Multiple Choice QuestionsIn 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
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 -5 /6
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!/(4!)3
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 areÂ
12 kg and 13 kg
5 kg and 5 kg
5 kg and 12 kg
5 kg and 12 kg
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
– 1/2 ≤ k ≤ 1/2
The differential equation of all circles passing through the origin and having their centres on the x-axis is
If p and q are positive real numbers such that p2 + q2 = 1, then the maximum value of (p + q) is
2
1/2
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
The sum of the series 20C0 – 20C1 + 20C2 – 20C3 + …… - ….. + 20C10 is-
– 20C10
0
0
B.
(1 + x)20 = 20C0 + 20C1x + … + 20C10x10 + …+ 20C20x20
put x = − 1,
0 = 20C0 − 20C1 + … − 20C9 + 20C10 − 20C11 + … + 20C20
0 = 2 (20C0 − 20C1 + … − 20C9) + 20C10
⇒ 20C0 − 20C1 + … + 20C10 =
 then the value of x
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