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1

### WB JEE 2022

English
Bengali

The value of a for which the sum of the squares of the roots of the equation $${x^2} - (a - 2)x - a - 1 = 0$$ assumes the least value is

A
0
B
1
C
2
D
3

$${x^2} - (a - 2)x - a - 1 = 0$$ সমীকরণের বীজদ্বয়ের বর্গের সমষ্টির মান ন্যূনতম করতে হলে a এর মান হবে

A
0
B
1
C
2
D
3
2

### WB JEE 2022

English
Bengali

If a, b are odd integers, then the roots of the equation $$2a{x^2} + (2a + b)x + b = 0$$, $$a \ne 0$$ are

A
rational
B
irrational
C
non-real
D
equal

যদি a, b অযুগ্ম পূর্ণসংখ্যা হয়, তবে $$2a{x^2} + (2a + b)x + b = 0$$, $$a \ne 0$$ সমীকরণের বীজদ্বয়

A
মূলদ হবে
B
অমূলদ হবে
C
বাস্তব হবে না
D
সমান হবে
3

### WB JEE 2021

English
Bengali
Let $$\alpha$$, $$\beta$$ be the roots of the equation x2 $$-$$ 6x $$-$$ 2 = 0 with $$\alpha$$ > $$\beta$$. If an = $$\alpha$$n $$-$$ $$\beta$$n for n $$\ge$$ 1, then the value of $${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$ is
A
1
B
2
C
3
D
4

## Explanation

Given, x2 $$-$$ 6x $$-$$ 2 = 0

$$\Rightarrow$$ xn $$-$$ 2(x2 $$-$$ 6x $$-$$ 2) = 0

$$\Rightarrow$$ xn $$-$$ 6xn $$-$$ 1 $$-$$ 2xn $$-$$ 2 = 0

When, n = 10,

Then, x10 $$-$$ 6x9 $$-$$ 2x8 = 0

$$\Rightarrow$$ x10 $$-$$ 2x8 = 6x9

Since, $$\alpha$$ and $$\beta$$ are roots of (i).

$$\therefore$$ $$\alpha$$10 $$-$$ 2$$\alpha$$8 = 6$$\alpha$$9

and $$\beta$$10 $$-$$ 2$$\beta$$8 = 6$$\beta$$9

$$\therefore$$ ($$\alpha$$10 $$-$$ $$\beta$$10) $$-$$ 2 ($$\alpha$$8 $$-$$ $$\beta$$8) = 6($$\alpha$$9 $$-$$ $$\beta$$9)

$$\Rightarrow$$ a10 $$-$$ 2a8 = 6a9 ($$\because$$ an = $$\alpha$$n $$-$$ $$\beta$$n)

$$\Rightarrow$$ $${{{a_{10}} - 2{a_8}} \over {2{a_9}}} = 3$$
x2 $$-$$ 6x $$-$$ 2 = 0 সমীকরণের বীজদ্বয় $$\alpha$$ ও $$\beta$$, $$\alpha$$ > $$\beta$$ । যদি an = $$\alpha$$n $$-$$ $$\beta$$n, n $$\ge$$ 1, তবে $${{{a_{10}} - 2{a_8}} \over {2{a_9}}}$$ হবে
A
1
B
2
C
3
D
4

## Explanation

দেওয়া, x2 $$-$$ 6x $$-$$ 2 = 0

$$\Rightarrow$$ xn $$-$$ 2(x2 $$-$$ 6x $$-$$ 2) = 0

$$\Rightarrow$$ xn $$-$$ 6xn $$-$$ 1 $$-$$ 2xn $$-$$ 2 = 0

যখন, n = 10,

তারপর, x10 $$-$$ 6x9 $$-$$ 2x8 = 0

$$\Rightarrow$$ x10 $$-$$ 2x8 = 6x9

যেহেতু, $$\alpha$$ এবং $$\beta$$ হল (i)-এর বীজদ্বয়।

$$\therefore$$ $$\alpha$$10 $$-$$ 2$$\alpha$$8 = 6$$\alpha$$9

এবং $$\beta$$10 $$-$$ 2$$\beta$$8 = 6$$\beta$$9

$$\therefore$$ ($$\alpha$$10 $$-$$ $$\beta$$10) $$-$$ 2 ($$\alpha$$8 $$-$$ $$\beta$$8) = 6($$\alpha$$9 $$-$$ $$\beta$$9)

$$\Rightarrow$$ a10 $$-$$ 2a8 = 6a9 ($$\because$$ an = $$\alpha$$n $$-$$ $$\beta$$n)

$$\Rightarrow$$ $${{{a_{10}} - 2{a_8}} \over {2{a_9}}} = 3$$
4

### WB JEE 2020

English
Bengali
If P(x) = ax2 + bx + c and Q(x) = $$-$$ax2 + dx + c, where ac $$\ne$$ 0 [a, b, c, d are all real], then P(x).Q(x) = 0 has
A
at least two real roots
B
two real roots
C
four real roots
D
no real root

## Explanation

Given,

P(x) = ax2 + bx + c, Q(x) = $$-$$ax2 + dx + c

Discriminant for P(x) = b2 $$-$$ 4ac

and Discriminant for Q(x) = d2 + 4ac

If ac > 0, Q(x) has real roots and ac < 0, P(x) has real roots.

$$\therefore$$P(x).Q(x) = 0 has at least two real roots.

যদি $$P(x) = a{x^2} + bx + c$$ এবং $$Q(x) = - a{x^2} + dx + c$$, $$ac \ne 0$$ হয় [a, b, c, d বাস্তব], সেক্ষেত্রে $$P(x).Q(x) = 0$$ এর

A
কমপক্ষে দুটি বাস্তব বীজ থাকবে
B
দুটি বাস্তব বীজ থাকবে
C
চারটি বাস্তব বীজ থাকবে
D
কোন বাস্তব বীজ থাকবে না

## Explanation

$$f(x) = P(x).Q(x)$$

$$= - {a^2}{x^4} + (ad - ab){x^3} + bd\,{x^2} + (bc + cd)x + {c^2}$$

$$f( - x) = - {a^2}{x^4} - (ad - ab){x^3} + bd\,{x^2} - (bc + cd)x + {c^2}$$

চিহ্নের পরিবর্তন

$$-$$ $$-$$ + $$-$$ +

$$\therefore$$ 3 বার চিহ্নের পরিবর্তন।

$$\therefore$$ কমপক্ষে দুটি বাস্তব বীজ থাকবে।

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