Correct Answer: D. 1 and 4

Explanation

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

This movement is against gravity, so it is negative.

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

\(s = 20m, u = 25ms^{-1}\)

\(20 = 25t - \frac{10t^{2}}{2} \implies 20 = 25t - 5t^{2}\)

\(5t^{2} - 25t + 20 = 0 \)

\(5t^{2} - 5t - 20t + 20 = 0 \implies 5t(t - 1) - 20(t - 1) = 0\)

\(5t - 20 = \text{0 or t - 1 = 0}\)

\(t = \text{1 or 4}\)

Correct Answer: D. \(\frac{32}{729}\)

Explanation

ar = \(\frac{2}{9}\) .....(i)

ar\(^3\) = \(\frac{8}{81}\) ......(ii)

\(\frac{ar3}{ar} = \frac{8}{81} \times \frac{9}{2}\)

r\(^2 = \frac{4}{9}\)

r = \(\sqrt{\frac{4}{9}}\)

= \(\frac{2}{3}\)

ar = \(\frac{2}{9}\)

a(\(\frac{2}{3}\)) = \(\frac{2}{9}\)

a = (\(\frac{2}{3}\)) = \(\frac{2}{9}\)

a = \(\frac{2}{9} \times \frac{3}{2}\)

a = \(\frac{1}{3}\)

T\(_r\) = ar\(^5\) = (\(\frac{1}{3}\))(\(\frac{2}{5}\))\(^5\)

= \(\frac{32}{729}\)

Correct Answer: D. -7/6

Explanation

\(∫^0_{-1}\) (x + 1)(x - 2) dx

= \(∫^0_{-1}\) \(x^2 - x - 2\) dx

Integrated \(x^2 - x - 2\) = \(\frac{x^3}{3} - \frac{x^2}{2} -2\)

Correct Answer: D. -0.9

Explanation

\(y = 4 - 9x\)

\(\frac{\mathrm d y}{\mathrm d x} = \frac{\Delta y}{\Delta x} = -9\)

\(\frac{\Delta y}{0.1} = -9 \implies \Delta y = -9 \times 0.1 = -0.9\)

Correct Answer: B. 60

Explanation

No of ways = 4 x 5 x 3 = 60 ways

Correct Answer: B. \(x \geq 3\) and \(x \leq -1\)

Explanation

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

\(x^{2} + x - 3x - 3 = (x + 1)(x - 3) \geq 0\)

\(x = -1 ; x = 3\)

Check: \(x = -1 : (-1)^{2} - 2(-1) = 1 + 2 \geq 3\) (satisfied)

\(-1 < x < 3 : 0^{2} - 2(0) = 0 \geq 3\) (not satisfied)

\(x < -1 : (-2)^{2} - 2(-2) = 4 + 4 = 8 \geq 3\) (satisfied)

\(x = 3 : 3^{2} - 2(3) = 9 - 6 = 3 \geq 3\) (satisfied)

\(x > 3 : 4^{2} - 2(4) = 16 - 8 = 8 \geq 3\) (satisfied)

\(\therefore x^{2} - 2x \geq \text{3 is satisfied in the region x} \leq \text{-1 and x} \geq 3\)

Correct Answer: A. \(2x - 3\)

Correct Answer: D. \(x^{2} + y^{2} - x - y - 8 = 0\)

Explanation

Given the endpoints of the diameter |EF|, the midpoint is the centre of the circle

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

The radius is the distance from the centre to any point on the circle. Using \((\frac{1}{2}, \frac{1}{2})\) and \((3, 2)\);

\(r^{2} = (3 - \frac{1}{2})^{2} + (2 - \frac{1}{2})^{2} = \frac{25}{4} + \frac{9}{4}\)

\(r^{2} = \frac{34}{4}\)

The equation of a circle is given as:

\((x - a)^{2} + (y - b)^{2} = r^{2}\), (a, b) as the centre of the circle.

\(= (x - \frac{1}{2})^{2} + (y - \frac{1}{2})^{2} = \frac{34}{4}\)

\(x^{2} - x + \frac{1}{4} + y^{2} - y + \frac{1}{4} = \frac{17}{2}\)

= \(x^{2} - y^{2} - x - y - 8 = 0\)

Explanation

f:x → x\(^2\) + 2 and g:x → \(\frac{1}{x+2}\).

(g?f)\(^{-1}\)

(g?f)\(^{x^2+2}\)

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

(g?f) = \(\frac{1}{(x^2+4}\)

let y = \(\frac{1}{(x^2+4}\)

y\(((x^2+4)\) = 1

\(yx^2+4y = 1\)

x\(^2\) = \(\frac{1-4y}{y}\)

x = \(\sqrt \frac{1-4y}{y}\)

(g?f)\(^{-1}\) = \(\sqrt \frac{1-4x}{x}\)

Explanation

They form 3 consecutive integers

let the number of orange be x, apple = x+1 and avocado pear = x+2

x + x+1 + x+2 = 12

3x + 3 = 12

3x = 9

x = 3

apple = 4

avocado pear = 5

probability that orange is drawn = \(\frac{3}{12}\)= \(\frac{1}{4}\)

probability that apple is drawn = \(\frac{4}{12}\)= \(\frac{1}{3}\)

probability that avocado pear is drawn = \(\frac{5}{12}\)

probability that the first is orange and the second is an avocado pear = \(\frac{3}{12}\) * \(\frac{5}{11}\) = \(\frac{5}{44}\)

ii.

both are of same fruit, they are both Orange or apple or Avocado pear

p(O∩O) + p(A∩A) + p(P∩P) = \(\frac{1}{4} * \frac{2}{11} + \frac{1}{3} * \frac{3}{11} + \frac{5}{12} * \frac{4}{11}\)

= \(\frac{1}{22} + \frac{1}{11} + \frac{5}{33}\)

= \(\frac{19}{66}\)

iii.

at least one is an apple = p(A∩O) or p(O∩A) or p(A∩P) or p(P∩A) or p(A∩A)

p(A∩O) + p(O∩A) + p(A∩P) + p(P∩A) + p(A∩A) =

= \(\frac{4}{12} * \frac{3}{11} + \frac{3}{11} * \frac{4}{12} + \frac{4}{12} * \frac{5}{11} + \frac{5}{12} * \frac{4}{11} + \frac{4}{12} * \frac{3}{11} \)

= \(\frac{12}{132}\) + \(\frac{12}{132}\) + \(\frac{20}{132}\) + \(\frac{20}{132}\) + \(\frac{12}{132}\)

= \(\frac{76}{132}\)

= \(\frac{19}{33}\)