Quant Marathon by Gaurav Sharma - Set 6


  • Director, Genius Tutorials, Karnal ( Haryana ) & Delhi | MSc (Mathematics)


    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


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