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1

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
2

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
3

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
$$a = - 2,b \in R - \{ 0\} ,c = 0$$
4

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
The $$\mathop {\lim }\limits_{x \to \infty } {\left( {{{3x - 1} \over {3x + 1}}} \right)^{4x}}$$ equals
A
1
B
0
C
e$$-$$8/3
D
e$$-$$4/9

Explanation

Let $$L = \mathop {\lim }\limits_{x \to \infty } {\left( {{{3x - 1} \over {3x + 1}}} \right)^{4x}}$$

$$L = {e^{\mathop {\lim }\limits_{x \to \infty } \left( {{{3x - 1} \over {3x + 1}} - 1} \right)4x}}$$

$$L = {e^{\mathop {\lim }\limits_{x \to \infty } \left( {{{3x - 1 - 3x - 1} \over {3x + 1}}} \right)4x}}$$

$$L = {e^{\mathop {\lim }\limits_{x \to \infty } {{ - 2 \times 4} \over {3 + {1 \over x}}}}}$$

$$L = {e^{ - 8/3}}$$
$$\mathop {\lim }\limits_{x \to \infty } {\left( {{{3x - 1} \over {3x + 1}}} \right)^{4x}}$$ হবে
A
1
B
0
C
e$$-$$8/3
D
e$$-$$4/9

Explanation

ধরা যাক $$L = \mathop {\lim }\limits_{x \to \infty } {\left( {{{3x - 1} \over {3x + 1}}} \right)^{4x}}$$

$$L = {e^{\mathop {\lim }\limits_{x \to \infty } \left( {{{3x - 1} \over {3x + 1}} - 1} \right)4x}}$$

$$L = {e^{\mathop {\lim }\limits_{x \to \infty } \left( {{{3x - 1 - 3x - 1} \over {3x + 1}}} \right)4x}}$$

$$L = {e^{\mathop {\lim }\limits_{x \to \infty } {{ - 2 \times 4} \over {3 + {1 \over x}}}}}$$

$$L = {e^{ - 8/3}}$$

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