Long Question 5

Posted on May 18, 2020 by Myhometuition

Question 5 (10 marks):

\begin{array}{l} (a) Prove \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) = \cos 3 x \\ (b) Hence, \\ (i) solve the equation s i n (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2} for 0 \leq x \leq 2 π \\ and give your answer in the simplest fraction form in terms of π radian . \\ (i i) sketch the graph of y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} for 0 \leq x \leq π . \end{array}

Solution:

\begin{array}{l} (a) Left hand side, \\ \sin (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) \\ = [\sin 3 x \cos \frac{π}{6} + \cos 3 x \sin \frac{π}{6}] - [\sin 3 x \cos \frac{π}{6} - \cos 3 x \sin \frac{π}{6}] \\ = 2 [\cos 3 x \sin \frac{π}{6}] \\ = 2 [\cos 3 x (\frac{1}{2})] \\ = \cos 3 x (right hand side) \end{array}

\begin{array}{l} (b) (i) \\ \sin (\frac{3 x}{2} + \frac{π}{6}) - \sin (\frac{3 x}{2} - \frac{π}{6}) = \frac{1}{2}, 0 \leq x \leq 2 π \\ \cos \frac{3 x}{2} = \frac{1}{2} \\ \frac{3 x}{2} = \frac{π}{3}, (2 π - \frac{π}{3}), (2 π + \frac{π}{3}) \\ \frac{3 x}{2} = \frac{π}{3}, \frac{5 π}{3}, \frac{7 π}{3} \\ x = \frac{2 π}{9}, \frac{10 π}{9}, \frac{14 π}{9} \end{array}

\begin{array}{l} (b) (i i) \\ y = s i n (3 x + \frac{π}{6}) - \sin (3 x - \frac{π}{6}) - \frac{1}{2} for 0 \leq x \leq π . \\ y = \cos 3 x - \frac{1}{2} \end{array}

Long Questions (Question 12)

Posted on May 18, 2020 by Myhometuition

Question 12 (10 marks):
(a) The mass of honeydews produced in a plantation is normally distributed with a mean of 0.8 kg and a standard deviation of 0.25 kg. The honeydews are being classified into three grades A, B and C according to their masses:

Grade A > Grade B > Grade C

(i) The minimum mass of a grade A honeydew is 1.2 kg.
If a honeydew is picked at random from the plantation, find the probability that the honeydew is of grade A.

(ii) Find the minimum mass, in kg, of grade B honeydew if 20% of the honeydews are of grade C.

(b) At the Shoot the Duck game booth at an amusement park, the probability of winning is 25%.
Jacky bought tickets to play n games. The probability for Jacky to win once is 10 times the probability of losing all games.

(i) Find the value of n.

(ii) Calculate the standard deviation of the number of wins.

Solution:
μ = 0.8 kg, σ = 0.25 kg

(a)(i)

\begin{array}{l} P (grade A) = P (X > 1.2) \\ = P (Z > \frac{1.2 - 0.8}{0.25}) \\ = P (Z > 1.6) \\ = 0.0548 \end{array}

(a)(ii)

\begin{array}{l} P (grade C) = 0.2 \\ P (X < m) = 0.2 \\ P (Z < \frac{m - 0.8}{0.25}) = 0.2 \\ P (Z < - 0.842) = 0.2 \\ \frac{m - 0.8}{0.25} = - 0.842 \\ m - 0.8 = - 0.2105 \\ m = 0.5895 \\ Minimum mass of grade B honeydew \\ is the same as the maximum mass of \\ grade C honeydew . \\ Minimum mass of grade B = 0.5895 kg \end{array}

(b)

\begin{array}{l} p = 0.25, X = B (n, 0.25) \\ P (X = r) = {}^{n}C_{r} p^{r} q^{n - r} \\ = {}^{n}C_{r} {(0.25)}^{r} {(0.75)}^{n - r} \end{array}

(b)(i)

\begin{array}{l} P (X = 1) = 10 P (X = 0) \\ {}^{n}C_{r} {(0.25)}^{1} {(0.75)}^{n - r} = 10 \times {}^{n}C_{0} {(0.25)}^{0} {(0.75)}^{n} \\ n \times 0.25 \times {(0.75)}^{n - r} = 10 \times 1 \times 1 \times {(0.75)}^{n} \\ 0.25 n \times \frac{{(0.75)}^{n - 1}}{{0.75}^{n}} = 10 \\ 0.25 n \times {0.75}^{- 1} = 10 \\ \frac{1}{4} n (\frac{4}{3}) = 10 \\ \frac{1}{3} n = 10 \\ n = 30 \end{array}

(b)(ii)

\begin{array}{l} n = 30, p = 0.25 \\ Standard deviation \\ = \sqrt{n p (1 - p)} \\ = \sqrt{30 \times 0.25 \times 0.75} \\ = 2.372 \end{array}

Long Questions (Question 11)

Posted on May 18, 2020 by Myhometuition

Question 11 (10 marks):
A study shows that the credit card balance of the customers is normally distributed as shown in Diagram 6.

(a)(i) Find the standard deviation.
(ii) If 30 customers are chosen at random, find the number of customers who have a credit card balance between RM1800 and RM3000.

(b) It is found that 25% of the customers have a credit card balance less than RM y.
Find the value of y.

Solution:
(a)(i)

\begin{array}{l} μ = 2870, x = 3770 \\ P (X > 3770) = 15.87 % \\ P (Z > \frac{3770 - 2870}{σ}) = 0.1587 \\ P (Z > 1.0) = 0.1587 \\ \frac{3770 - 2870}{σ} = 1.0 \\ σ = 900 \end{array}

(a)(ii)

\begin{array}{l} P (1800 < X < 3000) \\ = P (\frac{1800 - 2870}{900} < Z < \frac{3000 - 2870}{900}) \\ = P (- 1.189 < Z < 0.144) \\ = 1 - P (Z \leq - 1.189) - P (Z \geq 0.144) \\ = 1 - 0.1172 - 0.4427 \\ = 0.4401 \\ Number of customers = 0.4401 \times 30 \\ = 14 \end{array}

(b)

\begin{array}{l} μ = 2870, x = y \\ P (x < y) = 25 % \\ P (Z < \frac{y - 2870}{900}) = 0.25 \\ \frac{y - 2870}{900} = - 0.674 \\ y = 2263.40 \end{array}

Long Question 10

Posted on May 17, 2020 by Myhometuition

Question 10 (10 marks):
Diagram shows a curve y = 2x² – 18 and the straight line PQ which is a tangent to the curve at point K.

It is given that the gradient of the straight line PQ is 4.
(a) Find the coordinates of point K
(b) Calculate the area of the shaded region.

(c) When the region bounded by the curve, the x-axis and the straight line y = h is rotated through 180^o about the y-axis, the volume generated is 65π unit³.
Find the value of h.

Solution:
(a)

\begin{array}{l} y = 2 x^{2} - 18 \\ \frac{d y}{d x} = 4 x \\ Gradient of straight line P Q = 4 \\ 4 x = 4 \\ x = 1 \\ When x = 1, \\ y = 2 {(1)}^{2} - 18 = - 16 \\ Coordinates of K = (1, - 16) . \end{array}

(b)

\begin{array}{l} At x -axis, y = 0 \\ ∴ 2 x^{2} - 18 = 0 \\ 2 x^{2} = 18 \\ x^{2} = 9 \\ x = \pm 3 \\ The curve cuts the x -axis at (- 3, 0) and (3, 0) . \\ Area of shaded region \\ = Area of triangle - Area under the curve \\ = \frac{1}{2} (5 - 1) (16) - \int_{1}^{3} y d x \\ = 32 - \int_{1}^{3} (2 x^{2} - 18) d x \\ = 32 - | {[\frac{2 x^{3}}{3} - 18 x]}_{1}^{3} | \\ = 32 - | (\frac{2 {(3)}^{3}}{3} - 18 (3)) - (\frac{2 {(1)}^{3}}{3} - 18 (1)) | \\ = 32 - | (18 - 54 - \frac{2}{3} + 18) | \\ = 32 - | - 18 \frac{2}{3} | \\ = 32 - 18 \frac{2}{3} \\ = 13 \frac{1}{3} {units}^{2} \end{array}

(c)

\begin{array}{l} Volume generated = 65 π \\ π \int_{h}^{0} x^{2} d y = 65 π \\ π \int_{h}^{0} (\frac{y}{2} + 9) d x = 65 π \leftarrow \begin{array}{l} y = 2 x^{2} - 18 \\ x^{2} = \frac{y}{2} + 9 \end{array} \\ {[\frac{y^{2}}{4} + 9 y]}_{h}^{0} = 65 \\ 0 - (\frac{h^{2}}{4} + 9 h) = 65 \\ - \frac{h^{2}}{4} - 9 h = 65 \\ h^{2} + 36 h + 260 = 0 \\ (h + 10) (h + 26) = 0 \\ h = - 10 or h = - 26 (rejected) \\ Thus, h = - 10 \end{array}

Long Question 9

Posted on May 17, 2020 by Myhometuition

Question 9 (10 marks):
Diagram shows the straight line 4y = x – 2 touches the curve x = y² + 6 at point P.

Find
(a) the coordinates of P,
(b) the area of the shaded region,
(c) the volume of revolution, in terms of π, when the region bounded by the curve and the straight line x = 8 is revolved through 180^o about the x-axis.

Solution:
(a)

\begin{array}{l} 4 y = x - 2......... (1) \\ x = y^{2} + 6......... (2) \\ Substitute (2) into (1): \\ 4 y = (y^{2} + 6) - 2 \\ y^{2} - 4 y + 4 = 0 \\ (y - 2) (y - 2) = 0 \\ y - 2 = 0 \\ y = 2 \\ Substitute y = 2 into (2): \\ x = {(2)}^{2} + 6 \\ x = 10 \\ Thus, P = (10, 2) . \end{array}

(b)

\begin{array}{l} At x-axis, y = 0 \\ ∴ 4 y = x - 2 \\ 0 = x - 2 \\ x = 2 \\ Area of shaded region \\ = Area of triangle - Area under the curve \\ = \frac{1}{2} (10 - 2) (2) - \int_{6}^{10} y d x \\ = 8 - \int_{6}^{10} \sqrt{x - 6} d x \leftarrow \begin{array}{l} x = y^{2} + 6 \\ y = \sqrt{x - 6} \end{array} \\ = 8 - \int_{6}^{10} {(x - 6)}^{\frac{1}{2}} d x \\ = 8 - {[\frac{{(x - 6)}^{\frac{1}{2} + 1}}{\frac{1}{2} + 1}]}_{6}^{10} \\ = 8 - {[\frac{2 {(x - 6)}^{\frac{3}{2}}}{3}]}_{6}^{10} \\ = 8 - [\frac{2 {(10 - 6)}^{\frac{3}{2}}}{3} - \frac{2 {(6 - 6)}^{\frac{3}{2}}}{3}] \\ = 8 - \frac{16}{3} \\ = \frac{8}{3} {units}^{2} \end{array}

(c)

\begin{array}{l} Volume of revolution \\ = π \int_{6}^{8} y^{2} d x \\ = π \int_{6}^{8} (x - 6) d x \leftarrow \begin{array}{l} x = y^{2} + 6 \\ y^{2} = x - 6 \end{array} \\ = π {[\frac{x^{2}}{2} - 6 x]}_{6}^{8} \\ = π [(32 - 48) - (18 - 36)] \\ = 2 π {units}^{3} \end{array}

SPM Practice 3 (Linear Law) – Question 3

Posted on May 17, 2020 by Myhometuition

Question 3
The table below shows the corresponding values of two variables, x and y, that are related by the equation

y = 5 h x^{2} + \frac{k}{h} x

, where h and k are constants.

(a) Using a scale of 2 cm to 1 unit on the x - axis and 2 cm to 0.2 units on the

\frac{y}{x}

– axis, plot the graph of

\frac{y}{x}

against x . Hence, draw the line of best fit.

(b) Use your graph in (a) to find the values of
(i) h,
(ii) k,
(iii) y when x = 6.

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

(b) (iii) find the value of y when x = 6.

SPM Practice 3 (Linear Law) – Question 2

Posted on May 17, 2020 by Myhometuition

Question 2 (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 2(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}

Long Question 10

Posted on May 17, 2020 by Myhometuition

Question 10 (10 marks):
Diagram shows a triangle ABC. The straight line AE intersects with the straight line BC at point D. Point V lies on the straight line AE.

\begin{array}{l} It is given that \vec{B D} = \frac{1}{3} \vec{B C}, \vec{A C} = 6 \underset{˜}{x} and \vec{A B} = 9 \underset{˜}{y} . \\ (a) Express in terms of \underset{˜}{x} and / or \underset{˜}{y} : \\ (i) \vec{B C}, \\ (i i) \vec{A D} . \\ (b) It is given that \vec{A V} = m \vec{A D} and \vec{B V} = n (\underset{˜}{x} - 9 \underset{˜}{y}), where m and n are constants . \\ Find the value of m and of n . \\ (c) Given \vec{A E} = h \underset{˜}{x} + 9 \underset{˜}{y}, where h is a constant, find the value of h . \end{array}

Solution:
(a)(i)

\begin{array}{l} \vec{B C} = \vec{B A} + \vec{A C} \\ = - 9 \underset{˜}{y} + 6 \underset{˜}{x} \\ = 6 \underset{˜}{x} - 9 \underset{˜}{y} \end{array}

(a)(ii)

\begin{array}{l} \vec{A D} = \vec{A B} + \vec{B D} \\ = 9 \underset{˜}{y} + \frac{1}{3} \vec{B C} \\ = 9 \underset{˜}{y} + \frac{1}{3} (6 \underset{˜}{x} - 9 \underset{˜}{y}) \\ = 9 \underset{˜}{y} + 2 \underset{˜}{x} - 3 \underset{˜}{y} \\ = 2 \underset{˜}{x} + 6 \underset{˜}{y} \end{array}

(b)

\begin{array}{l} Given \vec{A V} = m \vec{A D} \\ = m (2 \underset{˜}{x} + 6 \underset{˜}{y}) \\ = 2 m \underset{˜}{x} + 6 m \underset{˜}{y} \\ \vec{A V} = \vec{A B} + \vec{B V} \\ = 9 \underset{˜}{y} + n (\underset{˜}{x} - 9 \underset{˜}{y}) \\ = 9 \underset{˜}{y} + n \underset{˜}{x} - 9 n \underset{˜}{y} \\ = n \underset{˜}{x} + (9 - 9 n) \underset{˜}{y} \\ By equating the coefficients of \underset{˜}{x} and \underset{˜}{y}, \\ 2 m \underset{˜}{x} + 6 m \underset{˜}{y} = n \underset{˜}{x} + (9 - 9 n) \underset{˜}{y} \\ 2 m = n \\ n = 2 m ............. (1) \\ 6 m = 9 - 9 n ............. (2) \\ Substitute (1) into (2), \\ 6 m = 9 - 9 (2 m) \\ 6 m = 9 - 18 m \\ 24 m = 9 \\ m = \frac{9}{24} = \frac{3}{8} \\ From (1) : \\ n = 2 (\frac{3}{8}) = \frac{3}{4} \end{array}

(c)

\begin{array}{l} A, D and E are collinear . \\ \vec{A D} = k (\vec{A E}) \\ \vec{A D} = k (h \underset{˜}{x} + 9 \underset{˜}{y}) \\ 2 \underset{˜}{x} + 6 \underset{˜}{y} = k h \underset{˜}{x} + 9 k \underset{˜}{y} \\ Equating the coefficients of \underset{˜}{y} : \\ 9 k = 6 \\ k = \frac{6}{9} \\ k = \frac{2}{3} \\ Equating the coefficients of \underset{˜}{x} : \\ k h = 2 \\ (\frac{2}{3}) h = 2 \\ h = 2 \times \frac{3}{2} \\ h = 3 \end{array}

Long Question 9

Posted on May 17, 2020 by Myhometuition

Question 9 (10 marks):
Diagram 5 shows triangles OAQ and OPB where point P lies on OA and point Q lies on OB. The straight lines AQ and PB intersect at point R.

\begin{array}{l} It is given that \vec{O A} = 18 \underset{˜}{x}, \vec{O B} = 16 \underset{˜}{y}, O P : P A = 1 : 2, O Q : Q B = 3 : 1, \\ \vec{P R} = m \vec{P B} and \vec{Q R} = n \vec{Q A}, where m and n are constants . \\ (a) Express \vec{O R} in terms of \\ (i) m, \underset{˜}{x} and \underset{˜}{y}, \\ (i i) n, \underset{˜}{x} and \underset{˜}{y}, \\ (b) Hence, find the value of m and of n . \\ (c) Given | \underset{˜}{x} | = 2 units, | \underset{˜}{y} | = 1 unit and O A is perpendicular to O B calculate | \vec{P R} | . \end{array}

Solution:
(a)(i)

\begin{array}{l} \vec{O R} = \vec{O P} + \vec{P R} \\ = \frac{1}{3} \vec{O A} + m \vec{P B} \\ = \frac{1}{3} (18 \underset{˜}{x}) + m (\vec{P O} + \vec{O B}) \\ = 6 \underset{˜}{x} + m (- 6 \underset{˜}{x} + 16 \underset{˜}{y}) \end{array}

(a)(ii)

\begin{array}{l} \vec{O R} = \vec{O Q} + \vec{Q R} \\ = \frac{3}{4} \vec{O B} + n \vec{Q A} \\ = \frac{3}{4} (16 \underset{˜}{y}) + n (\vec{Q O} + \vec{O A}) \\ = 12 \underset{˜}{y} + n (- 12 \underset{˜}{y} + 18 \underset{˜}{x}) \\ = (12 - 12 n) \underset{˜}{y} + 18 n \underset{˜}{x} \end{array}

(b)

\begin{array}{l} 6 \underset{˜}{x} + m (- 6 \underset{˜}{x} + 16 \underset{˜}{y}) = (12 - 12 n) \underset{˜}{y} + 18 n \underset{˜}{x} \\ 6 \underset{˜}{x} - 6 m \underset{˜}{x} + 16 m \underset{˜}{y} = 18 n \underset{˜}{x} + 12 \underset{˜}{y} - 12 n \underset{˜}{y} \\ by comparison; \\ 6 - 6 m = 18 n \\ 1 - m = 3 n \\ m = 1 - 3 n .............. (1) \\ 16 m = 12 - 12 n \\ 4 m = 3 - 3 n .............. (2) \\ Substitute (1) into (2), \\ 4 (1 - 3 n) = 3 - 3 n \\ 4 - 12 n = 3 - 3 n \\ 9 n = 1 \\ n = \frac{1}{9} \\ Substitute n = \frac{1}{9} into (1), \\ m = 1 - 3 (\frac{1}{9}) \\ m = \frac{2}{3} \end{array}

[adinserter block="3"]

(c)

\begin{array}{l} | \underset{˜}{x} | = 2, | \underset{˜}{y} | = 1 \\ \vec{P R} = \frac{2}{3} \vec{P B} \\ = \frac{2}{3} (- 6 \underset{˜}{x} + 16 \underset{˜}{y}) \\ = - 4 \underset{˜}{x} + \frac{32}{3} \underset{˜}{y} \\ | \vec{P R} | = \sqrt{{[- 4 (2)]}^{2} + {[\frac{32}{3} (1)]}^{2}} \\ = \sqrt{\frac{1600}{9}} \\ = \frac{40}{3} units \end{array}

Long Question (Question 10)

Posted on May 17, 2020 by Myhometuition

Question 10 (5 marks):
Solve the following simultaneous equations:
x – 3y = 1,
x²+ 3xy + 9y²= 7

Solution:

\begin{array}{l} x - 3 y = 1................... (1) \\ x^{2} + 3 x y + 9 y^{2} = 7................... (2) \\ From (1) : x = 3 y + 1................... (3) \\ Substitute (3) into (2) : \\ {(3 y + 1)}^{2} + 3 (3 y + 1) y + 9 y^{2} = 7 \\ 9 y^{2} + 6 y + 1 + 9 y^{2} + 3 y + 9 y^{2} - 7 = 0 \\ 27 y^{2} + 9 y - 6 = 0 \\ 9 y^{2} + 3 y - 2 = 0 \\ (3 y - 1) (3 y + 2) = 0 \\ y = \frac{1}{3} or - \frac{2}{3} \\ Substitute y into (3) : \\ When y = \frac{1}{3} \\ x = 3 (\frac{1}{3}) + 1 = 2 \\ When y = - \frac{2}{3} \\ x = 3 (- \frac{2}{3}) + 1 = - 1 \\ Hence, the solutions are x = 2, y = \frac{1}{3} or x = - 1, y = - \frac{2}{3} . \end{array}