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Mathematics Test - 28
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Mathematics Test - 28
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  • Question 1/10
    1 / -0

     Find θ[π2,3π2] satisfying 2cos2θ+sinθ2
    Solutions
    2cos2θ+sinθ2
    cos2θ+sin2θ=1
    =2(1sin2θ)+sinθ2
    =22sin2θ+sinθ2
    =2sin2θsinθ0
    =2t2t0
    let sinθ=t
    t(2t1)0
    (t)(2t1)0
    t(,0] and [1/2,)
    t=sinθ and sinθϵ[1,1]
    so
    sinθ[1,0] and [1/2,1]
    Hence , the correct option is B
  • Question 2/10
    1 / -0

    The locus of the midpoints of the chords of the circle x2y2=16 which are tangents to the hyperbola 9x216y2=144 is

    Solutions
    Let (h,k) be the middle point of a chord of the circle x2+y2=16
    Then its equation is hx+ky16=h2+k216i.e.
    hx+ky=h2+k2 (i)
    Let (i) touch the hyperbola
    9x216y2=144 i.e., ... (ii)
    At the point (ii) say, then (i) is identical with xα16yβ9=1
    (iii)
    Thus α16h=β9k=1h2+k2
    α=16hh2+k2 and β=9kh2+k2
    since (i) lies on the hyperbola (ii),
    116(16hh2+k2)219(9h2+k2)2=1
    16h29k2=(h2+k2)2
    Hence the required locus of (h,k) is (x2+y2)2=16x29y2
    Hence, the correct option is (A)
  • Question 3/10
    1 / -0

    In an acute - angled triangle ABC, points D, E and F are the feet of the perpendiculars from A, B and C onto BC, AC and AB, respectively. H is the orthocentre of ∆ABC . If sin A = 3/5 and BC = 39, then the length of AH is:

    Solutions

  • Question 4/10
    1 / -0

    lf Z=x+iy is a complex number then |x|+|y|≤ ?

    Solutions
    z=x+iy=rcosθ+irsinθ where |z|=r
    Now, |x|+|y|=|rcosθ|+|rsinθ|
    =|r|{cosθ|+|sinθ}
    But 2<cossθ+sinθ<2
    |x|+|y|2|r|
    2|z|
    Hence, the correct option is (A)
  • Question 5/10
    1 / -0

    Let ω be a complex number such that 2ω+1=z where z=3, if |1111ω21ω21ω2ω7|=3k, then k ie equal to.
    Solutions
    2w+1=z
    2w+1=3i
    w=1+3i2
    w2=13i2=w1
    Now, |1111ω21ω21ω2ω7|=3k...Given
    |1111ww21w2w|=3k
    1(w2w4)1(ww2)+1(w2w)=3k
    3w23w=3k
    3(w2w)=3k
    k=w2w
    k=12w
    =1(1+3i)
    =3i
    =z
    Hence, the correct option is (A)
  • Question 6/10
    1 / -0

    Let ω be a cube root of unity not equal to 1. Then the maximum possible value of |a+bω+cω2| where a, b, c ϵ {+1,-1} is:

    Solutions
    ω3=1
    ω31=0 or (ω1)(ω2+ω+1)=0
    since ω1,ω=1+i32,ω2=1i32
    |a+bω+αω2|=|a(b+c)2+i(32)(bc)|
    Square of the modulus is given by (2abc2)2+(3(bc)2)2
    This can be maximised by substituting a=1,b=1,c=1 or a=1,b=1,c=1
    Both the substitutions yield the same result, of 4 maximum modulus =2
    Hence, the correct option is (B)
  • Question 7/10
    1 / -0

    Number of solutions of the equation |z|2+7z=0 is:

    Solutions
    Let z=x+iy Then, |z|2+7z=x2+y2+7(x+iy)=0
    x2+y2+7x=0 \& 7y=0
    x2+7x=0 and y=0
    x=7 or x=0 and y=0
    So, solutions are z=7 or z=0 Hence, two solutions.
    Hence, the correct option is (B)
  • Question 8/10
    1 / -0

    If ω is a cube root of unity but not equal to 1, then minimum value of ∣a+b+cω2∣ (where a,b,ca,b,c are integers but not all equal) is:

    Solutions
    Let y=|a+bw+cw2|
    For y to be minimum y2 must be minimum y2=|a+bw+cw2|2
    =(a+bw+cw2)(a+bw+cw2)
    =12[(ab)2+(bc)2+(ca)2]
    since, a,b,c are not equal at a time. So, minimum value of y2 occurs when any two are same and third is differ by 1. Thus minimum of y=1 (as a,b,c are integers).
    Hence, the correct option is (C)
  • Question 9/10
    1 / -0

    If |z1|=1,|z2|=2,|z3|=3 and |9z1z2+4z1z3+z2z3|=12, then the value of |z1+z2+z3|
    is
    Solutions
    Given |z1|=1,|z2|=2,|z3|=3 and |9z1z2+4z1z3+z2z3|=12
    |9z1z2+4z1z3+z2z3|=12
    |z1||z2||z3||9z3+4z2+1z1|=12
    6|9z¯3|z3|2+4z¯2|z2|2+z¯1|z1|2|=12
    6|z1+z2+z3|=12
    6|z1+z2+z3|=12
    |z1+z2+z3|=2
    Hence, the correct option is (C)
  • Question 10/10
    1 / -0

    Let z=x+iy be a complex number where x and y are integers. Then the area of the
    rectangle whose vertices are the roots of the equation z¯z3+zz¯3=350 is:
    Solutions
    z¯z3+zz¯3=350
    zz¯(z¯2+z2)=350
    Put z=x+iy
    (x2+y2)(x2y2)=175
    since x and y are integer so only possibility is x2+y2=25
    and
    x2y2=7
    so solving these equation we get
    x=±4,y=±3
    Thus required area =8×6=48 sq. unit.
    Hence, the correct option is (A)
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