If f(0) = 0, f(1) = 1, f (2) = 2 and f(x) = fx - 2 + fx - 3 for x = 3, 4, 5, . . . , then f(9) = ?
12
13
14
10
D.
f(0) = 0, f(1) = 1, f (2) = 2 Given, f(x) = fx - 2 + fx - 3, x = 3, 4, 5, . . . The given function is known as "Reccurrence function".put x = 3, f3 = f1 + f0 = 1 + 0 = 1put x = 4, f4 = f2 + f1 = 2+ 1 ⇒ 3put x = 5, f5 = f3 + f2 = 1 + 2 ⇒ 3put x = 6, f6 = f4 + f3 = 3 + 1 ⇒ 4put x = 7, f7 = f5 + f4 = 3 + 3 ⇒ 6put x = 8, f8 = f6 + f5 = 3 + 4 ⇒ 7put x = 9, f9 = f7 + f6 = 6 + 6 ⇒ 10Hence, f(9) = 10
log42 - log82 + log162 - . . = ?
e2
loge2
1 + loge3
1 - loge2
For x ∈ R, the least value of x2 - 6x + 5x2 + 2x + 1 is
- 1
- 12
- 14
- 13
x ∈ R : 14xx + 1 - 9x - 30x - 4 < 0 = ?
(- 1, 4)
1, 4 ∪ 5, 7
(1, 7)
- 1, 1 ∪ 4, 6
If a, b and n are natural numbers, then a2n - 1 + b2n - 1 is divisible by
a + b
a - b
a3 + b3
a2 + b2
x2 + x + 1x - 1x - 2x - 3 = Ax - 1 + Bx - 2 + Cx - 3⇒ A + C =
4
5
6
8
Let f : R → R be defined byf(x) = α + sinxx, if x> 02, if x = 0β + sinx - xx3, if x < 0where, [x] denotes the integral part of x.If f continuous at x = 0, then β - α is equal to
1
0
2
If a, b, c and d ∈ R such that a2 + b2 = 4 and c2 + d2 = 4and if (a + ib) = (c + id)2 (x + iy), then x2 + y2 is equal to
3
If fx = p - xn1n, p > 0 and n is a positive integer, then ffx = ?
x
xn
p1/n
p - xn
If R is the set of all real numbers and f : R - {2} → R is defined by fx = 2 + x2 - x for x ∈ R - 2
R - {- 2}
R
R - {1}
R - {- 1}