Find the value of sin 45° - cos 30°
\(\frac{2+\sqrt{3}}{4}\)
\(\frac{\sqrt{2}+\sqrt{3}}{4}\)
\(\frac{\sqrt{2}+\sqrt{3}}{2}\)
\(\frac{\sqrt{2}-\sqrt{3}}{2}\)
Correct answer is D
\(Sin 45 - cos 30\\
\frac{\sqrt{2}}{2}-\frac{\sqrt{3}}{2}\\
=\frac{\sqrt{2}-\sqrt{3}}{2}\)
If y = 3 cos 4x, dy/dx equals?
6 sin 8x
-24 sin 4x
12 sin 4x
-12 sin 4x
Correct answer is D
y = 3 cos 4x
Applying product rule:
\(\frac{dy}{dx}\) = u\(\frac{dv}{dx}\) + v\(\frac{du}{dx}\)
Where u = 3 and v = cos 4x
\(\frac{du}{dx}\) of 3 = 0 and
\(\frac{dv}{dx}\) of cos 4x = -4sin 4x
\(\frac{dy}{dx}\) = 3 \(\times\) -4 sin 4x + cos 4x \(\times\) 0
\(\frac{dy}{dx}\) = -12 sin 4x
What is the value of r if the distance between the point (4,2) and (1,r) is 3 units?
1
2
3
4
Correct answer is B
A(4,2) and B(1,r), AB = 3 units
3 = √ (x2- x1)2 + (y2-y1)2
3 = √ (4-1)2 + (2-r)2
3 = √ 32 + (2-r)2
3 = √ 9 + 4 – 4r + r2
3 = √ r2 - 4r + 13
9 = r2 - 4r + 13 By squaring both sides
r2 - 4r + 4 = 0
(r-2)(r-2) = 0
r = 2
What is the value of p if the gradient of the line joining (-1,p) and (p, 4) is
| 2 |
| 3 |
-2
-1
1
2
Correct answer is D
| Gradient = | y2 - y1 |
| x2 - x1 |
| = | 4-p | ||
| p-(-1) |
| = | 4-p | ||
| p+1 |
| p = | 10 |
| 5 |
A circle of radius 4 cm and the center 0
The perpendicular bisector of the chords
A straight line passing through 0
A circle of radius 6 cm and with center 0
Correct answer is A
No explanation has been provided for this answer.