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:

SPM Practice Question 3

Posted on May 16, 2020 by Myhometuition

Question 3:
An arithmetic progression has 16 terms. The sum of the 16 terms is 188, and the sum of the even terms is 96. Find
(a) the first term and the common difference,
(b) the last term.

Solution:
(a) Let the first term = a
Common difference = d

\begin{array}{l} Given S_{16} = 188 \\ Thus, \frac{16}{2} [2 a + 15 d] = 188 \\ 8 [2 a + 15 d] = 188 \\ 2 a + 15 d = \frac{188}{8} \\ 2 a + 15 d = 23.5 - - - (1) \end{array}

Given the sum of the even terms = 96

\begin{array}{l} T_{2} + T_{4} + T_{6} + ..... + T_{16} = 96 \\ (a + d) + (a + 3 d) + (a + 5 d) + ..... + (a + 15 d) = 96 \\ \frac{8}{2} [(a + d) + (a + 15 d)] = 96 \\ 4 [2 a + 16 d] = 96 \\ 2 a + 16 d = 24 - - - (2) \end{array}

(2) – (1):
16d – 15d = 24 – 23.5
d = 0.5
Substitute d = 0.5 into (2):
2a + 16 (0.5) = 24
2a + 8 = 24
2a = 16
a = 8
Therefore, first term = 8 and common difference = 0.5.

(b) Last term
= T₂
= 8 + 15 (0.5)
= 8 +7.5
= 15.5

(Long Questions) – Question 8

Posted on May 16, 2020 by Myhometuition

Question 8:
Diagram below shows a cyclic quadrilateral PQRS.

(a) Calculate
(i) the length, in cm, of PR,
(ii) ∠PRQ.
(b) Find
(i) the area, in cm², of ∆ PRS,
(ii) the short distance, in cm, from point S to PR.

Solution:
(a)(i)

\begin{array}{l} P R^{2} = 7^{2} + 8^{2} - 2 (7) (8) \cos 80^{o} \\ P R^{2} = 113 - 19.4486 \\ P R = \sqrt{93.5514} \\ P R = 9.6722 cm \end{array}

(a)(ii)

\begin{array}{l} In cyclic quadrilateral \\ ∠ P Q R + ∠ P S R = 180 \\ ∠ P Q R + 80 = 180 \\ ∠ P Q R = 100^{o} \\ \frac{\sin ∠ Q P R}{3} = \frac{\sin 100}{9.6722} \\ \sin ∠ Q P R = 0.3055 \\ ∠ Q P R = 17^{o} 47' \\ ∠ P R Q = 180^{o} - 100^{o} - 17^{o} 47' \\ = 62^{o} 13' \end{array}

(b)(i)

\begin{array}{l} Area of △ P R S \\ = \frac{1}{2} \times 7 \times 8 \sin 80^{o} \\ = 27.5746 {cm}^{2} \end{array}

(b)(ii)

\begin{array}{l} Area of △ P R S = 27.5746 \\ \frac{1}{2} \times 9.6722 \times h = 27.5746 \\ h = \frac{27.5746 \times 2}{9.6722} \\ = 5.7018 cm \\ Shortest distance = 5.7018 cm \end{array}

(Long Questions) – Question 7

Posted on May 16, 2020 by Myhometuition

Question 7:
Diagram below shows a quadrilateral ABCD where the sides AB and CD are parallel. ∠BAC is an obtuse angle.

Given that AB = 14 cm, BC = 27 cm, ∠ACB = 30^o and AB : DC = 7 : 3.
Calculate
(a) ∠BAC.
(b) the length, in cm, of diagonal BD.
(c) the area, in cm², of quadrilateral ABCD.

Solution:
(a)

\begin{array}{l} \frac{\sin ∠ B A C}{27} = \frac{\sin 30^{o}}{14} \\ \sin ∠ B A C = \frac{\sin 30^{o}}{14} \times 27 \\ \sin ∠ B A C = 0.9643 \\ ∠ B A C = {74.64}^{o} \\ ∠ B A C (obtuse) = 180^{o} - {74.64}^{o} \\ = {105.36}^{o} \end{array}

(b)

\begin{array}{l} A B parallel with D C \\ ∠ B A C = ∠ A C D \\ ∠ B C D = {105.36}^{o} + 30^{o} \\ = {135.36}^{o} \\ \frac{D C}{A B} = \frac{3}{7} \\ D C = \frac{3}{7} \times 14 cm \\ = 6 cm \\ B D^{2} = 27^{2} + 6^{2} - 2 (27) (6) \cos ∠ B C D \\ B D^{2} = 765 - 324 \cos {135.36}^{o} \\ B D^{2} = 995.54 \\ B D = 31.55 cm \end{array}

(c)

\begin{array}{l} ∠ A B C = 180^{o} - 30^{o} - {105.36}^{o} \\ = {44.64}^{o} \\ A C^{2} = 27^{2} + 14^{2} - 2 (27) (14) \cos ∠ A B C \\ A C^{2} = 925 - 756 \cos {44.64}^{o} \\ A C^{2} = 387.08 \\ A C = 19.67 cm \\ Area △ A B C \\ = \frac{1}{2} (14) (27) \sin {44.64}^{o} \\ = 132.80 {cm}^{2} \\ Area △ A C D \\ = \frac{1}{2} (19.67) (6) \sin {105.36}^{o} \\ = 56.90 {cm}^{2} \\ Area of quadrilateral A B C D \\ = 132.80 + 56.90 \\ = 189.7 {cm}^{2} \end{array}

Short Questions (Question 17 – 19)

Posted on May 16, 2020 by Myhometuition

Question 17:

In the diagram above, the straight line PR is normal to the curve

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

at Q. Find the value of k.

Solution:

\begin{array}{l} y = \frac{x^{2}}{2} + 1 \\ \frac{d y}{d x} = x \\ At point Q, x -coordinate = 2, \\ Gradient of the curve, \frac{d y}{d x} = 2 \\ Hence, gradient of normal to the curve, P R = - \frac{1}{2} \\ \frac{3 - 0}{2 - k} = - \frac{1}{2} \\ 6 = - 2 + k \\ k = 8 \end{array}

Question 18:
The normal to the curve y = x² + 3x at the point P is parallel to the straight line
y = –x + 12. Find the equation of the normal to the curve at the point P.

Solution:

Given normal to the curve at point P is parallel to the straight line y = –x + 12.

Hence, gradient of normal to the curve = –1.

As a result, gradient of tangent to the curve = 1

y = x² + 3x

$\frac{d y}{d x}$ = 2x + 3

2x + 3 = 1

2x = –2

x = –1

y = (–1)²+ 3(–1)

y = –2

Point P = (–1, –2).

Equation of the normal to the curve at point P is,

y – (–2) = –1 (x – (–1))

y + 2 = – x – 1

y = – x– 3

Question 19:

Given that

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

, find the approximate change in x which will cause y to decrease from 48 to 47.7.

Solution:

\begin{array}{l} y = \frac{3}{4} x^{2} \\ \frac{d y}{d x} = (2) \frac{3}{4} x = \frac{3}{2} x \\ δ y = 47.7 - 48 = - 0.3 \\ Approximate change in x to y \\ \frac{δ x}{δ y} \approx \frac{d x}{d y} \\ δ x = \frac{d x}{d y} \times δ y \\ δ x = \frac{2}{3 x} \times (- 0.3) \\ δ x = \frac{2}{3 (8)} \times (- 0.3) \leftarrow \begin{array}{l} y = 48 \\ \frac{3}{4} x^{2} = 48 \\ x^{2} = 64 \\ x = 8 \end{array} \\ δ x = - 0.025 \end{array}