If : R→R and g : R→R are defined by fx = x and gx = x - 3 for x ∈ R, then gfx : - 85 < x < 85 is equal to
{0, 1}
{1, 2}
{- 3, - 2}
{2, 3}
f : [-6, 6] R is defined by f(x) = x2 - 3for x ∈ R, then(fofof) (-1) + (fofof) (0) + (fofof)(1) is equal to
f42
f32
f22
f2
A.
Given, fx = x2 - 3Now, f- 1 = - 12 - 3 = - 2⇒ fof- 1 = f- 2 = - 22 - 3 = 1⇒ fofof- 1 = f1 = 12 - 3 = - 2Now, f0 = 02 - 3 = - 3⇒ fof0 = f- 3 = - 32 - 3 = 6⇒ fofof0 = f6 = 62 - 3 = 33Again, f1 = 12 - 3 = - 2⇒ fof1 = f- 2 = - 22 - 3 = 1⇒ fofof1 = 12 - 3 = - 2Now,f42 = 422 - 3 = 32 - 3 = 29Hence, option a is correct
Given that a, b ∈ 0, 1, 2, . . . , 9 witha + b ≠ 0 and that a + b10x = ab + b100y = 1000. Then,1x - 1y is equal to
1
12
13
14
If x = 127 + 17, then x2 - 1x - x2 - 1 is equal to
2
3
4
If x2 + x + 1x2 + 2x + 1 = A + Bx + 1 + Cx + 12, then A - B is equal to
4C
4C + 1
3C
2C
∑k = 1∞1k!∑n = 1k2n - 1 is equal to
e
e2 + e
e2
e2 - e
If f : 2, 3→ R is defined by f (x) = x3 + 3x - 2, then the range f(x) iscontained in the interval
[1, 12]
[12, 34]
[35, 50]
[- 12, 12]
x ∈ R : 2x - 1x3 + 4x2 + 3x ∈ R equals
R - {0}
R - {0, 1, 3}
R - {0, - 1, - 3}
R - 0, - 1, - 3, +12
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) = ?
10
log42 - log82 + log162 - . . = ?
loge2
1 + loge3
1 - loge2