Correct Answer: C. \(\frac{4}{5}\)

Explanation

PQ = \(\begin {pmatrix} 2 & 3\\ -4 & 1\end {pmatrix}\) = \(\begin {pmatrix} 6 \\ 8 \end {pmatrix}\) = \(\begin {pmatrix} 36 \\ - 16\end {pmatrix}\)

\(\begin {pmatrix} 36 \\ -16 \end {pmatrix} \) = k = \(\begin {pmatrix} 45 \\ -20 \end {pmatrix} \)

k = \(\frac{36}{45}\)

= \(\frac{4}{5}\)

Correct Answer: C. \(\frac{1}{8}\)

Explanation

The remainder theorem states that if f(x) is divided by (x - a), the remainder is f(a).

\(f(x) = x^{3} - 2x + m\) divided by (x - 1), so that a = 1.

Remainder = \(f(1) = 1^3 - 2(1) + m = -1 + m\)

\(f(x) = 2x^{3} + x - m\) divided by (2x + 1), so that a = \(\frac{-1}{2}\)

\(f(\frac{-1}{2}) = 2(\frac{-1}{2}^{3}) + (\frac{-1}{2}) - m = \frac{-3}{4} - m\)

\(\implies m - 1 = \frac{-3}{4} - m\), collecting like terms,

\(2m = \frac{1}{4} \therefore m = \frac{1}{8}\)

Correct Answer: A. \(\frac{25}{4} - m\)

Explanation

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

\(a = 2, b = -5, c = m\)

\(\alpha + \beta = \frac{-b}{a} = \frac{5}{2}\)

\(\alpha \beta = \frac{c}{a} = \frac{m}{2}\)

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

= \((\frac{5}{2})^{2} - 2(\frac{m}{2}) \)

= \(\frac{25}{4} - m\)

Correct Answer: C. (1, - \(\frac{3}{2}\))

Explanation

\(\frac{3x^2}{3} + \frac{3y^2}{3} - \frac{6x}{3} + \frac{9y}{3} - \frac{5}{3}\) = 0

2gx = -2x, 2 fy = 3y

g = -1, f = \(\frac{3}{2}\)

Centre (1, - \(\frac{3}{2}\))

Explanation

(a) \(f(x) = \frac{x - 3}{2x - 1}, x \neq \frac{1}{2}\)

\(g(x) = \frac{x - 1}{x + 1}, x \neq -1\)

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

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

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

= \(\frac{- x - 2}{3x - 4}\) or \(\frac{-(x + 2)}{3x - 4}\).

(b)(i) \(y = 9x - x^{3}\)

If x = 0, y = 0.

If y = 0, \(9x - x^{3} = 0\)

\(x(9 - x^{2}) = x (3 - x)(3 + x) = 0\)

\(\implies x = 0, -3, 3\)

Turning points : \(\frac{\mathrm d y}{\mathrm d x} = 9 - 3x^{2}\)

\(9 - 3x^{2} = 0 \implies 9 = 3x^{2}\)

\(x^{2} = 3 \implies x = \pm \sqrt{3} = \pm 1.73\)

When x = 1.73, \(y = 9(1.73) - (1.73)^{3} = 15.57 - 5.18 = 10.39\)

Point is (1.73, 10.39).

When x = -1.73, \(y = 9(-1.73) - (-1.73)^{3} = -15.57 + 5.18 = -10.39\)

Point is (-1.73, -10.39).

Maximum or minimum ?

\(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = - 6x^{2}\)

At x = 1.73, \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} < 0 \therefore (1.73, 10.39) \text{ is a maximum point}\)

At x = -1.73, \(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} > 0 \therefore (-1.73, -10.39) \text{ is a maximum point}\).

Areas between x = -3 to x = 0 and x = 0 and x = 3 are equal.

\(\therefore Area = 2 \int_{0} ^{3} (9x - x^{3}) \mathrm {d} x = 2[\frac{9x^{2}}{2} - \frac{x^{4}}{4}]_{0} ^{3}\)

= \(81 - 40\frac{1}{2} = 40\frac{1}{2} sq. units\)

Correct Answer: D. 9s

Correct Answer: A. (1, 5)

Explanation

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

\(\frac{\mathrm d y}{\mathrm d x} = 6x - 1 = 5\)

\(6x = 5 + 1 = 6 \implies x = 1\)

\(y = 3(1^{2}) - 1 + 3 = 3 - 1 + 3 = 5\)

\(P = (1, 5)\)

Correct Answer: D. 3, \(\frac{-1}{3}\), 3

Explanation

log\(_5\) (\(\frac{125x^3}{\sqrt[3] {y}}\))

= \(\log_5 125 x^3 - \log _1 x^3 - log_5 y\frac{1}{3}\)

= \(3 log_5 5 + 3 log_5 x - \frac{1}{3} log _5 y\)

= 3, - \(\frac{1}{3}\), 3

Correct Answer: C. \(\frac{5\sqrt{2}}{2}\)

Explanation

\(\frac{\sqrt{3} + \sqrt{48}}{\sqrt{6}} = \frac{\sqrt{3} + 4\sqrt{3}}{\sqrt{6}}\)

\(\frac{5\sqrt{3}}{\sqrt{6} = \frac{5\sqrt{3} \times \sqrt{6}}{\sqrt{6} \times \sqrt{6}}\)

\(\frac{5\sqrt{18}{6} = \frac{15\sqrt{2}}{6}\)

= \(\frac{5\sqrt{2}}{2}\)

Explanation

he will be late to office = 2/5

he will not be late to office is = 1 - 2/5 = 3/5

If he will be late for 3 days only, he will also not be late for 3 days

\(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\)

p = 0.0138

(b) not be late to office is = 1 - 2/5 = 3/5

\(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\) * \(\frac{3}{5}\)

p = 0.0467

(c) be late to office = 2/5

\(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\) * \(\frac{2}{5}\)

p = 0.0041