Quant Marathon by Gaurav Sharma  Set 6

If n is an integer from 1 to 96, what is the probability for n x ( n + 1) x (n + 2) being divisible by 8 ?
 25%
 50%
 62.5%
 72.5%
All the even numbers substituted in n x (n + 1) x (n + 2) are divisible by 8 and there are 48 even numbers till 96.
The odd numbers that product multiples of 8, for example n = 7 gives a product of 7 x 8 x 9 and there are 12 such odd numbers
So the probability will be ( 48 + 12 ) / 96 = 0.625 = 62.5%
The product of the roots of the equation ( x – 2)^2 – 3 x – 2 + 2 = 0 is
 2
 – 4
 0
 None of these
We have (x – 2)^2 – 3 x – 2 + 2 = 0
x – 2^2 – 3 x – 2 + 2 = 0
(x – 2  2) (x – 2  1) = 0
x – 2 = 1, 2
x – 2 = +/ 1, +/ 2
x = 3, 1, 4, 0
Product of roots = 1 x 3 x 4 x 0 = 0
Solve the equation 2x^{log43} + 3^{log4x} = 27
2x^{log43} + 3^{log4x} = 27
2 * 3^{log4x} + 3^{log4x} = 27
3 * 3^{log4x} = 27
3^{log4x} = 9 = 3^2
log_{4}x = 2
x = 4^2 = 16
The number of real solutions of the equation x^2 / ( 1 – x – 5) = 1 is
 4
 2
 1
 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.
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.
 ½
 31/50
 17/41
 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
If the equation x^4 – 4x^3 + ax^2 + bx + 1 = 0 has four positive roots then
 a = 6, b = 4
 a = 4, b = 6
 a = 6, b = 4
 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
If the sum upto infinity of the series 1 + 4x + 7x^2 + 10x^3 + … is 35/16 then find x
 ½
 1/5
 1/7
 ¼
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
Find the sum of all rational terms in the expansion of ( 3^1/5 + 2^1/3)^15
T(r + 1) = 15Cr (3^{1/5})^(15 – r) (2^{1/3})^r
T (r + 1) = 15Cr (3)^( 3 – r/5) 2^(r/3)
For the rational terms r = 0, 15
Then T ( 0 + 1 ) = 15C0 3^3 2^0 = 27
T(15 + 1) = 15C15 3^0 2^5 = 32
Sum of rational terms = T1 + T16
= 27 + 32 = 59
If P is real root of 2x^3 – 3x^2 + 6x + 6 = 0 then [ P ] where [ . ] denotes the greatest integer function is equal to
 0
 1
 1
 2
f(x) = 2x^3 – 3x^2 + 6x + 6
Then d/dx [f(x)] = 6x^2 – 6x + 6 = 6 (x^2 – x + 1) > 0 for all Real number x
f(x) is increasing on R
f(x) has only one real root
Now f(0) = 6 and f( 1) = 5
Real root of f(x) lies between 0 and 1
[ P ] =  1
The number of solutions of the equation log_{3}(3 + root(x)) + log_{3} ( 1 + x^2) = 0 is
 0
 1
 2
 More than 2
log_{3}(3 + root(x)) + log_{3} (1 + x^2) = 0
x ≥ 0
Also 3 + root(x) > 3 and 1 + x^2 > 1 for all x > 0
Log_{3}(3 + root(x)) > 1 and log_{3}( 1 + x^2 ) > 0 for all x > 0
Log_{3}(3 + root(x)) > 1 and log_{3}( 1 + x^2 ) > 0 for all x ≥ 0
Hence the given equation has no solution