Correct Answer: B. \(7x^{2} - 4x + 5 = 0\)

Explanation

Let the roots of the equation be \(\alpha\) and \(\beta\).

\(\alpha + \beta = \frac{-b}{a} = \frac{4}{7}\)

\(\alpha \beta = \frac{c}{a} = \frac{5}{7}\)

\(\implies a = 7, b = -4, c = 5\)

Equation: \(ax^{2} + bx + c = 0 \)

= \(7x^{2} - 4x + 5 = 0\)

Correct Answer: A. 13%

Explanation

Let the original volume be V with radius r.

\(V = \frac{4}{3}\pi r^{3}\)

45% increased volume = 145%V.

Let the %age increase in radius = m%r

\(\frac{145}{100}V = \frac{4}{3}\pi (\frac{mr}{100})^{3}\)

\(1.45V = (\frac{4}{3}\pi r^{3})(\frac{m}{100})^{3}\)

\(1.45V = V(\frac{m}{100})^{3}\)

\(\implies 1.45 \times 10^{6} = m^{3}\)

\(m = \sqrt[3]{1.45 \times 10^{6}} = 113.2%\)

\(\therefore \text{%age increase =} 113.2 - 100 = 13.2%\)

\(\approxeq 13%\)

Correct Answer: B. 6.5

Explanation

\(f(x) = (x^{2} + 3)^{2}\)

Using the chain rule, \(\frac{\mathrm d y}{\mathrm d x} = \frac{\mathrm d y}{\mathrm d u} \times \frac{\mathrm d u}{\mathrm d x}\)

Let \(u = x^{2} + 3\) so that \(y = u^{2}\)

\(\frac{\mathrm d y}{\mathrm d u} = 2u\)

\(\frac{\mathrm d u}{\mathrm d x} = 2x\)

\(\therefore \frac{\mathrm d y}{\mathrm d x} = 2u(2x) = 4xu\)

But \(u = x^{2} + 3\),

\(\frac{\mathrm d y}{\mathrm d x} = 4x(x^{2} + 3)\)

At \(x = \frac{1}{2}, \frac{\mathrm d y}{\mathrm d x} = 4(\frac{1}{2})((\frac{1}{2})^{2} + 3)\)

= \(2 \times \frac{13}{4} = \frac{13}{2} = 6.5\)

Correct Answer: A. \(f : x \to 4x - 1\)

Explanation

The inverse of the inverse of a function gives the function

i.e \(f^{-1}(f^{-1}(x)) = f(x)\)

\(f^{-1}(x) = \frac{x + 1}{4}\)

Take y = x, so

\(f^{-1}(y) = \frac{y + 1}{4}\)

Let \(x = f^{-1}(y)\),

\(x = \frac{y + 1}{4} \implies 4x = y + 1\)

\(y = f(x) = 4x - 1\)

Explanation

(a) \(\frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}}\)

\(\frac{^{9}P_{3}}{^{15}C_{3}} = \frac{9!}{(9 - 3)!} \times \frac{3! 12!}{15!}\)

= \(504 \times \frac{1}{455} = 1.10769 \)

\(\frac{^{5}C_{3}}{^{3}P_{2}} = \frac{5!}{2! 3!} \times \frac{1}{3!}\)

= \(10 \times \frac{1}{6} = 1.66667\)

\(\therefore \frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}} = 1.10769 + 1.66667 = 2.77436\)

\(\approxeq 2.77\) (2 d.p)

(b) 6 tutors and 10 pupils to select 2 tutors and 5 pupils.

Select one tutor, one more tutor is needed = \(^{5}C_{1} = 5 ways\)

Keep aside 2 students, we have \(^{8}C_{5} = 56 ways\)

= \(5 \times 56 = \text{280 ways}\).

Correct Answer: A. -1

Explanation

\(\lim \limits_{x \to 3} \frac{x + 3}{x^{2} - x - 12} = \frac{3 + 3}{3^{2} - 3 - 12}\)

= \(\frac{6}{-6} = -1\)

Explanation

\(2x^{2} - 7x + 4 = 0\)

\(\alpha + \beta = \frac{7}{2}\)

\(\alpha \beta = \frac{4}{2} = 2\)

\(\frac{\alpha}{\beta} + \frac{\beta}{\alpha} = \frac{\alpha^{2} + \beta^{2}}{\alpha \beta}\)

= \(\frac{(\alpha + \beta)^{2} - 2\alpha \beta}{\alpha \beta}\)

= \(\frac{(\frac{7}{2})^{2} - 2(2)}{2}\)

= \(\frac{(\frac{49}{4} - \frac{16}{4})}{2}\)

= \(\frac{33}{8}\)

\(\frac{\alpha}{\beta} \times \frac{\beta}{\alpha} = 1\)

\(\therefore Equation : x^{2} - \frac{33}{8}x + 1 = 0\)

Correct Answer: D. 3y + x + 14 = 0

Explanation

\(3x - y + 11 = 0 \implies y = 3x + 11\)

\(Gradient = 3\)

Gradient of perpendicular line = \(\frac{-1}{3}\)

\(\therefore \frac{y - (-5)}{x - 1} = \frac{-1}{3}\)

\(3(y + 5) = 1 - x\)

\(3y + x + 15 - 1 = 3y + x + 14 = 0\)

Correct Answer: A. 100N

Explanation

\(F = ma = m(\frac{v - u}{t})\)

\(F = 25(\frac{35 - 15}{5}) = 25 \times 4 = 100N\)

Correct Answer: B. \(-1\)

Explanation

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

\(f(-3) = \frac{3(-3) + 1}{(-3)^{2} - 1} = \frac{-8}{8} = -1\)