Brain Teaser 6: Differentiate y = Ae^2x + Be^-x, where x ≥ 0,…

A curve has equation y = Ae^2x + Be^-x, where x ≥ 0. At the point where x = 0, y = 50 and dy/dx = -20.

i)             Show that A = 10 and find the value of B.                                             (5)

ii)           Using the values of A and B found in part (i), find the coordinates of the stationary point of the curve.                                                                 (4)

{My additional question: Express the exact values of the coordinates of this stationary point as     (C ln 2, D(2)^(2/3) ) and show that C = 1/3 and D = 30)

iii)          Determine the nature of the stationary point, giving a reason for your answer.     (2)

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In an earlier post, I have shown that e^{ln 2} = 2, 10^{lg 3} = 3 and generally b^log_b x = x. My additional question in part (ii) may require the use of this knowledge in order to get the exact value of the y-coordinate of the stationary point mentioned therein.

(Background to this Q: In the course of preparing a student to sit for her IGCSE Cambridge Add-Maths 0606 next Tuesday (10/06/2014), I came across the above Q as one of the past-year questions on "Calculus". After teaching her how to get the answers, I additionally asked her to get the exact values of the coordinates of the stationary point instead of just (0.231, 47.6). Thus, I have added an additional question to part (ii). :) )

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Answers:

i)                    A = 10; B = 40

ii)                   Stationary point = (0.231, 47.6) [or, ((1/3)ln 2, 30(2)^(2/3)); C = 1/3: D=30]

iii)                 d²y/dx² = 40e^2x + 40e^-x = +ve for all values of x – thus, minimum point.

Working leading to the answers will be posted in a week or so (please see below on "My Proposed Solutions"). You may post yours by way of comments if this question is of interest to you too. J

(Posted on 3/06/2014 from PJ SS2 Starbucks)

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My Proposed Solutions:

(i)             Given :    y = Ae^2x + Be^-x, where x ≥ 0

              At x = 0, y = 50 and dy/dx = -20

At x = 0, y = 50, therefore:          50 = Ae^o + Be^o

                                                   50 = A + B          ~~~ Eqn (1)

At x = 0, dy/dx = -20, therefore:  dy/dx = 2Ae^2x – Be^-x

                                                   -20  =  2Ae²⁽⁰⁾ – Be^-(0)

                                                   -20 = 2A – B      ~~~ Eqn (2)

Eqn (1) + Eqn (2):                       3A = 30,          hence, A = 10 (Shown)

Use Eqn (1):                               50 = 10 + B,    hence, B = 40 (Answer)

(ii)           Since,                 A = 10 and b = 40

therefore:            dy/dx = 2Ae^2x – Be^-x = 20e^2x – 40e^-x

At stationary pt, dy/dx = 0

therefore:            0 = 20e^2x – 40e^-x

x (e^x):                  0 = 20e^2x(e^x) – 40e^-x(e^x)

                           0 = 20e^3x – 40e^o

                           0 = 20e^3x – 40

                           e^3x = 40/20 = 2

ln both sides:       3x = ln 2.         Therefore,        x = (1/3)(ln 2) ~~ exact value

                                                                           x = 0.231 (3 sf) ~~ (Answer)

Put x = (1/3)(ln 2) into:   y = 10e^2x + 40e^-x

therefore,                        y = 10e^{2(1/3)(ln 2)} + 40e^{-(1/3)(ln 2)}

                                       y = 10e^{(ln 2^(2/3))} + 40e ^{(ln 2^(-1/3))}

[Use ln e^x = x:                 y = 10(2^(2/3)) + 40(2^(-1/3))]

All in terms of 2^(2/3):        y = 10(2^(2/3)) + 40(2^(-1/3))(2/2)

                                       y = 10(2^(2/3)) + [40(2^(-1/3))(2)]÷2

                                       y = 10(2^(2/3)) + [40(2^(1-(1/3)))] ÷2

                                       y = 10(2^(2/3)) + 20(2^(2/3)) = 30(2^(2/3))      ~~ exact value

                                       y = 47.6 (to 3 sf)               ~~ (Answer)                                       

(iii)          d²y/dx² = 40e^2x + 40e^-x = +ve value for all x ≥ 0

So, stationary pt is minimum.                                ~~ (Answer)

Please drop your comments if you have any view to express on the above. Have fun!

(Posted on 18/06/2014 from PJ SS2 Starbucks)

Brain Teaser 5: Number Sequence

A sequence of numbers U₁, U₂,…, U_n,…is given by the formula

       U_n = 3(2/3)ⁿ – 1 where n is a positive integer.

a)      Find the value of U₁, U₂ and U₃

b)      Show that sigma or summation of U_n (for n = 1 to 15) = -9.014 to 4 significant figures

c)      Prove that U_{(n + 1)} = 2(2/3)ⁿ – 1

Was at Dataran Sunway (Kota D’sara) Starbucks with Angela and this Q cropped up and we solved it pretty fast! What about you?

If before I put up the answers in a week or so, you are the 1st to post the correct answers by way of comments (with brief intro of yourself), then you win a Starbucks drink from me J.

Have fun with maths!

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From Borders' Starbucks @ The Curve, 17/5/2014 (Sat, 1.10pm):

My Proposed Solutions 

a)      To find the value of U₁, U₂ and U₃ – This is straight forward!

For U1, put n = 1 into U_n = 3(2/3)ⁿ – 1      

Therefore,                   U₁ = 3(2/3)¹ – 1 = 2 - 1

                                    U₁ = 1                            (Answer)

For U_2, put n = 2 into U_n = 3(2/3)ⁿ – 1

Therefore,                   U₂ = 3(2/3)² – 1 = 4/3 - 1

                                    U₂ = 1/3                        (Answer)

For U_3, put n = 3 into U_n = 3(2/3)ⁿ – 1

Therefore,                   U₃ = 3(2/3)³ – 1 = 8/9 - 1

                                    U₃ = -1/9                      (Answer)

b)      To show that sigma or summation of U_n (from n = 1 to 15) = -9.014 to 4 significant figures

If you use conventional approach, you will discover:

No common difference: U₂ – U₁ ≠ U₃ - U₂ (-2/3 ≠ - 4/9)

No common ratio: U₂/U₁ ≠ U₃/U₂ (1/3 ≠ - 1/3)

However, when you take a good look at U_n = 3(2/3)ⁿ – 1,

You will see that:     U_n = Geometric Term – 1…(GT = ar^(n-1) = (a/r)rⁿ)

Therefore,                 U_n = 3(2/3)ⁿ – 1  ≡ (a/r)(rⁿ) – 1

Hence,                          r = 2/3

                                  a/r = 3 or, a = 3r = 3(2/3) = 2

Therefore, sigma or summation of U_n = a(1-rⁿ)/(1-r) – n

Hence, sigma or summation first 15  U_n terms (from n=1 to 15)

                                 =  a(1-rⁿ)/(1-r) – n

Where,                  a = 2 (1^st of the GT – pl see above)

                              r = 2/3 (common ratio of the geometric sequence)

                             n = 15

Hence, sigma or summation first 15  U_n terms (from n=1 to 15)

                              =  a(1-rⁿ)/(1-r) – n

                              = 2(1 – (2/3)¹⁵)/(1 – 2/3) – 15

                              = 2(0.997716341)/(1/3) – 15

                              = 6(0.997716341) – 15

                              = 5.98629805 -15

                              = - 9.01370195

                              = - 9.014 (to 4 sf) (Shown)

c)      To prove that U_{(n + 1)} = 2(2/3)ⁿ – 1

Put n = (n+1) into U_n = 3(2/3)ⁿ – 1

Therefore,             U_{(n + 1)} = 3(2/3)^{(n + 1)} – 1

                              U_{(n + 1)} = 3(2/3)ⁿ(2/3)¹ – 1…Law of Index: b^{(m + n)} = b^m x bⁿ

                              U_{(n + 1)} = 2(2/3)ⁿ – 1 (Shown)

No one wins a Starbucks drink from me...ok, got to meet up with Karina of Garden Intl School later...

Brain Teaser 4: Find the centre and the radius of the circle with the equation x2 + y2 – 4x – 6y – 12 = 0 (or, x2 + y2 = ax + by + c, generally)

If you are given the centre of a circle as (a, b) and the radius as r (directly or indirectly), you can easily use Pythagoras Theorem to derive its equation as follows:

(x – a)² + (y – b)² = r²

If you are given equation of a circle not in the above format but in already simplified form, say

x2 + y2 – 4x – 6y – 12 = 0, would you be able to find:

1) its centre (a, b); and 

2) its radius, r?

Answer for Brain Teaser 4 will be out in a week or so. Let you try first.

Have fun with maths!

Notes:

All brain teasers in this blog are questions given to students to try and they took longer than usual to come up with solutions, and in some instances, even after 15 minutes, the student(s) just could not solve them.

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My Proposed Solution

Use: Completing-the-Square (CTS) Method:

Since, the eqn of the circle given is:  x² + y² – 4x – 6y – 12 = 0

Rearrange eqn for CTS:                   (x² – 4x) + (y² – 6y) = 12

Now, do CTS:                                  (x – 4/2)² + (y – 6/2)² = 12 + (-4/2)² +  (-6/2)²

Simplify:                                          (x – 2)² + (y – 3)² = 25

(x – a)² + (y – b)² = r² format:          (x – 2)² + (y – 3)² = 5²

(Since equation of a circle with radius r and centre (a, b) on Cartesian plane can be expressed by using Pythagoras Theorem as:   (x – a)² + (y – b)² = r² 

Therefore: centre is (2, 3) and radius is 5 units.

Brain Teaser 3: Solve sin(3 – z) = 0.8 for 0 < z < π radians.

I have gone through this question with my students. You should try it too. You may post your answer by way of comments to this post with brief intro of yourself.

Either I or one of my students will put up the answer in a week or so. Have fun with brain teasers of this blog! :)

(Acknowledgment: The above question was taken from IGCSE Cambridge Maths 0606, Yr 2012 .... Q 9(b))

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My Proposed Solution:

For 0 < z < π rad, solve sin(3 – z) = 0.8

Get domain for (3 – z):

Since 0 < z < π rad, … x (-1) to get -z, reverse inequality sign

Therefore, 0 > -z > -π rad

+ 3:           3 > (3 – z) > 3 – π

                 3 > (3 – z) > 0.142…domain for principal angle obtained

Since, sin(3 – z) = 0.8,  and sin^-1(0.8) = 0.9272…(> 0.142)

therefore, (3 – z) is in 1^st quadrant (qd) or 2^nd qd:

1^st qd:      3 – z = sin^-1(0.8)

                     z = 3 - sin^-1(0.8) = 2.072704782

                     z = 2.07 (to 3 sf) (Answer)

2^nd qd:     3 – z = π - sin^-1(0.8) (< 3)

                      z = 3 - π + sin^-1(0.8) = 0.7857…

                      z = 0.786 (to 3 sf) (Answer)

Brain Teaser 2: Find the value of e^x, where x = 4 + ln 2

If you can solve the above, you should be able to solve these too (without using calculator):

Find the value of:
(a) 3^x, where x = 4 + log₃ 2. (My question)
(b) 2^z, where z = 5 + log₂ 3  (IGCSE Cambridge Add-Maths 0606, Yr 2010/Summer/p22, Q10(b))

I will post their answers in a week or so. Meanwhile, you may post your answers by way of comments to this post. Have fun with maths!

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29.04.2014

My Proposed Solutions to These Questions:

Find the value of:

 (Lead Question): e^x, where x = 4 + ln 2
(a) 3^x, where x = 4 + log₃ 2. (My question)
(b) 2^z, where z = 5 + log₂ 3  (IGCSE Cambridge Add-Maths 0606, Yr 2010/Summer/p22, Q10(b))

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For the lead question:

Since            x = 4 + ln 2,

therefore:     e^x = e^{(4 + ln 2)} = e⁴ x e^{ln 2} ..............Law of index: b^{(m + n)} = b^m x bⁿ

Now:           let y = e^{ln 2}

ln both sides: ln y = ln 2

therefore:          y = 2 = e^{ln 2} (Hence, remember: e^{ln a} = a; 10^{lg b} = b; etc.)

Hence, e^x = e^{(4 + ln 2)} = e⁴ x e^{ln 2} = 2e⁴ (Answer)

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(a)    Since           x = 4 + log₃ 2

therefore: 3^x = 3^(4 + log₃ 2) = 3⁴ x 3^log₃ 2 = 3⁴ x 2 (see above)

                  3^x = 81 x 2 = 162 (Answer)

(b)   Since         z = 5 + log₂ 3

therefore:  2^z = 2^(5 + log₂ 3) = 2⁵ x 2^log₂3 = 2⁵ x 3

                2^z   = 32 x 3 = 96 (Answer)

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So, remember:

·        e^{ln a} = a

·        10^{lg b} = b

·        b^log_b c = c

Have fun with maths!

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(1.5.2014)

Brain Teaser 2B (requires slightly more skill than Brain Teaser 2)

Find the exact value of:

1) 3^x , where x = 2 + log₉ 3

2) 2^y , where y = 3 + log₈ 2

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Brain Teaser 1: Trigonometric Inequality

"For 0^o ≤ x ≤ 180^o, find to 3 significant figures the range of values of x for: 5 sin x > cos x."

In an earlier post, I have blogged about "A Negative Number Without The Negative Sign". In that post, I blogged about log x which is -ve but is not showing the -ve sign (log x, where 0 < x < 1). And, at the end of that post, I invited readers to think about other type(s) of -ve numbers without the -ve signs.

Have you thought about it? That depends on whether you are 'passionate' about that question, right? Any way, let me continue: Trigonometric ratios are, among others, such numbers: -ve numbers which may not show -ve signs.

Depending on the quadrants that the angles fall into, trigonometric ratios can be +ve or -ve. For examples:

• In the 1st quadrant (qd), ALL trigo ratios are +ve; ---------All        ~ Add
• In the 2nd qd, only sines (and thus, cosec) are +ve; --------Sine      ~ Sugar
• In the 3rd qd, only tangents (and thus, cot) are +ve; and ---Tangent ~ To
• In the 4th qd, only cosines (and, sec) are +ve --------------Cosine  ~ Coffee 
• My mnemonic for students to remember the above: Add Sugar To Coffee (since this post is written from TTDI Starbucks, :))

If you are trying to solve the above inequality algebraically, you can approach it quadrant by qaudrant. Dividing by cosine in the 2nd qd is dividing by a -ve number and, the inequality sign must be reversed!

However, you can also solve the above inequality by sketching the graphs of y = 5 sin x and y = cos x to identify y of the 1st graph that is above (>) the y of the 2nd graph, then use 5 sin x = cos x to find the root to determine bound of the range of values of x -- much in the same way as you would do for quadratic inequalities that are formed from the nature of the quadratic roots and the discriminant (2 equal roots means b² – 4ac = 0; 2 unequal roots means b² – 4ac > 0; or, no real roots means b² – 4ac < 0).

Okay, you may compare your answer with mine by posting a comment to this post.

Have a productive day!                                               (My answer: 11.3^o < x ≤ 180^o)

A Negative Number Without the Negative Sign!

       What kind of negative number does not show -ve sign? And, why is it important for you to        recognise them when you see them - particularly in inequalities?         

        Now, you know that in an inequality, say, 

                                 2 < 3

         when you x or ÷ both sides of the inequality by a -ve number, say, by -1,

         it becomes:    -2 (sign) -3, and

         the inequality sign is the reverse of the earlier sign because:

      

                                -2 > -3

                          

                         This is basic O-Maths:

                          ---------

                          Thus, for SPM 2013 Modern Maths (1449/1) Paper 1 Q 24:

                          Q 24: Find the solution for (x - 6)/(-3) < 5

                          You can easily solve Q24 by multiplying both sides of the inequality by -3 

                          and, you reverse the inequality sign to finally get the answer x > -9 (Answer C)

                          ---------

                          

                          Now, in Add. Maths:

                          

                          You may come across inequality like this (in solving GP question):

                                              n log 0.25 > log 0.0005 

                                              for you to find the greatest integer value of n.

                                                  

                              To solve this Q, you must know that: log (0 < x < 1) = negative numbers, although they                             do not carry the -ve signs with them. Let me explain:

         For y = log_b x:   When 0 < x < 1, y = log_b x = -ve number. (pl see here. or use calculator to verify)

         Thus, all Add-Maths students must know this:

                               log (0 < x < 1) = negative numbers 

                              (even though the -ve sign is not shown)

                                

           Thus, log 0.00001, log 0.25, log (1/5), log 0.99999 (and the likes, where 0 < x < 1)
           are all -ve numbers! 
                   
           So, in the inequality:  n log 0.25 > log 0.0005,
                            
            when you divide both sides of the inequality by log 0.25, to isolate and find n;
            you have to reverse the > sign since log 0.25 is a negative number (-0.602059991),
            Thus,        n < log 0.0005/log 0.25
                            n < 5.482892142
            Therefore, the greatest integer value of n = 5 
            
     (This skill is tested in Q9(f) of Edexcel 4PMO 2013 Jan Paper 2 on "Series" (pl see below). So, all A-Maths students - SPM and IGCSE Cambridge alike - beware! You may face a logarithmic inequality question of the above nature in time to come if you have not already been tested this way! By the way, there are also other negative numbers without the negative signs - can you name some of them? :)) 
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Edexcel PMO 2013 Jan Paper 2 Q9

9) The third and fifth terms of a geometric series S are 48 and 768 respectively. Find
(a) the two possible values of the common ratio of S,          (3)
(b) the first term of S.                                                         (1)

Given that the sum of the first 5 terms of S is 615
(c) find the sum of the first 9 terms of S.                            (4)

Another geometric series T has the same first term as S. The common ratio of T is 1
r where r is one of the values obtained in part (a). The nth term of T is tn.
Given that t2 > t3
(d) find the common ratio of T.                                    (1)

The sum of the first n terms of T is Tn
(e) Writing down all the numbers on your calculator display, find T9.      (2)

The sum to infinity of T is T
Given that T – Tn > 0.002
(f) find the greatest value of n.                                   (5)

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Past-Year Questions on Cosine Formula/Rule, Sine Formula/Rule (including ambiguous case)...

Edexcel Past-Year Questions (involving Sine Formula, Cosine Formula, Etc.)

A) Given 2 sides and 1 angle of Δ:

1)        In ΔABC, AB = 7 cm, AC = 6 cm, angle ABC = 42^o and angle ACB = Ө^o.

Find, to 1 decimal place, the 2 possible values of Ө.           (24/3/2004 P1 Q1)

2)        In ΔABC, AB = 5.7 cm, BC = 8.4 cm, angle ACB = 42^o.

Find, to the nearest 0.1^o, the 2 possible sizes of angle BAC.  (15/01/2010 P1 Q1)

3)    In triangle ABC, AB = 4.6 cm, AC = 5.7 cm and angle C = 52^o. Angle B is acute.

Calculate, to the nearest 0.1^o, the size of angle B.   (14/01/2011 P1 Q2)

4)    In ΔABC, BC = 12 cm, AC = 10 cm, angle ACB = 48^o.

Find, to 3 significant figures,

(a)    the length of AB,

(b)   the size of angle BAC                                      (27/05/2004 P1 Q3)

5)        In ΔLMN, LM = 5.6 cm, LN = 8.2 cm, angle MLN = 57^o.

Find, to 3 significant figures,

a)      the length of MN,

b)      the size of angle LMN                                      (17/05/2007 P2 Q3)

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B) Given 2 angles and 1 side of Δ:

6)        In ΔABC, angle A = 45^o and B = 60^o and AB = 7 cm.

Calculate to 3 significant figures, the length of BC.   (26/05/2006 P2 Q1)

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C) Given 3 sides of Δ

7)        A triangle has sides of lengths 4.6 cm, 5.3 cm and 6.5 cm.

Find, to the nearest degree, the size of the largest angle of the triangle.

                                                                                         (23/01/2007 P2 Q1)

8)        Triangle LMN has LM = 5 cm, LN = 8.2 cm and MN = 6.4 cm

Calculate, in degrees to the nearest 0.1^o, the size of angle LMN.

                                                                                         (21/01/2008 P1 Q1)

9)        The lengths of the sides of a triangle are 4 cm, 5 cm and 6 cm.

Find, in degrees to 1 decimal place, the size of the largest angle of the Δ.

                                                                                         (12/05/2008 P1 Q1)

10)    The lengths of the sides of a triangle are 5 cm, 6 cm and 8 cm.

(a)    Find, in degrees to 1 decimal place, the size of the smallest angle of the Δ.

(b)   Find, to the nearest cm², the area of the Δ.                   (13/05/2010 P1 Q4)

11)    In ΔABC, AB = 5 cm, BC = 8.3 cm and AC = 6.9 cm

(            a)    Find, in degrees to the nearest 0.1^o, the size of angle ACB.

(            b)   Find, in cm², to 3 significant figures, the area of ΔABC.

(18/01/2010 P2 Q2)

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Q12 to Q14 have to be scanned and shown below because the blogging software cannot produce the drawn triangles as they are - perhaps certain plug-in needed:

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SPM (2005 - 2013) Past Year Questions on Trigonometric Function Graphs

SPM Past-Yr Questions on Trigonometric Function Graphs (Yrs: 2005 – 2013):

(with a peek at similar questions from IGCSE Cambridge Add-Maths 0606 and Edexcel 4PMO (Further Pure Maths)

(Note: SPM 2013 P2 Q4 on trigonometric functions does not involve graphs unlike the years before)

The SPM Past Year Questions:

1)        2012 P2 Q6:

(a)    Prove that 2/(cos 2x + 1) = sec² x.                                          (2)

(b)   (i) Sketch the graph of y = cos 2x + 1 for 0 ≤ x ≤ 2π.               (3)

(ii) Hence, use the same axes, sketch a suitable straight line to find the number of solutions to the equation 2/sec² x = x/4π + 1 for 0 ≤ x ≤ 2π.

State the number of solutions.                                          (3)

2)        2011 P2 Q6:

(a)   Sketch the graph of y = -3 sin (3/2)x for 0 ≤ x ≤ 2π.               (4)

(b)   Hence, using the same axes, sketch a suitable graph to find the number of solutions to the equation π/x + 3 sin (3/2)x = 0 for 0 ≤ x ≤ 2π.      

State the number of solutions.                                                 (3)

3)        2010 P2 Q2:

(a)    Sketch the graph of y = 1 + 3 cos x for 0 ≤ x ≤ 2π.                  (4)

(b)   Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation 6π cos x = 4π – 3x for 0 ≤ x ≤ 2π.

State the number of solutions.                                                   (3)

4)        2009 P2 Q4:

(a)    Sketch the graph of y = (3/2) cos 2x for 0 ≤ x ≤ (3/2)π.            (3)

(b)   Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation (4/3π)x – cos 2x = (3/2) for 0 ≤ x ≤ (3/2)π.

State the number of solutions.                                                   (3)

5)        2008 P2 Q4:

(            a)    Prove that (2 tan x)/(2 - sec² x) = tan 2x.                                              (2)

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

(ii) Hence, using the same axes, sketch a suitable straight line to find the number of solutions to the equation 3x/π + (2 tan x)/(2 - sec² x) = 0

for 0 ≤ x ≤ π.

State the number of solutions.                                            (6)

6)        2007 P2 Q3:

(a)      Sketch the graph of y = ׀3 cos 2x׀ for 0 ≤ x ≤ 2π.                    (4)

(b)      Hence, using the same axes, sketch a suitable line to find the number of solutions to the equation 2 - ׀3 cos 2x׀ = x/2π for 0 ≤ x ≤ 2π.

State the number of solutions.                                                   (3)

7)        2006 P2 Q4:

(a)      Sketch the graph of y = -2 cos x for 0 ≤ x ≤ 2π.                        (4)

(b)      Hence, using the same axes, sketch a suitable graph to find the number of solutions to the equation π/x + 2 cos x = 0 for 0 ≤ x ≤ 2π.                   

State the number of solutions.                                                    (3)

8)        2005 P2 Q5:

(a)      Prove that cosec² x – 2 sin² x – cot² x  = cos 2x.                      (2)

(c)      (i) Sketch the graph of y = cos 2x for 0 ≤ x ≤ 2π.       

(ii) Hence, using the same axes, draw a suitable straight line to find the number of solutions to the equation 3 (cosec² x – 2 sin² x – cot² x) = (x/π) – 1 for 0 ≤ x ≤ 2π.

State the number of solutions.                                              (6)

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Try this question of mine before you proceed further:

My Q. For 0^o ≤ x ≤ 180^o, find to 3 significant figures the range of values of x for

           5 sin x > cos x.                     (Answer given at the end of this post)

                                   

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Let's take a look at a Cambridge IGCSE question on trigonometric function graphs:

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Let's take a look at a Edexcel IGCSE Add-Maths (4PMO) question on trigonometric function graphs:

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(Ans to My Q: 11.3^o < x ≤ 180^o - posed after the SPM Past-Yr Qs)

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Let's take a break and go jogging:

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P/S: If You Need Help In Physics, Chemistry, Add-Maths and/or O-Maths (IGCSE Yr 10/11 or SPM or CIE AS/A2 Chemistry 9701)

Help is Available in or around Petaling Jaya, Selangor:
1) Just email tutortan1@gmail.com; or WhatsApp: 018-3722 482. Act early to avoid disappointment! Currently in May 2016, existing students are about to finish their Summer exams. Their slots are up for grab starting now! All slots are normally taken up by end-June. Act now to avoid disappointment!

2) Those far-away can follow my student-friendly blog(s) for free. :).

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