Correct Answer: B. 11

Explanation

g(-3) = 5 - 2(-3)

= 5 + 6 = 11

Correct Answer: A. \(\frac{-1}{2}, 8\)

Explanation

U1 = x - 4

U2 = x + 2

U3 = 3x + 1

\(\frac{u_2}{u_1} = \frac{u_3}{u_2}\)

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

(x+2)(x+2) = (x-4)(3x+1)

x\(^2\) + 4x + 4 = 3x2 - 11x - 4

collecting like terms

2x\(^2\) - 15x - 8 =0

2x\(^2\) + x - 16x - 8 = 0

x(2x + 1) - 8(2x + 1) = 0

(x-8)(2x+1) = 0

x = (\(\frac{-1}{2}, 8\))

Explanation

(a) For P, Q and R to lie on a straight line,the gradient of any two of PQ, PR and QR must be equal.

Gradient of PR : \(\frac{6 - 1}{2 - 11} = -\frac{5}{9}\)

Gradient of QR : \(\frac{6 - \frac{13}{3}}{2 - 5} = \frac{\frac{5}{3}}{-3}\)

= \(- \frac{5}{9}\)

The gradients are the same therefore P, Q and R are on the same line.

(b) \(\overrightarrow{PQ} = (5i + \frac{13}{3}j) - (11i + j)\)

= \(-6i + \frac{10}{3}j\)

\(|\overrightarrow{PQ}| = \sqrt{(-6)^{2} + (\frac{10}{3})^{2}}\)

= \(\sqrt{36 + \frac{100}{9}}\)

= \(\sqrt{\frac{424}{9}}\)

\(\overrightarrow{QR} = (2i + 6j) - (5i + \frac{13}{3}j)\)

= \(-3i + \frac{5}{3}j\)

\(|\overrightarrow{QR}| = \sqrt{(-3)^{2} + (\frac{5}{3})^{2}}\)

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

= \(\sqrt{\frac{106}{9}}\)

\(\frac{|\overrightarrow{PQ}|}{|\overrightarrow{QR}|} = \frac{\sqrt{\frac{424}{9}}}{\sqrt{\frac{106}{9}}}\)

= \(\frac{4}{1}\)

\(\therefore |\overrightarrow{PQ}| : |\overrightarrow{QR}| = 4 : 1\)

Explanation

(a) 10% of bulbs are faulty, 90% are not faulty.

P(a bulb is flawless) = 90% = 0.9

(i) p(first sample is flawless and accepted) = \((0.9)^{5} = 0.59049 \approxeq 0.590\)

(ii) P(first sample of 5 bulbs contains at least one faulty bulb) = 1 - 0.59049 = 0.40951\)

Probability that the second sample of size 5 contains flawless bulbs = \((0.9)^{5} = 0.59049\)

\(\therefore\) probability that the first batch is accepted = \(0.4095 \times 0.5905 = 0.2418\)

(b) 3 black, 12 non-black. Total = 15.

p(black marble) = \(\frac{3}{15} = p = \frac{1}{5}\)

p(non- black marble) = \(\frac{12}{15} = q = \frac{4}{5}\)

\((p + q)^{10} = p^{10} + 10p^{9} q + 45p^{8} q^{2} + 120p^{7} q^{3} + 210p^{6} q^{4} + 252p^{5} q^{5} + 210p^{4} q^{6} + 120p^{3} q^{7} + 45p^{2} q^{8} + 10p q^{9} + q^{10}\)

(i) p(no black) = \(q^{10} = (\frac{4}{5})^{10} = \frac{1048576}{9765625}\)

(ii) p(black marble at most 3 times) = p(0 black marble) + p(1 black marble) + p(2 black marbles) + p(3 black marbles)

= \(\frac{1048576}{9765625} + 10(\frac{1}{5})(\frac{4}{5})^{9} + 45(\frac{1}{5})^{2} (\frac{4}{5})^{8} + 120(\frac{1}{5})^{3} (\frac{4}{5})^{7}\)

Correct Answer: C. \(k < \frac{25}{8}\)

Correct Answer: B. -4

Explanation

\((\sqrt{3} - 5\sqrt{2})(\sqrt{3} + \sqrt{2}) = 3 + \sqrt{6} - 5\sqrt{6} - 10\)

= \(-7 - 4\sqrt{6} = p + q\sqrt{6}\)

\(\therefore q = -4\)

Correct Answer: D. x\(^2\) - 6x + 7

Explanation

dy/dx = 2x - 6

y = ∫ 2x - 6

y = \(\frac{2x^2}{2} - 6 + c\)

y = x\(^2\) - 6x + c

passes through (1,2)

2 = 1\(^2\) - 6(1) + c

2 = 1 - 6 + c

c = 7

y = x\(^2\) - 6x + c

y = x\(^2\) - 6x + 7

Correct Answer: C. 0.8

Explanation

\(\begin{pmatrix} 2 & 1 \\ 4 & 3 \end{pmatrix}\)\(\begin{pmatrix} 5 \\ 4 \end{pmatrix}\) = k\(\begin{pmatrix} 17.5 \\ 40.0 \end{pmatrix}\)

\(\begin{pmatrix} 10 + 4 \\ 20 + 12\end{pmatrix}\) = \(\begin{pmatrix} 14 \\ 32\end{pmatrix}\) = k\(\begin{pmatrix} 17.5 \\ 40.0 \end{pmatrix}\)

k = \(\frac{14}{17.5} = \frac{32}{40} = 0.8\)

Correct Answer: B. \(6(-\sqrt{3} i - j)\)

Explanation

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

\(r = (12, 210°) = 12 \cos 210 i + 12 \sin 210 j\)

= \(- 6\sqrt{3} i - 6j\)

= \(6(-\sqrt{3} i - j)\)

Explanation

(a)(i) \(f(x) = \frac{2x - 3}{(x^{2} - 1)(x + 2)}\)

f(x) is undefined at \((x^{2} - 1)(x + 2) = 0\).

Either \((x^{2} - 1) = \text{0 or (}x + 2) = 0\)

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

\((x^{2} - 1) = 0 \implies x^{2} = 1 \)

\(x = \pm 1\)

At the points x = -1, 1 or 2, f(x) is undefined.

(ii) \(\frac{2x - 3}{(x^{2} - 1)(x + 2)} = \frac{A}{x + 1} + \frac{B}{x - 1} + \frac{C}{x + 2}\)

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

When x = -1,

\(-2A = 2(-1) - 3 \implies -2A = -5\)

\(A = \frac{5}{2}\)

When x = 1,

\(6B = 2(1) - 3 = -1\)

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

When x = -2,

\(3C = 2(-2) - 3 = -7\)

\(C = \frac{-7}{3}\)

\(\frac{2x - 3}{(x^{2} - 1)(x + 2)} = \frac{5}{2(x + 1)} - \frac{1}{6(x - 1)} - \frac{7}{3(x + 2)}\)

(c) Equation of circle : \((x + 3)^{2} + (y - 1)^{2} = r^{2}\)

Circle passes through (3,1) :

\(\therefore (3 + 3)^{2} + (1 - 1)^{2} = r^{2}\)

\(6^{2} = r^{2} \implies r = 6\)

Equation becomes :

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

\(x^{2} + 6x + 9 + y^{2} - 2y + 1 - 36 = 0\)

\(x^{2} + 6x + y^{2} - 2y - 26 = 0\)

\(x^{2} + y^{2} + 6x - 2y - 26 = 0\)