Explanation

If (x + 1) and (x - 2) are factors of the polynomial \(g(x)\), then \(g(-1) & g(2) = 0\)

i.e. \(g(-1) = (-1)^{4} + a(-1)^{3} + b(-1)^{2} - 16(-1) - 12 = 0\)

\(1 - a + b + 16 - 12 = 0 \implies a - b = 5 ... (1)\)

\(g(2) = (2)^{4} + a(2)^{3} + b(2)^{2} - 16(2) - 12 = 0\)

\(16 + 8a + 4b - 32 - 12 = 0 \implies 8a + 4b = 28 ... (2)\)

From (1), \(a = 5 + b\).

(2) becomes : \(8(5 + b) + 4b = 28 \implies 40 + 8b + 4b = 28\)

\(40 + 12b = 28 \implies 12b = -12\)

\(b = -1\)

\(a = 5 + b = 5 + (-1) = 4\)

\((a, b) = (4, -1)\)

Explanation

(a)(i) \((1 + x)^{4} = 1 + 4(x) + 6(x)^{2} + 4(x)^{3} + x^{4}\)

= \(1 + 4x + 6x^{2} + 4x^{3} + x^{4}\)

(ii) \((\frac{5}{4})^{4} = (1 + \frac{1}{4})^{4}\)

= \(1 + 4(\frac{1}{4}) + 6(\frac{1}{4})^{2} + 4(\frac{1}{4})^{3} + (\frac{1}{4})^{4}\)

= \(1 + 1 + 0.375 + 0.0625 + 0.00390625\)

= \(2 + 0.44140625\)

\(\approxeq 2.441\)

(b) \(T_{1} = a = 7\) (first term of an A.P)

\(T_{2} = a + d = 7 + d .... (1)\)

\(T_{5} = a + 4d = 7 + 4d .... (2)\)

\(T_{1} = 7 = a\) (first term of G.P)

\(T_{2} = ar = 7r\)

\(T_{3} = ar^{2} = 7r^{2}\)

\(\implies 7r = 7 + d\)

\(7r^{2} = 7 + 4d\)

\((1) \times 4 : 28r = 28 + 4d ... (3)\)

\((3) - (2) : 28r - 7r^{2} = 28 - 7 = 21\)

\(7r^{2} - 28r + 21 = 0\)

\(7r^{2} - 21r - 7r + 21 = 0\)

\(7r(r - 3) - 7(r - 3) = 0\)

\((7r - 7)(r - 3) = 0\)

\(\text{r = 1 or 3}\)

From (1), if r = 1,

\(7 = 7 + d \implies d = 0\) (This cannot be the case)

r = 3,

\(7(3) = 7 + d \implies d = 21 - 7 = 14\)

The common difference = 14.

Correct Answer: B. \(\frac{1}{2}\)

Explanation

x -5 -2 4 8

-5 25 10 -20 -40

-2 10 4 -8 -16

4 10 4 -8 -16

4 -20 -8 16 32

8 -40 -16 32 64

Pr(Two numbers is positive) = \(\frac{8}{16}\)

= \(\frac{1}{2}\)

Correct Answer: C. \(\frac{4}{3} seconds\)

Explanation

Given \(a(t) = 3t - 2\), \(v(t) = \int (a(t)) \mathrm {d} t\)

= \(\int (3t - 2) \mathrm {d} t = \frac{3}{2}t^{2} - 2t\)

\(\frac{3}{2}t^{2} - 2t = 0 \implies t(\frac{3}{2}t - 2) = 0\)

\(t = \text{0 or t} = \frac{3}{2}t - 2 = 0 \implies t = \frac{4}{3}\)

The time t = 0 was the starting point, The next time v = 0 m/s is at \(t = \frac{4}{3} seconds\).

Correct Answer: D. 2

Explanation

x?y = \(\sqrt{x+y - \frac{xy}{4}}\)

4?3 = \(\sqrt{4+3 - \frac{4*3}{4}}\)

= \(\sqrt{4+3-3}\)

= \(\sqrt{4}\)

= 2.

Explanation

(a) To find mean; = 67 + \(\frac{10}{100}\) = 67.1

(b) Substitute and find the standard deviation as \(\sqrt{\frac{7500}{100} - (\frac{10}{100})^2}\) = \(\sqrt{74.99}\)

= 8.6597

= 8.7, correct to one decimal place

Explanation

A.P ; \(S_{12} = 168 ; T_{3} = 7\)

\(S_{n} = \frac{n}{2}(2a + (n - 1) d)\)

\(T_{n} = a + (n - 1)d\)

\(168 = \frac{12}{2}(2a + (12 - 1)d \implies 168 = 6(2a + 11d)\)

\(168 = 12a + 66d \implies 84 = 6a + 33d ..... (1)\)

\(7 = a + 2d .... (2)\)

From (2), a = 7 - 2d. Put into (1), we have

\(84 = 6(7 - 2d) + 33d \)

\(84 = 42 - 12d + 33d \implies 84 - 42 = 21d\)

\(21d = 42 \implies d = 2\)

\(a = 7 - 2(2) = 7 - 4 = 3\)

\(a(\text{first term}) = 3 ; d(\text{common difference}) = 2\)

Correct Answer: B. 504

Correct Answer: A. (1, 0)

Explanation

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

\(\frac{\mathrm d y}{\mathrm d x} = 6x - 6x^{2} = 0 \implies 6x(1 - x) = 0\)

\(x = 0, 1\), when x = 1, y = 0.

when x = 1, y = 0.

The stationary point is (1, 0)

Correct Answer: B. \(\frac{4}{7}\)

Explanation

\(P(X?Y)=\frac{5}{8}\)

\(P(X?Y)=P(X)\times P(Y)\)

Since X and Y are independent events, the probability of their union (X ? Y) can be calculated as:

\(P(X?Y)=P(X)+P(Y)-P(X?Y)\)

\(=\frac{5}{8}=\frac{1}{8}+P(Y)-\frac{1}{8}\times P(Y)\)

\(=\frac{5}{8}-\frac{1}{8}=P(Y)-\frac{1}{8}\times P(Y)\)

\(=\frac{1}{2}=P(Y)(1-\frac{1}{8})\)

\(=\frac{1}{2}=P(Y)(\frac{7}{8})\)

\(=P(Y)=\frac{1}{2}÷\frac{7}{8}\)

\(∴P(Y)=\frac{1}{2}x\frac{8}{7}=\frac{4}{7}\)