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1

WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

Under which of the following condition(s) does(do) the system of equations $$\left( {\matrix{ 1 & 2 & 4 \cr 2 & 1 & 2 \cr 1 & 2 & {(a - 4)} \cr } } \right)\left( {\matrix{ x \cr y \cr z \cr } } \right) = \left( {\matrix{ 6 \cr 4 \cr a \cr } } \right)$$ possesses(possess) unique solution ?

A
$$\forall$$ a $$\in$$ R
B
a = 8
C
for all integral values of a
D
a $$\ne$$ 8

নিম্নলিখিত কোন শর্তাবলীর অধীনে

$$\left( {\matrix{ 1 & 2 & 4 \cr 2 & 1 & 2 \cr 1 & 2 & {(a - 4)} \cr } } \right)\left( {\matrix{ x \cr y \cr z \cr } } \right) = \left( {\matrix{ 6 \cr 4 \cr a \cr } } \right)$$ সমীকরণগুচ্ছের অনন্য সমাধান থাকবে ?

A
$$\forall$$a $$\in$$ R
B
a = 8
C
a-এর সকল পূর্ণসংখ্যা মানের জন্য
D
a $$\ne$$ 8
2

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
The determinant $$\left| {\matrix{ {{a^2} + 10} & {ab} & {ac} \cr {ab} & {{b^2} + 10} & {bc} \cr {ac} & {bc} & {{c^2} + 10} \cr } } \right|$$ is
A
divisible by 10 but not by 100
B
divisible by 100
C
not divisible by 100
D
not divisible by 10

Explanation

We have,

$$\left| {\matrix{ {{a^2} + 10} & {ab} & {ac} \cr {ab} & {{b^2} + 10} & {bc} \cr {ac} & {bc} & {{c^2} + 10} \cr } } \right|$$

$$ = {1 \over {abc}}\left| {\matrix{ {a({a^2} + 10)} & {a{b^2}} & {a{c^2}} \cr {{a^2}b} & {b({b^2} + 0)} & {b{c^2}} \cr {{a^2}c} & {{b^2}c} & {c({c^2} + 10)} \cr } } \right|$$

Taking common a, b, c from R1, R2 and R3 respectively

$$ = {{abc} \over {abc}}\left| {\matrix{ {{a^2} + 10} & {{b^2}} & {{c^2}} \cr {{a^2}} & {{b^2} + 10} & {{c^2}} \cr {{a^2}} & {{b^2}} & {{c^2} + 10} \cr } } \right|$$

Applying C1 $$\to$$ C1 + C2 + C3, we get

$$ = \left| {\matrix{ {{a^2} + {b^2} + {c^2} + 10} & {{b^2}} & {{c^2}} \cr {{a^2} + {b^2} + {c^2} + 10} & {{b^2} + 10} & {{c^2}} \cr {{a^2} + {b^2} + {c^2} + 0} & {{b^2}} & {{c^2} + 10} \cr } } \right|$$

$$ = ({a^2} + {b^2} + {c^2} + 10)\left| {\matrix{ 1 & {{b^2}} & {{c^2}} \cr 1 & {{b^2} + 10} & {{c^2}} \cr 1 & {{b^2}} & {{c^2} + 10} \cr } } \right|$$

R2 $$\to$$ R2 $$-$$ R1 and R3 $$\to$$ R3 $$-$$ R1

$$ = ({a^2} + {b^2} + {c^2} + 10)\left| {\matrix{ 1 & {{b^2}} & {{c^2}} \cr 0 & {10} & 0 \cr 0 & 0 & {10} \cr } } \right|$$

$$ = ({a^2} + {b^2} + {c^2} + 10)\,(100)$$

$$\because$$ If is divisible by 100.
$$\left| {\matrix{ {{a^2} + 10} & {ab} & {ac} \cr {ab} & {{b^2} + 10} & {bc} \cr {ac} & {bc} & {{c^2} + 10} \cr } } \right|$$ নির্ণায়কটি
A
10 দ্বারা বিভাজ্য কিন্তু 100 দ্বারা বিভাজ্য নয়
B
100 দ্বারা বিভাজ্য
C
100 দ্বারা বিভাজ্য নয়
D
10 দ্বারা বিভাজ্য নয়

Explanation

আমাদের কাছে,

$$\left| {\matrix{ {{a^2} + 10} & {ab} & {ac} \cr {ab} & {{b^2} + 10} & {bc} \cr {ac} & {bc} & {{c^2} + 10} \cr } } \right|$$

$$ = {1 \over {abc}}\left| {\matrix{ {a({a^2} + 10)} & {a{b^2}} & {a{c^2}} \cr {{a^2}b} & {b({b^2} + 0)} & {b{c^2}} \cr {{a^2}c} & {{b^2}c} & {c({c^2} + 10)} \cr } } \right|$$

যথাক্রমে R1, R2 এবং R3 থেকে কমন a, b, c নেওয়া

$$ = {{abc} \over {abc}}\left| {\matrix{ {{a^2} + 10} & {{b^2}} & {{c^2}} \cr {{a^2}} & {{b^2} + 10} & {{c^2}} \cr {{a^2}} & {{b^2}} & {{c^2} + 10} \cr } } \right|$$

C1 $$\to$$ C1 + C2 + C3, প্রয়োগ করলে আমরা পাই

$$ = \left| {\matrix{ {{a^2} + {b^2} + {c^2} + 10} & {{b^2}} & {{c^2}} \cr {{a^2} + {b^2} + {c^2} + 10} & {{b^2} + 10} & {{c^2}} \cr {{a^2} + {b^2} + {c^2} + 0} & {{b^2}} & {{c^2} + 10} \cr } } \right|$$

$$ = ({a^2} + {b^2} + {c^2} + 10)\left| {\matrix{ 1 & {{b^2}} & {{c^2}} \cr 1 & {{b^2} + 10} & {{c^2}} \cr 1 & {{b^2}} & {{c^2} + 10} \cr } } \right|$$

R2 $$\to$$ R2 $$-$$ R1 এবং R3 $$\to$$ R3 $$-$$ R1

$$ = ({a^2} + {b^2} + {c^2} + 10)\left| {\matrix{ 1 & {{b^2}} & {{c^2}} \cr 0 & {10} & 0 \cr 0 & 0 & {10} \cr } } \right|$$

$$ = ({a^2} + {b^2} + {c^2} + 10)\,(100)$$

$$\because$$ যদি 100 দ্বারা বিভাজ্য হয়।
3

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
Let T and U be the set of all orthogonal matrices of order 3 over R and the set of all non-singular matrices of order 3 over R respectively. Let A = {$$-$$1, 0, 1}, then
A
there exists bijective mapping between A and T, U.
B
there does not exist bijective mapping between A and T, U.
C
there exists bijective mapping between A and T but not between A and U.
D
there exists bijective mapping between A and U but not between A and T.

Explanation

Since,

n(A) $$\ne$$ n(T)

and n(A) $$\ne$$ n(U)

$$\therefore$$ There does not exist bijective mapping between A and T, U.
মনে কর T ও U, R-এর উপর তিনমাত্রার যথাক্রমে লম্ব ম্যাট্রিক্সের সেট ও অবিশিষ্ট ম্যাট্রিক্সের সেট এবং A = {1,0,1}, সেক্ষেত্রে
A
A ও T, U -এর মধ্যে একৈক উপরিচিত্ৰণ বিদ্যমান
B
A ও T, U-এর মধ্যে একৈক উপরিচিত্রণের অস্তিত্ব নেই
C
A ও T -এর মধ্যে একৈক উপরিচিত্রণ বিদ্যমান কিন্তু A ও U -এর মধ্যে নেই।
D
A ও U -এর মধ্যে একৈক উপরিচিত্ৰণ বিদ্যমান কিন্তু A ও T -এর মধ্যে নেই।

Explanation

যেহেতু,

n(A) $$\ne$$ n(T)

এবং n(A) $$\ne$$ n(U)

$$\therefore$$ A এবং T, U এর মধ্যে দ্বিমুখী ম্যাপিং নেই।
4

WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
If an (> 0) be the nth term of a G.P. then

$$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$$ is equal to
A
1
B
2
C
$$-$$2
D
0

Explanation

We have, an = arn $$-$$ 1

$$\Rightarrow$$ log an = log(arn $$-$$1)

$$\Rightarrow$$ log an = log a + (n $$-$$ 1) log r

Now, $$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$$

$$ = \left| {\matrix{ {\log a + (n - 1)\log r} & {\log a + n\log r} & {\log a + (n + 1)\log r} \cr {\log a + (n + 2)\log r} & {\log a + (n + 3)\log r} & {\log a + (n + 4)\log r} \cr {\log a + (n + 5) + \log r} & {\log a + (n + 6)\log r} & {\log a + (n + 7)\log r} \cr } } \right|$$

On applying R2 $$\to$$ R2 $$-$$ R1 R3 $$\to$$ R3 $$-$$ R1

$$\left| {\matrix{ {\log a + (n - 1)\log r} & {\log a + n\log r} & {\log a + (n + 1)\log r} \cr {3\log r} & {3\log r} & {3\log r} \cr {6\log r} & {6\log r} & {6\log r} \cr } } \right|$$

= 0 (since R2 and R3 are proportional)
যদি গুণােত্তর প্রগতির n তম পদ an (> 0) হয়, তবে

$$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$$ -এর মান হবে
A
1
B
2
C
$$-$$2
D
0

Explanation

আমাদের আছে, an = arn $$-$$ 1

$$\Rightarrow$$ log an = log(arn $$-$$1)

$$\Rightarrow$$ log an = log a + (n $$-$$ 1) log r

এখন, $$\left| {\matrix{ {\log {a_n}} & {\log {a_{n + 1}}} & {\log {a_{n + 2}}} \cr {\log {a_{n + 3}}} & {\log {a_{n + 4}}} & {\log {a_{n + 5}}} \cr {\log {a_{n + 6}}} & {\log {a_{n + 7}}} & {\log {a_{n + 8}}} \cr } } \right|$$

$$ = \left| {\matrix{ {\log a + (n - 1)\log r} & {\log a + n\log r} & {\log a + (n + 1)\log r} \cr {\log a + (n + 2)\log r} & {\log a + (n + 3)\log r} & {\log a + (n + 4)\log r} \cr {\log a + (n + 5) + \log r} & {\log a + (n + 6)\log r} & {\log a + (n + 7)\log r} \cr } } \right|$$

প্রয়োজক করে, R2 $$\to$$ R2 $$-$$ R1 R3 $$\to$$ R3 $$-$$ R1

$$\left| {\matrix{ {\log a + (n - 1)\log r} & {\log a + n\log r} & {\log a + (n + 1)\log r} \cr {3\log r} & {3\log r} & {3\log r} \cr {6\log r} & {6\log r} & {6\log r} \cr } } \right|$$

= 0 (যেহেতু R2 এবং R3 সমানুপাতিক)

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