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CBSE

Subject

Mathematics

Class

JEE Class 12

JEE Mathematics 2004 Exam Questions

Multiple Choice Questions

1.

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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2.

If |z2-1|=|z|2+1, then z lies on

  • the real axis

  • an ellipse

  • a circle

  • the imaginary axis


B.

an ellipse

Given that
|z2- 1| = |z|2+ 2
|z2 + (-1)| = |z2| + |-1|
It shows that the origin, -1 and z2 lies on a line and z2 and -1 lies on one side of the origin, therefore
z2 is a negative number. Hence z will be purely imaginary. So we can say that z lies on y-axis.

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3.

Let S(K) = 1 +3+5+..... (2K-1) = 3+K2. Then which of the following is true?

  • S(1) is correct

  • Principle of mathematical induction can be used to prove the formula

  • S(K) ≠S(K+1)

  • S(K)⇒ S(K+1)


D.

S(K)⇒ S(K+1)

S(K) = 1 + 3 + 5 + ...... + (2K - 1) = 3 + K2
Put K = 1 in both sides
∴ L.H.S = 1 and R.H.S. = 3 + 1 = 4 ⇒ L.H.S. ≠ R.H.S.
Put (K + 1) on both sides in the place of K L.H.S. = 1 + 3 + 5 + .... + (2K - 1) + (2K + 1)
R.H.S. = 3 + (K + 1)2 = 3 + K2 + 2K + 1
Let L.H.S. = R.H.S.
1 + 3 + 5 + ....... + (2K - 1) + (2K + 1) = 3 + K2 + 2K + 1
⇒ 1 + 3 + 5 + ........ + (2K - 1) = 3 + K2 If S(K) is true, then S(K + 1) is also true. Hence, S(K) ⇒ S(K + 1)

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4.

How many ways are there to arrange the letters in the word GARDEN with the vowels in alphabetical order?

  • 120

  • 480

  • 360

  • 240


C.

360

A total number of ways in which all letters can be arranged in alphabetical order = 6! There are two vowels in the word GARDEN. A total number of ways in which these two vowels can be arranged = 2!

∴ Total number of required ways 

∴ Total number of required waysspace equals space fraction numerator 6 factorial over denominator 2 factorial end fraction space equals space 360

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5.

If a1, a2, a3 , ....,an , .... are in G.P., then the value of the determinant open square brackets table row cell log space straight a subscript straight n end cell cell log subscript straight n plus 1 end subscript end cell cell log subscript straight n plus 2 end subscript end cell row cell log space straight a subscript straight n plus 3 end subscript end cell cell log space straight a subscript straight n plus 4 end subscript end cell cell log space straight a subscript straight n plus 5 end subscript end cell row cell log space straight a subscript straight n plus 6 end subscript end cell cell log space straight a subscript straight n plus 7 end subscript end cell cell log space straight a subscript straight n plus 8 end subscript end cell end table close square brackets is

  • 0

  • -2

  • 1

  • 2


A.

0

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6.

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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7.

The number of ways of distributing 8 identical balls in 3 distinct boxes so that none of the boxes is empty is

  • 5

  • 8C3
  • 38

  • 21


D.

21

The required number of ways
space equals space to the power of 8 minus 1 end exponent straight C subscript 3 minus 1 end subscript
space equals space to the power of 7 straight C subscript 2 space equals space fraction numerator 7 factorial over denominator 2 factorial 5 factorial end fraction
space equals space to the power of 7 straight C subscript 2 space equals space fraction numerator 7 space factorial over denominator 2 factorial space 5 factorial end fraction
space equals space fraction numerator 7.6 over denominator 2.1 end fraction space equals space 21

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8.

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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9.

Let two numbers have arithmetic mean 9 and geometric mean 4. Then these numbers are the roots of the quadratic equation

  • x2 + 18x +16 = 0

  • x2-18x-16 = 0

  • x2+18x-16 =0

  • x2-18x +16 =0


D.

x2-18x +16 =0

Let α and β be two numbers whose arithmetic mean is 9 and geometric mean is 4.
∴ α + β = 18 ........... (i)
and αβ =16 ........... (ii)
∴ Required equation is x2 - (α + β)x + (αβ) = 0 ⇒ x2 - 18x + 16 = 0 [using equation (i) and equation (ii)]

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10.

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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