If y = (1 - 2x)\(^3\), find the value of dy/dx at x = -1
57
27
-6
-54
Correct answer is D
y = (1 - 2x)\(^3\)
let u = 1-2x
du/dx = -2
∴y = u\(^3\)
dy/du = 3u\(^2\)
But dy/dx = du/dx * dy/du
= -2 * 3u\(^2\)
= -6u\(^2\)
= -6(1-2x)\(^2\)
At x = -1: dy/dx = -6(1 - 2(-1))\(^2\)
dy/dx = -6(1 + 2)\(^2\)
= -6 * 3\(^2\)
= -6 * 9
= -54
If sin θ = -1/2 for 0 < θ < 360o, the value of θ is
30o and 150o
150o and 210o
210o and 330o
150o and 330o
Correct answer is C
sin θ = -1/2
= -0.5
θ = sin-1 (0.5)
θ = 30o
Since θ is negative
θ = 180 + 30 = 210o
θ = 360 - 30 = 330o
∴θ = 210o and 330o
7 cm
14 cm
21 cm
28 cm
Correct answer is B
No explanation has been provided for this answer.
Find the equation of the perpendicular at the point (4,3) to the line y + 2x = 5
2y - x = 4
y + 2x = 3
y + 2x = 5
2y - x = 2
Correct answer is D
Gradient of the line y + 2x = 5 = -2
Gradient of the line perpendicular to
y + 2x + 5 = 1/2
equation of a line perpendicular to
y + 2x = 5 at the point (4, 3)
y - y1 = m(x - x1)
y - 3 = 1/2(x - 4)
2y - 6 = x -4
2y - x = 2
Two lines PQ and ST intersect at 75°. The locus of points equidistant from PQ and ST lies on the
perpendicular bisector of PQ
perpendicular bisector of ST
bisector of the angles between lines PQ and ST
bisector of the angles between lines PT and QS
Correct answer is C
No explanation has been provided for this answer.