Given that X : R → R is defined by x = y+15−y , y ∈ R, find the domain of x
{y : y ∈ R, y ≠ 0}
{y : y ∈ R, y ≠ 1}
{y : y ∈ R, y ≠ 5}
{y : y ∈ R, y ≠ 7}
Correct answer is C
No explanation has been provided for this answer.
Find the values of p and q
p = 5, q = 3
p = 5, q = -3
p = -5, q = -3
p = -5, q = 3
Correct answer is B
p + q = 2
p - q = 8
¯2p2=102
p = 5
from p + q = 2
5 + q = 2
q = 2 - 5
= -3
If ∫30(px2+16)dx = 129. Find the value of p
9
8
7
6
Correct answer is A
∫30(px2+16) = 129
px2+10+1+16x|30=129
px33+16x|30=129
(p(3)33+16(3)) - 0 = 129
9p + 48 = 129
9p = 129 - 48
9p9=819
p = 9
If cos x = -0.7133, find the values of x between 0o and 360o
44.5o , 224.5o
123.5o , 190.5o
135.5o , 213.5o
135.5o , 224.5o
Correct answer is D
cos x = -0.7133
x = cos −1(0.7133)
= 44.496o
x = 180 - 44.495o
x = 135.5o
and x = 180o + 44.495o
= 224.5o
(51−32)
(2−5−13)
(−52−13)
(5123)
Correct answer is B
Let A = (3512)
|A| = (3 x 2 - 5 x 1)
= 6 - 5
= 1
A−1 = (2−5−13)