Correct Answer: B. \(\frac{1}{4}\)

Explanation

The equation \(ax^{2} + bx + c\) is a perfect square if \(b^{2} = 4ac\).

\(x^{2} - x + p\)

\((-1)^{2} = 4(1)(p)\)

\(1 = 4p \implies p = \frac{1}{4}\)

Explanation

v = 0 m/s ; u = 30 m/s ; s = ? ; g = 10 m/s\(^{2}\).

(a) For maximum height, we use the formula

\(v^{2} = u^{2} - 2gs\)

\(s = \frac{u^{2} - v^{2}}{2g}\)

= \(\frac{30^{2} - 0^{2}}{2(10)}\)

= \(\frac{900}{20} = 45m\)

(b) Time to reach highest point:

\(v = u - gt \implies t = \frac{u - v}{g}\)

\(t = \frac{30 - 0}{10} = 3s\)

If the particle took 3s to get to the maximum height,

Time to reach the ground = 2(3) = 6s.

(c) From a height 40m above the ground, we use

\(s = ut - \frac{1}{2} gt^{2}\)

\(40 = 30t - 5t^{2}\)

\(5t^{2} - 30t + 40 = 0\)

\(5t^{2} - 20t - 10t + 40 = 0 \implies 5t(t - 4) - 10(t - 4) = 0\)

\((5t - 10)(t - 4) = 0 \implies \text{2s and 4s}\)

\(\therefore\) At times 2s and 4s, the height attained = 40m.

Correct Answer: B. 0

Explanation

To find the maximum value, we can use the second derivative test where, given \(f(x)\), the second derivative < 0, makes it a maximum value.

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

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} + 4x + 1 = 0\)

Solving, we have \( x = \frac{-1}{3}\) or \(-1\).

\(\frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 6x + 4\)

When \(x = \frac{-1}{3}, \frac{\mathrm d^{2} y}{\mathrm d x^{2}} = 2 > 0\)

When \(x = -1, \frac{\mathrm d^{2} y}{\mathrm d x^{2}} = -2 < 0\)

At maximum value of x being -1, \(y = -1(-1 + 1)^{2} = 0\)

Explanation

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

Let u = x\(^2\)

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

\(\frac{du}{dx} 2x\)

\(\frac{dv}{dx} = \frac{3}{2}\)(1 + x)\(^\frac{1}{2}\)

\(\frac{dy}{dx} = v\frac{du}{dx} + u\frac{dv}{dx}\)

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

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

(b)

2y - x = 3

when x = 0, y = \(\frac{3}{2}\)

when y = 0, x = -3

h = -3, k = \(\frac{3}{2}\)

Using

(x - h)\(^2\) + (y - k)\(^2\) = r\(^2\)

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

r(2, 3), x = 2, y = 3

(2 + 3)\(^2\) + (3 + \(\frac{3}{2}\))\(^2\) = r\(^2\)

\(\frac{109}{4}\) = r\(^2\)

The equation will give

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

= \(\frac{109}{4}\)

Correct Answer: A. 3.0 seconds

Explanation

\(s = ut + \frac{1}{2}gt^{2}\)

\(u = 0; s = 45m; g = 10ms^{-2}\)

\(45 = 0 + \frac{1}{2}(10)t^{2} = 5t^{2}\)

\(t^{2} = \frac{45}{5} = 9\)

\(t = 3.0s\)

Explanation

\(p(Ali) = p(A) = \frac{2}{3} ; p(Baba) = p(B) = \frac{3}{4} ; p(Katty) = p(K) = \frac{4}{5}\)

(a) \(p(\text{only Katty and Baba}) = p(K) \times p(B) \times p(A')\)

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

= \(\frac{1}{5}\)

(b) \(p(\text{none gains admission}) = p(A') \times p(K') \times p(B')\)

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

= \(\frac{1}{60}\).

(c) \(p(\text{at most two will gain admission}) = p(\text{none gains}) + p(\text{one gains}) + p(\text{two gains})\)

\(p(\text{one gains}) = p(A) \times p(B') \times p(K') + p(B) \times p(A') \times p(K') + p(K) \times p(A') \times p(B')\)

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

= \(\frac{2}{60} + \frac{3}{60} + \frac{4}{60}\)

= \(\frac{9}{60}\)

\(p(\text{two gains}) = p(A) \times p(B) \times p(K') + p(A) \times p(K) \times p(B') + p(B) \times p(K) \times p(A')\)

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

= \(\frac{6}{60} + \frac{8}{60} + \frac{12}{60}\)

= \(\frac{26}{60}\)

\(p(\text{at most two gains admission}) = \frac{1}{60} + \frac{9}{60} + \frac{26}{60}\)

= \(\frac{36}{60}\)

= \(\frac{3}{5}\).

Correct Answer: C. \(\frac{7}{15}\)

Explanation

\(P(\text{both bulbs are good}) = \frac{7}{10} \times \frac{6}{9}\)

= \(\frac{7}{15}\)

Correct Answer: C. 36

Explanation

Sample size = 6 x 6 = 36.

Explanation

(a)(i) Position vector of A, \(\overrightarrow{OA} = i + 5j\)

Position vector of B, \(\overrightarrow{OB} = 3i + 9j\)

Position vector of C, \(\overrightarrow{OC} = -i + j\)

The points A, B and C are collinear if \(|\overrightarrow{AB}| = k|\overrightarrow{BC}|\), where k is a constant.

If k > 0, then \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) are in the same direction.

If k < 0, then they are in opposite directions, in this case they are A, C, B.

\(\overrightarrow{AB} = \overrightarrow{AO} + \overrightarrow{OB} = - \overrightarrow{OA} + \overrightarrow{OB}\)

= \(- i - 5j + (3i + 9j) = 2i + 4j = 2(i + 2j)\)

\(\overrightarrow{BC} = \overrightarrow{BO} + \overrightarrow{OC} = - \overrightarrow{OB} + \overrightarrow{OC}\)

= \(- 3i - 9j + (-i + j) = - 4i - 8j = -4(i + 2j)\)

\(\overrightarrow{AB} = -\frac{1}{2} \overrightarrow{BC}\)

\(\therefore\) A, B and C are collinear with \(\overrightarrow{AB}\) and \(\overrightarrow{BC}\) in opposite directions.

(ii) \(|\overrightarrow{AB}| : |\overrightarrow{BC}| = 1 : 2\)

(b)

We take moments about X.

Clockwise moment = \(500 \times 8 + 100 \times 12 = 5200N\)

Anticlockwise moment = \(T_{2} \times 20\)

From the principles of moment,

\(20T_{2} = 5200N \implies T_{2} = 260N\)

For vertical equilibrium,

\(T_{1} + T_{2} = 600N\)

\(\therefore T_{1} = 600 - T_{2} = 600 - 260 =340N\)

The tensions in the strings are 340N and 260N respectively.

Correct Answer: D. (-2, -6)

Explanation

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

\(\frac{\mathrm d y}{\mathrm d x} = 2x + 4 = 0\)

\(2x = -4 \implies x = -2\)

\(y(-2) = (-2^{2}) + 4(-2) - 2 = 4 - 8 - 2 = -6\)

\(\therefore (x, y) = (-2, -6)\)