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CBSE

Subject

Mathematics

Class

JEE Class 12

JEE Mathematics 2007 Exam Questions

Multiple Choice Questions

1.

A body weighing 13 kg is suspended by two strings 5 m and 12 m long, their other ends being fastened to the extremities of a rod 13 m long. If the rod be so held that the body hangs immediately below the middle point. The tensions in the strings are 

  • 12 kg and 13 kg

  • 5 kg and 5 kg

  • 5 kg and 12 kg

  • 5 kg and 13 kg


C.

5 kg and 12 kg



straight T subscript 2 space cos space open parentheses straight pi over 2 minus straight theta close parentheses space equals space straight T subscript 1 space cosθ
rightwards double arrow space straight T subscript 1 space cos space straight theta space equals space straight T subscript 2 space sin space straight theta
straight T subscript 1 space sin space straight theta space plus space straight T subscript 2 space cos space straight theta space equals space 13
because space OC space equals space CA space space equals CB
rightwards double arrow space angle space AOC space equals space angle OAC space and space angle COB space equals space angle OBC
therefore space sin space straight theta space equals space sin space straight A space equals space 5 over 13 space and space cos space straight theta space equals 12 over 13
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 5 over 12 space rightwards double arrow space straight T subscript 1 space equals space 5 over 12 space and space cos space straight theta space equals space 12 over 13
rightwards double arrow space straight T subscript 1 over straight T subscript 2 space equals space 5 over 12 space rightwards double arrow space straight T subscript 1 space equals space 5 over 12 straight T subscript 2
straight T subscript 2 space open parentheses 5 over 12.5 over 13.12 over 13 close parentheses space equals space 13
straight T subscript 2 space open parentheses fraction numerator 169 over denominator 12.13 end fraction close parentheses space equals space 13
straight T subscript 2 space equals space 12 space kgs
rightwards double arrow space straight T subscript 1 space equals space 5 space kgs
106 Views

2.

In a geometric progression consisting of positive terms, each term equals the sum of the next two terms. Then the common ratio of this progression equals

  • 1 half left parenthesis 1 minus square root of 5 right parenthesis
  • 1 half square root of 5
  • square root of 5
  • 1 half left parenthesis square root of 5 minus 1 right parenthesis

D.

1 half left parenthesis square root of 5 minus 1 right parenthesis
Given space ar to the power of straight n minus 1 end exponent space equals ar to the power of straight n space plus space ar to the power of straight n plus 1 end exponent
rightwards double arrow space 1 space equals space straight r plus space straight r squared
therefore space straight r space equals space fraction numerator square root of 5 minus 1 over denominator 2 end fraction
330 Views

3.

A tower stands at the centre of a circular park. A and B are two points on the boundary of the park such that AB (= a) subtends an angle of 60º at the foot of the tower, and the angle of elevation of the top of the tower from A or B is 30º. The height of the tower is

  • 2 straight a divided by square root of 3
  • 2 straight a square root of 3
  • straight a divided by square root of 3
  • straight a square root of 3

C.

straight a divided by square root of 3


∆OAB is equilateral
∴ OA = OB = AB = a
Now space tan space 30 to the power of straight o space equals space straight h over straight a
therefore space straight h space equals space fraction numerator straight a over denominator square root of 3 end fraction
167 Views

4.

The sum of the series 20C020C1 + 20C220C3 + …… - ….. + 20C10 is-

  • 20C10

  • 1 half straight C presuperscript 20 subscript 10
  • 0

  • 20C10

B.

1 half straight C presuperscript 20 subscript 10

(1 + x)20 = 20C0 + 20C1x + … + 20C10x10 + …+ 20C20x20
put x = − 1,
0 = 20C0 − 20C1 + … − 20C9 + 20C1020C11 + … + 20C20
0 = 2 (20C020C1 + … − 20C9) + 20C10
20C020C1 + … + 20C10 =1 half straight C presuperscript 20 subscript 10

1417 Views

5.

Consider a family of circles which are passing through the point (-1, 1) and are tangent to x-axis. If (h, K) are the co-ordinates of the centre of the circles, then the set of values of k is given by the interva

  • 0 < k < 1/2

  • k ≥ 1/2

  • – 1/2 ≤ k ≤ 1/2

  • k ≤ ½


B.

k ≥ 1/2

Equation of circle (x − h)2+ (y − k)2 = k2
It is passing through (− 1, 1) then
(− 1 − h)2+ (1 − k)2= k2
h2+ 2h − 2k + 2 = 0
D ≥ 0
2k − 1 ≥ 0 ⇒ k ≥ 1/2.

169 Views

6.

The differential equation of all circles passing through the origin and having their centres on the x-axis is

  • straight x squared space equals straight y squared space plus space xy dy over dx
  • straight x squared space equals straight y squared space plus space 3 xy dy over dx
  • straight x squared space equals straight y squared space plus space 2 xy dy over dx
  • straight x squared space equals straight y squared space minus space 2 xy dy over dx

C.

straight x squared space equals straight y squared space plus space 2 xy dy over dx

General equation of all such circles is
x2+ y2 + 2gx = 0.
Differentiating, we get
2 straight x space plus 2 straight y space dy over dx space plus 2 straight g space equals space 0
therefore space Desired space equation space is
straight x squared space plus space straight y squared space plus space open parentheses negative 2 straight x minus 2 straight y dy over dx close parentheses space straight x space equals space 0
rightwards double arrow straight y squared space equals space straight x squared space plus space 2 xy dy over dx

121 Views

7.

If p and q are positive real numbers such that p2 + q2 = 1, then the maximum value of (p + q) is

  • 2

  • 1/2

  • fraction numerator 1 over denominator square root of 2 end fraction
  • square root of 2

D.

square root of 2
Using space straight A. straight M. space greater or equal than space straight G. straight M comma
fraction numerator straight p squared space plus straight q squared over denominator 2 end fraction space greater or equal than pq
rightwards double arrow space pq space less or equal than 1 half
left parenthesis straight p plus straight q right parenthesis squared space equals space straight p squared space plus straight q squared space plus 2 pq
rightwards double arrow space left parenthesis straight p plus straight q right parenthesis squared space equals straight p squared space plus straight q squared space plus 2 pq
rightwards double arrow space straight p plus straight q space less or equal than square root of 2
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8. If space sin to the power of negative 1 end exponent space open parentheses straight x over 5 close parentheses space plus space cosec to the power of negative 1 end exponent space open parentheses 5 over 4 close parentheses space equals space straight pi over 2 then the value of x
  • 1

  • 3

  • 4

  • 5


B.

3

sin to the power of negative 1 end exponent space straight x over 5 space plus space sin to the power of negative 1 end exponent space 4 over 5 space equals space straight pi over 2
rightwards double arrow space sin to the power of negative 1 end exponent space straight x over 5 space equals space cos to the power of negative 1 end exponent space 4 over 5
rightwards double arrow space sin to the power of negative 1 end exponent space straight x over 5
space equals space sin to the power of negative 1 end exponent space 3 over 5
therefore space straight x space equals space 3
104 Views

9.

The set S: {1, 2, 3, …, 12} is to be partitioned into three sets A, B, C of equal size. Thus, A ∪ B ∪ C = S, A ∩ B = B ∩ C = A ∩ C = φ. The number of ways to partition S is-

  • 12!/3!(4!)3

  • 12!/3!(3!)4

  • 12!/(4!)3

  • 12!/(3!)4


C.

12!/(4!)3

Number of ways

straight C presuperscript 12 subscript 4 space straight x space straight C presuperscript 8 subscript 4 space straight x space straight C presuperscript 4 subscript 4 space equals space fraction numerator 12 factorial over denominator left parenthesis 4 factorial right parenthesis cubed end fraction

181 Views

10.

In the binomial expansion of (a - b)n, n ≥ 5, the sum of 5th and 6th terms is zero, then
a/b equals

  • 5/n −4

  • 6 /n −5

  • n -5 /6

  • n -4 /5


D.

n -4 /5

straight C presuperscript straight n subscript 4 space straight a to the power of straight n minus 4 end exponent space left parenthesis negative straight b right parenthesis to the power of 4 space plus straight C presuperscript straight n subscript 5 straight a to the power of straight n minus 5 end exponent space left parenthesis negative straight b right parenthesis to the power of 5 space equals space 0
rightwards double arrow space open parentheses straight a over straight b close parentheses space equals space fraction numerator straight n minus 5 plus 1 over denominator 5 end fraction
159 Views

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