CBSE
Gujarat Board
Haryana Board
Class 10
Class 12
∑k = 1513 + 23 + .... + k31 + 3 + 5 + ... + 2k - 1 is equal to
22.5
24.5
28.5
32.5
A.
∑k = 1513 + 23 + .... + k31 + 3 + 5 + ... + 2k - 1 = ∑k = 15kk + 122k2 ∵ ∑n3 = nn + 122 and 1 + 3 + 5 + ... + 2k - 1 = k2= ∑k = 15k + 124 = 22 + 32 + 42 + 52 + 624= 4 + 9 + 16 + 25 + 364= 904 = 22.5
The function f: C → C defined, by fx = ax + dcx + d for x ∈ C where bd ≠ 0 reduces to a constant function, if
a = c
b = d
ad = bc
ab = cd
C.
fx = ax + dcx + d ... iNow take option ci.e. ab = cd = kFrom equation i fx = bkx + bdkx + d = bkx + 1dkx + 1 = bd = constant
For any integer n > 1, the number of positive divisors of n is denoted by d(n). Then, for a prime P, d (d (d(P)7)) is equal to
1
2
3
p
Since, dnrepresents number of the divisors of n.∴ dp7 = 8 d8 = d23 = 4 d4 = d22 = 3 dddp7 = 3
If f : R → R is defined by f(x) = x5 for x ∈ R, where [y] denotes the greatest integer not exceeding y, then fx : x < 71 is equal to
- 14, - 13, . . . 0, . . . 13, 14
- 14, - 13, . . . 0, . . . 14, 15
- 15, - 14, . . . 0, . . . 14, 15
- 15, - 14, . . . 0, . . . 13, 14
D.
Given, f : R → R and f(x) = x5Also,f(x) : x < 71 = f(x) : - 71 < x < 71= x5 : - 71 < x < 71= - 70 - 0.15, . . . ,- 705, . . . , 05, . . . ,655, . . . , 705, . . . ,70 + 0.999h= - 15, - 14, . . . , 0, . . . , 13, 14