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1

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

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

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

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

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

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

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
Let f : D $$\to$$ R where D = [$$-$$0, 1] $$\cup$$ [2, 4] be defined by

$$f(x) = \left\{ {\matrix{ {x,} & {if} & {x \in [0,1]} \cr {4 - x,} & {if} & {x \in [2,4]} \cr } } \right.$$ Then,
A
Rolle's theorem is applicable to f in D.
B
Rolle's theorem is not applicable to f in D.
C
there exists $$\xi $$$$\in$$D for which f'($$\xi $$) = 0 but Rolle's theorem is not applicable.
D
f is not continuous in D.

Explanation

We have, f : D $$\to$$ R where D = [0, 1] $$\cup$$ [2, 4]

$$f(x) = \left\{ {\matrix{ {x,} & {if} & {x \in [0,1]} \cr {4 - x,} & {if} & {x \in [2,4]} \cr } } \right.$$


Clearly f(x) is increasing in [0, 1] and decreasing in [2, 4]

$$\therefore$$ f(x) is neither continuous and nor differentiable in D.

$$\therefore$$ Rolles theorem is not applicable to f in D.
মনে কর f : D $$\to$$ R যেখানে D = [$$-$$0, 1] $$\cup$$ [2, 4] এভাবে সংজ্ঞায়িত আছে যে

$$f(x) = \left\{ {\matrix{ {x,} & \text{যখন} & {x \in [0,1]} \cr {4 - x,} & \text{যখন} & {x \in [2,4]} \cr } } \right.$$ সেক্ষেত্রে,
A
D সেটে রােলের উপপাদ্যটি f অপেক্ষকে প্রযুক্ত হবে
B
D সেটে রােলের উপপাদ্যটি f অপেক্ষকে প্রযুক্ত হবে না
C
$$\xi \in$$D-এর অস্তিত্ব আছে যার জন্য f'($$\xi $$) = 0 হবে কিন্তু রােলের উপপাদ্যটি প্রযুক্ত হবে না
D
f, D সেটে সন্তত নয়

Explanation

আমাদের কাছে, f : D $$\to$$ R যেখানে D = [0, 1] $$\cup$$ [2, 4]

$$f(x) = \left\{ {\matrix{ {x,} & \text{যখন} & {x \in [0,1]} \cr {4 - x,} & \text{যখন} & {x \in [2,4]} \cr } } \right.$$


স্পষ্টতই f(x) [0, 1] এ বাড়ছে এবং [2, 4] এ কমছে

$$\therefore$$ f(x) D-এ সন্তত এবং অন্তরকলন যোগ্য নয়।

$$\therefore$$ D সেটে রােলের উপপাদ্যটি f অপেক্ষকে প্রযুক্ত হবে না।
4

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
Let $${S_n} = {\cot ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....$$ to nth term. Then $$\mathop {\lim }\limits_{n \to \infty } {S_n}$$ is
A
$${\pi \over 3}$$
B
$${\pi \over 4}$$
C
$${\pi \over 6}$$
D
$${\pi \over 8}$$

Explanation

Given,

$${S_n} = {\cot ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....$$ to nth term.

$$ = {\cot ^{ - 1}}2 \times {1^2} + {\cot ^{ - 1}}2 \times {2^2} + {\cot ^{ - 1}}2 \times {3^2} + {\cot ^{ - 1}}2 \times {4^2} + ...$$

$$\therefore$$ nth term = cot$$-$$1 2n2

$$ \Rightarrow {t_n} = {\tan ^{ - 1}}\left( {{1 \over {2{n^2}}}} \right)$$

$$ = {\tan ^{ - 1}}\left( {{{(2n + 1) - (2n - 1)} \over {1 + (2n + 1)(2n - 1)}}} \right)$$

$$ = {\tan ^{ - 1}}(2n + 1) - {\tan ^{ - 1}}(2n - 1)$$

$$\therefore$$ $${S_n} = {\tan ^{ - 1}}(2n + 1) - {\tan ^{ - 1}}(1)$$

$$\therefore$$ $$\mathop {\lim }\limits_{n \to \infty } {S_n} = {\tan ^{ - 1}}(\infty ) - {\tan ^{ - 1}}(1)$$

$$ = {\pi \over 2} - {\pi \over 4} = {\pi \over 4}$$
মনে কর $${S_n} = {\cot ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....$$ n তম পদ পর্যন্ত. সেক্ষেত্রে $$\mathop {\lim }\limits_{n \to \infty } {S_n}$$ হবে
A
$${\pi \over 3}$$
B
$${\pi \over 4}$$
C
$${\pi \over 6}$$
D
$${\pi \over 8}$$

Explanation

দেওয়া,

$${S_n} = {\cot ^{ - 1}}2 + {\cot ^{ - 1}}8 + {\cot ^{ - 1}}18 + {\cot ^{ - 1}}32 + ....$$ n তম পদ পর্যন্ত.

$$ = {\cot ^{ - 1}}2 \times {1^2} + {\cot ^{ - 1}}2 \times {2^2} + {\cot ^{ - 1}}2 \times {3^2} + {\cot ^{ - 1}}2 \times {4^2} + ...$$

$$\therefore$$ n তম পদ = cot$$-$$1 2n2

$$ \Rightarrow {t_n} = {\tan ^{ - 1}}\left( {{1 \over {2{n^2}}}} \right)$$

$$ = {\tan ^{ - 1}}\left( {{{(2n + 1) - (2n - 1)} \over {1 + (2n + 1)(2n - 1)}}} \right)$$

$$ = {\tan ^{ - 1}}(2n + 1) - {\tan ^{ - 1}}(2n - 1)$$

$$\therefore$$ $${S_n} = {\tan ^{ - 1}}(2n + 1) - {\tan ^{ - 1}}(1)$$

$$\therefore$$ $$\mathop {\lim }\limits_{n \to \infty } {S_n} = {\tan ^{ - 1}}(\infty ) - {\tan ^{ - 1}}(1)$$

$$ = {\pi \over 2} - {\pi \over 4} = {\pi \over 4}$$

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