Long Question 3

Posted on May 17, 2020 by Myhometuition

Question 3:

(a) Prove that

\frac{2 \tan x}{2 - \sec^{2} x} = \tan 2 x .

(b)(i) Sketch the graph of y = – tan 2x for 0 ≤ x ≤ π .

(b)(ii) Hence, by drawing a suitable straight line on the same axes, find the number of solutions satisfying the equation

\frac{3 x}{π} + \frac{2 \tan x}{2 - \sec^{2} x} = 0

for 0 ≤ x ≤ π .

State the number of solutions.

Solution:

(a)

\begin{array}{l} \frac{2 \tan x}{2 - \sec^{2} x} = \tan 2 x \\ L H S : \\ \frac{2 \tan x}{2 - \sec^{2} x} = \frac{2 \tan x}{2 - (1 + \tan^{2} x)} \\ = \frac{2 \tan x}{2 - \tan^{2} x} \\ = \tan 2 x (RHS) \end{array}

(b)(i)

(b)(ii)

\begin{array}{l} \frac{3 x}{π} + \frac{2 \tan x}{2 - \sec^{2} x} = 0 \\ \frac{3 x}{π} + \tan 2 x = 0 \leftarrow from part (a) \\ - \tan 2 x = \frac{3 x}{π} \\ y = \frac{3 x}{π} \\ The suitable straight line to sketch is y = \frac{3 x}{π} . \end{array}

When x = 0, y = 0.

When x = π, y = 3.

Number of solutions = 3

Long Question 8

Posted on May 16, 2020 by Myhometuition

Question 8:
Diagram below shows quadrilateral OPQR. The straight line PR intersects the straight line OQ at point S.

\begin{array}{l} It is given that \vec{O P} = 7 \underset{˜}{x}, \vec{O R} = 5 \underset{˜}{y}, P S : S R = 3 : 1 and \vec{O R} is parallel to \vec{P Q} . \\ (a) Express in terms of \underset{˜}{x} and \underset{˜}{y}, \\ (i) \vec{P R} \\ (ii) \vec{O S} \\ (b) Using \vec{P Q} = m \vec{O R} and \vec{S Q} = n \vec{O S}, where m and n are constants, \\ Find the value of m and of n . \\ (c) Given that | \underset{˜}{y} | = 4 units and the area of O R S {is 50 cm}^{2}, find the \\ perpendicular distance from point S to O R . \end{array}

Solution:
(a)(i)

\begin{array}{l} \vec{P R} = \vec{P O} + \vec{O R} \\ = - 7 \underset{˜}{x} + 5 \underset{˜}{y} \end{array}

(a)(ii)

\begin{array}{l} \vec{O S} = \vec{O P} + \vec{P S} \\ = 7 \underset{˜}{x} + \frac{3}{4} \vec{P R} \\ = 7 \underset{˜}{x} + \frac{3}{4} (- 7 \underset{˜}{x} + 5 \underset{˜}{y}) \\ = 7 \underset{˜}{x} - \frac{21}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} \\ = \frac{7}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} \end{array}

(b)

\begin{array}{l} \vec{P S} = \vec{P Q} - \vec{S Q} \\ \frac{3}{4} \vec{P R} = m \vec{O R} - n \vec{O S} \\ \frac{3}{4} (- 7 \underset{˜}{x} + 5 \underset{˜}{y}) = m (5 \underset{˜}{y}) - n (\frac{7}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y}) \\ - \frac{21}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} = 5 m \underset{˜}{y} - \frac{7}{4} n \underset{˜}{x} - \frac{15}{4} m \underset{˜}{y} \\ - \frac{21}{4} \underset{˜}{x} + \frac{15}{4} \underset{˜}{y} = - \frac{7}{4} n \underset{˜}{x} + \frac{5}{4} m \underset{˜}{y} \\ - 21 \underset{˜}{x} + 15 \underset{˜}{y} = - 7 n \underset{˜}{x} + 5 m \underset{˜}{y} \\ 7 n = 21 \\ n = 3 \\ 5 m = 15 \\ m = 3 \end{array}

(c)

\begin{array}{l} Area of Δ O R S = 50 \\ \frac{1}{2} \times (5 \underset{˜}{y}) \times t = 50 \\ \frac{1}{2} \times 5 (4) \times t = 50 \\ 10 t = 50 \\ t = 5 \\ ∴ Perpendicular distance from point S to O R = 5 units . \end{array}

Long Question 7

Posted on May 16, 2020 by Myhometuition

Question 7:
Diagram below shows quadrilateral OPBC. The straight line AC intersects the straight line PQ at point B.

\begin{array}{l} It is given that \vec{O P} = \underset{˜}{a}, \vec{O Q} = \underset{˜}{b}, \vec{O A} = 4 \vec{A P}, \vec{O C} = 3 \vec{O Q}, \vec{P B} = h \vec{P Q} and \\ \vec{A B} = k \vec{A C} . \\ (a) Express \vec{O B} in terms of h, \underset{˜}{a} and \underset{˜}{b} . \\ (b) Express \vec{O B} in terms of k, \underset{˜}{a} and \underset{˜}{b} . \\ (c)(i) Find the value of h and of k . \\ (ii) Hence, state \vec{O B} in terms of \underset{˜}{a} and \underset{˜}{b} . \end{array}

Solution:
(a)

\begin{array}{l} \vec{O B} = \vec{O P} + \vec{P B} \\ = \underset{˜}{a} + h \vec{P Q} \\ = \underset{˜}{a} + h (\vec{P O} + \vec{O Q}) \\ = \underset{˜}{a} + h (- \underset{˜}{a} + \underset{˜}{b}) \\ = \underset{˜}{a} - h \underset{˜}{a} + h \underset{˜}{b} \\ \vec{O B} = (1 - h) \underset{˜}{a} + h \underset{˜}{b} \end{array}

(b)

\begin{array}{l} \vec{O B} = \vec{O P} + \vec{P B} \\ = \underset{˜}{a} + \vec{P A} + \vec{A B} \\ = \underset{˜}{a} + (- \frac{1}{5} \vec{O P}) + k \vec{A C} \\ = \underset{˜}{a} + (- \frac{1}{5} \underset{˜}{a}) + k (\vec{A O} + \vec{O C}) \\ = \frac{4}{5} \underset{˜}{a} + k (- \frac{4}{5} \vec{O P} + 3 \vec{O Q}) \\ = \frac{4}{5} \underset{˜}{a} + k (- \frac{4}{5} \underset{˜}{a} + 3 \underset{˜}{b}) \\ = \frac{4}{5} \underset{˜}{a} - \frac{4}{5} k \underset{˜}{a} + 3 k \underset{˜}{b} \\ \vec{O B} = \frac{4}{5} (1 - k) \underset{˜}{a} + 3 k \underset{˜}{b} \end{array}

(c)(i)

\begin{array}{l} (1 - h) \underset{˜}{a} + h \underset{˜}{b} = \frac{4}{5} (1 - k) \underset{˜}{a} + 3 k \underset{˜}{b} \\ 1 - h = \frac{4}{5} - \frac{4}{5} k .......... (1) \\ h = 3 k .......... (2) \\ Substitute (2) into the (1) \\ 1 - 3 k = \frac{4}{5} - \frac{4}{5} k \\ 5 - 15 k = 4 - 4 k \\ 11 k = 1 \\ k = \frac{1}{11} \\ Substitute k = \frac{1}{11} into (2) \\ h = 3 (\frac{1}{11}) \\ = \frac{3}{11} \end{array}

(c)(ii)

\begin{array}{l} \vec{O B} = (1 - h) \underset{˜}{a} + h \underset{˜}{b} \\ when h = \frac{3}{11} \\ = (1 - \frac{3}{11}) \underset{˜}{a} + (\frac{3}{11}) \underset{˜}{b} \\ = \frac{8}{11} \underset{˜}{a} + \frac{3}{11} \underset{˜}{b} \end{array}

Long Question 6

Posted on May 16, 2020 by Myhometuition

Question 6:
Diagram below shows a trapezium OABC and point D lies on AC.

\begin{array}{l} It is given that \vec{O C} = 18 \underset{˜}{b}, \vec{O A} = 6 \underset{˜}{a} and \vec{O C} = 2 \vec{A B} . \\ (a) Express in terms of \underset{˜}{a} and \underset{˜}{b}, \\ (i) \vec{A C} \\ (ii) \vec{O B} \\ (b) It is given that \vec{A D} = k \vec{A C}, where k is a constant . \\ Find the value of k if the points O, D and B are collinear . \end{array}

Solution:
(a)(i)

\begin{array}{l} \vec{A C} = \vec{A O} + \vec{O C} \\ = - 6 \underset{˜}{a} + 18 \underset{˜}{b} \\ = 18 \underset{˜}{b} - 6 \underset{˜}{a} \end{array}

(a)(ii)

\begin{array}{l} \vec{O C} = 2 \vec{A B} \\ 18 \underset{˜}{b} = 2 (\vec{A O} + \vec{O B}) \\ 18 \underset{˜}{b} = 2 (- 6 \underset{˜}{a} + \vec{O B}) \\ 18 \underset{˜}{b} = - 12 \underset{˜}{a} + 2 \vec{O B} \\ \vec{O B} = 6 \underset{˜}{a} + 9 \underset{˜}{b} \end{array}

(b)

\begin{array}{l} \vec{O D} = h \vec{O B} \\ = h (6 \underset{˜}{a} + 9 \underset{˜}{b}) \\ = 6 h \underset{˜}{a} + 9 h \underset{˜}{b} \\ \vec{A D} = \vec{O D} - \vec{O A} \\ = 6 h \underset{˜}{a} + 9 h \underset{˜}{b} - 6 \underset{˜}{a} \\ = \underset{˜}{a} (6 h - 6) + 9 h \underset{˜}{b} \\ \vec{A D} = k \vec{A C} \\ \underset{˜}{a} (6 h - 6) + 9 h \underset{˜}{b} = k (18 \underset{˜}{b} - 6 \underset{˜}{a}) \\ \underset{˜}{a} (6 h - 6) + 9 h \underset{˜}{b} = - 6 k \underset{˜}{a} + 18 k \underset{˜}{b} \\ 6 h - 6 = - 6 k \\ h - 1 = - k \\ h = 1 - k .......... (1) \\ 9 h = 18 k \\ h = 2 k \\ From (1), \\ 1 - k = 2 k \\ 3 k = 1 \\ k = \frac{1}{3} \end{array}

Long Question 8

Posted on May 16, 2020 by Myhometuition

Question 8:
Diagram below shows the curve

y = \frac{4}{x^{2}}

and the straight line y = mx + c. The straight line y = mx + c is a tangent to the curve at (2, 1).

(a) Find the value of m and of c.

(b) Calculate the area of the shaded region.

(c) It is given that the volume generated when the region bounded by the curve, the x–axis and the straight lines x = 2 and x = h is revolved through 360^o about the x-axis is

\frac{38 π}{81} {unit}^{3} .

Find the value of h, such that h > 2.

Solution:
(a)

\begin{array}{l} y = \frac{4}{x^{2}} = 4 x^{- 2} \\ \frac{d y}{d x} = - 8 x^{- 3} \\ = - \frac{8}{x^{3}} \\ At x = 2, \\ \frac{d y}{d x} = - \frac{8}{2^{3}} \\ = - 1 \\ Equation of tangent: \\ y - y_{1} = m (x - x_{1}) \\ y - 1 = - 1 (x - 2) \\ y = - x + 2 + 1 \\ y = - x + 3 \\ m = - 1, c = 3 \end{array}

(b)

\begin{array}{l} At x -axis, y = 0 \\ From the straight line y = - x + 3, x = 3 \\ Area of the shaded region \\ = Area under the curve - Area of triangle \\ = \int_{2}^{4} y d x - \frac{1}{2} \times 1 \times 1 \\ = \int_{2}^{4} (4 x^{- 2}) d x - \frac{1}{2} \\ = {[\frac{4 x^{- 1}}{- 1}]}_{2}^{4} - \frac{1}{2} \\ = {[- \frac{4}{x}]}_{2}^{4} - \frac{1}{2} \\ = [- \frac{4}{4} - (- \frac{4}{2})] - \frac{1}{2} \\ = \frac{1}{2} {unit}^{2} \end{array}

(c)

\begin{array}{l} Volume generated = \frac{38 π}{81} \\ π \int_{2}^{h} y^{2} d x = \frac{38 π}{81} \\ \int_{2}^{h} {(\frac{4}{x^{2}})}^{2} d x = \frac{38}{81} \\ \int_{2}^{h} (\frac{16}{x^{4}}) d x = \frac{38}{81} \\ \int_{2}^{h} (16 x^{- 4}) d x = \frac{38}{81} \\ {[\frac{16 x^{- 3}}{- 3}]}_{2}^{h} = \frac{38}{81} \\ {[- \frac{16}{3 x^{3}}]}_{2}^{h} = \frac{38}{81} \\ - \frac{16}{3 h^{3}} - (- \frac{16}{3 {(2)}^{3}}) = \frac{38}{81} \\ \frac{16}{3 h^{3}} = \frac{16}{24} - \frac{38}{81} \\ \frac{16}{3 h^{3}} = \frac{16}{81} \\ 3 h^{3} = 81 \\ h^{3} = 27 \\ h = 3 \end{array}

Long Question 7

Posted on May 16, 2020 by Myhometuition

Question 7:
Diagram below shows a curve

y = \frac{1}{4} x^{2} + 3

which intersects the straight line y = x + 6 at point A.

(a) Find the coordinates of A.
(b) Calculate
(i) the area of the shaded region M,
(ii) the volume generated, in terms of π, when the shaded region N is revolved 360^o about the y-axis.

Solution:
(a)

\begin{array}{l} y = \frac{1}{4} x^{2} + 3.......... (1) \\ y = x + 6.......... (2) \\ Substitute (2) into (1), \\ x + 6 = \frac{1}{4} x^{2} + 3 \\ 4 x + 24 = x^{2} + 12 \\ x^{2} - 4 x - 12 = 0 \\ (x + 2) (x - 6) = 0 \\ x = - 2 or x = 6 (rejected) \\ When x = - 2 \\ y = - 2 + 6 = 4 \\ Therefore, A = (- 2, 4) . \end{array}

(b)(i)

\begin{array}{l} At x -axis, y = 0 \\ From y = x + 6, x = - 6 \\ Area of region M \\ = Area of triangle + Area under the curve \\ = \frac{1}{2} \times (6 - 2) \times 4 + \int_{- 2}^{0} y d x \\ = 8 + \int_{- 2}^{0} (\frac{1}{4} x^{2} + 3) d x \\ = 8 + {[\frac{x^{3}}{4 (3)} + 3 x]}_{- 2}^{0} \\ = 8 + [0 - (\frac{{(- 2)}^{3}}{12} + 3 (- 2))] \\ = 8 + [0 - (- \frac{8}{12} - 6)] \\ = 8 + [0 - (- \frac{20}{3})] \\ = 14 \frac{2}{3} {unit}^{2} \end{array}

(b)(ii)

\begin{array}{l} At y -axis, x = 0, \\ y = \frac{1}{4} (0) + 3 \\ y = 3 \\ y = \frac{1}{4} x^{2} + 3 \\ 4 y = x^{2} + 12 \\ x^{2} = 4 y - 12 \\ Volume of N \\ = π \int_{3}^{4} x^{2} d y \\ = π \int_{3}^{4} (4 y - 12) d y \\ = π \int_{3}^{4} (2 y^{2} - 12 y) d y \\ = π {[(2 y^{2} - 12 y)]}_{3}^{4} \\ = π [(2 {(4)}^{2} - 12 (4)) - (2 {(3)}^{2} - 12 (3))] \\ = π (- 16 + 18) \\ = 2 π {unit}^{3} \end{array}

Long Question 4

Posted on May 16, 2020 by Myhometuition

Question 4:

Diagram below shows a curve x = y² – 1 which intersects the straight line 3y = 2x at point Q.

Calculate the volume generated when the shaded region is revolved 360^o about the y-axis.

Solution:

\begin{array}{l} x = y^{2} - 1 \to (1) \\ 3 y = 2 x \\ x = \frac{3}{2} y \to (2) \\ Substitute (2) into (1), \\ \frac{3}{2} y = y^{2} - 1 \\ 2 y^{2} - 3 y - 2 = 0 \\ (2 y + 1) (y - 2) = 0 \\ y = - \frac{1}{2} or y = 2 \end{array}

\begin{array}{l} When y = 2, x = \frac{3}{2} (2) = 3, Q = (3, 2) \\ I_{1} (Volume of cone) \\ = \frac{1}{3} π r^{2} h = \frac{1}{3} π {(3)}^{2} (2) \\ = 6 π {unit}^{3} \\ I_{2} (Volume of the curve) \\ = π \int_{1}^{2} x^{2} d y \\ = π \int_{1}^{2} {(y^{2} - 1)}^{2} d y \\ = π \int_{1}^{2} (y^{4} - 2 y^{2} + 1) d y \\ = π {[\frac{y^{5}}{5} - \frac{2 y^{3}}{3} + y]}_{1}^{2} \\ = π [(\frac{2^{5}}{5} - \frac{2 {(2)}^{3}}{3} + 2) - (\frac{1^{5}}{5} - \frac{2 {(1)}^{3}}{3} + 1)] \\ = π (\frac{46}{15} - \frac{8}{15}) \\ = \frac{38}{15} π {unit}^{3} \\ ∴ Volume generated = I_{1} - I_{2} \\ = 6 π - \frac{38}{15} π \\ = \frac{52}{15} π {unit}^{3} \end{array}

SPM Practice Question 13 & 14

Posted on May 16, 2020 by Myhometuition

Question 13:
The sum of the first n term of a geometry progression is given by 6(1 – 0.5ⁿ ). Find
(a) The fourth term of the progression,
(b) The sum to infinity of the progression.

Solution:

Question 14:
A gardener has the task of digging an area of 800 m². On the first day he digs an area of 10 m². On each successive day he digs an area of 1.2 times the area that he dug the previous day, until the day when the task is completed. Find the number of days needed to complete the task.

Solution:

SPM Practice Question 4 – 6

Posted on May 16, 2020 by Myhometuition

Question 4:
The sum of the first n terms of the geometric progression 5, 15, 75, … is 5465.
Find the value of n.

Solution:

Question 5:
The first three terms of a geometric progression are 5k + 6, 2k, k – 2.
Find
(a) the positive value of k,
(b) the sum from the third term to the sixth term, using the value of k obtained in (a)

Solution:

Question 6:
The first three terms of a sequence are 2, x, 18. Find the positive value of x so that the sequence is
(a) an arithmetic progression,
(b) a geometric progression.

Solution:

SPM Practice Question 4

Posted on May 16, 2020 by Myhometuition

Question 4:
Mr Choong started working for a company on 1 January 2008 with an initial annual salary of RM28800. Every January, the company increased his salary by 5% of the previous year’s salary.

Calculate
(a) his annual salary, to the nearest RM on 1 January 2013,
(b) the minimum value of n such that his annual salary in the nth year will exceed RM40 000,
(c) the total salary, to the nearest RM, paid to him by the company, for the years 2008 to 2013.

Solution: