Correct Answer: D. 4

Explanation

\(F = ma\)

\(32 = 8m \implies m = 4 kg\)

Correct Answer: A. \(\frac{1}{3}\)

Explanation

Differentiate distance twice to get the acceleration and then equate to get p.

\(s = 7 + pt^{3} + t^{2}\)

\(\frac{\mathrm d s}{\mathrm d t} = v(t) = 3pt^{2} + 2t\)

\(\frac{\mathrm d v}{\mathrm d t} = a(t) = 6pt + 2\)

\(a(3) = 6p(3) + 2 = 8 \implies 18p = 8 - 2 = 6\)

\(p = \frac{6}{18} = \frac{1}{3}\)

Explanation

(ii)

bi) Median 1/2 of N = 1/2 x 400 = 200

Cumulative frequency level From the curve, it is 118.5cm

(ii) Inter quartile range = upper quartile - lower quartile upper quartile

= 3/4 of N

= 3/4 x 400 = 300

Cumulative frequency level From the curve, it is 126.5cm

lower quartile = 1/4 of N

= 1/4 x 400 = 100

Cumulative frequency level From the curve, it is 111.0cm

Inter quartile range = 126.5cm -111.0cm = 15.5cm

(ii) From the curve, students whose heights are at most 130cm are 240.

Percentage óf students whose heights are at most 130cm = \(\frac{240}{400}\) x 100

= 60%

Correct Answer: B. \(\frac{65}{12}\)

Explanation

\({1_0^∫} x^2(x^3+2)^3\)dx

let \( u = x^3 + 2, du = 3x^2dx\)

when x = 1, u = 3

when x = 0, u = 2

dx = \(\frac{du}{3x^2}\)

\({3_2^∫}\) \(\frac{x^2[u]^3}{3x^2}\)

\({3_2^∫}\) \(\frac{u^3}{3}\) du

= \(\frac{u^4}{3*4}\)\(_2\)3

\(\frac{1}{12} [u^4]\)\(_2\)3

\(\frac{1}{12} [3^4 - 2^4]\)

\(\frac{1}{12}[81 - 16]\)

\(\frac{65}{12}\)

Explanation

\(3x^{2} - 6x + 10\)

\(3(x^{2} - 2x) + 10\)

\(3[x^{2} - 2x + (\frac{1}{2}(-2))^{2}] + 10 - 3\)

= \(3(x - 1)^{2} + 7\)

The minimum value is 7 and this occurs at x - 1 = 0 ==> x = 1.

Correct Answer: C. 180

Explanation

No. of ways = \(\frac{6!}{2!2!}\)

= \(\frac{6 \times 5 \times 4 \times 3 \times 2!}{2 \times 1 \times 2!}\)

= 180 ways

Correct Answer: C. (\(\frac{2}{3}, \frac{-4}{3}\))

Explanation

The equation for a circle with centre coordinates (a, b) and radius r is

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

Expanding the above equation, we have

\(x^{2} - 2ax +a^{2} + y^{2} - 2by + b^{2} - r^{2} = 0\) so that

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

Taking the original equation given, \(3x^{2} + 3y^{2} - 4x + 8y = 2\) and making the coefficients of \(x^{2}\) and \(y^{2}\) = 1,

\(x^{2} + y^{2} - \frac{4x}{3} + \frac{8y}{3} = \frac{2}{3}\), comparing, we have

\(2a = \frac{4}{3}; 2b = \frac{-8}{3}\)

\(\implies a = \frac{2}{3}; b = \frac{-4}{3}\)

Explanation

Note that P(X not hitting the target) = \(\frac{2}{3}\)

P(Y not hittiing the target) = \(\frac{4}{5}\) and P(Z not hitting the target) = \(\frac{3}{4}\)

So that, P(only one) = (\(\frac{1}{3} \times \frac{4}{5} \times \frac{3}{4}\)) + (\(\frac{2}{3} \times \frac{1}{5} \times \frac{3}{4}\)) + (\(\frac{2}{3} \times \frac{4}{5} \times \frac{1}{4}\)) = \(\frac{12}{60} + \frac{6}{60} = \frac{13}{30}\)

= 0.43, correct to two decimal places.

Correct Answer: A. -5

Explanation

Let \(f(x) = y\), then we have

\(y = \frac{x+1}{2} \implies 2y = x+1; x = 2y-1\)

Let \(f^{1}(x) = x; x = 2y-1\), replacing y with x,

\(f^{1}(x) = 2x - 1 \implies f^{1}(-2) = 2(-2) -1= -5\)

Correct Answer: D. \(4 + 2\sqrt{3}\)