসিয়ে] এবং y-অক্ষের উপর B বিন্দুর স্থানাঙ্ক (0, y $$-$$ xy') [x = 0 বসিয়ে]

$$PA:PB = 3:1$$

$$\therefore$$ P বিন্দুর স্থানাঙ্ক (x, y) হলে

$$x = {{3.0 + 1.\left( {x{y \over {y'}}} \right)} \over 4}$$

বা, $$4x = x - {y \over {y'}}$$

বা, $$3xy' = - y \Rightarrow 3x{{dy} \over {dx}} + y = 0$$

আবার $$3{{dy} \over y} + {{dx} \over x} = 0$$

বা, $$3\log y + \log x = k$$ বক্রের সমীকরণ।

ইহা (1, 1) বিন্দুগামী।

$$\therefore$$ k = 0

$$\therefore$$ $$\log (x{y^3}) = 0$$

বা, $$x{y^3} = 1$$

ইহা $$\left( {{1 \over 8},2} \right)$$ বিন্দুগামী।

4

WB JEE 2020

MCQ (More than One Correct Answer)
English
Bengali
A particle is projected vertically upwards. If it has to stay above the ground for 12 sec, then
A
velocity of projection is 192 ft/sec
B
greatest height attained is 600 ft
C
velocity of projection is 196 ft/sec
D
greatest height attained is 576 ft

Explanation

We have, $$v = u - gt$$

at t = 6, v = 0

$$ = u - gt = 0$$

$$ \Rightarrow $$ u = 6g = 192 ft/sec ($$ \because $$ g = 32 ft/sec2)

Now, $$h = ut - {1 \over 2}g{t^2}$$

$$ = 192.6 - {1 \over 2} \times 32 \times 6 \times 6$$

= 576 ft

একটি বস্তুকণা উল্লম্ব অভিমুখে প্রক্ষিপ্ত হল। কণাটি ভূমির উপরে 12 সেকেন্ড অতিবাহিত করলে

A
প্রক্ষেপের গতিবেগ হবে 192 ft/sec
B
সর্বোচ্চ অর্জিত উচ্চতা হবে 600 ft
C
প্রক্ষেপের গতিবেগ হবে 196 ft/sec
D
সর্বোচ্চ অর্জিত উচ্চতা হবে 576 ft

Explanation

6 সেকেন্ডে সর্বোচ্চ উচ্চতায় উঠবে।

$$\therefore$$ $$0 = u - 32 \times 6 \Rightarrow u = 192$$ ft/sec

$$0 = {u^2} - 2 \times 32 \times H$$

বা, $$H = {{{u^2}} \over {64}} = 576$$ ft.

Joint Entrance Examination

JEE Main JEE Advanced WB JEE

Graduate Aptitude Test in Engineering

GATE CSE GATE ECE GATE EE GATE ME GATE CE GATE PI GATE IN

Medical

NEET

CBSE

Class 12