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1

JEE Advanced 2020 Paper 1 Offline

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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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