Correct Answer: B. 3

Explanation

\(\frac{\sqrt{3}}{\sqrt{3} - 1} + \frac{\sqrt{3}}{\sqrt{3} + 1}\)

= \(\frac{\sqrt{3}(\sqrt{3} + 1) + \sqrt{3}(\sqrt{3} - 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)}\)

= \(\frac{3 + \sqrt{3} + 3 - \sqrt{3}}{3 + \sqrt{3} - \sqrt{3} - 1}\)

= \(\frac{6}{2} = 3\)

Correct Answer: D. \(\frac{x^3}{3} - 2x - \frac{1}{x} + c\)

Explanation

\((x - \frac{1}{x})^{2} = x^2 - 2 + \frac{1}{x^2}\)

\(\int (x^2 + \frac{1}{x^2} - 2) \mathrm {d} x\)

= \(\int (x^2 + x^{-2} - 2) \mathrm {d} x\)

= \(\frac{x^3}{3} - 2x - \frac{1}{x}\)

Correct Answer: C. 15

Explanation

f(x) = x\(^3\) + px + q

f(3) 3\(^2\) + 3pf + q = 6

f\(^{1}\)(x) = 2x + p

f\(^1\)(3) = 2(3) + p = 0

p = -6

from 9 + 3p + q = 6

9 + 3(-6) + q = 6

9 - 18 + q = 6

-9 + q = 6

Q = 6 + 9

= 15

Correct Answer: B. 11.7 N

Explanation

\(F = F \cos \theta i + F \sin \theta j\) (Resolving F into its components)

\(F_{1} = (10 N, 090°) = 10 \cos 90 i + 10 \sin 90 j\)

= \(10 j\)

\(F_{2} = (6 N, 180°) = 6 \cos 180 i + 6 \sin 180 j\)

= \(-6 i\)

\(R = - 6 i + 10 j\)

\(|R| = \sqrt{(-6)^{2} + (10)^{2}}\)

= \(\sqrt{136}\)

= 11.7 N

Correct Answer: C. \(\frac{1}{8} (2x + 1)^4 + k\)

Explanation

\(\int(2x + 1)^3 dx\)

Using substitution method, Let \(u = 2x + 1\)

\(\frac{du}{dx}=2==>du=2dx==>dx=\frac{du}{2}\)

=\(\int\frac{u^3}{2} du = \frac{1}{2}\int u^3 du\)

=\(\frac{1}{2}(\frac{u^4}{4})=\frac{u^4}{8}\)

\(\therefore\frac{1}{8} (2x + 1)^4 + k\)

Correct Answer: C. \(a = 9b^{3}\)

Explanation

\(\log_{3}a - 2 = 3\log_{3}b\)

Using the laws of logarithm, we know that \( 2 = 2\log_{3}3 = \log_{3}3^{2}\)

\(\therefore \log_{3}a - \log_{3}3^{2} = \log_{3}b^{3}\)

= \(\log_{3}(\frac{a}{3^{2}}) = \log_{3}b^{3} \implies \frac{a}{9} = b^{3}\)

\(\implies a = 9b^{3}\)

Correct Answer: B. \(-\frac{2}{5}\)

Explanation

\(S_{\infty} = \frac{a}{1 - r}\) (Sum to infinity of a GP)

\(250 = \frac{350}{1 - r} \implies 250(1 - r) = 350\)

\(350 = 250 - 250r \implies 350 - 250 = -250r\)

\(250r = -100 \implies r = \frac{-100}{250} = -\frac{2}{5}\)

Correct Answer: B. \(x \leq 2\)

Explanation

\(f : x \to \sqrt{4 -2x}\) defined on the set of real numbers,R, which has range from \((-\infty, \infty)\) but because of the root sign, it is defined from \([0, \infty)\).

This is because the root of numbers only has real number values from 0 and upwards.

\(\sqrt{4-2x} \geq 0 \implies 4-2x \geq 0\)

\(-2x \geq -4; x \leq 2\)

Correct Answer: B. 17.55j - 13.78i

Explanation

Converting the forces to their rectangular forms

F = (10N, 060º)

Fx = 10cos60 = 5i

fy = 10sin60 = 8.66j

F = 5i + 8.66j

P = (15N, 120º)

Px = 15cos120 = -7.5i

py = 15sin120 = 12.99j

P = -7.5i + 12.99j

Q = (12N, 200º)

Qx = 12cos200 = -11.28i

Qy = 12sin200 = -4.1j

Q = -11.28i -4.1j

The resultant force = F + P + Q

R = 5i + 8.66j + (-7.5i + 12.99j) + (-11.28i -4.1j)

R = -13.78i + 17.55j

Correct Answer: D. 0.97

Explanation

(0.9 x 0.7) + (0.1 x 0.7) = (0.9 x 0.3) = 0.63 + 0.07 + 0.27 = 0.97