# Quant Boosters - Gaurav Sharma - Set 2

• Q26) Solve the equation 2 * x^log3 + 3^logx = 27 (base of log is 4)

• 2 * x^log3 + 3^logx = 27
2 * 3^logx + 3^logx = 27
3 * 3^logx = 27
3^logx = 9 = 3^2
logx = 2
x = 4^2 = 16 (as log base is 4)

• Q27) The number of real solutions of the equation x^2 / (1 – |x – 5|) = 1 is
a) 4
b) 2
c) 1
d) None of these

• We have x^2 / ( 1 – |x – 5|) = 1
x^2 = 1 - |x – 5|
x^2 – 1 = - |x – 5|
Total number of real solutions of this equation is equal to the number of points of intersection of the curves y = x^ - 1 and y = - | x – 5| Clearly these two graphs do not intersect and hence no solution

• Q28) If two numbers a and b are chosen at random from the set {1, 2, 3, … , 10} with replacement. Then, what is the probability that the roots of the equation x^2 + ax + b = 0 are real.
a) 1/2
b) 31/50
c) 17/41
d) 7/11

• For real roots, the discriminant of the equation x^2 + ax + b = 0 should be greater than or equal to 0.
a^2 – 4b ≥ 0
a^2 ≥ 4b
(a/2)^2 ≥ b

Pairs that satisfy the inequality are
a = 2, b =1; a = 3, b = 1, 2; a = 4, b = 1, 2, 3, 4;
a = 5, b = 1, 2, 3, 4, 5, 6; a = 6, b = 1, 2, 3, 4, 5, 6, 7, 8, 9
a = 7, b = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
a = 8, b = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
a = 9, b = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10
a = 10, b = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10

So 62 pairs out of 100, hence probability is 62/100 = 31/50

• Q29) If the equation x^4 – 4x^3 + ax^2 + bx + 1 = 0 has four positive roots then
a) a = 6, b = -4
b) a = -4, b = 6
c) a = -6, b = 4
d) a = -6, b = -4

• Let p, q, r and s be the four positive roots of the equation. Then
p + q + r + s = 4 and pqrs = 1
( p + q + r + s ) / 4 = 1 and (pqrs)^1/4 = 1
AM = GM
p = q = r = s = 1
x^4 – 4x^3 + ax^2 + bx + 1 = (x – 1)^4
x^4 – 4x^3 + ax^2 + bx + 1 = x^4 – 4x^3 + 6x^2 - 4x + 1
a = 6 and b = - 4

• Q30) If the sum upto infinity of the series 1 + 4x + 7x^2 + 10x^3 + … is 35/16 then find x
a) 1/2
b) 1/5
c) 1/7
d) 1/4

• S = 1 + 4x + 7x^2 + 10x^3 + …
Sx = x + 4x^2 + 7x^3 + 10x^4 + …
S (1 – x) = 1 + 3x + 3x^2 + 3x^3 + …
S (1 – x) = 1 + 3x/( 1 – x)
35/16(1 – x) = (1 – x + 3x) / ( 1 – x) = ( 1 + 2x ) / ( 1 – x)
35 (1 – x) ^2 = 16 (1 + 2x )
35x^2 – 102x + 19 = 0
(7x – 19) (5x – 1) = 0
x = 19/7, 1/5
But x # 19/7 (because 19/7 > 1)
So x = 1/5

61

63

61

61

67

64

55

65