Long Question 6

Posted on April 22, 2020 by user

Question 6:
Diagram below shows part of the curve

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

which passes through B (1, 2).

(a) Find the equation of the tangent to the curve at the point B.
(b) A region is bounded by the curve, the x-axis and the straight lines x = 2 and x = 3.
(i) Find the area of the region.
(ii) The region is revolved through 360° about the x–axis. Find the volume generated, in terms of p.

Solution:
(a)

\begin{array}{l} y = \frac{2}{{(3 x - 2)}^{2}} = 2 {(3 x - 2)}^{- 2} \\ \frac{d y}{d x} = - 4 {(3 x - 2)}^{- 3} (3) \\ \frac{d y}{d x} = \frac{- 12}{{(3 x - 2)}^{3}} \\ \frac{d y}{d x} = \frac{- 12}{{(3 (1) - 2)}^{3}}, x = 1 \\ \frac{d y}{d x} = - 12 \\ y - 2 = - 12 (x - 1) \\ y - 2 = - 12 x + 12 \\ y = - 12 x + 14 \end{array}

(b)(i)

\begin{array}{l} Area = \int_{2}^{3} y d x \\ = \int_{2}^{3} \frac{2}{{(3 x - 2)}^{2}} d x = \int_{2}^{3} 2 {(3 x - 2)}^{- 2} d x \\ = {[\frac{2 {(3 x - 2)}^{- 1}}{- 1 (3)}]}_{2}^{3} \\ = {[- \frac{2}{3 (3 x - 2)}]}_{2}^{3} \\ = [- \frac{2}{3 [3 (3) - 2]}] - [- \frac{2}{3 [3 (2) - 2]}] \\ = - \frac{2}{21} + \frac{1}{6} \\ = \frac{1}{14} {unit}^{2} \end{array}

(b)(ii)

\begin{array}{l} Volume generated \\ = π \int y^{2} d x \\ = π \int_{2}^{3} \frac{4}{{(3 x - 2)}^{4}} d x = π \int_{2}^{3} 4 {(3 x - 2)}^{- 4} d x \\ = π {[\frac{4 {(3 x - 2)}^{- 3}}{- 3 (3)}]}_{2}^{3} = π {[- \frac{4}{9 {(3 x - 2)}^{3}}]}_{2}^{3} \\ = π [- \frac{4}{9 {[3 (3) - 2]}^{3}}] - [- \frac{4}{9 {[3 (2) - 2]}^{3}}] \\ = π (- \frac{4}{3087} + \frac{4}{576}) \\ = \frac{31}{5488} π {unit}^{3} \end{array}

Gradient Of A Straight Line

Posted on April 22, 2020 by user

(B) Gradient of a Straight Line
If the points

A (x_{1}, y_{1})

and

B (x_{2}, y_{2})

lie on the straight line

y = m x + c

, then gradient of ,the straight line

m = \frac{y_{2} - y_{1}}{x_{2} - x_{1}} o r \frac{y_{1} - y_{2}}{x_{1} - x_{2}}

Steps To Draw The Line Of Best Fit

Posted on April 22, 2020 by user

Steps to draw a line of Best Fit
(i) Select suitable scales for the x-axis and the y-axis, make sure the points plotted accurately and the graph produced is large enough on the graph paper,
(ii) Mark the points correctly,
(iii) Use a long and transparent ruler to draw the line of best fit.

Step 1 : Select the suitable scale on x and y axis
(the graph produced must be more than 50% of the graph paper)

Step 2 : Mark the points correctly

Step 3 : Draw the Line of Best Fit

* Note

-the line passes through four points

-one point is above the line

-one point is below the line

SPM Practice 2 (Question 1 – 3)

Posted on April 22, 2020 by user

Question 1:
Reduce non-linear relation, $y = p x^{n - 1}$ , where k and n are constants, to linear equation. State the gradient and vertical intercept for the linear equation obtained.
[Note : Reduce No-linear function to linear function]

Solution:

Question 2:
The diagram shows a line of best fit by plotting a graph of $y^{2}$ against $\sqrt{x}$ .

Find the equation of the line of best fit.
Determine the value of
1. x when y = 4,
2. y when x = 25.

Solution:

Question 3:
The diagram shows part of the straight line graph obtained by plotting $\sqrt{y}$ against $x^{2}$ .

Express y in terms of x.

Solution:

SPM Practice 3 (Linear Law) – Question 1

Posted on April 22, 2020 by user

Question 1 (10 marks):
Use a graph to answer this question.
Table 1 shows the values of two variables, x and y, obtained from an experiment.
The variables x and y are related by the equation

y - \sqrt{h} = \frac{h k}{x}

, where h and k are constants.

(a) Plot xy against x, using a scale of 2 cm to 1 unit on the x-axis and 2 cm to 5 units on the xy-axis.
Hence, draw the line of best fit.

(b) Using the graph in 9(a), find
(i) the value of h and of k,
(ii) the correct value of y if one of the values of y has been wrongly recorded during the experiment.

Solution:
(a)

(b)

\begin{array}{l} y - \sqrt{h} = \frac{h k}{x} \\ x y - \sqrt{h} x = h k \\ x y = \sqrt{h} x + h k \\ Y = m X + C \\ Y = x y, m = \sqrt{h}, C = h k \end{array}

(b)(i)

\begin{array}{l} m = \frac{36.5}{5.1} \\ \sqrt{h} = \frac{36.5}{5.1} \\ \sqrt{h} = 7.157 \\ h = 51.22 \\ C = - 4 \\ h k = - 4 \\ k = \frac{- 4}{h} \\ k = - \frac{4}{51.22} \\ k = - 0.0781 \end{array}

(b)(ii)

\begin{array}{l} x y = 21 \\ 3.5 y = 21 \\ y = \frac{21}{3.5} = 6.0 \\ Correct value of y is 6 .0 . \end{array}

Tips To Reduce Non-Linear Function To Linear Function

Posted on April 22, 2020 by user

Tips:
(1) The equation must have one constant (without x and y).
(2) X and Y cannot have constant, but can have the variables (for example x and y).
(3) m and c can only have the constant (for example a and b), cannot have the variables x and y.

Examples
(1)
X and Y cannot have constant, but can have the variables (for example x and y)

(2)
m and c can only have the constant (for example a and b), cannot have the variables x and y

SPM Practice 2 (Linear Law) – Question 4

Posted on April 22, 2020 by user

Question 4
The table below shows the corresponding values of two variables, x and y, that are related by the equation

y = q x + \frac{p}{q x}

, where p and q are constants.

One of the values of y is incorrectly recorded.
(a) Using scale of 2 cm to 5 units on the both axis, plot the graph of xy against

x^{2}

. Hence, draw the line of best fit

(b) Use your graph in (a) to answer the following questions:
(i) State the values of y which is incorrectly recorded and determine its actual value.
(ii) Find the value of p and of q.

Solution
Step 1 : Construct a table consisting X and Y.

Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit

Steps to draw line of best fit - Click here

(b) (i) State the values of y which is incorrectly recorded and determine its actual value.

Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph

Step 4 : Rewrite the original equation given and reduce it to linear form

Step 5 : Compare with the values of m and c obtained, find the values of the unknown required

Distance Between Two Points

Posted on April 22, 2020 by user

Distance between point $A (x_{1}, y_{1})$ and point is $B (x_{2}, y_{2})$ given by

$\sqrt{{(x_{1} - x_{2})}^{2} + {(y_{1} - y_{2})}^{2}}$

Reduce Non-Linear Function To Linear Function – Examples (G) To (L)

Posted on April 22, 2020 by user

Examples:
Reduce each of the following equations to the linear form. Hence, state the gradient and the Y-intercept of the linear equations in terms of a and b.

(g)

k x^{2} + t y^{2} = x

(h)

y = \frac{x}{p + q x}

(i)

h y = x + \frac{k}{x}

(j)

y = a b^{x}

(k)

y = a x^{b}

(l)

y = a b^{x + 1}

[Note :
X and Y cannot have constant, but can have the variables (for example x and y)
m and c can only have the constant (for example a and b), cannot have the variables x and y]

Solution:

SPM Practice 3 (Linear Law) – Question 3

Posted on April 22, 2020 by user

Question 3
The table below shows the values of two variables, x and y, obtained from an experiment. The variables x and y are related by the equation

\frac{a}{y} = \frac{b}{x} + 1

, where k and p are constants.

(a) Based on the table above, construct a table for the values of

\frac{1}{x}

and

\frac{1}{y}

. Plot

\frac{1}{y}

against

\frac{1}{x}

, using a scale of 2 cm to 0.1 unit on the

\frac{1}{x}

- axis and 2 cm to 0.2 unit on the

\frac{1}{y}

- axis. Hence, draw the line of best fit.
(b) Use the graph from (b) to find the value of
(i) a,
(ii) b.

Solution
Step 1 : Construct a table consisting X and Y.

Step 2 : Plot a graph of Y against X, using the scale given and draw a line of best fit

Steps to draw line of best fit - Click here

Step 3 : Calculate the gradient, m, and the Y-intercept, c, from the graph

Step 4 : Rewrite the original equation given and reduce it to linear form

Step 5 : Compare with the values of m and c obtained, find the values of the unknown required