Loading....

Press & Hold to Drag Around |
|||

Click Here to Close |

**Question 1**
**Report**

Find the point (x, y) on the Euclidean plane where the curve y = 2x2 - 2x + 3 has 2 as gradient

**Answer Details**

We know that the gradient of a curve is given by its derivative. Therefore, we need to find the derivative of the given curve and equate it to 2 to find the point where the gradient is 2. y = 2x^2 - 2x + 3 dy/dx = 4x - 2 Equating dy/dx to 2, we get: 4x - 2 = 2 4x = 4 x = 1 Substituting x = 1 in the original equation, we get: y = 2(1)^2 - 2(1) + 3 y = 3 Therefore, the point where the curve has a gradient of 2 is (1, 3). So, the correct option is: (1, 3).

**Question 2**
**Report**

Solve for r in the following equation 1r−1 $\frac{1}{r-1}$ + 2r+1 $\frac{2}{r+1}$ = 3r

**Answer Details**

1r−1
$\frac{1}{r-1}$ + 2r+1
$\frac{2}{r+1}$ = 3r
$\frac{3}{r}$

Multiply through by r(r -1) which is the LCM

= (r)(r + 1) + 2(r)(r - 1)

= 3(r - 1)(r + 1)

= r2 + r + 2r2 - 2r

3r2 - 3 = 3r2

r = 3r2 - 3

-r = -3

∴ r = 3

**Question 3**
**Report**

An open rectangular box is made of wood 2cm thick. If the internal dimensions of the box are 50cm long, 36cm wide and 20cm deep, the box volume of wood in the box is

**Answer Details**

Internal dimension are 50cm, 36cm and 20cm

internal volume = 50 x 36 x 20cm3

1000 x 36cm3

= 36000cm3

External dimension are 54cm x 40cm x 22cm

= 2160cm2 x 22cm = 47520cm3

Volume of wood = Ext. volume - Int. volume

= 47,520cm3 - 36,000cm3

= 11,520cm3

**Question 4**
**Report**

x12345f21212 $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ f& 2& 1& 2& 1& 2\end{array}$

Find the variance of the frequency distribution above

**Answer Details**

xffx¯x?x(¯x?x)2f(¯x?x)2122?248212?111326000414111221024882418
$\begin{array}{cccccc}x& f& fx& \stackrel{\xaf}{x}?x& (\stackrel{\xaf}{x}?x{)}^{2}& f(\stackrel{\xaf}{x}?x{)}^{2}\\ 1& 2& 2& ?2& 4& 8\\ 2& 1& 2& ?1& 1& 1\\ 3& 2& 6& 0& 0& 0\\ 4& 1& 4& 1& 1& 1\\ 2& 2& 10& 2& 4& 8\\ & 8& 24& & & 18\end{array}$

x = ?fx?f
$\frac{?fx}{?f}$

= 248
$\frac{24}{8}$

= 3

Variance (62) = ?f(¯x?x)2?f
$\frac{?f(\stackrel{\xaf}{x}?x{)}^{2}}{?f}$

= 188
$\frac{18}{8}$

= 94

**Question 5**
**Report**

Find the inequality which represents the shaded portion in the diagram

**Answer Details**

The shaded area in the diagram represents the region below the line passing through the points (1, 0) and (0, 2). To find the equation of the line, we first need to find its slope: slope = (change in y) / (change in x) slope = (0 - 2) / (1 - 0) slope = -2 Next, we use the point-slope form of the equation of a line to find the equation of the line: y - 0 = -2(x - 1) y = -2x + 2 Now we can test each inequality option to see which one represents the shaded region. We can do this by picking a point in the shaded region, plugging in its coordinates into the inequality, and checking if the inequality is true. For example, the point (0, 0) is in the shaded region, so we plug in x=0 and y=0 into each inequality: - 2(0) - 0 - 2 ≥ 0 is false - 2(0) - 0 - 2 ≤ 0 is true - 2(0) - 0 - 2 < 0 is true - 2(0) - 0 - 2 > 0 is false Therefore, the inequality that represents the shaded portion in the diagram is 2x - y - 2 ≤ 0.

**Question 6**
**Report**

Evaluate log5(0.04)log318−log32

**Answer Details**

To evaluate the expression, we need to use the properties of logarithms. First, we can simplify the expression inside the parentheses of the logarithm: log5(0.04) = log5(4/100) = log5(4) - log5(100) = log5(2^2) - log5(10^2) = 2log5(2) - 2 Next, we can simplify the second part of the expression: log318 - log32 = log3(18/32) = log3(9/16) = log3(3^2) - log3(2^4) = 2log3(3) - 4 Substituting these simplified expressions back into the original expression, we get: 2log5(2) - 2 - (2log3(3) - 4) Simplifying further, we get: 2log5(2) - 2 - 2log3(3) + 4 Combining like terms, we get: 2log5(2) - 2log3(3) + 2 Now, we can plug in the values of log5(2) and log3(3) using a calculator: log5(2) ≈ 0.4307 and log3(3) = 1 Substituting these values, we get: 2(0.4307) - 2(1) + 2 ≈ 0.8614 - 2 + 2 ≈ -0.1386 Therefore, the answer is -1, since it is the only option that is negative.

**Question 7**
**Report**

Class Interval1−56−1011−1516−2021−25Frequency6152072 $\begin{array}{cccccc}\text{Class Interval}& 1-5& 6-10& 11-15& 16-20& 21-25\\ Frequency& 6& 15& 20& 7& 2\end{array}$

Estimate the median of the frequency distribution above

**Answer Details**

Median = L + [N2−ffm
$\frac{\frac{N}{2}-f}{fm}$]h

N = Sum of frequencies

L = lower class boundary of median class

f = sum of all frequencies below L

fm = frequency of modal class and

h = class width of median class

Median = 11 + [502−2120
$\frac{\frac{50}{2}-21}{20}$]5

= 11 + (25−2120
$\frac{25-21}{20}$)5

= 11 + ((4)20
$\frac{(4)}{20}$)

11 + 1 = 12

**Question 8**
**Report**

In the frustum of the cone, the top diagram is twice the bottom diameter. If the height of the frustum is h centimeters, find he height of the cone.

**Answer Details**

xr $\frac{x}{r}$ = x+h2r $\frac{x+h}{2r}$

2 x r = r (x + h)

Total height of cone = x + h

but x = h

total height = 2h

**Question 9**
**Report**

Find the value of log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63

**Answer Details**

log10 r + log10 r2 + log10 r4 + log10 r8 + log10 r16 + log10 r32 = 63

log10r63 = 63

63 = 1063

∴ r = 10

**Question 10**
**Report**

Find P if x?3(1?x)(x+2) $\frac{x?3}{(1?x)(x+2)}$ = p1?x $\frac{p}{1?x}$ + Qx+2

**Answer Details**

x−3(1−x)(x+2)
$\frac{x-3}{(1-x)(x+2)}$ = p1−x
$\frac{p}{1-x}$ + Qx+2
$\frac{Q}{x+2}$

Multiply both sides by LCM i.e. (1 - x(x + 2))

∴ x - 3 = p(x + 2) + Q(1 - x)

When x = +1

(+1) - 3 = p(+1 + 2) + Q(1 - 1)

-2 = 3p + 0(Q)

3p = -2

∴ p = −23

**Question 11**
**Report**

In the diagram, PTS is a tangent to the circle TQR at T. Calculate < RTS

**Answer Details**

RTS = RQT (angle between a tangent and a chord at the point of contact is equal to the angle in the alternate segment) But R = Q + T = 180

RQT = 180? $?$ - (50 + 60)

= 180? $?$ - 110? $?$

= 70? $?$

Since RQT = RTS = 70?

**Question 12**
**Report**

Calculate the perimeter, in cm, of a sector of a circle of radius 8cm and angle 45o

**Answer Details**

Perimeter = OP + OQ + PQ

= 8 + 8 + PQ

length PQ = θ360×2πr
$\frac{\theta}{360\times 2\pi r}$

= 45360
$\frac{45}{360}$ x 2 x π
$\pi $ x 8

= 2π
$\pi $

Perimeter of sector 2r + L

Where l = length of arc and r = radius

∴ P = 2(8) + 2π
$\pi $

= 16 + 2π

**Question 13**
**Report**

Evaluate 13÷[57(910?1+34)]

**Answer Details**

13÷[57(910?1+34)]
$\frac{1}{3}\xf7[\frac{5}{7}(\frac{9}{10}?1+\frac{3}{4})]$

13÷[57(910?1010+34)]
$\frac{1}{3}\xf7[\frac{5}{7}(\frac{9}{10}?\frac{10}{10}+\frac{3}{4})]$

= 13÷[57(?110+34)]
$\frac{1}{3}\xf7[\frac{5}{7}(\frac{?1}{10}+\frac{3}{4})]$

= 13÷[57(?2+1520)]
$\frac{1}{3}\xf7[\frac{5}{7}(\frac{?2+15}{20})]$

= 13÷[57×1320]
$\frac{1}{3}\xf7[\frac{5}{7}\times \frac{13}{20}]$

13+[1328]
$\frac{1}{3}+[\frac{13}{28}]$ = 13×2813
$\frac{1}{3}\times \frac{28}{13}$

= 2839

**Question 14**
**Report**

In the diagram. Find h

**Answer Details**

A△ $\u25b3$ = √S(S−a)(S−b)(S−c) $\sqrt{S(S-a)(S-b)(S-c)}$ (Hero's Formula)

S = a+b+c2 $\frac{a+b+c}{2}$ = 5+6+72 $\frac{5+6+7}{2}$

182=9 $\frac{18}{2}=9$

A△ $\u25b3$ √9×4×3×2 $\sqrt{9}\times 4\times 3\times 2$

√216=6√6cm3 $\sqrt{216}=6\sqrt{6}c{m}^{3}$

A△ $\u25b3$ = 12×6×h $\frac{1}{2}\times 6\times h$

6√6=12×7×h $\sqrt{6}=\frac{1}{2}\times 7\times h$

h = 12h√6

**Question 15**
**Report**

Given that for sets A and B, in a universal set E, A ? $?$ B then A ? $?$(A ? $?$ B)1 is

**Answer Details**

A ⊂
$\subset $ B means A is contained in B i.e. A is a subset of B(A ∩
$\cap $ B)1 = A1

A(A ∩
$\cap $ B)1 = A ∩
$\cap $ A1

The intersection of complement of a set P and P1 has no element

i.e. n(A ∩
$\cap $ A1) = ϕ

**Question 16**
**Report**

If M(4, q) is the mid-point of the line joining L(p, -2) and N(q, p). Find the values of p and q

**Answer Details**

To find the values of p and q, we can use the midpoint formula, which states that the midpoint of a line segment between two points (x1, y1) and (x2, y2) is ((x1+x2)/2, (y1+y2)/2). Here, we are given that M(4,q) is the midpoint of the line joining L(p,-2) and N(q,p). So we know that: - The x-coordinate of M is the average of the x-coordinates of L and N: (p+q)/2 = 4 - The y-coordinate of M is the average of the y-coordinates of L and N: (-2+p)/2 = q Simplifying the first equation, we get: (p+q)/2 = 4 p+q = 8 q = 8-p Substituting this into the second equation, we get: (-2+p)/2 = q (-2+p)/2 = 8-p -2+p = 16-2p 3p = 18 p = 6 Therefore, the values of p and q are: - p = 6 - q = 8-p = 2 So the correct option is p = 6, q = 2.

**Question 17**
**Report**

x12345fy+2y−22y−3y+43y−4 $\begin{array}{cccccc}x& 1& 2& 3& 4& 5\\ f& y+2& y-2& 2y-3& y+4& 3y-4\end{array}$

This table shows the frequency distribution of a data if the mean is 4314 $\frac{43}{14}$ find y

**Question 18**
**Report**

In the diagram, PQRS is a parallelogram. Find the value of < SQR

**Answer Details**

SQR + RQV + VQU = 18o angle on a straight line SP is parallel to QR and PV is parallel to TR

< STP = < RQV = 30o

But SQR + 30o + 50o = 180o

SQR = 180 - 80

= 100o

**Question 19**
**Report**

Simplify √48 $48$ - 9√3 $\frac{9}{\sqrt{3}}$ + √75

**Answer Details**

√48
$48$ - 9√3
$\frac{9}{\sqrt{3}}$ + √75
$75$

Rearrange = √48
$48$ + √75
$75$ - 9√3
$\frac{9}{\sqrt{3}}$

= (√16 x √3) + (√25 x √3) - 9√3
$\frac{9}{\sqrt{3}}$

=4√3 + 5√3 - 9√3
$\frac{9}{\sqrt{3}}$

Rationalize →
$\to $ 9√3 = 9√3
$\frac{9}{\sqrt{3}}$ x √3√3
$\frac{\sqrt{3}}{\sqrt{3}}$

= 9√3√9
$\frac{9\sqrt{3}}{\sqrt{9}}$ - 9√3√3
$\frac{9\sqrt{3}}{\sqrt{3}}$

= 3√3

**Question 20**
**Report**

Find the range of values of x for which 1x
$\frac{1}{x}$ > 2 is true

**Answer Details**

1x $\frac{1}{x}$ > 2 =