Correct Answer: b. #120

Correct Answer: B. \(\frac{3x - 7}{2x - 4}, x \neq 2\)

Explanation

\(h : x \to 2 - \frac{1}{2x - 3}\)

\(h(x) = \frac{2(2x - 3) - 1}{2x - 3} = \frac{4x - 7}{2x - 3}\)

Let x = h(y)

\(x = \frac{4y - 7}{2y - 3}\)

\(x(2y - 3) = 4y - 7 \implies 2xy - 4y = 3x - 7\)

\(y = \frac{3x - 7}{2x - 4}\)

\(h^{-1}(x) = \frac{3x - 7}{2x - 4}\)

Correct Answer: A. \(^{-1}/_3\)

Explanation

\(T_2 = \frac{-2}{3};S_\infty \frac {3}{2}\)

\(T_n = ar^n - 1\)

∴ \(T_2 = ar = \frac{-2}{3}\)---eqn.(i)

\(S_\infty = \frac{a}{1 - r} = \frac{3}{2}\)---eqn.(ii)

= 2a = 3(1 - r)

= 2a = 3 - 3r

∴ a = \(\frac{3 - 3r}{2}\)

Substitute \(\frac{3 - 3r}{2}\) for a in eqn.(i)

= \(\frac{3 - 3r}{2} \times r = \frac{-2}{3}\)

= \(\frac{3r - 3r^2}{2} = \frac{-2}{3}\)

= 3(3r - 3r\(^2\)) = -4

= 9r - 9r\(^2\) = -4

= 9r\(^2\) - 9r - 4 = 0

= 9r\(^2\) - 12r + 3r - 4 = 0

= 3r(3r - 4) + 1(3r - 4) = 0

= (3r - 4)(3r + 1) = 0

∴ r = \(\frac{4}{3} or - \frac{1}{3}\)

For a geometric series to go to infinity, the absolute value of its common ratio must be less than 1 i.e. |r| < 1.

∴ r = -\(^1/_3\) (since |-\(^1/_3\)| < 1)

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

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

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

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

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

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

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

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

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

Correct Answer: D. 399000

Correct Answer: C. 1¹¹/₆₀

Correct Answer: D. 4x⁴

Explanation

\(56x^{-4} \div 14x^{-8}\)

= \(4x^{- 4 - (-8)}\)

= \(4x^{4}\)

Correct Answer: C. 45

Explanation

101101₂ = 1 x 2⁵ + 0 x 2⁴ + 1 x 2³ + 1 x 2² + 0 x 2¹ + 1 x 2⁰ = 32 + 0 + 8 + 4 + 0 + 1 = 45₁₀

Correct Answer: A. 1

Explanation

Acceleration, \(a = (4 - 2t) ms^{-2}\)

\(\therefore v = \int a \mathrm {d} t \)

= \(\int (4 - 2t) \mathrm {d} t = (4t - t^{2} + c)\)

When t = 0, v = 5 m/s

\(5 = 0 - 0 + c \implies c = 5\)

\(\therefore v = (4t - t^{2} + 5) ms^{-1}\)

Distance, \(s = \int (4t - t^{2} + 5) \mathrm {d} t\)

= \(2t^{2} - \frac{t^{3}}{3} + 5t + c\)

When t = 0, s = 18 m.

\(0 - 0 + 0 + c = 18 \implies c = 18\)

\(\therefore s = 2t^{2} - \frac{t^{3}}{3} + 5t + 18\)

(a) Velocity is greatest when a = 0.

\(4 - 2t = 0 \implies t = 2s\)

(b)(i) Particle is momentarily at rest when v = 0 m/s.

\(4t - t^{2} + 5 = 0 \implies t^{2} - 4t - 5 = 0\)

\(t(t - 5) + 1(t - 5) = 0 \implies t = -1 ; t = 5\)

Since time cannot be negative, t = 5s.

(ii) Distance when t = 5s.

\(s = 2(5^{2}) - \frac{1}{3} (5^{3}) + 5(5) + 18\)

= \(50 - \frac{125}{3} + 25 + 18\)

= \(\frac{154}{3}\)

= \(51\frac{1}{3} m\)