Correct Answer: B. \(8\sqrt{3}\)

Explanation

\((1 + 2\sqrt{3})^{2} = 1 + 4\sqrt{3} + 12 = 13 + 4\sqrt{3}\)

\((1 - 2\sqrt{3})^{2} = 1 - 4\sqrt{3} + 12 = 13 - 4\sqrt{3}\)

\(13 + 4\sqrt{3} - (13 - 4\sqrt{3}) = 13 + 4\sqrt{3} - 13 + 4\sqrt{3}\)

= \(8\sqrt{3}\)

Correct Answer: C. \(\frac{13}{40}\)

Correct Answer: A. 3.6 cm

Explanation

Length of arc given angle in radians = \(r \theta\)

= \(4 \times 0.9 = 3.6 cm\)

Correct Answer: D. \(\frac{-7}{6}\)

Explanation

Expanding \((x+1)(x-2) = x^{2} - 2x + x - 2 = x^{2} - x - 2\)

\(\int_{-1}^{0} (x^{2} - x - 2) \mathrm{d}x = [\frac{x^{3}}{3} - \frac{x^{2}}{2} - 2x]_{-1}^{0}\)

= \([\frac{0}{3} - \frac{0}{2} - 2\times0 - (\frac{-1^{3}}{3} - \frac{-1^{2}}{2} - 2\times-1)]\)

= \(0 + \frac{1}{3} + \frac{1}{2} - 2 = \frac{-7}{6}\)

Note: This can also be solved using integration by parts.

\(\int uv \mathrm{d}x = u\int v \mathrm{d}x - \int u'(\int v \mathrm{d}x)\mathrm{d}x\).

Correct Answer: B. \(f(x) = x^{3} + \frac{1}{x^{4}} + 2\)

Explanation

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - \frac{4}{x^5} = 3x^{2} - 4x^{-5}\)

\(y = \int (3x^{2} - 4x^{-5}) \mathrm {d} x \)

\(y = x^{3} + \frac{1}{x^{4}} + c\)

f(1) = 4; \(4 = 1^{3} + \frac{1}{1^{4}} + c \implies 4 = 2 + c\)

\(c = 2\)

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

Correct Answer: B. 192

Explanation

\(S_{n} = \frac{n}{2}(2a + (n - 1)d)\) ( for an arithmetic progression)

\(S_{3} = 18 = \frac{3}{2}(2(4) + (3 - 1) d) \)

\(18 = \frac{3}{2} (8 + 2d)\)

\(18 = 12 + 3d \implies 3d = 6\)

\(d = 2\)

\(\therefore T_{1} = 4 \implies T_{2} = 4 + 2 = 6; T_{3} = 6 + 2 = 8\)

Their product = \(4 \times 6 \times 8 = 192\)

Explanation

\(\over{PS}\) = \(\over{OS}\) - \(\over{OP}\) = (\(^{-2}_3\)) - (\(^{-1}_4\)) = (\(^{-1}_{-1}\))

Similarly \(\over{QR}\) = \(\over{OR}\) - \(\over{OQ}\) = (\(^x_y\)) - (\(^2_3\)) = (\(^{x - 2}_{y - 3}\))

Since \(\over{PS}\) = \(\over{QR}\), it flows that (\(^{x - 2}_{y - 3}\)) = (\(^{-1}_{-1}\)). Therefore, x - 2 = -1 which solved will yield x = 1.

Also, y - 3 = -1 which implies that y = 2

(b)(i), Since the particle starts from rest, the initial velocity (u) = 0 and substituting for final velocity (v) and distance (s) into the formula v\(^2\) = u\(^2\) + 2as to obtain 20\(^2\) = 0 + 2a(8) and when simplifies and solved for acceleration (a), to get a = 25ms\(^{-2}\)

(b)(ii), S = ut + \(\frac{1}{2}\)at\(^2\), where s = 40, u = 0 and a = 25

Substituting these values into the equation we obtain 40 = 0 + \(\frac{1}{2}\) x 25t\(^2\) and solving for t gave t = 1.79 seconds.

Explanation

(a)(i) If repetition is allowed, the numbers can be formed as follows:

Number of ways = \(5^{5} = 3125\)

(ii) If no repetition = \(5!\) ways

= 120 ways

(b) No repetition and the numbers odd

The 5th position is filled by 5 and 7 for it to be odd ie 2 ways

Ways = \(4 \times 3 \times 2 \times 1 \times 2 = 48\)

Correct Answer: D. 9

Explanation

4x\(^2\) - 12x + k;

a = 4, b = -12, c = k.

For perfect square b\(^2\) = 4ac

(-12)\(^2\) = 4 * 4 * k;

144 = 16k

k = 144\16

k = 9

Correct Answer: B. \(\frac{1}{\sqrt{29}}(-2i + 5j)\)

Explanation

The unit vector, \(\hat{n}\) is given by \(\hat{n} = \frac{\overrightarrow{r}}{|\overrightarrow{r}|}\)

= \(\frac{(-2i + 5j)}{\sqrt{(-2)^{2} + 5^{2}}} = \frac{(-2i + 5j)}{\sqrt{29}}\)

= \(\frac{1}{\sqrt{29}}(-2i + 5j)\)