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|

    0_1510554155208_2362da66-47f6-48f5-94f7-df222cd7387b-qm61.png

    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


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