Correct Answer: C. 2 square units

Explanation

\(\int^0_{-2} x + 1\) x\(^2 + x|^0_{-2}\)

0 - (2\(^2 - 2\))

= 2 square units

Correct Answer: C. -5< x ≤ 7/3

Correct Answer: A. 6

Explanation

\(log2 = 0.3010\hspace{1mm}given\\

log2^y = 1.8062\\

∴ ylog2 = 1.8062\\

y=\frac{1.8062}{logy}=\frac{1.8062}{0.3010}=6\)

Correct Answer: B. 135°

Explanation

(a) 10% of bulbs are faulty, 90% are not faulty.

P(a bulb is flawless) = 90% = 0.9

(i) p(first sample is flawless and accepted) = \((0.9)^{5} = 0.59049 \approxeq 0.590\)

(ii) P(first sample of 5 bulbs contains at least one faulty bulb) = 1 - 0.59049 = 0.40951\)

Probability that the second sample of size 5 contains flawless bulbs = \((0.9)^{5} = 0.59049\)

\(\therefore\) probability that the first batch is accepted = \(0.4095 \times 0.5905 = 0.2418\)

(b) 3 black, 12 non-black. Total = 15.

p(black marble) = \(\frac{3}{15} = p = \frac{1}{5}\)

p(non- black marble) = \(\frac{12}{15} = q = \frac{4}{5}\)

\((p + q)^{10} = p^{10} + 10p^{9} q + 45p^{8} q^{2} + 120p^{7} q^{3} + 210p^{6} q^{4} + 252p^{5} q^{5} + 210p^{4} q^{6} + 120p^{3} q^{7} + 45p^{2} q^{8} + 10p q^{9} + q^{10}\)

(i) p(no black) = \(q^{10} = (\frac{4}{5})^{10} = \frac{1048576}{9765625}\)

(ii) p(black marble at most 3 times) = p(0 black marble) + p(1 black marble) + p(2 black marbles) + p(3 black marbles)

= \(\frac{1048576}{9765625} + 10(\frac{1}{5})(\frac{4}{5})^{9} + 45(\frac{1}{5})^{2} (\frac{4}{5})^{8} + 120(\frac{1}{5})^{3} (\frac{4}{5})^{7}\)

Correct Answer: D. -9,2/3

Explanation

3x\(^2\) + 25x - 18 = 0

3x\(^2\) + 27x - 2x - 18 = 0

3x(x + 9) - 2(x + 9) = 0

(3x - 2)(x + 9) = 0

x = \(\frac{2}{3}\) or x = -9.

Explanation

\((x - a)^{2} + (y - b)^{2} = r^{2}\)

\(x^{2} - 2ax + a^{2} + y^{2} - 2by + b^{2} = r^{2}\)

Using A(3, 2) :

\(9 - 6a + a^{2} + 4 - 4b + b^{2} = r^{2}\)

\(13 - 6a - 4b + a^{2} + b^{2} = r^{2} .... (1)\)

Using B(-1, -2) :

\(1 + 2a + a^{2} + 4 + 4b + b^{2} = r^{2}\)

\(5 + 2a + 4b + a^{2} + b^{2} = r^{2} .... (2)\)

Using C(5, -4) :

\(25 - 10a + a^{2} + 16 + 8b + b^{2} = r^{2}\)

\(41 - 10a + 8b + a^{2} + b^{2} = r^{2} ... (3)\)

(1) - (2) :

\(8 - 8a - 8b = 0 \implies 1 - a - b = 0 .... (4)\)

(3) - (2) :

\(24 - 12a + 12 + 4b = 0 \implies 36 - 12a + 4b = 0 \)

\(\equiv 9 - 3a + b = 0... (5)\)

Solving (4) and (5), we have :

\(a = 2\frac{1}{2} ; b = -1\frac{1}{2}\)

Coordinates of the centre : \(C(2\frac{1}{2}, -1\frac{1}{2})\).

(b) To get the radius, we can make use of any point on the circle as well as the centre.

Using point \(A(3, 2) & C(2\frac{1}{2}, -1\frac{1}{2})\)

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

= \(\sqrt{\frac{1}{4} + \frac{49}{4}}\)

= \(\sqrt{\frac{50}{4}}\)

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

(c) Equation of the circle :

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

\(x^{2} - 11x + \frac{25}{4} + y^{2} + 3y + \frac{9}{4} = \frac{50}{4}\)

\(x^{2} + y^{2} - 11x + 3y + \frac{34}{4} - \frac{50}{4} = 0\)

\(x^{2} + y^{2} - 11x + 3y - 4 = 0\)

Explanation

(a)

Difference in longitudes = 124° + 56° = 180°

Distance PQ = \(\frac{180°}{360°} \times 2\pi r\)

where \(r = R \cos \theta\)

\(/PQ/ = \frac{180}{360} \times 2 \times 3.142 \times R \cos 68.7°\)

\(/PQ/ = 3.142 \times 6400 \times 0.3633\)

= \(7305.53 km\)

\(\approxeq 7300 km\) (3 sig. figs)

(b) Red balls = 4, white = 3 ; green = 5

Total = 4 + 3 + 5 = 12 balls.

(i) P(green) = \(\frac{5}{12}\)

P(not green) = \(1 - \frac{5}{12} = \frac{7}{12}\)

(ii) P(picking red and white) = P( red then white) or P(white then red)

= \(\frac{4}{12} \times \frac{3}{11} + \frac{3}{12} \times \frac{4}{11}\)

= \(\frac{2}{11}\)

Explanation

Frequency Distribution Table

Correct Answer: A. 29