Correct Answer: A. \(-1\)

Explanation

\(16^{3x} = \frac{1}{4}(32^{x - 1})\)

\((2^{4})^{3x} = (2^{-2})((2^{5})^{x - 1})\)

\(2^{12x} = 2^{-2 + 5x - 5}\)

\(12x = -7 + 5x\)

\(7x = -7 \implies x = -1\)

Correct Answer: B. -5

Explanation

\(\begin{vmatrix} 2 & -3 \\ 1 & 4 \end{vmatrix} \begin{vmatrix} -6 \\ k \end{vmatrix} \begin{vmatrix} 3 \\ -26 \end{vmatrix} = 15\)

\(\begin{vmatrix} 2[-6] & - 3k \\ 1[-6] & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)

\(\begin{vmatrix} -12 & - 3k \\ -6 & + 4k \end{vmatrix} = \begin{vmatrix} 3 \\ -26 \end{vmatrix}\)

-12 - 3k = 3

-3k = 3 + 12

k = \(\frac{15}{-3}\)

k = -5

Correct Answer: A. 8.25

Explanation

Mean \(\bar{x}\) = \(\frac{3 + 7 + 6 + 2 + 8 + 5 + 9 + 1 + 4 + 10}{10} \)

= \(\frac{55}{10} = 5.50\)

Variance = \(\frac{\sum (x - \bar{x})^{2}}{n}\)

= \(\frac{82.5}{10}\)

= 8.25

Correct Answer: A. \(\frac{3}{5}\)

Explanation

5+4+x+9+2x+1 = 40

19+3x = 40

3x = 21

x = 7

The probability that the hunter covered at least 6km, means the hunter covered either 6km or 7km, or 8km.

24 hunters covered at least 6km

Explanation

(a) Time taken to reach the ground

Using S = ut + 1/2gt\(^2\)

(u= 0 for a body falling from rest)

50 = 0 x t + 1/2 x 10t\(^2\)

5t\(^2\) = 50;

t\(^2\) =50/5 = 10

t = √10 =3.16

Distance when it strikes the ground This journey is independent of gravity

d =vt =30 x 3.16 = 94.8m

(b) Using Lami's rule,

\(\frac{20}{sin 34}\) = \(\frac{T_1}{sin 90}\) = \(\frac{T_2}{sin 56}\)

= T\(_1\) \(\frac{20.sin 90}{sin 34}\) = T\(_2\) \(\frac{20.sin 56}{sin 34}\)

T\(_1\) = 35.79N and T\(_2\) = 29.65N

Explanation

nC\(_4\) = \(\frac{n!}{[n-4]! 4!}\) → \(\frac{n[n-1][n-2][n-3][n-4]!}{[n-4]!4!}\)

nC\(_5\) = \(\frac{n!}{[n-5]! 5!}\) → \(\frac{n[n-1][n-2][n-3][n-4][n-5]!}{[n-5]!5!}\)

and nC\(_6\) = \(\frac{n!}{[n-6]! 6!}\) → \(\frac{n[n-1][n-2][n-3][n-4][n-5][n-6]!}{[n-6]!6!]}\)

d = U2 - U1 = U3 - U2

nC\(_5\) - nC\(_4\) = \(\frac{n(n-1)(n-2)(n-3)(n-9)}{5!}\)

nC\(_6\) - nC\(_5\) = \(\frac{n(n-1)(n-2)(n-3)(n-4)(n-11)}{5! 6}\)

nC\(_5\) - nC\(_4\) = nC\(_6\) - nC\(_5\)

\(\frac{n(n-1)(n-2)(n-3)(n-9)}{5!}\) = \(\frac{n(n-1)(n-2)(n-3)(n-4)(n-11)}{5! 6}\)

divide both sides by n(n-1)(n-2)(n-3)

\(\frac{n-9}{5!}\) = \(\frac{[n-4][n-11]}{5! 6}\)

multiply both sides by 5! * 6

6(n-9) = (n-4)(n-11)

6n - 54 = n\(^2\) -11n - 4n + 44

6n - 54 = n\(^2\) - 15n + 44

n\(^2\) - 21n + 98 = 0

n\(^2\) - 7n - 14n + 98 = 0

n(n - 7) -14(n - 7) = 0

(n-7)(n-14) = 0

n = 7 or 14

ii.

d = U2 - U1

when n = 7

d = 7C\(_5\) - 7C\(_4\)

d = \(\frac{7*6*5!}{2! * 5!}\) - \(\frac{7*6*5*4!}{3! * 4!}\)

d = 21 - 35

d = -14

when n = 14

d = 14C\(_5\) - 14C\(_4\)

d = \(\frac{14*13*12*11*10*9!}{9! * 5!}\) - \(\frac{14*13*12*11*10!}{10! * 4!}\)

d = 2002 - 1001

d = 1001

Correct Answer: A. \(f^{-1} : x \to \frac{2x + 3}{x - 1}, x \neq 1\)

Explanation

\(f(x) = \frac{x + 3}{x - 2}\)

\(f(y) = \frac{y + 3}{y - 2}\)

Let f(y) = x,

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

\(x(y - 2) = y + 3\)

\(xy - y = 2x + 3 \implies y(x - 1) = 2x + 3\)

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

Correct Answer: A. \(\frac{-1}{2}\)

Explanation

Given the formula for p * q as: \(p + q + 2pq\) and its identity element is 0, such that if, say, t is the inverse of p, then

\(p * t = 0\), then \(p + t + 2pt = 0 \therefore p + (1 + 2p)t = 0\)

\(t = \frac{-1}{1 + 2p}\) is the formula for the inverse of p and is undefined on R when

\(1 + 2p) = 0\) i.e when \(2p = -1; p = \frac{-1}{2}\).

Correct Answer: A. 2.9.seconds

Explanation

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

40 = \(\frac{1}{2} \times 9.8 \times t^2\)

\(\frac{80}{9.8} = t^2\)

t = \(\sqrt{8.16326}\)

t≈ 2.9 secs

Explanation

Curve : \(y = x(3 - x^{2})\)

\(y = 3x - x^{3}\)

Gradient function : \(\frac{\mathrm d y}{\mathrm d x} = 3 - 3x^{2}\)

When x = 2, y = \(3(2) - (2^{3}) = -2\)

Gradient of tangent : \(3 - 3(-2)^{2} = -9\)

Gradient of normal : \(\frac{1}{9}\)

Equation of normal through (2, -2)

\(\frac{y + 2}{x - 2} = \frac{1}{9}\)

\(9y + 18 = x - 2 \implies x - 9y - 20 = 0\)