Evaluate \(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
3\(\frac{2}{3}\)
4
4\(\frac{1}{3}\)
4\(\frac{2}{3}\)
Correct answer is D
\(\int^{1}_{-1}(2x + 1)^2 \mathrm d x\)
= \(\int^{1}_{-1}(4x^2 + 4x + 1) \mathrm d x\)
= \(\int^{1}_{-1}\)[\(\frac{4x^3}{3} + 2x^2 + x]\)
= [\(\frac{4}{3}\) + 2 + 1] - [\(\frac{-4}{3}\)+2 -1]
= \(\frac{13}{3}\) + \(\frac{1}{3}\)
= \(\frac{14}{3}\)
= \(4 \frac{2}{3}\)
Integrate \(\frac{1 - x}{x^3}\) with respect to x
\(\frac{x - x^2}{x^4}\) + k
\(\frac{4}{x^4} - \frac{3 + k}{x^3}\)
\(\frac{1}{x} - \frac{1}{2x^2}\) + k
\(\frac{1}{3x^2} - \frac{1}{2x}\) + k
Correct answer is C
\(\int \frac{1 - x}{x^3}\)
= \(\int^{1}_{x^3} - \int^{x}_{x^3}\)
= x-3 dx - x-2dx
= \(\frac{1}{2x^2} + \frac{1}{x}\)
Find the point (x, y) on the Euclidean plane where the curve y = 2x2 - 2x + 3 has 2 as gradient
(1, 3)
(2, 7)
(0, 3)
(3, 15)
Correct answer is A
Equation of curve;
y = 2x2 - 2x + 3
gradient of curve;
\(\frac{dy}{dx}\) = differential coefficient
\(\frac{dy}{dx}\) = 4x - 2, for gradient to be 2
∴ \(\frac{dy}{dx}\) = 2
4x - 2 = 2
4x = 4
∴ x = 1
When x = 1, y = 2(1)2 - 2(1) + 3
= 2 - 2 + 3
= 5 - 2
= 3
coordinate of the point where the curve; y = 2x2 - 2x + 3 has gradient equal to 2 is (1, 3)
If y = 3t3 + 2t2 - 7t + 3, find \(\frac{dy}{dt}\) at t = -1
-1
1
-2
2
Correct answer is C
y = 3t3 + 2t2 - 7t + 3
\(\frac{dy}{dt}\) = 9t2 + 4t - 7
When t = -1
\(\frac{dy}{dt}\) = 9(-1)2 + 4(-1) - 7
= 9 - 4 -7
= 9 - 11
= -2
What is the value of sin(-690)?
\(\frac{\sqrt{3}}{2}\)
-\(\frac{\sqrt{3}}{2}\)
\(\frac{-1}{2}\)
\(\frac{1}{2}\)
Correct answer is D
Sin(-690o) = Sin(-360 -3300
sin - 360 = sin 0
∴ sin(-690o) = sin(330o)
Negative angles are measured in clockwise direction
The acute angle equivalent of sin(-330o) = sin(30o)
sin(-330o) = sin(30o)
= \(\frac{1}{2}\)
= 0.5