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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

$$\mathop {\lim }\limits_{x \to \infty } \left( {{{{x^2} + 1} \over {x + 1}} - ax - b} \right),(a,b \in R)$$ = 0. Then

A
a = 0, b = 1
B
a = 1, b = $$-$$1
C
a = $$-$$1, b = 1
D
a = 0, b = 0

$$\mathop {\lim }\limits_{x \to \infty } \left( {{{{x^2} + 1} \over {x + 1}} - ax - b} \right),(a,b \in R)$$ -এর মান 0 দেওয়া আছে। সেক্ষেত্রে

A
a = 0, b = 1
B
a = 1, b = $$-$$1
C
a = $$-$$1, b = 1
D
a = 0, b = 0
2

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
3

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
4

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

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