Correct Answer: B. \(\frac{-3}{(5 - x)^4}\)

Explanation

y = (5 - x)\(^{-3}\)

\(\frac{dy}{dx}\) = -3(5 - x)\(^{-4}\)

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

Correct Answer: A. N = X\(^{\frac{v}{p}}\) - M

Explanation

\(\frac{v}{p} = \frac{p}{p} log _x(M + N)\)

\(\log_x(M + N) = \frac{v}{p}\)

\(x^{\frac{v}{p}} = M + N\)

N = X\(^{\frac{v}{p}}\) - M

Correct Answer: D. \(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\)

Explanation

\(\begin{pmatrix} 1 & -2 \\ 3 & 4 \end{pmatrix} \(\begin{pmatrix} -2 & 3 \\ 1 & 0 \end{pmatrix}\)

=\(\begin{pmatrix} (1 \times -2) + (-2 \times 1) & (1 \times 3) + (-2 \times 0) \\ (3 \times -2) + (4 \times 1) & (3 \times 3) + (4 \times 0) \end{pmatrix}\)

= \(\begin{pmatrix} -4 & 3 \\ -2 & 9 \end{pmatrix}\)

Explanation

(a) \(\frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}}\)

\(\frac{^{9}P_{3}}{^{15}C_{3}} = \frac{9!}{(9 - 3)!} \times \frac{3! 12!}{15!}\)

= \(504 \times \frac{1}{455} = 1.10769 \)

\(\frac{^{5}C_{3}}{^{3}P_{2}} = \frac{5!}{2! 3!} \times \frac{1}{3!}\)

= \(10 \times \frac{1}{6} = 1.66667\)

\(\therefore \frac{^{9}P_{3}}{^{15}C_{3}} + \frac{^{5}C_{3}}{^{3}P_{2}} = 1.10769 + 1.66667 = 2.77436\)

\(\approxeq 2.77\) (2 d.p)

(b) 6 tutors and 10 pupils to select 2 tutors and 5 pupils.

Select one tutor, one more tutor is needed = \(^{5}C_{1} = 5 ways\)

Keep aside 2 students, we have \(^{8}C_{5} = 56 ways\)

= \(5 \times 56 = \text{280 ways}\).

Correct Answer: B. (\(\frac{1}{4}\), \(\frac{3}{4}\))

Explanation

2x\(^2\) + 2y\(^2\) - x - 3y - 41

standard equation of circle

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

General form of equation of a circle.

x\(^2\) + y\(^2\) + 2gx + 2fy + c = 0

a = -g, b = -f., r2 = g2 + f2 - c

the centre of the circle is (a,b)

comparing the equation with the general form of equation of circle.

2x\(^2\) + 2y\(^2\) - x - 3y - 41

= x\(^2\) + y\(^2\) + 2gx + 2fy + c

2x\(^2\) + 2y\(^2\) - x - 3y - 41 = 0

divide through by 2

g = \(\frac{-1}{4}\) ; 2g = \(\frac{-1}{2}\)

f = \(\frac{-3}{4}\) ; 2f = \(\frac{-3}{2}\)

a = -g → - \(\frac{-1}{4}\) ; = \(\frac{1}{4}\)

b = -f → - (\frac{-3}{4}\) = (\frac{3}{4}\)

therefore the centre is (\(\frac{1}{4}\), \(\frac{3}{4}\))

Explanation

(a) Using long division

(x + 2)(2x\(^2\) + 7x - 15)

(x + 2)(2x - 3)(x - 5)

zeros of g (x) are;

x + 2 =0, 2x - 3 = 0

x + 5 = 0, x = -2

x = \(\frac{5}{2}\) and x = -5

(b) 3(2^\(^x\)) + 3\(^{y - 2}\) = 25 .......x1

2\(^x\) - 3\(^{y + 1}\) = -19 .........x3

3(2\(^x\)) + 3\(^{y - 2}\) = 25

3\(^{y - 2}\) + 3(3\(^{y + 1}\)) = 82

\(\frac{3^y}{9} + \frac{3^y}{1}\) = 82

\(\frac{3^y + 9(3^y)}{9}\) = 82

3\(^y\) + 9(3\(^y)\)) = 82 x 9

let p = 3y

p + qp = 738

\(\frac{10p}{10} = \frac{738}{10}\)

p = 73.8, 3\(^y\) = 73.8

log3\(^y\) = log 73.8

\(\frac{y log 3}{log 3} = \frac{log 73.8}{\log 3}\)

y = 3.9

from

2\(^x\) + 3\(^{y + 1}\) = - 19

2\(^x\) + 3\(^{3.9 + 1}\) = - 19

2\(^x\) + 3\(^{4 - 9}\) = - 19

2\(^x\) - 217.7 = - 19

2\(^x\) = 198.71

log 2\(^x\) = log 198.71

\(\frac{x \log 2}{log 2} = \frac{log 198.71}{log 2}\)

x = 7.6

Correct Answer: D. \(y = x^{3} - 2x^{2} + 16\)

Explanation

The gradient of a curve is gotten by differentiating the equation of the curve. Therefore, given the gradient, integrate to get the equation of the curve back.

\(\frac{\mathrm d y}{\mathrm d x} = 3x^{2} - 4x\)

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

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

To find c (the constant of integration), when x = -2, y = 0

\(0 = (-2^{3}) - 2(-2^{2}) + c\)

\(0 = -8 - 8 + c \implies c = 16\)

\(\therefore y = x^{3} - 2x^{2} + 16\)

Correct Answer: B. \(x^2-2x-15=0\)

Explanation

\(x^2-\)(sum of roots)\(x+\)(product of roots) = \(0\)

\(4x^2-4x-15=0\)

Divide through by 4

\(=x^2-x-\frac{15}{4}=0\)

\(=x^2-x+(-\frac{15}{4})=0\)

\(=x^2-(1)x+(-\frac{15}{4})=0\)

sum of roots =1

= m + (m + 4) = 1

=2m+4=1

=2m=-3

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

The equation whose roots are 2m and 2m+8

2m=2×-\(\frac{3}{2}=-3\)and \(2m+8=2×-\frac{3}{2}+8=5\)

\(=x^2-(-3+5)x+(-3)(5)=0\)

\(=x^2-2x+(-15)=0\)

\(∴x^2-2x-15=0\)

Explanation

4 girls and 5 boys

(a) p(only boys) = \(\frac{^{5}C_{3}}{^{9}C_{3}}\)

= \(\frac{5!}{2!3!} \times \frac{3!6!}{9!}\)

= \(\frac{10 \times 1}{84}\)

= \(\frac{5}{42}\)

(b) p(more girls than boys) = p(3 girls and no boys) or p(2 girls and 1 boy)

Number of ways of selecting = \(^{4}C_{3} \times ^{5}C_{0} + ^{4}C_{2} \times ^{5}C_{1}\)

= \(4 \times 1 + 6 \times 5\)

= 34 ways.

Without restrictions, selection can be made in \(^{9}C_{3}\) ways.

= \(\frac{9!}{6! 3!}\)

= 84 ways.

\(\therefore\) p(more girls than boys) = \(\frac{34}{84}\)

= \(\frac{17}{42}\).

Explanation

(a) \(F_{1} = \begin{pmatrix} 35 \cos 210° \\ 35 \sin 210° \end{pmatrix}\)

\(F_{2} = \begin{pmatrix} 35\sqrt{3} \cos 300° \\ 35\sqrt{3} \sin 300° \end{pmatrix}\)

\(R = F_{1} + F_{2}\)

= \(\begin{pmatrix} -35 \cos 30 \\ -35 \sin 30 \end{pmatrix} + \begin{pmatrix} 35\sqrt{3} \cos 60 \\ -35\sqrt{3} \sin 60 \end{pmatrix}\)

= \(\begin{pmatrix} -30.31 \\ -17.5 \end{pmatrix} + \begin{pmatrix} 30.31 \\ -52.5 \end{pmatrix}\)

= \(\begin{pmatrix} 0 \\ -70 \end{pmatrix}\)

Magnitude of resultant of \(F_{1}\) and \(F_{2}\) = 70N.

(b) Resultant acting on the particle = resultant of \(F_{1}\) and \(F_{2}\) less resistance.

= 70N - 40N = 30N.