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 Multiple Choice QuestionsMultiple Choice Questions

1.

If roots of the equation x2 – bx + c = 0 be two consectutive integers, then b2 – 4c equals

  • – 2

  • 3

  • 2

  • 1


D.

1

Let α, α + 1 be roots
α + α + 1 = b
α(α + 1) = c
∴ b2 – 4c = (2α + 1)2 - 4α(α + 1) = 1.

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2.

The value of α for which the sum of the squares of the roots of the equation x2 – (a – 2)x – a – 1 = 0 assume the least value is

  • 1

  • 0

  • 3

  • 2


A.

1

x2 – (a – 2)x – a – 1 = 0
⇒ α + β = a – 2
α β = –(a + 1)
α2 + β2 = (α + β)2 - 2αβ
= a2 – 2a + 6 = (a – 1)2 + 5
⇒ a = 1

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3.

If z = x – i y and z1/3 = p+ iq , then fraction numerator begin display style open parentheses straight x over straight p plus straight y over straight q close parentheses end style over denominator left parenthesis straight p squared plus straight q squared right parenthesis end fraction is equal to 

  • 1

  • -2

  • 2

  • -1


B.

-2

D.

-1

straight z to the power of 1 divided by 3 end exponent space equals space straight p space plus space iq
left parenthesis straight x minus iy right parenthesis to the power of 1 divided by 3 end exponent space equals space left parenthesis straight p plus iq right parenthesis      therefore,(Qz = x − iy)
(x - iy) = (p + iq)3
⇒ (x - iy) = p3 +(iq)3 + 3p2qi + 3pq2i2
⇒ (x - iy) = p3 - iq3 + 3p2qi - 3pq2
⇒ (x - iy) = (p3 - 3pq2 ) + i (3p2 q - q3 ) On comparing both sides, we get
⇒ x = (p3 - 3pq2) and - y = 3p2 q - q3
⇒ x = p(p2 - 3q2 ) and y = q(q2 - 3p2 )
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4.

If one root of the equation x2+px+12 =0 is 4, while the equation x2 +px +q = 0 has equal roots, then the value of 'q' is

  • 49/3

  • 4

  • 3

  • 12


A.

49/3

Since 4 is one of the roots of equation x2 + px + 12 = 0. So it must satisfied the equation.
∴ 16 + 4p + 12 = 0
⇒ 4p = -28
⇒ p = -7
The other equation is x2 - 7x + q = 0 whose roots are equal. Let roots are α and α of above equation

therefore space sum space of space roots space equals straight alpha space plus straight alpha space equals 7 over 1

⇒ 2α = 7 ⇒ α = 7/ 2 and product of roots α.α = q ⇒ α2 = q

(7/2)2 = q
q =49/4

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5.

The coefficient of xn in expansion of (1+x)(1-x)n is

  • (n-1)

  • (-1)n(1-n)

  • (-1)n-1(n-1)2

  • (-1)n-1n


B.

(-1)n(1-n)

The coefficient of xn in expansion of (1+x)(1-x)n is = coefficient of xn + coefficient of xn-1

left parenthesis negative 1 right parenthesis to the power of straight n space fraction numerator straight n factorial over denominator straight n factorial 0 factorial end fraction space minus fraction numerator straight n factorial over denominator 1 factorial left parenthesis straight n minus 1 right parenthesis factorial end fraction
space equals space left parenthesis negative 1 right parenthesis to the power of straight n space open square brackets fraction numerator straight n factorial over denominator straight n factorial 0 factorial end fraction minus fraction numerator straight n factorial over denominator 1 factorial left parenthesis straight n minus 1 right parenthesis factorial end fraction close square brackets
equals space minus space left parenthesis 1 right parenthesis to the power of straight n space left square bracket 1 minus straight n right square bracket

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6.

If x = ω – ω2 – 2. Then the value of (x4 + 3x3 + 2x2 – 11x – 6) is

  • 0

  • -1

  • 1

  • -3


C.

1

If (x + 2)2 = (ω – ω2 )
2 x2 + 4 + 4x = ω2 + ω4 – 2ω3
x2 + 4 + 4x = ω2 + ω –2 (x2 + 4x + 7) = 0 ...(i)
x4 + 3x3 + 2x2 – 11x – 6
= x2 (x2 + 4x + 7) –x(x2 + 4x + 7) – (x2 + 4x + 7) +1
= x 2 (0) – x(0) – 0 + 1 By (i)
= 1

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7.

Let z, w be complex numbers such that z iw + = 0 and arg zw = π. Then arg z equals

  • π/4

  • 5π/4

  • 3π/4

  • π/2


C.

3π/4

Since z + iw = 0 ⇒ z = −iw
⇒ z = iw
⇒ w = -iz
Also arg(zw) = π
⇒ arg (-iz2) = π
⇒ arg (-i) + 2 arg(z) = π
negative straight pi over 2 space plus space 2 arg space left parenthesis straight z right parenthesis space equals space straight pi space space left parenthesis because space arg space left parenthesis negative straight i right parenthesis space equals space minus straight pi divided by 2 right parenthesis
2 space arg space left parenthesis straight z right parenthesis space equals space fraction numerator 3 straight pi over denominator 2 end fraction
arg space left parenthesis straight z right parenthesis space equals space fraction numerator 3 straight pi over denominator 4 end fraction

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8.

If 2a + 3b + 6c =0, then at least one root of the equation ax+ bx+ c = 0  lies in the interval

  • (0,1)

  • (1,2)

  • (2,3)

  • (1,3)


A.

(0,1)

Let space straight f apostrophe left parenthesis straight x right parenthesis space equals space ax squared space plus bx space plus straight c
rightwards double arrow space straight f left parenthesis straight x right parenthesis space equals space ax cubed over 3 plus bx squared over 2 space plus cx space plus straight d
straight f left parenthesis straight x right parenthesis space equals space fraction numerator 2 ax cubed space plus 3 bx squared space plus 6 cx space plus 6 straight d over denominator 6 end fraction
straight f left parenthesis 1 right parenthesis space equals space fraction numerator 2 straight a space plus 3 straight b space plus 6 straight c space plus 6 straight d over denominator 6 end fraction space equals fraction numerator 6 straight d over denominator 6 end fraction space equals space straight d
straight f left parenthesis 1 right parenthesis space space equals fraction numerator 6 straight d over denominator 6 end fraction space equals straight d
straight f left parenthesis 0 right parenthesis space equals space straight f left parenthesis 1 right parenthesis
straight f apostrophe left parenthesis straight x right parenthesis space equals space 0

∴ One of the roots of ax2 + bx + c = 0 lies between 0 and 1.
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9.

If (1 – p) is a root of quadratic equation x2 +px + (1-p)=0 , then its roots are

  • 0, 1

  • -1, 2

  • 0, -1

  • -1, 1


C.

0, -1

Since (1 - p) is the root of quadratic equation
x2 + px + (1 - p) = 0 ........ (i)
So, (1 - p) satisfied the above equation
∴ (1 - p)2 + p(1 - p) + (1 - p) = 0
(1 - p)[1 - p + p + 1] = 0 (1 - p)(2) = 0
⇒ p = 1 On putting this value of p in equation (i)
x2 + x = 0
⇒ x(x + 1) = 0 ⇒ x = 0, -1

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10.

If the cube roots of unity are 1, ω, ω2 then the roots of the equation (x – 1)+ 8 = 0, are

  • -1 , - 1 + 2ω, - 1 - 2ω2

  • -1 , -1, - 1

  • -1 , 1 - 2ω, 1 - 2ω2

  • -1 , 1 + 2ω, 1 + 2ω2


C.

-1 , 1 - 2ω, 1 - 2ω2

(x – 1)3 + 8 = 0
⇒ (x – 1) = (-2) (1)1/3
⇒ x – 1 = -2 or -2ω or -2ω2 or
n = -1 or 1 – 2ω or 1 – 2ω2 .

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