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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

If $$(\cot {\alpha _1})(\cot {\alpha _2})\,......\,(\cot {\alpha _n}) = 1,0 < {\alpha _1},{\alpha _2},....\,{\alpha _n} < \pi /2$$, then the maximum value of $$(\cos {\alpha _1})(\cos {\alpha _2}).....(\cos {\alpha _n})$$ is given by

A
$${1 \over {{2^{n/2}}}}$$
B
$${1 \over {{2^n}}}$$
C
$${1 \over {2n}}$$
D
1

$$(\cos {\alpha _1})(\cos {\alpha _2}).....(\cot {\alpha _n}) = 1,0 < {\alpha _1},{\alpha _2},....{\alpha _n} < \pi /2$$ হলে $$(\cos {\alpha _1})(\cos {\alpha _2}).....(\cos {\alpha _n})$$ এর সর্বোচ্চ মান হবে

A
$${1 \over {{2^{n/2}}}}$$
B
$${1 \over {{2^n}}}$$
C
$${1 \over {2n}}$$
D
1
2

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
The equation 6x + 8x = 10x has
A
no real root
B
infinitely many rational roots
C
exactly one real root
D
two distinct real roots

Explanation

Given, equation, 6x + 8x = 10x

$$ \Rightarrow {\left( {{6 \over {10}}} \right)^x} + {\left( {{8 \over {10}}} \right)^x} = 1$$

$$ \Rightarrow {\left( {{3 \over 5}} \right)^x} + {\left( {{4 \over 5}} \right)^x} = 1$$

Let $${3 \over 5} = \sin \theta \Rightarrow {4 \over 5} = \cos \theta $$

$$\therefore$$ $${(\sin \theta )^x} + {(\cos \theta )^x} = 1$$

It is possible only x = 2

$$\therefore$$ given equation has exactly one real root.
6x + 8x = 10x সমীকরণের
A
কোন বাস্তব বীজ নেই
B
সমীকরণটির অসংখ্য মূলদ বীজ আছে
C
শুধুমাত্র একটি বাস্তব বীজ আছে
D
দুটি পৃথক বাস্তব বীজ আছে

Explanation

দেওয়া, সমীকরণ, 6x + 8x = 10x

$$ \Rightarrow {\left( {{6 \over {10}}} \right)^x} + {\left( {{8 \over {10}}} \right)^x} = 1$$

$$ \Rightarrow {\left( {{3 \over 5}} \right)^x} + {\left( {{4 \over 5}} \right)^x} = 1$$

ধরা যাক $${3 \over 5} = \sin \theta \Rightarrow {4 \over 5} = \cos \theta $$

$$\therefore$$ $${(\sin \theta )^x} + {(\cos \theta )^x} = 1$$

এটা সম্ভব শুধুমাত্র x = 2

$$\therefore$$ প্রদত্ত সমীকরণের ঠিক একটি বাস্তব বীজ রয়েছে।
3

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
Let f(x) = sin x + cos ax be periodic function. Then,
A
a is any real number
B
a is any irrational number
C
a is rational number
D
a = 0

Explanation

Given, f(x) = sin x + cos ax

sin x is periodic with 2$$\pi $$.

f(x) is periodic only on sin x and cos x both are periodic.

$$ \therefore $$ cos ax is periodic at $${{2\pi } \over a}$$.

LCM of $${2\pi }$$ and $${{2\pi } \over a}$$ is possible only a is rational number.

Note :
(1) When $$n$$ is odd then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}\theta $$, $${\sec ^n}\theta $$ = $$2\pi $$

(2) When $$n$$ is even then the period of $${\sin ^n}\theta $$, $${\cos ^n}\theta $$, $${\csc ^n}\theta $$, $${\sec ^n}\theta $$ = $$\pi $$

(3) When $$n$$ is even/odd then the period of $${\tan ^n}\theta $$, $${\cot ^n}\theta $$ = $$\pi $$

(3) When $$n$$ is even/odd then the period of $$\left| {{{\sin }^n}\theta } \right|$$, $$\left| {{{\cos }^n}\theta } \right|$$, $$\left| {{{\csc }^n}\theta } \right|$$, $$\left| {{{\sec }^n}\theta } \right|$$, $$\left| {{{\tan }^n}\theta } \right|$$, $$\left| {{{\cot }^n}\theta } \right|$$ = $$\pi $$

দেওয়া আছে যে $$f(x) = \sin x + \cos ax$$ পর্যায়বৃত্ত অপেক্ষক। সেক্ষেত্রে

A
'a' যেকোনো বাস্তব সংখ্যা
B
'a' যেকোনো অমূলদ সংখ্যা
C
'a' মূলদ সংখ্যা
D
a = 0

Explanation

ধরি t হল f(x)-এর পর্যায়।

$$\therefore$$ $$f(x + t) = f(x)$$

$$ \Rightarrow \cos a(x + t) + \sin (x + t) = \cos ax + \sin x$$

x = 0 এবং x = $$-$$t বসিয়ে পাই

$$\cos at + \sin t = 1$$ এবং $$1 = \cos at - \sin t$$

$$\therefore$$ সমাধান করে পাই

$$\cos at = 1$$ এবং $$\sin t = 0$$

$$\therefore$$ $$at = 2m\pi $$ এবং $$t = n\pi $$

যেখানে m ও n পূর্ণসংখ্যা

এবং m $$\ne$$ 0, n $$\ne$$ 0

$$\therefore$$ $${{at} \over t} = {{2m\pi } \over {n\pi }} \Rightarrow a = {{2m} \over n}$$ [$$\because$$ t $$\ne$$ 0]

$$\therefore$$ a একটি মূলদ সংখ্যা।

4

WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
$$\cos (2x + 7) = a(2 - \sin x)$$ can have a real solution for
A
all real values of a
B
$$a \in [2,6]$$
C
$$a \in ( - \infty ,2)\backslash \{ 0\} $$
D
$$a \in (0,\infty )$$

$$\cos (2x + 7) = a(2 - \sin x)$$ এর বাস্তব সমাধান থাকবে

A
a-এর সকল বাস্তব মানের জন্য
B
$$a \in [2,6]$$-এর জন্য
C
$$a \in ( - \infty ,2)/\{ 0\} $$-এর জন্য
D
$$a \in (0,\infty )$$-এর জন্য

Explanation

$$\cos (2x + 7) = a(2 - \sin x)$$

এখানে $$ - 1 \le a(2 - \sin x) \le 1$$

বা, $$ - {1 \over a} \le 2 - \sin x \le {1 \over a}$$ [যখন $$a > 0$$]

বা, $$ - \left( {2 + {1 \over a}} \right) \le - \sin x \le - 2 + {1 \over a}$$

$$\therefore$$ $$2 + {1 \over a} \le 1$$ & $$2 - {1 \over a} \ge - 1$$

$$ \Rightarrow {1 \over a} \le - 1$$ & $$ - {1 \over a} \ge - 3$$

$$ \Rightarrow a \ge - 1$$ & $$a \le 3$$

$$\because$$ $$a > 0$$ $$\therefore$$ $$0 < a \le 3$$

প্রদত্ত উভয়ের কোনটি সঠিক নয়।

$$\therefore$$ প্রদত্ত সম্ভাব্য উত্তরগুলি ভুল।

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