# Quant Marathon by Gaurav Sharma - Set 6

Gaurav Sharma, Director of Genius Classes Of Mathematics (GCM), Karnal ( Haryana ), completed his graduation from Kurukshetra University in Electronics. Being inclined towards Mathematics, he pursued his Masters in Mathematics and also completed his L.L.B. He believes that, when a man quits learning he will be more senile onwards. He is a mentor right from his college days and loves teaching and inspiring students. Team Genius Classes, along with the regular classes, organizes workshops, events and seminars time to time. You can find lot of good stuffs in his Facebook group - CAT Mausi

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 ?

1. 25%
2. 50%
3. 62.5%
4. 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

1. 2
2. – 4
3. 0
4. 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 2xlog43 + 3log4x = 27

2xlog43 + 3log4x = 27

2 * 3log4x + 3log4x = 27

3 * 3log4x = 27

3log4x = 9 = 3^2

log4x = 2

x = 4^2 = 16

The number of real solutions of the equation x^2 / ( 1 – |x – 5|) = 1 is

1. 4
2. 2
3. 1
4. 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.

1. ½
2. 31/50
3. 17/41
4. 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

1. a = 6, b = -4
2. a = -4, b = 6
3. a = -6, b = 4
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

1. ½
2. 1/5
3. 1/7
4. 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 (31/5)^(15 – r) (21/3)^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

1. 0
2. -1
3. 1
4. -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 log3(3 + root(x)) + log3 ( 1 + x^2) = 0 is

1. 0
2. 1
3. 2
4. More than 2

log3(3 + root(x)) + log3 (1 + x^2) = 0

x ≥ 0

Also 3 + root(x) > 3 and 1 + x^2 > 1 for all x > 0

Log3(3 + root(x)) > 1 and log3( 1 + x^2 ) > 0 for all x > 0

Log3(3 + root(x)) > 1 and log3( 1 + x^2 ) > 0 for all x ≥ 0

Hence the given equation has no solution