Explanation

\(\log_{3} x - 3\log_{x} 3 + 2 = 0\)

\(\log_{3} x - 3\log_{x} 3 + 2 = 0\)

Let \(\log_{3} x = a\), then \(\log_{x} 3 = \frac{1}{a}\).

\(a - \frac{3}{a} + 2 = 0\)

\(a^{2} + 2a - 3 = 0\)

\(a^{2} - a + 3a - 3 = 0\)

\(a(a - 1) + 3(a - 1) = 0\)

\((a - 1)(a + 3) = 0\)

\(\text{a = 1 or -3}\)

\(\log_{3} x = 1 \implies x = 3^{1} = 3\)

\(\log_{3} x = -3 \implies x = 3^{-3} = \frac{1}{27}\)

Explanation

\(\tan (2x - 15)° - 1 = 0 \implies \tan (2x - 15)° = 1\)

\((2x - 15)° = \tan^{-1} (1) = \frac{n \pi}{4}, n = 1, 2, ...\)

\(2x - 15 = 45°, 225°, 405°, 585°, 0° \leq x \leq 360°\)

\(2x = 60°, 240°, 420°, 600°\)

\(x = 30°, 120°, 210°, 300°\)

Explanation

Using Lami's theorem, \(\frac{150}{\sin 80} = \frac{T_{1}}{\sin 40} = \frac{T_{2}}{\sin 60}\)

\(T_{1} = \frac{150 \sin 40}{\sin 80} = \frac{150 \times 0.6428}{0.9848} = 97.41 N\)

\(T_{2} = \frac{150 \sin 60}{\sin 80} = \frac{150 \times 0.8660}{0.9848} = 131. 90 N\)

(b) \(h = 2 + 20t + kt^{2}\)

\(\frac{\mathrm d h}{\mathrm d t} = v = 20 + 2kt\)

At the highest point, \(v = 0 ms^{-1}; t = 2 secs\)

\(\implies 20 + 2k(2) = 0\)

\(20 = -4k \implies k = -5\)

(ii) Its highest point

\(h = 2 + 20t - 5t^{2}\)

At highest point, t = 2 secs

\(h(2) = 2 + 20(2) - 5(2^{2}) = 2 + 40 - 20 = 22 m\)

Correct Answer: A. \(2^{-n}\)

Explanation

\(8^{n} \times 2^{2n} \div 4^{3n} = 2^{3n} \times 2^{2n} \div 2^{6n}\)

\(2^{2n + 3n - 6n}\)

= \(2^{-n}\)

Explanation

(a) \(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A}{(x - 1)}+\frac{B}{2x + 3}+\frac{C}{(2x + 3)^2}\)

\(\frac{8x^2 + 8x + 9}{(x - 1)(2x + 3)^2}≡\frac{A(2x + 3)2+B(x - 1)(2x + 3)+C(x - 1)}{(x - 1)(2x + 3)^2}\)

\(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)

Put \(x=1\)

\(8(1)^2+8(1)+9=A(2(1)+3)^2+B(1-1)(2(1)+3)+C(1-1)\)

⇒25=25A

=A=\(\frac{25}{25}=1\)

Put \(x=-\frac{3}{2}\)

\(8(-\frac{3}{2})^2+8(-\frac{3}{2})+9=A(2(-\frac{3}{2})+3)^2+B(-\frac{3}{2}-1)(2(-\frac{3}{2})+3)+C(-\frac{3}{2}-1)\)

⇒15=-2.5C

\(=C=-\frac{15}{2.5}=-6\)

Since \(8x^2+8x+9=A(2x+3)^2+B(x-1)(2x+3)+C(x-1)\)

\(⇒8x^2+8x+9=A(4x^2+12x+9)+B(2x^2+x-3)+C(x-1)\)

\(=8x^2+8x+9=4Ax^2+12A+9A+2Bx^2+Bx-3B+Cx-C\)

\(=8x^2+8x+9=4Ax^2+2Bx^2+12Ax+Bx+Cx+9A-3B-C\)

\(=8x^2+8x+9=(4A+2B)x^2+(12A+B+C)x+9A-3B-C\)

By comparing the coefficient of \(x, 8=12A+B+c\)

=8=12(1)+B-6

=8=12+B-6

=8=6+B

=8-6=B

=\(\therefore \frac{8x^2+8x+9}{(x-1)(2x+3)^2}=\frac{1}{x-1}+\frac{2}{2x+3}-\frac{6}{(2x+3)^2}\)

(b) Equation of a circle =\((x - a)^2 + (y - b)^2 = r^2\)

Where "a" and "b" are the coordinate of the center and "r" is the radius

2πr = 6π (given)

∴ r = 3 units

=\( (x - (-2))^2 + (y - 5)^2 = 3^2\)

= \((x + 2)^2 + (y - 5)^2 = 9\)

= \(x^2 + 4x + 4 + y^2 - 10y + 25 = 9\)

= \(x^2 + y^2 + 4x - 10y + 29 - 9 = 0\)

∴ \(x^2 + y^2 + 4x - 10y + 20 = 0\)

Correct Answer: B. - 22

Explanation

\(\frac{6x + k}{2x^2 + 7x - 15} = \frac{4}{x + 5} - \frac{2}{2x - 3}\)

6x + k = 4 (2x - 3) - 2(x + 5)

6x + k = 8x - 12 - 2x - 10

6x + k = 6x - 22

k = - 22

Correct Answer: D. 0.3

Explanation

P(X \(\cup\) Y) = P(X)P(Y)

0.15 = 0.5m

m = \(\frac{0.015}{0.5}\)

= 0.3

Correct Answer: C. \(\frac{2}{3}\)\(\sqrt{5}\) + \(\frac{5}{6}\sqrt{2}\) + \(\frac{1}{2}\sqrt{10}\) + 2

Explanation

\(\frac{(\sqrt{5} + 3)(4 + \sqrt{10})}{(4 - \sqrt{10})(4 + \sqrt{10})}\)

= \(\frac{4\sqrt{5} + \sqrt{50} + 12 + 3\sqrt{10}}{4^2 - (\sqrt{10})^2}\)

= \(\frac{4\sqrt{5} + 5\sqrt{2} + 12 + 3\sqrt{10}}{16 - 10}\)

= \(\frac{4 \sqrt{5}}{6} + \frac{5 \sqrt{2}}{6} + \frac{12}{6} + \frac{3\sqrt{10}}{6}\)

= \(\frac{2}{3}\)\(\sqrt{5}\) + \(\frac{5}{6}\sqrt{2}\) + \(\frac{1}{2}\sqrt{10}\) + 2

Correct Answer: D. 64

Explanation

\(\log_{2} y^{\frac{1}{2}} = \log_{5} 125\)

\(\log_{2} y^{\frac{1}{2}} = \log_{5} 5^{3} = 3\log_{5} 5 = 3\)

\(\log_{2} y^{\frac{1}{2}} = 3 \implies y^{\frac{1}{2}} = 2^{3} = 8\)

\(y = 8^{2} = 64\)

Correct Answer: B. 4.606

Explanation

(0,3)\(^x\) = (0.5)\(^8\)

xlog 0,3 = 9 log 0.5

x = \(\frac{8 \log 0.5}{\log 0.3}\)

= 4.606