Correct Answer: A. \(\frac{2}{3}\)

Explanation

For independent events \(P(X \cap Y) = P(X) \times P(Y)\)

\(\frac{2}{15} = \frac{1}{5} \times P(Y)\)

\(P(Y) = \frac{2}{15} ÷ \frac{1}{5} = \frac{2}{3}\)

Correct Answer: B. \(\frac{3x^{2}}{\sqrt[3]{(3x^{3} + 1)^{2}}}\)

Explanation

\(y = \sqrt[3]{3x^{3} + 1} = (3x^{3} + 1)^{\frac{1}{3}}\)

Let u = \(3x^{3} + 1\); y = \(u^{\frac{1}{3}}\)

\(\frac{\mathrm d y}{\mathrm d x} = (\frac{\mathrm d y}{\mathrm d u})(\frac{\mathrm d u}{\mathrm d x})\)

\(\frac{\mathrm d y}{\mathrm d u} = \frac{1}{3}u^{\frac{-2}{3}}\)

\(\frac{\mathrm d u}{\mathrm d x} = 9x^{2}\)

\(\frac{\mathrm d y}{\mathrm d x} = (\frac{1}{3}(3x^{3} + 1)^{\frac{-2}{3}})(9x^{2})\)

= \(\frac{3x^{2}}{\sqrt[3]{(3x^{3} + 1)^{2}}}\)

Explanation

\(y = 2x^{2} + x - 3\)

\(\frac{\mathrm d y}{\mathrm d x} = 4x + 1\)

At (2, 7), the gradient is \(4(2) + 1 = 9\)

The gradient of normal = \(-1 \div 9 = \frac{-1}{9}\)

Equation of the normal = \(y - 7 = \frac{-1}{9}(x - 2)\)

\(9y - 63 = 2 - x\)

At the point of meeting the x- axis, y = 0

\(0 - 63 = 2 - x \implies x = 65\)

The coordinate of P = (65, 0).

Correct Answer: B. \(\frac{-\sqrt{3}}{2}\)

Explanation

\(\cos (x + y) = \cos x \cos y - \sin x \sin y\)

\(\cos (\frac{\pi}{2} + \frac{\pi}{3}) = \cos \frac{\pi}{2} \cos \frac{\pi}{3} - \sin \frac{\pi}{2} \sin \frac{\pi}{3}\)

= \((0 \times \frac{1}{2}) - (1 \times \frac{\sqrt{3}}{2})\)

= \(0 - \frac{\sqrt{3}}{2} = -\frac{\sqrt{3}}{2}\)

Correct Answer: D. 8.66 N

Explanation

\(F_{1} = (5 N, 060°) = 5 \cos 60 i + 5 \sin 60 j = 2.5 i + \frac{5\sqrt{3}}{2}j\)

\(F_{2} = (10 N, 180°) = 10 \cos 180 i + 10 \sin 180 j = -10i\)

\(R = F_{1} + F_{2} = -7.5 i + \frac{5\sqrt{3}}{2}j\)

\(|R| = \sqrt{(-7.5)^{2} + (\frac{5\sqrt{3}}{2})^{2}}\)

= \(\sqrt{56.25 + 18.75} = \sqrt{75}\)

= 8.66 N

Explanation

To find the determinant of \(\begin{pmatrix} 5 & -6 & 4\\ 7 & 4 & -3\\ 2 & 1 & 6 \end{pmatrix}\) which is 5(24 + 3) + 6(4 + 6) + 4(7 - 8) = 135 + 288 - 4 = 419.

Then, substitude the column of the coefficient of x by the right - hand side of the equations and find the determinant of \(\begin{pmatrix} 5 & -6 & 4\\ 7 & 4 & -3\\ 46 & 1 & 6 \end{pmatrix}\) = 15(24 + 3) + 6(114 + 138) + 4(19 - 184) = 405 + 1512 - 660 = 1257.

Similarly, to find the coefficient of y, find the determinant of \(\begin{pmatrix} 5 & 15 & 4\\ 7 & 19 & -3\\ 2 & 46 & 6 \end{pmatrix}\)

= 5(114 + 138) - 13(42 + 6) - 4(322 - 38) = 1260 - 720 + 1136 = 1676.

Finally, for the coefficient of z find the determinant of \(\begin{pmatrix} 5 & -6 & 15\\ 7 & 19 & -3\\ 2 & 46 & 6 \end{pmatrix}\)

= 5(184 - 19) + 6(322 - 38) + 15(7 - 8) = 825 + 1704 - 15 = 2514

Therefore, x = \(\frac{1257}{419}\) = 3, y = \(\frac{1676}{419}\) = 4

and z = \(\frac{2514}{419}\) = 6

Correct Answer: C. 24.5

Explanation

Modal class 20 - 29

Modal class mark = \(\frac{20+29}{2}\)

= \(\frac{49}{2}\)

= 24.5

Explanation

(a)

∑ clockwise moments = ∑ anti-clockwise moments

= 3 x mJ + 4 x 15 = 60 x 2

= 3mJ + 60 = 120

= 3mJ = 120 - 60

= 3mJ = 60

\(∴mJ=\frac{60}{3}=20kg\)

(bi)

\(∑f_y=0==>N-mgcosθ=0\)

\(=N-12(10)cos30^o=0\)

\(=N=120cos 30^o\)

\(=N=60√3N\)

At the point of moving down,

\(∑f_x=0==>μN-P-mgsinθ=0\)

\(=\frac{2}{3}(60√3)-P-12(10)sin30^o=0\)

\(=40√3-P-60=0\)

\(=P=40√3-60\)

\(∴P=9.28N\)

(bii)

\(∑f_y=0==>N-mgcosθ=0\)

\(=N-12(10)cos 30^o=0\)

\(=N=120 cos 30^o\)

\(=N=60√3N\)

At the point of moving up,

\(∑f_x=0==>P-μN-mgsinθ=0\)

\(=P-\frac{2}{3}(60√3)-12(10)sin30^o=0\)

\(=P-40√3-60=0\)

\(=P=40√3+60\)

\(∴P=129.28N\)

Correct Answer: A. -1

Explanation

\(\lim \limits_{x \to 3} \frac{x + 3}{x^{2} - x - 12} = \frac{3 + 3}{3^{2} - 3 - 12}\)

= \(\frac{6}{-6} = -1\)

Correct Answer: B. p = 5, q = -3

Explanation

If \(\begin{pmatrix} p+q & 1\\ 0 & p-q \end {pmatrix}\) = \(\begin{pmatrix} 2 & 1 \\ 0 & 8 \end{pmatrix}\)

Find the values of p and q

If \(\begin{pmatrix} p+q & 1\\ 0 & p-q \end {pmatrix}\) = \(\begin{pmatrix} 2 & 1 \\ 0 & 8 \end{pmatrix}\)

Find the values of p and q

p + q = 2

p - q = 8

\(\overline{\frac{2p}{2} = \frac{10}{2}}\)

p = 5

from p + q = 2

5 + q = 2

q = 2 - 5

= -3