Answer: Option A. -> [3 – (x + y + z)]/2
:
A
Sol:- a = 0, b = 1, c = -1. x = 1, y = 2, c = 2. ab + bc + ca = -1. Only option a) satisfies
Alternatively: -
Given
x = 1 + a², =>a²= x - 1
y = 1 + b²,=>b²= y - 1
z = 1 + c²=>c²= z - 1and(a + b + c)²= 0
As we know,(a + b + c)²= a² + b² + c² + 2(ab + bc + ca)
Put the given values.
0 = (x - 1) + (y - 1) + (z - 1) +2(ab + bc + ca)
(ab + bc + ca) =[3 – (x + y + z)]/2

Answer: Option A. -> [3 – (x + y + z)]/2
:
D
1st method: -
Sum of the roots (a + b) = p and Product of the roots (ab) = r
for other equation, a/2 + 2b = q and (ab) = r
From above equations,
a = 2/3 (2p - q) and b = (2q - p)/3
The value of r is

2^nd method: - x² – 3x + 2 = 0 and α = 2 and β = 1 and α/2 = 1 and 2β = 2 forms the same expression. P = q = 3 and r = 2. Only d) gives r = 2

Answer: Option C. -> Real
:
C
Solve using assumption
Let a = 0, b = 1, c = 2. The expression becomes 3x² -6x + 2 = 0. The discriminantof this exp is positive. You can try with some other values. You will always get real roots.
Alternatively: -
(x-b)(x-c) + (x-a)(x-c) + (x-a)(x-b) = 0
x² - cx - bx + bc + x² - cx - ax + ac + x² - bx - ax + ab = 0
3x² - 2(a + b + c)x + (ab + bc + ca) = 0
Discriminant = b² - 4ac
4(a + b + c)² - 4 x 3 x(ab + bc + ca)
4{a² + b² + c² + 2(ab + bc + ca) - 3(ab + bc + ca)}
4{a²+ b²+ c²- (ab + bc + ca)}
Thediscriminantof this exp is positive.You will always get real roots.

Answer: Option C. -> -6  
:
C
Let roots of the given equation are a and a².
Sum of the roots (a + a²) = - P/3 .........................(1)
Product of the roots (a)³ = 1 => a = 1
From equation (1),
1 + 1 = - P/3 => P = - 6
Alternatively: - 1, 1 be the roots. Eqn. becomes x² - 2x + 1 = 0 or 3 x² - 6x + 3 = 0.

Answer: Option A. -> at least one root in (0,1)
:
A
These questions areflexible. So, we can assume values of a, b, and c to make an other quardatic equation.
Use Assumption
Let a = 0, b = 1, c = -1. Equation becomes 2x -1 = 0, x = ½.
Now, we can eleminate options (b) and (c). It satisfies option (a)

Answer: Option B. -> 42
:
B
S = I + 2II + 3IIIX = I+ II +IIIwhere I = Number of people who playone gameII = Number of people who play two games III=Number of people who playthree gamesGiven that II+III=24, and III=12, thus II= 12S= 25+25+28= 78S-X= II+2III=>X= 78-36= 42. Option (b)

Answer: Option A. -> 5
:
A

Let x be the number of children playing only cricket
a+b= 12-4=8
Thus 25= 12+8+x => x= 5. Option (a)

Answer: Option A. ->  
:
A
= ; = ; = .
+ = + = ; = ->3 - (1/3)
Thus, on similar lines,
+.......... =;

Answer: Option C. -> 62.33
:
C
Ans: (c)The terms of the given sequence are as follows :a₁ = 81.33 a₇ = a₁a₂ = -19 a₈ = a₂a₃ = a₂ – a₁ a₉ = a₃a₄ = - a₁ a₁₀ = a₄ = -a₁a₅ = -a₂a₁₁ = a₄ = - a₂a₆ = - a₂ + a₁ a₁₂ = a₆ = -a₃ and so on.The sum of the first six terms, the next six terms and so on is 0.The sum of the first 6002 terms can be written as the sum of first 6000 terms + 6001^st term + 6002^nd term. From the above explanation, the sum of the first 6000 terms is zero, 6001 term will be a1 and 6002^nd term will be a₂. The sum of the first 6002 terms will be a₁ + a₂ = 81.33 + (-19) = 62.33.

Answer: Option D. -> 70
:
D
M (M (A (M (x, y), S (y, x)), A (y, x)
M (M (A (6, 1), 2), A (3, 2))
M (M (7, 2), A (3, 2))
M (14, 5) = 70