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Mathematics Test 125
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Mathematics Test 125

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  • Question 1/10
    4 / -1

    Coefficient of x2 in the expansion of (x3 + 2x2 + x + 4)15 is -

    Solutions

    Let (x3 + 2x2 + x + 4)15 = a0 +a1x + a2x2+ ...
    Differentiating w.r.t. x
    15(x3 + 2x2 + x + 4)14 (3x2 + 4x + 1)
    = a1 +2a2x + 3a3x2+ ...
    Again different w.r.t. x
    15{(x3 + 2x2 + x + 4)14 (6x +4) + 14(x3 + 2x2 + x + 4)13(3x2 + 4x + 1)2}
    = 2a2 + 6a3x + ...
    Now put x = 0
    ⇒ 15(4.414 + 14.413) = 2a2
    ⇒ 2a2 = 15 × 413(16 + 14)
    ⇒ a2 = (15)2 × 226
    Which is a perfect square

     

  • Question 2/10
    4 / -1

    If r & s be the positive integers and ƒ(r, s) be the function which gives number of ordered quadraples (a, b, c, d) of positive integers such that 3r7s = LCM (a, b, c) = LCM (a, b, d) = LCM(a, c, d) = LCM (b, c, d), then

    Solutions

    To decide (r1, r2, r3, r4).
    Case-1 →  r1 = r2 = r3 = r4 = r
    Case-2 → Exactly 3 gets r & other can be filled with 0, 1, 2, ..... r–1.
                   4C3.r
    Case-3 → Exactly 2 gets r, 4C2.r2
    ⇒ Total number of ways = (6r2 + 4r + 1)
        Similarly (s1, s2, s3, s4) can be decided
        in (6s2 + 4s + 1) way.
    ⇒ Total number of quadruples
         = (6r2 + 4r + 1)(6s2 + 4s + 1) = 672.

     

  • Question 3/10
    4 / -1

    The number of ways in which the word 'PARALLEL' is arranged so that both A's are together and no two L are together is

    Solutions

    |P|AA|R|E|
    5C3 × 4!

     

  • Question 4/10
    4 / -1

    If  are three unit vectors such that 
     , then the value of n is

    Solutions

     

  • Question 5/10
    4 / -1

    Let   and . If  is a unit vector inclined at an angle 60º to both  and  such that , then which of the following is/are true

  • Question 6/10
    4 / -1

    Number of ways to choose 3 positive integers a < b < c so that abc = 30030 is not equal to

    Solutions

    30030 = 2 × 3 × 5 × 7 × 11 × 13
    6 primes 2, 3, 5, 7, 11 and 13 can be alloted to any of a, b, c in 36 ways.
    But it include the cases when all primes are alloted to either a or b or c.
    Hence  are number of ways such that a.b.c = 30030

     

  • Question 7/10
    4 / -1

    Coefficient of xaybzc in the expansion of (2x – y + 3z)13, where a,b,c are three distinct terms of a GP can be

    Solutions

     

  • Question 8/10
    4 / -1

    Given that vec a and  are two unit vectors such that angle between vec a and  is . If  be a vector in the plane of  vec a and  , such that  and  then

    Solutions

     

  • Question 9/10
    4 / -1

    Let  and . If D,E and F are the mid-points of BC,CA and AB respectively. P,Q and R are three points such that ,  and  then

    Area of polygon ARBPCQ is equal to

    Solutions

     So, AGBR is parallelogram as diagonals bisects each other.
     Hence, [AGB] = [ARB]
     Similarly, [AQC] = [AGC]
     and [BPC] = [BGC]
     So, [ARBPCQ] = 2[ABC]

     

  • Question 10/10
    4 / -1

    Let  and . If D,E and F are the mid-points of BC,CA and AB respectively. P,Q and R are three points such that ,  and  then

    If the vector A is at (1,1,1), then the equation of median through B is

    Solutions

     

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