Correct Answer: C. \(\frac{1}{33}\)(6 + ?3)

Explanation

( \(\frac{1}{2 - √3}\) + \(\frac{2}{2 + √3}\) )\(^{-1}\)

= \(\frac{2 + √3 + 2(2 - √3)}{(2 - √3)(2 + √3)}\)

= 2 + √3 + 4 - 2√3

= (6 - √3)\(^{-1}\)

= \(\frac{1}{6 - √3}\)

= \(\frac{6 + √3}{6 - √3 * 6 + √3}\)

= \(\frac{6 + √3}{33}\)

Correct Answer: D. \(\frac{10x}{(1-x^2)^6}\)

Explanation

\(y = \frac{1} {(1 - x^2)^5} = (1-x^2)^{-5}\)

Let u = \(1 - x^2; y = u ^{-5}\)

\(\frac{du}{ dx}=-2x;\frac{dy}{ du}=-5u^{-6}\)

Using chain rule:

\(\frac{dy}{dx}=\frac{dy}{du}\times\frac{du}{ dx}\)

\(\frac{dy}{dx}=-2x\times-5u^{-6}\)

\(\frac{dy}{dx}=-2x\times-5(1 - x^2)^{-6}\)

\(\therefore\frac{dy}{dx} = 10x(1-x^2)^{-6}=\frac{10x}{(1-x^2)^6}\)

Correct Answer: C. 3.50

Explanation

\(Mean (\bar {x}) = \frac{\sum fx}{\sum f} = \frac{75}{10} = 7.5\)

\(SD = \sqrt{\frac{\sum fd^{2}}{\sum f}}\)

= \(\sqrt{\frac{122.5}{10}}\)

= \(\sqrt{12.25} = 3.50\)

Correct Answer: C. k = 1, p = -2

Explanation

Un = kn\(^2\) + pn,

U\(_1\) = -1,

U\(_5\) = 15,

when n = 1

U1 = k(1)\(^2\)+ p(1) = -1

k + p = -1 --------eqn1

when n = 5

U\(_5\) = k(5)\(^2\)+ p(5) = 15

25k + 5p = 15 --------eqn2

multiply eqn1 by 5 and eqn2 by 1

5k + 5p = -5 -------eqn3

25k + 5p = 15 -------eqn4

eqn4 - eqn3

20k = 20

k = 1

sub for k in eqn1

1 + p = -1

p = -1 -1 = -2

Explanation

\(p(M) = 0.9 = \frac{9}{10} ; p(M') = 0.1 = \frac{1}{10}\)

\(p(P) = 0.75 = \frac{3}{4} ; p(P') = 0.25 = \frac{1}{4}\)

\(p(B) = 0.8 = \frac{4}{5} ; p(B') = 0.2 = \frac{1}{5} \)

p(at least two subjects) = p(Maths and Physics not Biology) or p(Maths and Biology not Physics) or p(Physics and Biology not Maths) or p(all three subjects)

= \((\frac{9}{10} \times \frac{3}{4} \times \frac{1}{5}) + (\frac{9}{10} \times \frac{4}{5} \times \frac{1}{4}) + (\frac{3}{4} \times \frac{4}{5} \times \frac{1}{10}) + (\frac{9}{10} \times \frac{3}{4} \times \frac{4}{5})\)

= \(\frac{27}{200} + \frac{36}{200} + \frac{12}{200} + \frac{108}{200}\)

= \(\frac{183}{200}\)

= \(0.915\)

Explanation

(a)

(b) Modal class: This is the class boundary with the highest frequency.

In this case, it is 139.5 and 149.5.

From the histogram, the modal income is 3 +139.5

= 142.5 or N142500 (in thousands of naira)

Correct Answer: B. \(3i + 3\sqrt{3}j\)

Explanation

\(V = V\cos \theta i + V\sin \theta j\)

\(6N = 6 \cos 60 i + 6 \sin 60 j\)

= \(3i + 3\sqrt{3} j\)

Explanation

(a) \(2x^{2} + 5x - 6 = 0\)

\(a = 2 ; b = 5 ; c = -6\)

\(\alpha \beta = \frac{c}{a} = \frac{-6}{2} = -3\)

\(\alpha + \beta = \frac{-b}{a} = \frac{-5}{2}\)

Product of new roots : \((\alpha - 2)(\beta - 2) = \alpha \beta - 2(\alpha + \beta) + 4\)

= \(-3 - 2(-\frac{5}{2}) + 4 = 6\)

Sum of new roots : \((\alpha - 2) + (\beta - 2) = \alpha + \beta - 4\)

= \(\frac{-5}{2} - 4 = -\frac{13}{2}\)

Equation : \(x^{2} - (sum) x + product = 0\)

\(x^{2} - (-\frac{13}{2} x) + 6 = 0\)

\(x^{2} + \frac{13}{2} x + 6 = 0 \implies 2x^{2} + 13x + 12 = 0\)

(b) \(\int_{0} ^{k} (x^{2} - 2x) \mathrm {d} x = \frac{4}{3}\)

\([\frac{x^{3}}{3} - x^{2}]_{0} ^{k} = \frac{4}{3}\)

\(\frac{k^{3}}{3} - k^{2} = \frac{4}{3}\)

\(k^{3} - 3k^{2} = 4 \implies k^{3} - 3k^{2} - 4 = 0\)

Correct Answer: C. 81°

Explanation

\(\overrightarrow{x} . \overrightarrow{y} = |\overrightarrow{x}||\overrightarrow{y}|\cos\theta\)

\(\overrightarrow{x} . \overrightarrow{y} = (i - 3j) . (6i + j) = 6 - 3 = 3\)

\(|\overrightarrow{x}| = \sqrt{1^{2} + (-3)^{2}} = \sqrt{10}\)

\(|\overrightarrow{y}| = \sqrt{6^{2} + 1^{2}} = \sqrt{37}\)

\(\therefore 3 = (\sqrt{10})(\sqrt{37})\cos \theta\)

\(\cos\theta = \frac{3}{\sqrt{370}} = 0.1559\)

\(\theta = \cos^{-1} 0.1559 \approxeq 81°\)

Correct Answer: D. 250

Explanation

The value of the sine of an angle is negative in the third and fourth quadrant. Hence options A and B are not the options.

\(\sin 250 = -\sin (250 - 180) = - \sin 70\)

\(\sin 290 = - \sin (360 - 290) = - \sin 70\)