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1

### JEE Advanced 2020 Paper 1 Offline

Suppose a, b denote the distinct real roots of the quadratic polynomial x2 + 20x $$-$$ 2020 and suppose c, d denote the distinct complex roots of the quadratic polynomial x2 $$-$$ 20x + 2020. Then the value of

ac(a $$-$$ c) + ad(a $$-$$ d) + bc(b $$-$$ c) + bd(b $$-$$ d) is
A
0
B
8000
C
8080
D
16000

## Explanation

Given quadratic polynomials, x2 + 20x $$-$$ 2020 and x2 $$-$$ 20x + 2020 having a, b distinct real and c, d distinct complex roots respectively.

So, a + b = $$-$$20, ab = $$-$$2020

and c + d = 20, cd = 2020

Now, ac(a $$-$$c) + ad(a $$-$$ d) + bc(b $$-$$c) + bd(b $$-$$ d)

= a2(c + d) $$-$$ a(c2 + d2) + b2(c + d) $$-$$ b(c2 + d2)

= (c + d) (a2 + b2) $$-$$ (c2 + d2) (a + b)

= (c + d)[(a + b)2 $$-$$ 2ab] $$-$$ (a + b)[(c + d)2 $$-$$ 2cd]

= 20[(20)2 + 4040] + 20[(20)2 $$-$$ 4040]

= 2 $$\times$$ 20 $$\times$$ (20)2 = 40 $$\times$$ 400 = 16000
2

### JEE Advanced 2017 Paper 2 Offline

Let p, q be integers and let $$\alpha$$, $$\beta$$ be the roots of the equation, x2 $$-$$ x $$-$$ 1 = 0 where $$\alpha$$ $$\ne$$ $$\beta$$. For n = 0, 1, 2, ........, let an = p$$\alpha$$n + q$$\beta$$n.

FACT : If a and b are rational numbers and a + b$$\sqrt 5$$ = 0, then a = 0 = b.
If a4 = 28, then p + 2q =
A
14
B
7
C
21
D
12

## Explanation

$$\alpha = {{1 + \sqrt 5 } \over 2}$$,

$$\beta = {{1 - \sqrt 5 } \over 2}$$

$${a_4} = {a_3} + {a_2}$$

$$= 2{a_2} + {a_1}$$

$$= 3{a_1} + 2{a_0}$$

$$28 = p(3\alpha + 2) + q(3\beta + 2)$$

$$28 = (p + q)\left( {{3 \over 2} + 2} \right) + (p - q)\left( {{{3\sqrt 5 } \over 2}} \right)$$

$$\therefore$$ p $$-$$ q = 0

and $$(p + q) \times {7 \over 2} = 28$$

$$\Rightarrow$$ p + q = 8

$$\Rightarrow$$ p = q = 4

$$\therefore$$ p + 2q = 12
3

### JEE Advanced 2017 Paper 2 Offline

Let p, q be integers and let $$\alpha$$, $$\beta$$ be the roots of the equation, x2 $$-$$ x $$-$$ 1 = 0 where $$\alpha$$ $$\ne$$ $$\beta$$. For n = 0, 1, 2, ........, let an = p$$\alpha$$n + q$$\beta$$n.

FACT : If a and b are rational numbers and a + b$$\sqrt 5$$ = 0, then a = 0 = b.
a12 = ?
A
a11 + 2a10
B
2a11 + a10
C
a11 $$-$$ a10
D
a11 + a10

## Explanation

$$\alpha$$2 = $$\alpha$$ + 1

$$\beta$$2 = $$\beta$$ + 1

an = p$$\alpha$$n + q$$\beta$$n

= p($$\alpha$$n$$-$$1 + $$\alpha$$n$$-$$2) + q($$\beta$$n$$-$$1 + $$\beta$$n$$-$$2)

= an$$-$$1 + an$$-$$2

$$\therefore$$ a12 = a11 + a10
4

### JEE Advanced 2016 Paper 1 Offline

Let $$- {\pi \over 6} < \theta < - {\pi \over {12}}.$$ Suppose $${\alpha _1}$$ and $${B_1}$$ are the roots of the equation $${x^2} - 2x\sec \alpha + 1 = 0$$ and $${\alpha _2}$$ and $${\beta _2}$$ are the roots of the equation $${x^2} + 2x\,\tan \theta - 1 = 0.$$ $$If\,{\alpha _1} > {\beta _1}$$ and $${\alpha _2} > {\beta _2},$$ then $${\alpha _1} + {\beta _2}$$ equals
A
$$2\left( {\sec \theta - \tan \theta } \right)$$
B
$$2\,\sec \,\theta$$
C
$$- 2\tan \theta$$
D
$$0$$

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