CBSE
If the number of terms in the expansion of is 28, then the sum of the coefficients of all the terms in this expansion is
64
2187
243
729
D.
729
Clearly, number of terms in the expansion of
If the 2^{nd}, 5^{th} and 9th terms of a non-constant A.P. are in G.P., then the common ratio of this G.P. is:
8/5
4/3
1
7/4
B.
4/3
Let a be the first term and d be a common difference. Then, we have a+d, a+4d, a+8d in GP,
ie. (a +4d)^{2} = (a+d)(a+8d)
⇒ a^{2} +16d^{2} +8ad = a^{2} +8ad +ad +8d^{2}
⇒ 8d^{2} = ad
8d =a [∴ d≠0]
Now, common ratio,
If
is an empty set
contains exactly one element.
contains exactly two elements.
contains more than two elements.
C.
contains exactly two elements.
We have,
Hence, S contains exactly two elements.
The sum of all real values of x satisfying the equation
is:
3
-4
6
5
A.
3
Given,
A man is walking towards a vertical pillar in a straight path, at a uniform speed. At a certain point A on the path, he observes that the angle of elevation of the top of the pillar is 30^{o}. After walking for 10 minutes from A in the same direction, at a point B, he observes that the angle of elevation of the top of the pillar is 60^{o}. Then the time taken (in minutes) by him, from B to reach the pillar, is:
6
10
20
5
D.
5
According to given information, we have the following figure:
Now, from ΔACD and ΔBCD, we have
∵ speed is uniform.
∵ distance y will be cover in 5 min
∵ distance x covered in 10 min
∵ Distance x/2 will be cover in 5 min
If the sum of the first ten terms of the series,
is 16/5 m, the m is equal to
102
101
100
99
B.
101
Let S_{10} be the sum of first ten terms of the series. Then, we have
A value of θ for which is purely imaginary is
π/3
π/6
D.
Let z = is purely imaginary. Then, we have Re (z) = 0
We have Re (z) = 0
Now, consider z =
If all the words (with or without meaning) having five letters, formed using the letters of the word SMALL and arranged as in a dictionary; then the position of the word SMALL is:
46^{th}
59^{th}
52^{nd}
58^{th}
D.
58^{th}
Clearly, number of words start with A = 4!/ 2! = 12
Number of words start with L = 4! = 24
A number of words start with M = 4!/2! = 12
Number of words start with SA = 3!/2! = 3
Number of words starts with SL = 3! = 6
Note that, next word will be 'SMALL'
Hence, the position of word 'SMALL' is 58th.
If A = and A adj A = AA^{T}, then 5a +b is equal to
-1
5
4
5
B.
5
Given, A = and A adj A = AA^{T}, Clearly, A (adj A) = |A|I_{n}|
Now, substituting the value of b in Eq. (iii) we get
5a = 2
Hence, 5a + b = 2 +3 = 5