Evaluate \(\int^2_1(x^2 - 4x)dx\)
\(\frac{11}{3}\)
\(\frac{3}{11}\)
\(-\frac{3}{11}\)
\(-\frac{11}{3}\)
Correct answer is D
\(\int^2_1(x^2 - 4x)dx\)
\((\frac{x^3}{3} - \frac{4x^2}{2})\)
substituting integrate values
\([\frac{8}{3} - \frac{4 \times 4}{2}] - [\frac{1}{2} - 2]\)
= \(-\frac{11}{3}\)
If y = x2 - \(\frac{1}{x}\), find \(\frac{\delta y}{\delta x}\)
2x - \(\frac{1}{x^2}\)
2x + x2
2x - x2
2x + \(\frac{1}{x^2}\)
Correct answer is D
y = x2 - \(\frac{1}{x}\)
y = x2 - x-1
\(\frac{\delta y}{\delta x}\) = 2x + x-2
\(\frac{\delta y}{\delta x}\) = 2x + \(\frac{1}{x^2}\)
If angle \(\theta\) is 135°, evaluate cos\(\theta\)
\(\frac{1}{2}\)
\(\frac{\sqrt{2}}{2}\)
\(-\frac{\sqrt{2}}{2}\)
\(-\frac{1}{2}\)
Correct answer is C
\(\theta\) = 135°
Cos 135° = Cos(90 + 45)°
= cos90° cos45° - sin90° sin45°
= 0cos45° - (1 x \(\frac{\sqrt{2}}{2}\))
= \(-\frac{\sqrt{2}}{2}\)
Find the equation of the line through the points (-2, 1) and (-\(\frac{1}{2}\), 4)
y = 2x - 3
y = 2x + 5
y = 3x - 2
y = 2x + 1
Correct answer is B
\(\frac{y - y_1}{x - x_1}\) = \(\frac{y_2 - y_1}{x_2 - x_1}\)
\(\frac{y - 1}{x - -2}\) = \(\frac{4 - 1}{-\frac{1}{2} + 2}\)
= \(\frac{y - 1}{x + 2}\) = \(\frac{3}{\frac{3}{2}}\)
y = 2x + 5
The distance between the point (4, 3) and the intersection of y = 2x + 4 and y = 7 - x is
\(\sqrt{13}\)
\(3\sqrt{2}\)
\(\sqrt{26}\)
\(10\sqrt{5}\)
Correct answer is B
P1 (4, 3), P2 (x, y)
y = 2x + 4 .....(1)
y = 7 - x .....(2)
Substitute (2) in (1)
7 - x = 2x + 4
7 - 4 = 2x + x
3 = 3x
x = 1
Substitute in eqn (2)
y = 7 - x
y = 7 - 1
y = 6
P2 (1, 6)
Distance between 2 points is given as
D = \(\sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}\)
D = \(\sqrt{(1 - 4)^2 + (6 - 3)^2}\)
D = \(\sqrt{(-3)^2 + (3)^2}\)
D = \(\sqrt{9 + 9}\)
D = \(\sqrt{18}\)
D = \(\sqrt{9 \times 2}\)
D = \(3\sqrt{2}\)