Quant Boosters by Shubham Ranka, IIM Calcutta


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    a:b = 3:7
    b:c = 4:9
    c:d = 6:7
    a:b:c:d=72:168:378:441
    TRICK:
    [a:b:c:d = 3 * 4 * 6: 7 * 4 * 6 : 7 * 9 * 6 : 7 * 9 * 7]
    a+b+c+d=52950
    d = 52950 * 441/1059 = 22050


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    Q 16) f(x)=ax^2+bx+c (b > a) & f(x) ≥ 0 for all x. Find the minimum value of (a+b+c)/(b-a)


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    a > 0, b^2 < = 4ac
    By symmetry:- a=c== > b=2a (D=0)
    4/1== > 4

    Explanation:

    We know, a > 0, b^2 < =4ac. So, c > 0. Now sice b > a == > a,b,c > 0
    Now, Considering the boundary case:- b^2 = 4ac. And, in this eqn: interchanging a and c does not change my final outcome. So, I safely assumed a=c to check for the boundary case. Now, it could either be minima or maxima. Check for other values to figure out that.


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    Q 17) a1 = p, b1=q.
    a(n)=pb(n-1) and b(n) = qb(n-1) for all even n > 1
    a(n)=pa(n-1) and b(n) = qa(n-1) for all odd n > 1
    Find sigma[a(n)+b(n)] for even n.


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    a1=p, b1=q
    a2=pq, b2 = q^2
    a3 = p^2q, b3 = pq^2
    a4 = p^2q^2, b4 = pq^3
    a5 = p^3q^2, b5 = p^2q^3
    a6 = p^3q^3, b6 = p^2q^4
    ...
    an = p^n/2.q^n/2,
    bn = sigma[a(n)+b(n)] = pq(1+pq+p^2q^2+....(pq)^n/2 -1) + q^2 (1+pq+p^2q^2 + ...(pq)^n/2 -1)
    = (pq+q^2)* pq^(n/2-1)/(pq-1)
    = q(p+q)[pq^(n/2)-1]/ (pq-1)


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    Q18 ) Find the minimum value of |x^2 - 2x + 5| + |2x^2 - 4x + 9| + |3x^2 - 6x + 7| .


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    |(x-1)^2 + 4 | + |2(x-1)^2 + 7| + |3(x-1)^2 + 4|
    min. @ x=1 --- > 15
    PS: Assuming x to be a real no.


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    Q 19) If a + b + c = 6 where a,b,c are positive then maximum value of (a+1)(b+2)(c+3) is


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    AM > = GM
    (a+1 + b+2 + c+3)/3 > = (a+1)(b+2)(c+3)^1/3
    4 > = (a+1)(b+2)(c+3)^1/3
    (a+1)(b+2)(c+3) < = 4^3
    (a+1)(b+2)(c+3) < = 64
    Maximum value is 64

    Alternate approach:
    a + 1 =x, b + 2 = y and c + 3 = z
    x + y + z = 12
    For maximum value equate the terms x = y = z = 4
    So product max is 4 x 4 x 4 = 64


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    Q 20) If 2x^2 + 3y^2 = 1. If M is the maximum value and N is the minimum value of x^2 + 3xy + 2y^2. What is M+N - MN


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    Let x^2 + 3xy + 2y^2 = k
    Given that 1 = 2x^2 + 3y^2
    So, x^2 + 3xy + 2y^2 = k = k*1 = k(2x^2 + 3y^2)
    == > x^2 (1-2k) + 3xy + y^2 (2-3k) = 0
    Dividing by y^2:-
    (x/y)^2 (1-2k) + (x/y)*3 + (2-3k) = 0
    For the above quadratic eqn in (x/y) to have a real solution:-
    D > = 0 thus
    9 - 4(1-2k)(2-3k) > = 0'
    so 24k^2 - 28k - 1 < = 0. Thus the interval is [N,M] so N and M are roots of the equation, thus M + N = 7/6 and MN = -1/24
    so M + N - MN= 7/6 - 1/24 = 27/24 = 9/8


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    Q 21) Find the value of the series 3/1! + 7/2! + 15/3! + 27/4! + 43/5! + ...


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    Taylor series expansion for e^x = 1 + x + x^2/2! + x^3/3! +...
    e-1 = 1+1/2!+1/3!+1/4!+....
    3,7,15,27,43,
    Numerator:-
    Tn = Tn-1 + 4(n-1), n > =2
    T1=3
    Tn = 3+ 4 (n(n-1)/2)
    == > Generic Tn = [3+2(n(n-1))]/n!
    = 3/n! + 2/(n-2)! (n > =2)
    = 3(e-1) + 2e
    = 5e-3


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    Q 22) In the year 1980, the age of a person is 1/89th of the year of his birth. What was the age of person in 2012?


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    Let x be the man's year of birth,
    Then x/89 = 1980 - x
    == > x = 1958
    == > In 2012 Age was: 54 years


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    Q23) Let a, b, c, d belongs to [1/2, 2]. Suppose abcd=1. Find the maximum and minimum value of (a + 1/b)(b + 1/c)(c + 1/d)(d + 1/a).


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    (ab+1)(bc+1)(cd+1)(ad+1)
    (ab+1)(cd+1) = (ab+1)(1/ab + 1)
    Now: (x+1)(1/x +1) = 1+x+1/x+1
    We know, that: x+1/x > =2 (2 @ x=1)
    == > (ab+1)(1/ab + 1) > 2+2 =4
    Similarly:- (bc+1)(ad+1) > =4
    Since, in the given domain: a=b=c=d=1 satisfies:
    == > Min. values= 4*4 = 16
    Put a=b=c=d=2 to get the max. value


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    Q24) In the series 3,6,9,12.. how many of the first 100 terms can be expressed as a difference of 2 perfect squares ?


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    (a+b)(a-b)=3k {k=1 to 100}
    Sum of factors and difference of factors =even . So, both even or both odd.
    For odd values of k, a-b=1, and take a=3k+1/2, b=3k-1/2. So, 50 values of k
    For even values of k, should be a multiple of 4. So, 25 values of k.
    Hence, a total of 75 terms can be expressed.


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    Q25) In the game of cricket , Runs scored in one ball can be 0(dot ball) , 1, 2, 3 , 4 or 6. What is the probability that there is at least one dot ball in an over in which 7 runs was scored. No extra deliveries were bowled.


 

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