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 (Single Correct Answer)
English
Bengali

Let f : [a, b] $$\to$$ R be continuous in [a, b], differentiable in (a, b) and f(a) = 0 = f(b). Then

A
there exists at least one point $$c \in (a,b)$$ for which $$f'(c) = f(c)$$
B
$$f'(x) = f(x)$$ does not hold at any point of (a, b)
C
at every point of $$(a,b),f'(x) > f(x)$$
D
at every point of $$(a,b),f'(x) < f(x)$$

f : [a, b] $$\to$$ R, [a, b]-তে সন্তত, (a, b)-তে অন্তরকলনযোগ্য এবং f(a) = 0 = f(b) । সেক্ষেত্রে

A
অন্তত একটি বিন্দু $$c \in (a,b)$$ এর অস্তিত্ব আছে যেক্ষেত্রে $$f'(c) = f(c)$$
B
$$(a,b)$$ এর কোন বিন্দুতেই $$f'(x) = f(x)$$ হবে না
C
$$(a,b)$$ এর প্রতিটি বিন্দুতে $$f'(x) > f(x)$$ হবে
D
$$(a,b)$$ এর প্রতিটি বিন্দুতে $$f'(x) < f(x)$$ হবে
2

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

$$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x}\ln \sqrt {{{1 + x} \over {1 - x}}} } \right)$$ is

A
$${1 \over 2}$$
B
0
C
1
D
does not exist

Explanation

$$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x}\ln \sqrt {{{1 + x} \over {1 - x}}} } \right)$$

Put $$x = \cos 2\theta $$

$$ \Rightarrow 2\theta = {\cos ^{ - 1}}(x)$$

$$ \Rightarrow \theta = {1 \over 2}{\cos ^{ - 1}}(x)$$

when $$x \to 0$$ then $$\theta \to {\pi \over 4}$$

$$\therefore$$ $$\mathop {\lim }\limits_{\theta \to {\pi \over 4}} \left( {{{\ln \sqrt {{{1 + \cos 2\theta } \over {1 - \cos 2\theta }}} } \over {\cos 2\theta }}} \right)$$

$$ = \mathop {\lim }\limits_{\theta \to {\pi \over 4}} \left( {{{\ln \sqrt {{{2{{\cos }^2}\theta } \over {2{{\sin }^2}\theta }}} } \over {\cos 2\theta }}} \right)$$

$$ = \mathop {\lim }\limits_{\theta \to {\pi \over 4}} \left( {{{\ln {{({{\cot }^2}\theta )}^{{1 \over 2}}}} \over {\cos 2\theta }}} \right)$$

$$ = \mathop {\lim }\limits_{\theta \to {\pi \over 4}} \left( {{{\ln |\cot \theta |} \over {\cos 2\theta }}} \right)$$

when $$\theta \to {\pi \over 4}$$ then $$\cot \theta > 0$$

$$\therefore$$ $$|\cot \theta | = \cot \theta $$

$$ = \mathop {\lim }\limits_{\theta \to {\pi \over 4}} {{\ln (\cot \theta )} \over {\cos 2\theta }}$$ ($${0 \over 0}$$ form)

Applying L' Hospital Rule,

$$ = \mathop {\lim }\limits_{\theta \to {\pi \over 4}} {{{1 \over {\cot \theta }} \times ( - \cos e{c^2}\theta )} \over { - \sin 2\theta \times 2}}$$

$$ = {{{1 \over {\cot {\pi \over 4}}} \times - \cos e{c^2}{\pi \over 4}} \over { - \sin {\pi \over 2} \times 2}}$$

$$ = {{{1 \over 1} \times - {{(\sqrt 2 )}^2}} \over { - 1 \times 2}} = 0$$

$$ = {2 \over 2} = 1$$

$$\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x}\ln \sqrt {{{1 + x} \over {1 - x}}} } \right)$$ =
A
$${1 \over 2}$$
B
0
C
1
D
এর অস্তিত্ব নেই
3

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

Let $$f(x) = {a_0} + {a_1}|x| + {a_2}|x{|^2} + {a_3}|x{|^3}$$, where $${a_0},{a_1},{a_2},{a_3}$$ are real constants. Then f(x) is differentiable at x = 0

A
whatever be $${a_0},{a_1},{a_2},{a_3}$$.
B
for no values of $${a_0},{a_1},{a_2},{a_3}$$.
C
only if $${a_1} = 0$$
D
only if $${a_1} = 0,{a_3} = 0$$

Explanation

Given,

$$f(x) = {a_0} + {a_1}|x| + {a_2}|x{|^2} + {a_3}|x{|^3}$$

$$\therefore$$ $$f(x) = \left\{ {\matrix{ {{a_0} + {a_1}x + {a_2}{x^2} + {a_3}{x^3},} & {x \ge 0} \cr {{a_0} - {a_1}x + {a_2}{x^2} - {a_3}{x^3},} & {x < 0} \cr } } \right.$$

$$ \Rightarrow f(x) = \left\{ {\matrix{ {{a_1} + 2{a_2}x + 3{a_3}{x^2},} & {x \ge 0} \cr { - {a_1} + 2{a_2}x - 3{a_3}{x^2},} & {x < 0} \cr } } \right.$$

$$f(x)$$ is differentiable at $$x = 0$$

$$\therefore$$ $$\mathrm{L.H.D = R.H.D}$$

$$ \Rightarrow - {a_1} + 0 + 0 = {a_1} + 0 + 0$$

$$ \Rightarrow 2{a_1} = 0$$

$$ \Rightarrow {a_1} = 0$$

মনে কর $$f(x) = {a_0} + {a_1}|x| + {a_2}|x{|^2} + {a_3}|x{|^3}$$, যেখানে $${a_0},{a_1},{a_2},{a_3}$$ বাস্তব ধ্রুবক। তবে f(x) অপেক্ষকটি $$x = 0$$ বিন্দুতে অন্তরকলনযোগ্য হবে

A
$${a_0},{a_1},{a_2},{a_3}$$-এর যে কোন মানের জন্য
B
$${a_0},{a_1},{a_2},{a_3}$$-এর কোন মানের জন্যই নয়
C
কেবলমাত্র যদি $${a_1} = 0$$ হয়
D
কেবলমাত্র যদি $${a_1} = 0,{a_3} = 0$$ হয়
4

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

The values of a, b, c for which the function $$f(x) = \left\{ \matrix{ {{\sin (a + 1)x + \sin x} \over x},x < 0 \hfill \cr c,x = 0 \hfill \cr {{{{(x + b{x^2})}^{{1 \over 2}}} - {x^{{1 \over 2}}}} \over {b{x^{{1 \over 2}}}}},x > 0 \hfill \cr} \right.$$ is continuous at x = 0, are

A
$$a = {3 \over 2},b = - {3 \over 2},c = {1 \over 2}$$
B
$$a = - {3 \over 2},c = {3 \over 2},b$$ is arbitrary non-zero real number.
C
$$a = - {5 \over 2},b = - {3 \over 2},c = {3 \over 2}$$
D
$$a = - 2,b \in R - \{ 0\} ,c = 0$$

Explanation

For continuous at x = 0,

$$\mathop {\lim }\limits_{x \to 0} f(x) = f(0)$$

$$RHL = \mathop {\lim }\limits_{x \to {0^ + }} {{{{(x + b{x^2})}^{{1 \over 2}}} - {x^{{1 \over 2}}}} \over {b{x^{{1 \over 2}}}}}$$

$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{x^{{1 \over 2}}}\left\{ {{{(1 + bx)}^{{1 \over 2}}} - 1} \right\}} \over {b{x^{{1 \over 2}}}}}$$

$$ = \mathop {\lim }\limits_{x \to {0^ + }} {{{{(1 + bx)}^{{1 \over 2}}} - 1} \over b}$$

= 0 (when b $$\ne$$ 0)

$$LHL = \mathop {\lim }\limits_{x \to {0^ - }} {{\sin (a + 1)x + \sin x} \over x}\left( {{0 \over 0}} \right)$$

$$ = \mathop {\lim }\limits_{x \to {0^ - }} {{(\cos (a + 1)x)(a + 1) + \cos x} \over 1}$$

$$ = a + 1 + 1$$

$$ = a + 2$$

And $$f(0) = c$$ (given)

We know, for continuous function

$$LHL = RHL = f(0)$$

$$\therefore$$ $$a + 2 = 0(b \ne 0) = c$$

$$\therefore$$ c = 0, a = $$-$$2

and b = R $$-$$ {0}

a, b, c -এর যেসব মানের জন্য অপেক্ষক $$f(x) = \left\{ \matrix{ {{\sin (a + 1)x + \sin x} \over x},x < 0 \hfill \cr c,x = 0 \hfill \cr {{{{(x + b{x^2})}^{{1 \over 2}}} - {x^{{1 \over 2}}}} \over {b{x^{{1 \over 2}}}}},x > 0 \hfill \cr} \right.$$ x = 0 বিন্দুতে সন্তত হবে, সেগুলি হল

A
$$a = {3 \over 2},b = - {3 \over 2},c = {1 \over 2}$$
B
$$a = - {3 \over 2},c = {3 \over 2},b$$ যদৃচ্ছ অশূণ্য বাস্তব সংখ্যা
C
$$a = - {5 \over 2},b = - {3 \over 2},c = {3 \over 2}$$
D
$$a = - 2,b \in R - \{ 0\} ,c = 0$$

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