Correct Answer: C. 21

Explanation

Let the sum of the 8 numbers = y.

\(\frac{y}{8} = 20 \implies y = 20 \times 8 = 160\)

\(160 - 17 = 143 (\text{the sum of the other 7 numbers})\)

\(mean = \frac{143 + 25}{8} = \frac{168}{8}\)

= 21

Correct Answer: A. ±12

Explanation

The general form of a quadratic equation is:

\(x^2\)-(sum of roots) = 0

For equal roots it's:\(x^2 - 2(α)+(α)^2=0\)

\(3x^2+px+12=0\)

Divide through by 3

= \(x^2+\frac{p}{3}x+4=0\)

=\(x^2-(-\frac{p}{3})x+4=0\)

\(So,α^2=4\)

= α = √4 = ±2

Also,-\(\frac{p}{3}= 2α\)

When α = 2

-\(\frac{p}{3} = 2(2)=4\)

=p=-12

When α =-2

-\(\frac{p}{3}=2(-2)=-4\)

= p = 12

∴ values of p = ±12

Correct Answer: C. \(\frac{-33}{65}\)

Explanation

From Pythagoras' theorem

when sin x = \(\frac{12}{13}\), cos x = \(\frac{5}{13}\)

Using cos (x + y) = cosx cosy - sinxsiny

\(\frac{5}{13}\) * \(\frac{3}{5}\) - \(\frac{12}{13}\) * \(\frac{4}{5}\)

= \(\frac{-33}{65}\)

Correct Answer: C. 14m

Explanation

\(a = \frac{v - u}{t} = \frac{9 - 5}{2} = 2 ms^{-2}\)

\(s = ut + \frac{1}{2}at^{2} = 5(2) + \frac{1}{2}(2 \times 2^{2})\)

= \(14m\)

Correct Answer: C. 15

Explanation

6C\(_2 = \frac{6!}{[6-2]![2!]}\)

\(\frac{6*5*4!}{4!*2!}\)

= \(\frac{6*5}{2}\)

= 15

Explanation

(a) \(\frac{1 + 4 + k + (k + 4) + 11}{5} = k + 1\)

\(\frac{2k + 20}{5} = k + 1\)

\(2k + 20 = 5(k + 1) \implies 2k + 20 = 5k + 5\)

\(20 - 5 = 5k - 2k \implies 15 = 3k\)

\(k = 5\)

(b) Mean \(\bar{x} = 6\)

\(\sigma = \sqrt{\frac{\sum fx^{2}}{\sum f} - (\frac{\sum fx}{\sum f})^{2}}\)

= \(\sqrt{\frac{244}{5} - (\frac{30}{5})^{2}}\)

= \(\sqrt{48.8 - 36}\)

= \(\sqrt{12.8} = 3.58\)

Correct Answer: D. 3

Explanation

\((\sqrt{x} + 1) * (\sqrt{x} - 1) = 4 \implies \frac{\sqrt{x} + 1}{\sqrt{x} - 1} + \frac{\sqrt{x} - 1}{\sqrt{x} + 1} = 4\)

\(\frac{(\sqrt{x} + 1)(\sqrt{x} + 1) + (\sqrt{x} - 1)(\sqrt{x} - 1)}{(\sqrt{x} - 1)(\sqrt{x} + 1)}\)

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

\(2x + 2 = 4x - 4 \therefore 4x - 2x = 2x = 2 + 4= 6\)

\(x = 3\)

Correct Answer: C. 3 + √6

Explanation

\(\frac{3}{3 - √3}\)

Rationalize

\(= \frac{3}{3 - √6} \times \frac{3 + √6}{3 + √6}\)

\(=\frac{3(3 + √6)}{(3 - √6)(3 + √6)}\)

\(=\frac{3(3 + √6)}{9 - 6}=\frac{3(3 + √6)}{3}\)

∴ 3 + √6

Explanation

(a) \(T_{n} = 4T_{n - 1} - 3\)

\(T_{3} = 4T_{2} - 3 ; T_{2} = 4T_{1} - 3\)

\(2T_{3} = 5T_{2} \implies T_{3} = \frac{5}{2} T_{2}\)

\(\frac{5}{2} T_{2} = 4T_{2} - 3 \implies 3 = \frac{3}{2} T_{2}\)

\(T_{2} = \frac{3}{\frac{3}{2}} = 2\)

\(2 = 4T_{1} - 3 \implies 5 = 4T_{1}\)

\(T_{1} = \frac{5}{4}\)

\(2T_{3} = 5(2) = 10\)

\(T_{3} = 5\)

The first three terms of the sequence are \(\frac{5}{4}, 2, 5\).

(b) \(\begin{pmatrix} 2 & 0 & 1 \\ 5 & -3 & 1 \\ 0 & 4 & 6 \end{pmatrix} \begin{pmatrix} 1 \\ m \\ r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\)

\(\begin{pmatrix} 2(1) + r \\ 5(1) - 3m + r \\ 4m + 6r \end{pmatrix} = \begin{pmatrix} k \\ 2 \\ 26 \end{pmatrix}\)

\(2 + r = k ... (1)\)

\(5 - 3m + r = 2 \implies 3m - r = 3 ... (2)\)

\(4m + 6r = 26 ... (3)\)

\((2) \times 6 : 18m - 6r = 18\)

\((2) + (3) : 22m = 44 \implies m = 2\)

\(4(2) + 6r = 26 \implies 6r = 18\)

\(r = 3\)

\(2 + r = k \implies 2 + 3 = 5\)

\((k, m, r) = (5, 2, 3)\).

Correct Answer: D. \((\frac{2}{5}, 4\frac{1}{5})\)