Correct Answer: C. 198

Explanation

T\(_3\) = 6

T\(_5\) = 12

S\(_{12}\) = ?

T\(_n\) = a + (n - 1)d

⇒ T\(_3\) = a + 2d = 6 ----- (i)

⇒ T\(_5\) = a + 4d = 12 ----- (ii)

Subtract equation (ii) from (i)

⇒ -2d = -6

⇒ d\(\frac{-6}{-2}\) = 3

Substitute 3 for d in equation (i)

⇒ a + 2(3) = 6

⇒ a + 6 = 6

⇒ a = 6 - 6 = 0

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

⇒ S\(_{12}\) = \(\frac{12(2 \times 0 + (12 - 1)3)}{2}\)

⇒ S\(_{12}\) = 6(0 + 11 x 3)

⇒ S\(_{12}\) = 6(33)

∴ S\(_{12}\) = 198

Correct Answer: C. 4

Correct Answer: C. 36°

Correct Answer: A. 136°

Correct Answer: D. \(\frac{1}{3x + 1}\) + \(\frac{3}{2x - 1}\)

Correct Answer: C. 2\(\frac{7}{20}\)

Explanation

Sin x = \(\frac{opp}{hyp}\)

sinx = \(\frac{3}{5}\)

using Pythagoras' theorem

hyp\(^2\) = opp\(^2\) + adj\(^2\)

adj\(^2\) = 5\(^2\) - 3\(^2\) = 25 - 9

adj\(^2\) = 16

adj =√ 16

adj = 4.

tanx = \(\frac{opp}{adj}\)

= \(\frac{3}{4}\)

cosx = \(\frac{adj}{hyp}\)

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

(tanx + 2cosx) = \(\frac{3}{4}\) + 2(\(\frac{4}{5}\))

= \(\frac{15 + 32}{20}\)

= \(\frac{47}{20}\) or

2 \(\frac{7}{20}\)

Correct Answer: C. \(\frac{23}{7}\)

Explanation

\(\frac{15 - 2x}{(x+4)(x-3)}\) = \(\frac{R}{(x+4)}\) + \(\frac{9}{7(x-3)}\)

15-2x = R(x - 3) +9(x +4)/7

Put x =-4, we have 15 -2(-4) = -7R

23 -7R;

R = 23/7

Correct Answer: B. 3(2¹⁸ - 1)

Explanation

Rank coefficient = I = \(\frac{6\sum d^2}{n(n^2 - 1)}\)

= 1 - \(\frac{6(141)}{10(10^2 - 1)}\)

= 1 - \(\frac{6(141)}{10 \times 99}\)

= 1 - \(\frac{846}{990}\)

= 1 - 0.85

= 0.15

(ii)

Rank coefficient = 1 - \(\frac{6(78)}{10(10^2 - 1)}\)

= 1 - \(\frac{468}{990}\)

= 1 - 0.472

= 0.53

(iii)

Rank coefficient = 1 - \(\frac{6(78)}{10(10^2 - 1)}\)

= 0.53

(b) Examiner II and III

Explanation

(a) \(a = \begin{pmatrix} 8 \\ 3 \end{pmatrix} ; b = \begin{pmatrix} 6 \\ -5 \end{pmatrix} ; c = \begin{pmatrix} 2 \\ -3 \end{pmatrix} ; |d| = \sqrt{41}\)

\(p = a + b - 2c\)

= \(\begin{pmatrix} 8 \\ 3 \end{pmatrix} + \begin{pmatrix} 6 \\ -5 \end{pmatrix} + 2\begin{pmatrix} 2 \\ -3 \end{pmatrix}\)

= \(\begin{pmatrix} 8 + 6 + 4 \\ 3 - 5 - 6 \end{pmatrix}\)

= \(\begin{pmatrix} 10 \\ -8 \end{pmatrix}\)

\(|p| = \sqrt{10^{2} + (-8)^{2}} = \sqrt{164}\)

\(|p| = 2\sqrt{41}\)

\(|d| = \frac{1}{2} |p|\)

\(\therefore d = \frac{1}{2} p = \frac{1}{2} \begin{pmatrix} 10 \\ -8 \end{pmatrix}\)

= \(\begin{pmatrix} 5 \\ -4 \end{pmatrix}\)

(b) A(3, 4) ; B(3, n).

\(OA = 3i + 4j ; OB = 3i + nj\)

\(OA \cdot OB = |OA||OB| \cos \theta\)

\((3i + 4j) \cdot (3i + nj) = |3i + 4j||3i + nj| \cos 30\)

\(9 + 4n = 5 (\sqrt{9 + n^{2}}) (\sqrt{3}{2})\)

\(2(9 + 4n) = 5(\sqrt{3(9 + n^{2})})\)

\(18 + 8n = 5\sqrt{27 + 3n^{2}}\)

Squaring both sides,

\(324 + 288n + 64n^{2} = 25(27 + 3n^{2})\)

\(324 + 288n + 64n^{2} = 675 + 75n^{2}\)

\(75n^{2} - 64n^{2} - 288n + 675 - 324 = 0\)

\(11n^{2} - 288n + 351 = 0\)

\(n = \frac{-(-288) \pm \sqrt{(-288)^{2} - 4(11)(351)}}{2(11)}\)

\(n = \frac{288 \pm \sqrt{82944 - 15444}}{22}\)

\(n = \frac{288 \pm \sqrt{67500}}{22}\)

\(n = \frac{288 \pm 259.81}{22}\)

\(n = \frac{288 + 259.81}{22} ; n = \frac{288 - 259.81}{22}\)

\(n = \frac{547.81}{22} ; n = \frac{28.19}{22}\)

\(n = 24.9 ; n = 1.28\)