Correct Answer: C. 20m

Correct Answer: A. 1931.25 m\(^2\) ≤ A < 2021.25 m\(^2\)

Explanation

The sides have been given to the nearest meter, so

51.5 m ≤ length < 52.5

37.5 m ≤ width < 38.5

Minimum area = 37.5 x 51.5 = 1931.25 m\(^2\)

Maximum area = 38.5 x 52.5 = 2021.25 m\(^2\)

∴ The range of the area = 1931.25 m\(^2\) ≤ A < 2021.25 m\(^2\)

Correct Answer: C. \(\frac{4y - x}{y}\)

Explanation

(\(\frac{x -y}{y}\)); \(\frac{3}{1} - \frac{x y}{y}\)

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

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

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

Correct Answer: B. \(\frac{7-q}{4}\)

Correct Answer: D. \({x : 2< x \leq 6}\)

Explanation

(a) \((\frac{4}{25})^{-\frac{1}{2}} \times 2^{4} \div (\frac{15}{2})^{-2}\)

= \((\frac{25}{4})^{\frac{1}{2}} \times 2^{4} \div (\frac{2}{15})^{2}\)

= \(\frac{5}{2} \times 16 \times \frac{225}{4} = 2250\)

(b) \(\log_{5} (\frac{3}{5}) + 3\log_{5} (\frac{15}{2}) - \log_{5} (\frac{81}{8})\)

= \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} 15 - 3\log_{5} 2 - \log_{5} 81 + \log_{5} 8\)

= \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} (3 \times 5) - 3\log_{5} 2 - \log_{5} 3^{4} + \log_{5} 2^{3}\)

= \(\log_{5} 3 - \log_{5} 5 + 3\log_{5} 3 + 3\log_{5} 5 - 3\log_{5} 2 - 4\log_{5} 3 + 3\log_{5} 2\)

= \(2\log_{5} 5 \)

= \(2\)

Correct Answer: C. -12 cos (-4x)

Explanation

(b)(i) m\(^2\) = 5\(^2\) + 9\(^2\)

m\(^2\) = 25 + 81

m\(^2\) = 106

m = \(\sqrt{106}\) = 10.296 km/hr

Speed = \(\frac{distance}{times}\)

distance = speed x time

= 10.296 x 2

= 21km

(b)(ii)

tan\(\theta\) = \(\frac{9km/hr}{5km/hr}\)

tan\(\theta\) = 1.8

\(\theta\) = tan\(^{-1}\)

\(\theta\) = 60.9\(^o\)

the bearing of cyclist x from y

90\(^o\) + 19.9\(^o\) = 109.9\(^o\)

= 110\(^o\)

(c) The average speed = \(\frac{distance }{time}\)

= \(\frac{21}{4}\)km

= 5km/her

Explanation

(ai) Let the quadratic equation be \(ax^2+bx+c\)

\(g(x)=ax^2+bx+c\)

Since \((2x+1)\) is a factor,\(x=-\frac{1}{2}\)is a root

\(∴g(-\frac{1}{2})=a(-\frac{1}{2})^2+b(-\frac{1}{2})+c=0\)

\(=a(\frac{1}{4})-b(\frac{1}{2})+c=0\)

Multiply through by 4

=a-2b+4c=0-----(i)

For(x-1)and(x-2),x=1 and x = 2 respectively

So,

\(g(1)=a(1)^2+b(1)+c=-6\)

=a+b+c=-6-----(ii)

and

\(g(2)=a(2)^2+b(2)+c=-5\)

= 4a+2b+c=-5

Adding equations (i) and (iii) gives;

=5a+5b=-5

Divide through by 5

=a+b=-1......(iv)

Adding equation (i) to 2 times equation (iv) gives:

3a+6c=-12.....(v)

Equation (v) minus three times equation (iv) gives:

=3c=-12-(-3)

=3c=-9

=c\(\frac{-9}{3}=-3\)

Substitute (-3) for c in equation (v)

=3a+6(-3)=-12

=3a-18=-12

=3a=-12+18

=3a=6

=a=\(\frac{6}{3}=2\)

Substitute 2 for a and (-3) for c in equation (ii)

=2+6-3=-6

=b-1=-6

=b=-6+1

=b=-5

\(\therefore g(x)=2x^2-5x-3\)

(aii) \(2\times2 - 5x - 3\)

\(= 2\times2 - 6x + x - 3\)

\(= 2x(x - 3) + 1(x - 3)\)

\(= (x - 3)(2x + 1)\)

zeros of \(g(x)\) are \(2x + 1 = 0\) and \(x - 3 = 0\)

∴ zeros of \(g(x)\) are -\frac{1}{2}\) and \(3\)

(b) \((\frac{x}{2}-1)^8\)

rth term is given as \(^nC_r-1 a^n-(r-1)b^r-1\)

\(a=\frac{x}{2},b=-1,n=8,r=3,r-1=2\)

3rd term = \(^8C_2(\frac{x}{2})^8-2(-1)^2\)

= \(28×(\frac{x}{2})^6\times1\)

∴3rd term = \(28\times \frac{x^6}{64}=\frac{7x^6}{16}\)

Correct Answer: A. \(\frac{a \pi^2}{3} (x + 3y) \)