Correct Answer: C. 40

Explanation

x\(^3\)y\(^2\) in (x-2y)\(^5\)

n = 5, r = 3, p = x, q = -2y

5C\(_3\) * x\(^3\) -2y\(^2\)

5C\(_3\) = \(\frac{5!}{[5-3]!3!}\)

\(\frac{5*4*3!}{2! 3!}\) → \(\frac{5*4}{2}\)

5C\(_3\) = 10

: 5C\(_3\) * x\(^3\) -2y\(^2\) = 10 * x\(^3\) 4y\(^2\)

40x\(^3\)y\(^2\)

the coefficient is 40

Correct Answer: A. \(\begin{pmatrix} -3 & 1 \end{pmatrix}\)

Correct Answer: B. 1.4

Explanation

n = 5

x? = \(\frac{∑x}{n} = \frac{85 + 84 + 83 + 86 + 87}{5} = \frac{425}{5} = 85\)

\(S.D = √\frac{∑(x - x? )}{n} = √\frac{10}{5}\)

∴ S.D = √2 =1.4

Correct Answer: A. \(\frac{63}{65}\)

Explanation

\(sin x = \frac{4}{5}\) and \(cos y = \frac{12}{13}\)

x is obtuse i.e sin x = + ve while cos x = + ve

\(cos x=\frac{3}{5}==>cos x=-\frac {3}{5}(obtuse)\)

\(sin y= \frac{5}{13}\)

\(sin (x-y) = sin x\) \(cos y - cos x\) \(sin y\)

\(sin(x-y) = \frac{4}{5}\times\frac{12}{13}-(-\frac{3}{5})\times\frac{5}{13}\)

\(sin(x-y) = \frac{48}{65}-(-\frac{3}{13})\)

\(\therefore sin (x-y) = \frac{48}{65} + \frac{3}{13} = \frac{63}{65}\)

Correct Answer: B. \(3\sqrt{13}\)

Explanation

Given \(a = i - 3j; b = -2i + 5j; c = 3i - j\)

\(a- b + c = (1 - (-2) + 3)i + (-3 - 5 + (-1))j = 6i - 9j\)

\(|a - b + c| = \sqrt{6^{2} + (-9)^{2}} = \sqrt{36 + 81} = \sqrt{117}\)

\(= \sqrt{9 \times 13} = 3\sqrt{13}\)

Explanation

(a) \(\frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x}\)

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

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

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

\(\implies \frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x} = \frac{A}{x} + \frac{B}{x - 1} + \frac{C}{x - 3}\)

\(\implies \frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x} = \frac{A(x - 1)(x - 3) + B(x)(x - 3) + C(x)(x - 1)}{x(x - 1)(x - 3)}\)

Equating, we have

\(2x^{2} - 5x + 1 = A(x - 1)(x - 3) + B(x)(x - 3) + C(x)(x - 1) \)

Let x = 1, we have

\(2 - 5 + 1 = -2B \implies B = 1\)

Let x = 3, we have

\(18 - 15 + 1 = 6C \implies C = \frac{4}{6} = \frac{2}{3}\)

Let x = 0, we have

\(0 - 0 + 1 = 3A \implies A = \frac{1}{3}\)

\(\therefore \frac{2x^{2} - 5x + 1}{x^{3} - 4x^{2} + 3x} = \frac{1}{3x} + \frac{1}{x - 1} + \frac{2}{3(x - 3)}\)

(b) \(\begin{vmatrix} x - 3 & -4 & 3 \\ 5 & 2 & 2 \\ 2 & -4 & 6 - x \end{vmatrix} = -24\)

\(\implies (x - 3)(12 - 2x + 8) - (-4)(30 - 5x - 4) + 3(-20 - 4) = -24\)

\((x - 3)(20 - 2x) + 4(26 - 5x) - 72 + 24 = 0\)

\(20x - 2x^{2} - 60 + 6x + 104 - 20x - 48 = 0\)

\(-2x^{2} + 6x - 4 = 0\)

\(2x^{2} - 6x + 4 = 0\)

\(2x^{2} - 4x - 2x + 4 = 0 \implies 2x(x - 2) - 2(x - 2) = 0\)

\((2x - 2)(x - 2) = 0 \implies \text{x = 1 or 2}\).

Correct Answer: B. \(2^{(n+2)}\sqrt2\)

Explanation

8√2, 16√2, 32√2, ..

\(a = 8\sqrt2; r =\frac{T_2}{T_1}=\frac{16\sqrt2}{8\sqrt2}=2\)

\(T_n=ar^{n-1}\)

\(T_n=8\sqrt2 \times 2^{n-1}\)

\(T_n=2^3\times2^{n-1}\times\sqrt2\)

\(T_n=2^{3+n-1}\times\sqrt2\)

\(\therefore T_n= 2^{(n+2)}\sqrt2\)

Explanation

(a)1) The sequence is an A.P: 500, 750, 1000 ..

with a = 500 and d=250; Tn=15000

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

15000 = 500+ (n - 1) x 250

15000 = 500 +250n - 250

250n = 14500 ; n = \(\frac{14500}{250}\)

= 58

The jogger will reach a distance of 15km in 58 days.

(ii) Finding total distance he would have run in the training

Using S\(_n\) =n/2 [2a + (n - 1)d] n= 58, d=250, a = 500

= 58/2 [2 x 500 + (58 -1) x 250]

= 29[ 1000 +(57 x 250)]

= 29 (1000+ 14250),

= 29 x 15250 = 442250

(b) 2nd term => ar = -3; a= -3/r

S ∞ = \(\frac{a}{1-r}\) = 25/2

\(\frac{-3}{r}\) x \(\frac{1}{1-r}\) = 25/2;

25r - 25r\(^2\) = -6;

25r\(^2\) - 25r - 6 = 0

(5r +1) (5r - 6) = 0

r= -1/5 and 6/5

Explanation

(a) \(\int_{0}^{3} \frac{\mathrm d x}{x^{2} + 1}\)

h = 0.5

\(y_{1} = 1; y_{2} = 0.8; y_{3} = 0.5; y_{4} = 0.308; y_{5} = 0.2; y_{6} = 0.138; y_{7} = 0.1\)

\(y_{1} + y_{7} = 1 + 0.1 = 1.1\)

\(2(y_{2} + ... + y_{6}) = 2(0.8 + 0.5 + 0.308 + 0.2 + 0.138) = 2(1.946) = 3.892\)

\(\int_{0}^{3} \frac{\mathrm d x}{x^{2} + 1} = \frac{1}{2} \times 0.5 (1.1 + 3.892) = 0.25(4.992)\)

= \(1.248 \approxeq 1.25 (\text{to 2 decimal place})\)

(b) -2x + y = 3

-x + 4y = 1

\(det = \begin{vmatrix} -2 & 1 \\ -1 & 4 \end{vmatrix} = -7\)

\(det_{1} = \begin{vmatrix} 3 & 1 \\ 1 & 4 \end{vmatrix} = 11\)

\(det_{2} = \begin{vmatrix} -2 & 3 \\ -1 & 1 \end{vmatrix} = 1\)

\(\therefore x = \frac{det_{1}}{det} = \frac{11}{-7} = -1\frac{4}{7}\)

\(\therefore y = \frac{det_{2}}{det} = \frac{1}{-7}\)

Explanation

(ii) Using Lami's theorem,

\(\frac{48}{\sin 150} = \frac{T}{\sin 120}\)

\(T = \frac{48 \sin 120}{\sin 150}\)

\(T = 83.14N\)

The tension in DB is 83.14N.

(iii) Using Lami's theorem,

\(\frac{10q}{\sin 90} = \frac{48}{\sin 150}\)

\(10q = \frac{48 \sin 90}{\sin 150}\)

\(10q = 96 \implies q = 9.6kg\)