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1

WB JEE 2021

MCQ (More than One Correct Answer)
English
Bengali
$$\mathop {\lim }\limits_{n \to \infty } \left\{ {{{\sqrt n } \over {\sqrt {{n^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 4)}^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 8)}^3}} }} + .... + {{\sqrt n } \over {\sqrt {{{[n + 4(n - 1)]}^3}} }}} \right\}$$ is
A
$${{5 - \sqrt 5 } \over {10}}$$
B
$${{5 + \sqrt 5 } \over {10}}$$
C
$${{2 + \sqrt 3 } \over {2}}$$
D
$${{2 - \sqrt 3 } \over {2}}$$

Explanation

Let $$I = \mathop {\lim }\limits_{n \to \infty } \left\{ {{{\sqrt n } \over {\sqrt {{n^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 4)}^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 8)}^3}} }} + .... + {{\sqrt n } \over {\sqrt {{{[n + 4(n - 1)]}^3}} }}} \right\}$$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{{\sqrt n } \over {\sqrt {{{(n + 4r)}^3}} }}} $$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{{n\sqrt n } \over {\sqrt {{{(n + 4r)}^3}} }}} $$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{1 \over {\sqrt {{{\left( {1 + {{4r} \over n}} \right)}^{3/2}}} }}} $$

$$ \Rightarrow I = \int_0^1 {{{dx} \over {{{(1 + 4x)}^{3/2}}}}} $$

$$ \Rightarrow I = \left[ {{{{{(1 + 4x)}^{ - 3/2 + 1}}} \over {4( - 3/2 + 1)}}} \right]_0^1$$

$$ \Rightarrow I = {{ - 1} \over 2}[{(1 + 4x)^{ - 1/2}}]_0^1$$

$$ \Rightarrow I = {{ - 1} \over 2}\left[ {{1 \over {\sqrt 5 }} - 1} \right]$$

$$ \Rightarrow I = {{ - 1} \over 2}\left[ {{{1 - \sqrt 5 } \over {\sqrt 5 }}} \right)$$

$$ \Rightarrow I = {{\sqrt 5 - 1} \over {2\sqrt 5 }} = {{\sqrt 5 (\sqrt 5 - 1)} \over {10}} = {{5 - \sqrt 5 } \over {10}}$$
$$\mathop {\lim }\limits_{n \to \infty } \left\{ {{{\sqrt n } \over {\sqrt {{n^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 4)}^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 8)}^3}} }} + .... + {{\sqrt n } \over {\sqrt {{{[n + 4(n - 1)]}^3}} }}} \right\}$$ -এর মান হল
A
$${{5 - \sqrt 5 } \over {10}}$$
B
$${{5 + \sqrt 5 } \over {10}}$$
C
$${{2 + \sqrt 3 } \over {2}}$$
D
$${{2 - \sqrt 3 } \over {2}}$$

Explanation

ধরা যাক $$I = \mathop {\lim }\limits_{n \to \infty } \left\{ {{{\sqrt n } \over {\sqrt {{n^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 4)}^3}} }} + {{\sqrt n } \over {\sqrt {{{(n + 8)}^3}} }} + .... + {{\sqrt n } \over {\sqrt {{{[n + 4(n - 1)]}^3}} }}} \right\}$$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } \sum\limits_{r = 0}^{n - 1} {{{\sqrt n } \over {\sqrt {{{(n + 4r)}^3}} }}} $$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{{n\sqrt n } \over {\sqrt {{{(n + 4r)}^3}} }}} $$

$$ \Rightarrow I = \mathop {\lim }\limits_{n \to \infty } {1 \over n}\sum\limits_{r = 0}^{n - 1} {{1 \over {\sqrt {{{\left( {1 + {{4r} \over n}} \right)}^{3/2}}} }}} $$

$$ \Rightarrow I = \int_0^1 {{{dx} \over {{{(1 + 4x)}^{3/2}}}}} $$

$$ \Rightarrow I = \left[ {{{{{(1 + 4x)}^{ - 3/2 + 1}}} \over {4( - 3/2 + 1)}}} \right]_0^1$$

$$ \Rightarrow I = {{ - 1} \over 2}[{(1 + 4x)^{ - 1/2}}]_0^1$$

$$ \Rightarrow I = {{ - 1} \over 2}\left[ {{1 \over {\sqrt 5 }} - 1} \right]$$

$$ \Rightarrow I = {{ - 1} \over 2}\left[ {{{1 - \sqrt 5 } \over {\sqrt 5 }}} \right)$$

$$ \Rightarrow I = {{\sqrt 5 - 1} \over {2\sqrt 5 }} = {{\sqrt 5 (\sqrt 5 - 1)} \over {10}} = {{5 - \sqrt 5 } \over {10}}$$
2

WB JEE 2020

MCQ (More than One Correct Answer)
English
Bengali
Let $$f(x) = {1 \over 3}x\sin x - (1 - \cos \,x)$$. The smallest positive integer k such that $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {{x^k}}} \ne 0$$ is
A
4
B
3
C
2
D
1

Explanation

Given, $$f(x) = {1 \over 3}x\sin x - (1 - \cos \,x)$$

$$ \therefore $$ $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {{x^k}}} = \mathop {\lim }\limits_{x \to 0} {{x\sin x - 3(1 - \cos \,x)} \over {3{x^k}}}$$

$$ = \mathop {\lim }\limits_{x \to 0} {{2x\sin {x \over 2}\cos {x \over 2} - 6{{\sin }^2}{x \over 2}} \over {3{x^k}}}$$

= $${1 \over 3}\mathop {\lim }\limits_{x \to 0} \left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right).\mathop {\lim }\limits_{x \to 0} \left( {{{2x\cos {x \over 2} - 6\sin {x \over 2}} \over {2{x^{k - 1}}}}} \right)$$

$$ = {1 \over 3}\mathop {\lim }\limits_{x \to 0} \left( {{{x\cos {x \over 2} - 3\sin {x \over 2}} \over {{x^{k - 1}}}}} \right)$$

For, $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {{x^k}}} \ne 0 \Rightarrow k - 1 = 1 \Rightarrow k = 2$$

দেওয়া আছে যে $$f(x) = {1 \over 3}x\sin x - (1 - \cos x)$$। $$\mathop {\lim }\limits_{x \to 0} {{f(x)} \over {{x^k}}} \ne 0$$ হলে k-এর সম্ভাব্য ক্ষুদ্রতম ধনাত্মক মান হবে

A
4
B
3
C
2
D
1

Explanation

$$f(x) = {1 \over 3}x\sin x - (1 - \cos x)$$

$$ = {1 \over 3}{x^2}.{{\sin x} \over x} - 2{\sin ^2}{x \over 2}$$

$$ = {1 \over 3}.{x^2}.{{\sin x} \over x} - 2.{\left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)^2}.{4 \over {{x^2}}}$$

$$ = {1 \over {{x^2}}}\left[ {{1 \over 3}.{{\sin x} \over x} - 8{{\left( {{{\sin {x \over 2}} \over {{x \over 2}}}} \right)}^2}} \right]$$

$$\therefore$$ $$k = 2$$

3

WB JEE 2019

MCQ (More than One Correct Answer)
English
Bengali
Consider the function $$f(x) = {{{x^3}} \over 4} - \sin \pi x + 3$$
A
f(x) does not attain value within the interval [$$-$$2, 2]
B
f(x) takes on the value $$2{1 \over 3}$$ in the interval [$$-$$2, 2]
C
f(x) takes on the value $$3{1 \over 4}$$ in the interval [$$-$$2, 2]
D
f(x) takes no value p, 1 < p < 5 in the interval [$$-$$2, 2].

Explanation

Given, $$f(x) = {{{x^3}} \over 4} - \sin \pi x + 3$$

differentiating w.r.t. x, we get

$$f'(x) = {{3{x^2}} \over 4} - \pi \cos (\pi x)$$

at $$x = - 2,f( - 2) = 1$$

and at $$x = 2,f(2) = 5$$

$$f(x) = {{{x^3}} \over 4} - \sin \pi x + 3$$ অপেক্ষকটি বিবেচনা করো।

A
অন্তরাল [$$-$$2, 2] -তে f(x) কোন মান পরিগ্রহ করে না
B
অন্তরাল [$$-$$2, 2] তে f(x), $$2{1 \over 3}$$ মানটি পরিগ্রহ করে
C
অন্তরাল [$$-$$2, 2] তে f(x), $$3{1 \over 4}$$ মানটি পরিগ্রহ করে
D
অন্তরাল [$$-$$2, 2] তে f(x) কোন মান p পরিগ্রহ করে না যেখানে 1 < p < 5

Explanation

$$f(x) = {{{x^3}} \over 4} - \sin \pi x + 3$$

$$f( - 2) = - 2 + 0 + 3 = 1$$

$$f(2) = 2 - 0 + 3 = 5$$

$$\therefore$$ $$f( - 2) \ne f(2)$$

আবার $$f(x)$$ অপেক্ষক $$[ - 2,2]$$ পরিসরে সন্তত।

$$f(x)$$ অপেক্ষকের সকল মান $$(1,5)$$ পরিসরে থাকবে।

4

WB JEE 2019

MCQ (More than One Correct Answer)
English
Bengali
Let $$f:[1,3] \to R$$ be a continuous function that is differentiable in (1, 3) an

f'(x) = | f(x) |2 + 4 for all x$$ \in $$ (1, 3). Then,
A
f(3) $$-$$ f(1) = 5 is true
B
f(3) $$-$$ f(1) = 5 is false
C
f(3) $$-$$ f(1) = 7 is false
D
f(3) $$-$$ f(1) > 0 only at one point of (1, 3)

Explanation

Given that f : [1, 3] $$ \to $$ R be a continuous and differentiable in (1, 3).

and f'(x) = | f(x) |2 + 4

By applying LMVT, there exist at least one point c$$ \in $$(1, 3) such that

$${{f(3) - f(1)} \over {3 - 1}} = f'(c)$$ $$ \because $$ $$\left[ {f'(c) = {{f(b) - f(a)} \over {b - a}}} \right]$$

$$ \Rightarrow {{f(3) - f(1)} \over 2} = f'(c)$$

$$ \Rightarrow {{f(3) - f(1)} \over 2} = [f|c{|^2} + 4]$$

($$ \because $$ $$f'(c) = |f(c){|^2} + 4$$)

$$ \Rightarrow f(3) - f(1) = 2\,.\,|f(c){|^2} + 8$$

$$ \Rightarrow f(3) - f(1) \ge 8$$

It is clear from the given options that (B) and (C) the correct.

মনে করো $$f:[1,3] \to R$$ সন্তত অপেক্ষক ও $$(1,3)$$ অন্তরালে অন্তরকলনযোগ্য এবং সকল $$x \in (1,3)$$ এর জন্য $$f'(x) = {\left| {f(x)} \right|^2} + 4$$ সেক্ষেত্রে -

A
$$f(3) - f(1) = 5$$ সত্য হবে
B
$$f(3) - f(1) = 5$$ প্রযোজ্য হবে না
C
$$f(3) - f(1) = 7$$ প্রযোজ্য হবে না
D
$$(1,3)$$ এর মাত্র একটি বিন্দুতে $$f(3) - f(1) < 0$$হবে

Explanation

LMVT অনুযায়ী $$f'(x) = {{f(3) - f(1)} \over {3 - 1}}$$

বা, $$f(3) - f(1) = 2f'(x)$$

বা, $$f(3) - f(1) = 2{\left| {f(x)} \right|^2} + 8$$

$$\because$$ $$2{\left| {f(x)} \right|^2} > 0$$

$$\therefore$$ $$f(3) - f(1) > 8$$

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