Correct Answer: C. 120°

Correct Answer: B. x > -6

Correct Answer: D. 5y - 2x - 12 = 0

Explanation

\(Line: 2y + 5x - 6 = 0\)

\(2y = 6 - 5x \implies y = 3 - \frac{5x}{2}\)

Gradient = \(\frac{-5}{2}\)

For the line perpendicular to the given line, Gradient = \(\frac{-1}{\frac{-5}{2}} = \frac{2}{5}\)

The midpoint of P(4, 3) and Q(-6, 1) = \((\frac{-6 + 4}{2}, \frac{3 + 1}{2})\)

= (-1, 2).

Therefore, the line = \(\frac{y - 2}{x + 1} = \frac{2}{5}\)

\(2(x + 1) = 5(y - 2) \implies 5y - 2x - 12 = 0\)

Explanation

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

\(y+dy=4(x+dx)-\frac{7}{(x + dx)^2}\)

\(y+dy=4(x+dx)-\frac{7}{(x + dx)^2}-(4x-\frac{7}{x^2})\)

\(dy=4dx-\frac{7}{(x + dx)^2}+\frac{7}{x^2}\)

\(dy=4dx-(\frac{7}{(x + dx)^2}-\frac{7}{x^2})\)

\(dy=4dx-\frac{7x^2 - 7(x + dx)^2}{x^2(x + dx)^2)}\)

\(dy=4dx-\frac{(7(x^2 - (x + dx)^2)}{x^2(x + dx)^2)}\)

\(a^a-b^2=(a+b)(a-b)\)

\(dy=4dx-(\frac{7((x+x+dx)(x-x-dx))}{x^2(x+dx)^2})\)

\(dy=4dx-(\frac{7((2x+dx)(-dx))}{x^2(x+dx)^2})\)

\(dy=dx(4-(\frac{7((2x+dx)(-1))}{x^2(x+dx)^2}))\)

\(dy=4-(\frac{7((2x+dx)(-1))}{x^2(x+dx)^2})\)

As \(dx \to 0 \frac{dy}{dx}=4-(\frac{7((2x)(-1))}{x^2(x)^2})\)

\(\frac{dy}{dx}=4-(\frac{-14x}{x^4}\)

\(\therefore \frac{dy}{dx}=4+\frac{14}{x^3}\)

(b) tan \(P= \frac{3}{x - 1},tan Q={2}{x + 1}\)

\(tan (P-Q)=\frac{tan P - tan Q}{1 + tan P tan Q}\)

tan P-tan Q=\(\frac{3}{x - 1}-\frac{2}{x + 1}=\frac{3(x + 1) - 2(x - 1)}{(x + 1)(x - 1)}\)

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

1+tan P tan Q =\(1+(\frac{3}{x - 1})(\frac{2}{x + 1})=1+\frac{6}{(x + 1)(x - 1)}\)

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

\(\frac{tan P - tan Q}{1 + tan Ptan Q}=\frac{x + 5}{(x + 1)(x - 1)}÷\frac{(x + 1)(x - 1)+6}{(x + 1)(x - 1)}\)

\(=\frac{x + 5}{(x + 1)(x - 1)}\times\frac{(x + 1)(x - 1)}{(x + 1)(x - 1) + 6}\)

\(=\frac{x + 5}{(x + 1)(x - 1) + 6}=\frac{x + 5}{x^2 - 1 + 6}\)

∴tan(P-Q)\(=\frac{x + 5}{x^2 + 5}\)

Correct Answer: C. 1

Explanation

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

\(8^{y} = 2^{y + 2} \implies (2^{3})^{y} = 2^{y + 2}\)

\(3y = y + 2 \implies 2y = 2\)

\(y = 1\)

Correct Answer: C. \(\sqrt{\frac{b}{b - a}}\)

Correct Answer: D. \(\frac{2}{3}\)

Explanation

\(\text{p(a head and a six)} = \text{p(a head)} + \text{p(a six)}\)

= \(\frac{1}{2} + \frac{1}{6} = \frac{2}{3}\).

Hint: Probability of A and B occurring should be greater than probability A or B happening.

Correct Answer: C. \(\frac{7}{12}\)

Correct Answer: D. 12

Explanation

Recall, the terms of a GP is given by \(T_{n} = ar^{n - 1}\)

Given, \(T_{1} = a = \frac{3}{4}\) and \(T_{2} \times T_{3} = ar \times ar^{2} = 972\)

\(\implies \frac{3}{4}r \times \frac{3}{4}r^{2} = \frac{9}{16}r^{3} = 972\)

\(r^{3} = \frac{972 \times 16}{9} = 1728 \implies r = \sqrt[3]{1728} = 12\)

Correct Answer: C. u = \(\frac{2x + 3}{2 - 3x}\)

Explanation

x =\(\frac{2u-3}{3u + 2}\)

cross multiply

x(3u + 2) = 2u - 3

3ux + 2x = 2u - 3

collect like terms of u

2x + 3 = 2u - 3ux

\(\frac{2x + 3}{2 - 3x}\) = u