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CBSE

Subject

Mathematics

Class

JEE Class 12

JEE Mathematics 2008 Exam Questions

Multiple Choice Questions

1.

The conjugate of a complex number is fraction numerator 1 over denominator straight i minus 1 end fraction. Then the complex number is

  • fraction numerator negative 1 over denominator straight i minus 1 end fraction
  • fraction numerator 1 over denominator straight i plus 1 end fraction
  • fraction numerator negative 1 over denominator 1 plus straight i end fraction
  • fraction numerator 1 over denominator 1 plus straight i end fraction

C.

fraction numerator negative 1 over denominator 1 plus straight i end fraction
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2.

AB is a vertical pole with B at the ground level and A at the top. A man finds that the angle of elevation of the point A from a certain point C on the ground is 60°. He moves away from the pole along the line BC to a point D such that CD = 7 m. From D the angle of elevation of the point A is 45°.
Then the height of the pole is

  • fraction numerator 7 square root of 3 over denominator 2 end fraction. fraction numerator 1 over denominator square root of 3 begin display style minus end style begin display style 1 end style end fraction straight m
  • fraction numerator 7 square root of 3 over denominator 2 end fraction. left parenthesis square root of 3 plus 1 right parenthesis space straight m
  • fraction numerator 7 square root of 3 over denominator 2 end fraction. left parenthesis square root of 3 minus 1 right parenthesis space straight m
  • fraction numerator 7 square root of 3 over denominator 2 end fraction. fraction numerator 1 over denominator square root of 3 begin display style plus end style begin display style 1 end style end fraction

B.

fraction numerator 7 square root of 3 over denominator 2 end fraction. left parenthesis square root of 3 plus 1 right parenthesis space straight m


BD = AB = 7 + x
Also AB = x tan 60° = x√3
∴ x√3 = 7+x
straight x space equals space fraction numerator 7 over denominator square root of 3 minus 1 end fraction

AB space equals space fraction numerator 7 square root of 3 over denominator 2 end fraction left parenthesis square root of 3 plus 1 right parenthesis

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3.

The mean of the numbers a, b, 8, 5, 10 is 6 and the variance is 6.80. Then which one of the following gives possible values of a and b?

  • a = 0, b = 7 

  • a = 5, b = 2

  • a = 1, b = 6

  • a = 3, b = 4 


D.

a = 3, b = 4 

Mean of a, b, 8, 5, 10 is 6

rightwards double arrow space fraction numerator straight a plus space straight b plus space 8 space plus space 5 plus space 10 over denominator 5 end fraction space equals space 6
rightwards double arrow space straight a space plus straight b space equals space 7 space space space space... space left parenthesis straight i right parenthesis
Given space that space variance space is space 6.8
therefore comma space Variance space equals space fraction numerator straight capital sigma space left parenthesis straight x subscript straight i space minus space straight A right parenthesis squared over denominator straight n end fraction
space equals space fraction numerator left parenthesis straight a minus 6 right parenthesis squared space plus space left parenthesis straight b minus 6 right parenthesis squared space plus space 4 plus 1 space plus 16 over denominator 5 end fraction space equals space 6.8
rightwards double arrow space straight a squared space plus straight b squared space equals space 25
straight a squared space plus space left parenthesis 7 minus straight a right parenthesis squared space equals space 25
rightwards double arrow space straight a squared minus 7 straight a space plus 12 space equals space 0
therefore space straight a space equals space 4 comma 3 space and space straight b space equals space 3 comma 4

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4.

A focus of an ellipse is at the origin. The directrix is the line x = 4 and the eccentricity is 1/2. Then the length of the semi−major axis is 

  • 8/3

  • 2/3

  • 5/3

  • 4/3


A.

8/3

Major axis is along x-axis.

straight a over straight e minus space ae space equals space 4
straight a space open parentheses 2 minus 1 half close parentheses space equals 4
straight a space equals 8 over 3

737 Views

5.

The perpendicular bisector of the line segment joining P (1, 4) and Q (k, 3) has y−intercept − 4. Then a possible value of k is

  • 1

  • 2

  • -2

  • -4


D.

-4

slope space of space bisector space equals space straight k minus 1
Middle space point space equals space open parentheses fraction numerator straight k plus 1 over denominator 2 end fraction comma 7 over 2 close parentheses
Equation space of space bisector space is
straight y space minus space 7 over 2 space equals space left parenthesis straight k minus 1 right parenthesis space open parentheses straight x minus fraction numerator left parenthesis straight k plus 1 right parenthesis over denominator 2 end fraction close parentheses
Put space straight x space equals space 0 space and space straight y space equals space minus space 4
rightwards double arrow space straight k space plus-or-minus 4
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6.

The solution of the differential equation dy over dx space equals space fraction numerator straight x plus straight y over denominator straight x end fraction   satisfying the condition y (1) = 1 is  

  • y = ln x + x

  • y = x ln x + x2

  •  y = xe(x−1)

  • y = x ln x + x


D.

y = x ln x + x

y = vx

dy over dx space equals space straight v space plus space straight x dv over dx
straight v space plus space straight x dv over dx space equals space 1 space plus space straight v
rightwards double arrow space dv space equals space dx over straight x
therefore space straight v space equals space log space straight x space plus space straight c
rightwards double arrow space straight y over straight x space equals space log space straight x space space plus straight c

Since, y (1) = 1, we have y = x log x + x

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7.

The point diametrically opposite to the point P (1, 0) on the circle x2+ y2 + 2x + 4y − 3 = 0 is 

  • (-3,4)

  • (-4,3)

  • (-3,-4)

  • (3,4)


C.

(-3,-4)

Centre (− 1, − 2)
Let (α, β) is the required point
fraction numerator straight alpha plus 1 over denominator 2 end fraction space equals space minus space 1 space and space fraction numerator straight beta plus 0 over denominator 2 end fraction space equals space minus space 2

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8.

Statement − 1: For every natural number n ≥ 2 fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space..... space plus space fraction numerator 1 over denominator square root of straight n end fraction space greater than space square root of straight n

Statement −2: For every natural number n ≥ 2,straight n greater or equal than 2 comma space square root of straight n left parenthesis straight n plus 1 right parenthesis space end root space less than space straight n plus 1

  • Statement −1 is false, Statement −2 is true

  • Statement −1 is true, Statement −2 is true, Statement −2 is a correct explanation for Statement −1

  • Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

  • Statement − 1 is true, Statement − 2 is false. 


C.

Statement −1 is true, Statement −2 is true; Statement −2 is not a correct explanation for Statement −1.

straight P space left parenthesis straight n right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space..... space plus space fraction numerator 1 over denominator square root of straight n end fraction
straight P space left parenthesis 2 right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space greater than space square root of 2
Let space us space assume space that space straight P space left parenthesis straight k right parenthesis space equals space fraction numerator 1 over denominator square root of 1 end fraction space plus space fraction numerator 1 over denominator square root of 2 end fraction space plus space...... fraction numerator 1 over denominator square root of straight k end fraction space plus fraction numerator 1 over denominator square root of straight k plus 1 end root end fraction space greater than square root of straight k plus 1 end root
has space to space be space true.
straight L. straight H. straight S greater than thin space square root of straight k space plus space fraction numerator 1 over denominator square root of straight k plus 1 end root end fraction space equals space fraction numerator square root of straight k space left parenthesis straight k plus 1 right parenthesis end root space plus 1 over denominator square root of straight k plus 1 end root end fraction
since space square root of straight k left parenthesis straight k plus 1 right parenthesis end root space greater than straight k space space left parenthesis for all space straight k greater or equal than 0 right parenthesis
therefore space fraction numerator square root of straight k space left parenthesis straight k plus 1 right parenthesis end root plus 1 over denominator square root of straight k plus 1 end root end fraction space greater than space fraction numerator straight k plus 1 over denominator square root of straight k plus 1 end root end fraction space equals space square root of straight k plus 1 end root
Let space straight p left parenthesis straight n right parenthesis space space equals space square root of straight n space left parenthesis straight n plus 1 right parenthesis end root space less than space straight n plus 1
State space minus 1 space is space correct.
straight P space left parenthesis 2 right parenthesis space space equals space square root of 2 space straight x space 3 end root space less than space 3
If space straight P space left parenthesis straight k right parenthesis space space equals space square root of straight k left parenthesis straight k plus 1 right parenthesis end root space less than space left parenthesis straight k plus 1 right parenthesis space is space true
Now space space straight P space left parenthesis straight k plus 1 right parenthesis space equals space square root of left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end root space less than space straight k plus 2 space has space to space be space true
square root of left parenthesis straight k plus 1 right parenthesis left parenthesis straight k plus 2 right parenthesis end root space thin space left parenthesis straight k plus 2 right parenthesis
Hence Statement −2 is not a correct explanation of Statement −1. 
125 Views

9.

A die is thrown. Let A be the event that the number obtained is greater than 3. Let B be the event that the number obtained is less than 5. Then P (A ∪ B) is 

  • 3/5

  • 0

  • 1

  • 2/5


C.

1

A = {4, 5, 6} , B = {1, 2, 3, 4} .
Obviously P (A ∪ B) = 1.

177 Views

10.

A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is at 

  • (0, 2)

  • (1, 0)

  • (0,1)

  • (2,0)


B.

(1, 0)

vertex (0,1) 

342 Views

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