Correct Answer: A. 432

Explanation

\((3x + 4)^{4} = ^{4}C_{0}(3x)^{0}(4)^{4} + ^{4}C_{1}(3x)^{1}(4)^{3} + ^{4}C_{2}(3x)^{2}(4)^{2} + ^{4}C_{3}(3x)^{3}(4)^{1} + ^{4}C_{4}(3x)^{4}(4)^{0}\)

\(x^{3} = ^{4}C_{3}(3x)^{3}(4) = \frac{4!}{3!1!} \times 3^{3} \times 4\)

= \(432x^{3}\)

Correct Answer: A. \({x : x \in R, x \neq 3}\)

Explanation

\(f(x) = \frac{x}{3 - x} \)

f(x) has a defined value except at x = 3 where the function is undefined.

Explanation

(a)(i) \(A(1 + r) + A(1 + r)^{2} + ... + A(1 + r)^{n}\)

\(S_{n} = \frac{a(r^{n} - 1)}{r - 1}\) (sum of terms in a G.P).

\(a = A(1 + r) ; r = (1 + r)\)

\(S_{n} = \(\frac{(A(1 + r))[(1 + r)^{n} - 1]}{(1 + r) - 1}\)

= \(\frac{(A(1 + r))[(1 + r)^{n} - 1]}{r}\)

(ii) Sum of the 6th to 10th term = \(S_{10} - S_{5}\)

\(a = 40(1 + 0.08) = 43.2 ; r = 0.08\)

\(S_{10} = \frac{43.2[(1 + 0.08)^{10} - 1]}{0.08} = 540[(1.08)^{10} - 1]\)

\(S_{5} = \frac{43.2[(1.08)^{5} - 1]}{0.08} = 540[(1.08)^{5} - 1]\)

\(S_{10} - S_{5} = 540[(1.08)^{10} - (1.08)^{5}]\)

= 371.8

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

\(\frac{\mathrm d y}{\mathrm d x} = -x^{-2}\)

When \(x = 2; y = \frac{1}{2}\), a point on the curve.

Gradient of tangent = \(-2^{-2} = -\frac{1}{4}\)

Equation : \(\frac{y - \frac{1}{2}}{x - 2} = -\frac{1}{4}\)

\(4y - 2 = 2 - x \implies x + 4y - 4 = 0\)

Explanation

\(|AB| = \sqrt{(-1 - 3)^{2} + (2 - 5)^{2}}\)

= \(\sqrt{16 + 9} = \sqrt{25}\)

= 5 units.

\(|AC| = \sqrt{(4 - 1)^{2} + (8 - 2)^{2}}\)

= \(\sqrt{25 + 36}\)

= \(\sqrt{61}\) units.

\(|BC| = \sqrt{(4 - 3)^{2} + (8 - 5)^{2}}\)

= \(\sqrt{1 + 9} \)

= \(\sqrt{10}\) units.

From the cosine rule,

\((\sqrt{61})^{2} = 5^{2} + (\sqrt{10})^{2} - 2(5)(\sqrt{10}) \cos <PBQ\)

\(\cos < PBQ = \frac{25 + 10 - 61}{2(5)(\sqrt{10})}\)

\(\cos < PBQ = \frac{-26}{10\sqrt{10}} = -0.4747\)

\(< PBQ = 118°, \beta = < PBQ = 118°\)

\(< BPS = \alpha = 180° - 118° = 62°\)

The magnitude of the resultant

\(|R|^{2} = 5^{2} + (3\sqrt{10})^{2} - 2(5)(3\sqrt{10}) \cos 62\)

= \(25 + 90 - (30\sqrt{10})(0.4695)\)

= \(115 - 44.538\)

= \(70.462\)

\(|R| = \sqrt{70.462} = 8.39N\)

Let \(\delta\) be the angle the resultant makes with the 5N force.

\(PS^{2} = BS^{2} + BP^{2} - 2(BS)(BP) \cos PBS\)

\((3\sqrt{10})^{2} = (\sqrt{70.462})^{2} + 5^{2} - 2(8.39)(5) \cos \delta\)

\(90 = 70.462 + 25 - 83.9 \cos \delta\)

\(\cos \delta = \frac{70.462 + 25 - 90}{83.9}\)

\(\cos \delta = 0.0651\)

\(\delta = \cos^{-1} (0.0651) = 86.27°\)

\(\approxeq 86°\).

The resultant makes an angle of 86° with the 5N force.

(b)(i) \(u = 0 ms^{-1} ; v = 20 ms^{-1} ; s = 8 m ; a = ?\)

\(v^{2} = u^{2} + 2as \)

\(20^{2} = 0^{2} + 2(8)a \implies 400 = 16a\)

\( a = 25 ms^{-2}\)

(ii) \(t = ?\)

\(s = ut + \frac{1}{2} at^{2}\)

\(40 = 0 + \frac{1}{2} (25)(t^{2})\)

\(80 = 25t^{2}\)

\(t^{2} = \frac{80}{25} = 3.2\)

\(t = \sqrt{3.2} = 1.79 s\)

It takes 1.79s to cover a distance of 40m.

Explanation

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

\(F_{1} = (10 N, 090°) = \begin{pmatrix} 10 \cos 90 \\ 10 \sin 90 \end{pmatrix} = \begin{pmatrix} 0 \\ 10 \end{pmatrix}\)

\(F_{2} = (20 N, 210°) = \begin{pmatrix} 20 \cos 210 \\ 20 \sin 210 \end{pmatrix} = \begin{pmatrix} -10\sqrt{3} \\ - 10 \end{pmatrix}\)

\(F_{3} = (4 N, 330°) = \begin{pmatrix} 4 \cos 330 \\ 4 \sin 330 \end{pmatrix} = \begin{pmatrix} 2\sqrt{3} \\ -2 \end{pmatrix}\)

\(F_{R} = \begin{pmatrix} 0 \\ 10 \end{pmatrix} + \begin{pmatrix} -10\sqrt{3} \\ -10 \end{pmatrix} + \begin{pmatrix} 2\sqrt{3} \\ -2 \end{pmatrix}\)

= \(\begin{pmatrix} -8\sqrt{3} \\ -2 \end{pmatrix}\)

\(|F_{R}| = \sqrt{(8\sqrt{3})^{2} + (-2)^{2}}\)

= \(\sqrt{192 + 4} \)

= \(\sqrt{196} = 14 N\)

Explanation

\(2^{(2y + 2)} - 9(2^{y}) = -2\)

= \((2^{2})(2^{2y}) - 9(2^{y}) = -2\)

= \(4(2^{y})^{2} - 9(2^{y}) = -2\)

Let \(2^{y} = x\)

\(\implies 4x^{2} - 9x = -2\)

\(4x^{2} - 9x + 2 = 0\)

\(4x^{2} - 8x - x + 2 = 0\)

\(4x(x - 2) - 1(x - 2) = 0\)

\((4x - 1)(x - 2) = 0 \implies 4x = 1; x = \frac{1}{4}\)

or \(x = 2\)

If \(x = 2^{y} = \frac{1}{4} = 2^{-2} \implies y = -2\)

If \(x = 2^{y} = 2^{1} \implies y = 1\)

\(\therefore y = -2 ; y = 1\)

Correct Answer: B. 90°

Explanation

\(a . b = |a||b|\cos \theta\)

\(\cos \theta = \frac{a . b}{|a||b|}\)

= \( \frac{(5i + 3j).(3i - 5j)}{(\sqrt{5^2 + 3^2})(\sqrt{3^{2} + (-5)^{2}})}\)

= \(\frac{0}{34} = 0\)

\(\theta = \cos^{-1} 0 = 90°\)

Correct Answer: A. 840

Explanation

= \(^{8}C_{5} \times ^{6}C_{4}\)

=\(56 \times 15\)

= 840

Correct Answer: B. \(7\sqrt{2} - \frac{17\sqrt{3}}{3}\)

Explanation

Given \(\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}}\),

first, we rationalise by multiplying through with \(2\sqrt{3} - 3\sqrt{2}\) (the inverse of the denominator).

\((\frac{8 - 3\sqrt{6}}{2\sqrt{3} + 3\sqrt{2}})(\frac{2\sqrt{3} - 3\sqrt{2}}{2\sqrt{3} - 3\sqrt{2}})\)

= \(\frac{16\sqrt{3} - 24\sqrt{2} - 18\sqrt{2} + 18\sqrt{3}}{4(3) - 6\sqrt{6} + 6\sqrt{6} - 9(2)}\)

= \(\frac{34\sqrt{3} - 42\sqrt{2}}{-6} = 7\sqrt{2} - \frac{17\sqrt{3}}{3}\)

Correct Answer: C. (x - 2)(x + 5)(y + 1)

Explanation

\(x^{2} + x^{2}y + 3x - 10y + 3xy -10\)

= \(x^{2} + x^{2}y + 3x + 3xy - 10y - 10 = x^{2}(1 + y) + 3x(1 + y) - 10(y + 1)\)

= \((x^{2} + 3x - 10)(y + 1)\)

= \((x^{2} + 3x - 10) = x^{2} - 2x + 5x - 10\)

= \(x(x - 2) + 5(x - 2) = (x - 2)(x +5)\)

\(\therefore x^{2} + x^{2}y + 3x - 10y + 3xy -10 = (x - 2)(x + 5)(y + 1)\).