Correct Answer: C. 29

Explanation

Using remainder theorem, the remainder when \(5x^{3} + 2x^{2} - 7x -5\) is divided by (x - 2) = f(2)

\(f(2) = 5(2^{3}) + 2(2^{2}) - 7(2) -5 = 40 + 8 - 14 - 5\)

= 29

Correct Answer: A. \(f^{-1} : x \to \frac{1+2x}{2-x}, x \neq 2\)

Explanation

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

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

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

\(xy - 2y = -1 - 2x \implies y = \frac{-1 - 2x}{x - 2}\)

\(f^{-1} : x \to \frac{1 + 2x}{2 - x} ; x \neq 2\)

Correct Answer: B. \(\text{1 and }\frac{-1}{3}\)

Explanation

For equal roots, \(b^{2} - 4ac = 0\)

From the equation, \(a = P, b = (P+1), c = P\)

\((P+1)^{2} - 4(P)(P) = P^{2} + 2P + 1 - 4P^{2} = 0\)

\(-3P^{2} + 2P + 1 = 0 \implies 3P^{2} - 2P - 1 = 0\)

\(3P^{2} - 3P + P - 1 = 0\)

\(3P(P - 1) + 1(P - 1) = 0\)

\(P = \text{1 or }\frac{-1}{3}\)

Correct Answer: C. 60°

Explanation

\(2\sin^{2}\theta = 1 + \cos \theta \implies 2(1 - \cos^{2}\theta) = 1 + \cos \theta\)

\(2 - 2\cos^{2}\theta = 1 + \cos \theta\)

\(2 - 2\cos^{2}\theta - 1 - \cos \theta = 0\)

\(2\cos^{2}\theta + \cos \theta - 1 = 0\)

\(2\cos^{2}\theta + 2\cos\theta - \cos \theta - 1 = 0 \implies 2\cos \theta(\cos \theta + 1) - 1(\cos \theta + 1) = 0\)

\((2\cos \theta - 1)(\cos \theta + 1) = 0 \implies \cos \theta = \frac{1}{2} \)

\(\theta = \cos^{-1} \frac{1}{2} = 60°\)

Explanation

(a) \(y = (2x + 3)^{7} + \frac{x - 1}{2x - 1}\)

\(\frac{\mathrm d y}{\mathrm d x} = 7(2x + 3)^{6}. 2 + \frac{(2x - 1).1 - (x - 1).2}{(2x - 1)^{2}}\)

= \(14(2x + 3)^{6} + \frac{2x - 1 - 2x + 2}{(2x - 1)^{2}}\)

= \(14(2x + 3)^{6} + \frac{1}{(2x - 1)^{2}}\)

At x = -1, \(\frac{\mathrm d y}{\mathrm d x} = 14(-1)^{6} + \frac{1}{(2(-1) - 1)^{2}}\)

= \(\frac{\mathrm d y}{\mathrm d x} = 14 + \frac{1}{9} = 14\frac{1}{9}\)

(b) \(\int_{1} ^{2} \frac{x - 1}{(x + 2)^{4}} \mathrm d x\)

\( u = x + 2 \implies \mathrm d u = \mathrm d x\)

If x = 1, u = 1 + 2 = 3; x = 2, u = 2 + 2 = 4.

\(\int_{1} ^{2} \frac{x - 1}{(x + 2)^{4}} \mathrm d x = \int_{3} ^{4} \frac{u - 3}{u^{4}}\)

= \(\int_{3} ^{4} (\frac{u}{u^{4}} - \frac{3}{u^{4}}) \mathrm d u\)

= \(\int_{3} ^{4} (u^{-3} - 3u^{-4}) \mathrm d u\)

= \([\frac{u^{-2}}{-2} + u^{-3}]_{3} ^{4}\)

= \([\frac{1}{u^{3}} - \frac{1}{2u^{2}}]_{3} ^{4}\)

= \((\frac{1}{4^{3}} - \frac{1}{2(4^{2})}) - (\frac{1}{3^{3}} - \frac{1}{2(3^{2})})\)

= \(\frac{1}{64} - \frac{1}{32} - \frac{1}{27} + \frac{1}{18}\)

= \(\frac{5}{1728}\).

Correct Answer: D. -2x + 3y - 6 = 0

Explanation

Recall:

Where 'a' and 'b' are the x and y intercept respectively.

2x-3y = -6

2x - 3y + 6 = 0 -------(1)

multiply through by -1

-2x + 3y - 6 = 0

Explanation

(a)(i) \(f(x) = x^{3} + px^{2} - 10x + q\)

\(x^{2} + x - 6 = 0 \implies x^{2} - 2x + 3x - 6 = 0\)

\(x(x - 2) + 3(x - 2) = 0 \implies (x - 2)(x + 3) = 0\)

\(x = 2 ; x = -3\)

\(f(-3) = (-3)^{3} + p(-3)^{2} - 10(-3) + q = 0\)

\(-27 + 9p + 30 + q = 0 \implies 9p + q = -3 .... (1)\)

\(f(2) = (2)^{3} + p(2)^{2} - 10(2) + q = 0\)

\(8 + 4p - 20 + q = 0 \implies 4p + q = 12 .... (2)\)

(2) - (1) :

\((4p + q) - (9p + q) = (12 - (-3)) \implies -5p = 15\)

\(p = \frac{15}{-5} = -3\)

Put p = -3 in (2), we have:

\(4(-3) + q = 12 \implies -12 + q = 12 \)

\(q = 12 + 12= 24\)

\(f(x) = x^{3} - 3x^{2} - 10x + 24\)

(ii) Using the method of long division,

\(\frac{x^{3} - 3x^{2} - 10x + 24}{x^{2} + x - 6}\)

we get the third factor = \(x - 4\).

(b) \(\frac{\mathrm d v}{\mathrm d t} = 2\frac{1}{2} cm^{3} s^{-1}\)

\(v = x^{3}\)

\(\frac{\mathrm d v}{\mathrm d x} = 3x^{2}\)

\(\frac{\mathrm d v}{\mathrm d t} = \frac{\mathrm d v}{\mathrm d x} \cdot \frac{\mathrm d x}{\mathrm d t}\)

\(\frac{5}{2} = 3x^{2} \cdot \frac{\mathrm d x}{\mathrm d t}\)

When x = 2cm, we have

\(\frac{5}{2} = 3(2)^{2} \cdot \frac{\mathrm d x}{\mathrm d t}\)

\(\frac{\mathrm d x}{\mathrm d t} = \frac{5}{2} \div 12 = \frac{5}{24} cm s^{-1}\)

Explanation

\(r = 4t i + (12 - 3t)j\)

(a) \(v = \frac{\mathrm d r}{\mathrm d t} = 4i - 3j\)

(b) When t = 0, r = 0i + 12j = 12j.

When t = 5, r = 20i - 3j.

Displacement = \(r_{t = 5} - r_{t = 0}\)

= \(20i - 3j - 12j\)

= \(20i - 15j\)

\(|Disp| = \sqrt{20^{2} + 15^{2}} = \sqrt{400 + 225}\)

= \(\sqrt{625} = 25m\)

Explanation

\(R_{1} = \begin{pmatrix} -20 \cos 60° \\ -20 \sin 60° \end{pmatrix} ; R_{2} = \begin{pmatrix} -10 \cos 60° \\ 10 \sin 60° \end{pmatrix} \)

\(R_{3} = \begin{pmatrix} 8\sqrt{2} \cos 45° \\ 8\sqrt{2} \sin 45° \end{pmatrix}\)

\(R = R_{1} + R_{2} + R_{3}\)

\(R = \begin{pmatrix} -20 \cos 60° \\ -20 \sin 60° \end{pmatrix} + \begin{pmatrix} -10 \cos 60° \\ 10 \sin 60° \end{pmatrix} + \begin{pmatrix} 8\sqrt{2} \cos 45° \\ 8\sqrt{2} \sin 45° \end{pmatrix}\)

\(R = \begin{pmatrix} -20 \times \frac{1}{2} \\ -20 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatrix} -10 \times \frac{1}{2} \\ 10 \times \frac{\sqrt{3}}{2} \end{pmatrix} + \begin{pmatrix} 8\sqrt{2} \times \frac{1}{\sqrt{2}} \\ 8\sqrt{2} \times \frac{1}{\sqrt{2}} \end{pmatrix}\)

= \(\begin{pmatrix} - 10 - 5 + 8 \\ -10\sqrt{3} + 5\sqrt{3} + 8 \end{pmatrix}\)

= \(\begin{pmatrix} -7 \\ 8 - 5\sqrt{3} \end{pmatrix} = \begin{pmatrix} -7 \\ -0.66 \end{pmatrix}\)

Let \(\theta\) be the direction from the horizontal.

\(\tan \theta = \frac{-0.66}{-7}\)

\(\tan \theta = 0.0943\)

\(\theta = \tan^{-1} (0.0943) = 5.39°\)

Correct Answer: A. 25

Explanation

Change in momentum = m (v - u)

= \(5 \times (\begin{pmatrix} 4 \\ 7 \end{pmatrix} - \begin{pmatrix} 1 \\ 3 \end{pmatrix})\)

= \(5 \times \begin{pmatrix} 3 \\ 4 \end{pmatrix}\)

= \(\begin{pmatrix} 15 \\ 20 \end{pmatrix}\)

\(|m(v - u)| = \sqrt{15^{2} + 20^{2}} = \sqrt{625} = 25\)