Correct Answer: A. 5

Correct Answer: C. x > 1

Explanation

2x - x\(^2\) > 0

0 > x\(^2\) - 2x

x\(^2\) - 2x < 0

x(x - 2) < 0

x > 1

Explanation

\(f ' ' (x) = \frac{\mathrm d ^{2} y}{\mathrm d ^{2} x} = 2\)

\(f ' (x) = \int 2 \mathrm {d} x\)

= \(2x + c\)

When x = 1, f'(x) = 0.

\(2(1) + c = 0 \implies c = -2\)

\(\therefore f ' (x) = 2x - 2\)

\(f(x) = \int (2x - 2) \mathrm {d} x\)

= \(x^{2} - 2x + c\)

When x = 0, f(x) = -8

\(0^{2} - 2(0) + c = -8\)

\(c = -8\)

\(\therefore f(x) = x^{2} - 2x - 8\)

Explanation

At y-intercept, 3(0) — 2y + 6 = 0 and solving for y, y = 3.

Therefore, the coordinates of y — intercept is (0, 3).

Then, r\(^2\) = (0 — 2)\(^2\) + (3 — 3)\(^2\) = 4. With the centre at (2, 3). the equation of the circle was (x — 2)\(^2\) + (y — 3)\(^2\) = 4 and which simplified to \(x^2\) - y\(^2\) — 4x — 6y + 9 = 0

Correct Answer: A. \(\frac{-(2x + y)}{x}\)

Explanation

\(\frac{\mathrm d}{\mathrm d x}(x^2 + xy - 5) = \frac{\mathrm d (x^{2})}{\mathrm d x} + \frac{\mathrm d (xy)}{\mathrm d x} - \frac{\mathrm d (5)}{\mathrm d x} = 0\)

= \(2x + x\frac{\mathrm d y}{\mathrm d x} + y = 0\)

\(\implies x\frac{\mathrm d y}{\mathrm d x} = -(2x + y)\)

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

Correct Answer: A. -2

Explanation

\(\int_{1}^{2} (2 + 2x - 3x^{2}) \mathrm {d} x\)

= \((2x + x^{2} - x^{3})|_{1}^{2}\)

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

= \(0 - 2 = -2\)

Correct Answer: B. 21

Explanation

\(Mean = \frac{sum of items}{total number of items}\)

\(20 = \frac{x}{8} \implies x = 160\)

The sum of the 7 nos = 160 - 17 = 143

Correct mean = \(\frac{143 + 25}{8} = \frac{168}{8} = 21\)

Correct Answer: B. 0

Explanation

\(\log_{10} (\frac{1}{3} + \frac{1}{4}) + 2\log_{10} 2 + \log_{10} (\frac{3}{7})\)

\(\frac{1}{3} + \frac{1}{4} = \frac{7}{12}\)

= \(\log_{10} (\frac{7}{12} \times 2^{2} \times \frac{3}{7})\)

= \(\log_{10} 1 = 0\)

Explanation

(a) 4 boys, 5 girls

P(both are boys) = P(1st is a boy and 2nd is a boy)

= \(\frac{4}{9} \times \frac{3}{8}\)

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

(b) If p is the probability of a standard transistors, then \(p = 0.8 = \frac{4}{5}\)

If q is the probability of a transistor below standard, \(q = 0.2 = \frac{1}{5}\)

For 6 transistors, we have

\((p + q)^{6} = p^{6} + 6p^{5}q + 15p^{4}q^{2} + 20p^{3}q^{3} + 15p^{2}q^{4} + 6pq^{5} + q^{6}\)

(i) p(exactly 2 standard transistors) = \(15p^{2}q^{4}\)

= \(15 \times (\frac{4}{5})^{2} \times (\frac{1}{5})^{4}\)

= \(\frac{48}{3125}\)

(ii) p(exactly 1 standard transistor) = \(6pq^{5}\)

= \(6 \times \frac{4}{5} \times (\frac{1}{5})^{5}\)

= \(\frac{24}{15625}\)

(iii) p(at least 2 standard transistors) = 1 - [p(0 standard transistor) + p(1 standard transistor)]

p(0 standard transistor) = \(q^{6}\)

= \((\frac{1}{5})^{6} = \frac{1}{15625}\)

p(at least 2 standard transistors) = \(1 - [\frac{1}{15625} + \frac{24}{15625}]\)

= \(\frac{15600}{15625} = \frac{624}{625}\)

(iv) p(at most 2 standard transistors) = p(0 standard transistor) + p(1 standard transistor) + p(2 standard transistors)

= \(\frac{1}{15625} + \frac{24}{15625} + \frac{48}{3125}\)

= \(\frac{265}{15625} = \frac{53}{3125}\)

Correct Answer: A. 3 sec

Explanation

\(u = 27 ms^{-1}; a = -9 ms^{-2}; v = 0; t = ?\)

\(v = u + at; t = \frac{v - u}{a}\)

\(t = \frac{0 - 27}{-9} = \frac{-27}{-9}\)

\(\therefore t = 3 sec\)