6
2
-2
-8
Correct answer is B
f (x) = x\(^2\) + px + q
\(\frac{dy}{dx}\) = 2x + p
2x + p = 0
2(-3) + p = 0
p = +6
f(-2) = -2\(^2\) + p(-2) + q
= -6 2p + q = -10
-2(-16) + q = -10
-12 + q = -10
q = -10 + 12
= 2
Evaluate \(\int^0_0 \sqrt{x} dx\)
3
9
18
27
Correct answer is C
\(\int^0_0 \sqrt{x} dx\)
\(\frac{x^{\frac{1}{2}} + 1}{\frac{1}{2} + 1}\) |\(^9_0\)
\(\frac{2x^{\frac{3}{2}}}{3}\)|\(^9_0\)
= \(\frac{2(9)^{\frac{3}{2}}}{3}\)
= \(\frac{2(3)^3}{3}\)
= \(\frac{2 \times 27}{3}\)
= 2 x 9
= 18
Find the coordinates of the centre of the circle 3x\(^2\) + 3y\(^2\) - 6x + 9y - 5 = 0
(-3. \(\frac{9}{2}\))
(-1. \(\frac{3}{2}\))
(1, - \(\frac{3}{2}\))
(3. -\(\frac{9}{2}\))
Correct answer is C
\(\frac{3x^2}{3} + \frac{3y^2}{3} - \frac{6x}{3} + \frac{9y}{3} - \frac{5}{3}\) = 0
2gx = -2x, 2 fy = 3y
g = -1, f = \(\frac{3}{2}\)
Centre (1, - \(\frac{3}{2}\))
\(\frac{27}{64}\)y\(^2\)
\(\frac{27}{8}\)y\(^2\)
8y\(^2\)
9y\(^2\)
Correct answer is B
(2x + \(\frac{3y}{4}\))\(^3\)
= 3(2x) (\(\frac{3y}{4}\))\(^2\)
= \(\frac{27}{8}\)y\(^2\)
Given that P and Q are non-empty subsets of the universal set, U. Find P \(\cap\) (Q U Q`).
p
P`
Q
Q`
Correct answer is A
P \(\cap\) (Q U Q`)
P \(\cap\) U = P