Masha Hamid Has a Question on Logarithm

Mashal Hamid <mashalhamid.xx@gmail.com>	8:08 PM (3 hours ago)
to me

Hello Mr.Tan 

I was wandering if you could give me the solution to this question 16 from past year paper on logarithmic functions and indices, that i found on your blog, as I am having troubles with part (c) .

I would really appreciate it. 

Thanks. 

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Hello Mashal,

My proposed solution to part (c) is as follows:

Use the given f(x) = 2xlog_x3 – 5log_x9 – x + 5 to factorise into:

                     f(x) = (x – 5)(alog_x3 – b),

then the values of a and of b should be obvious as follows:

f(x) = 2xlog_x3 – 5log_x9 – x + 5

       = 2xlog_x3 – 5log_x3² – x + 5

       = 2xlog_x3 – 10log_x3 – x + 5

       = (2xlog_x3 – x) – (10log_x3 + 5)        ~ rearrange to factorise

       = x(2log_x3 – 1) – 5(2log_x3 - 1)

       = (x – 5)(2log_x3 – 1)

      [≡(x – 5)(alog_x3 – b)]

Hence, a = 2 and b = 1.

Hope the above helps. Have a good day!

J

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Mashal Hamid	Sep 19 (2 days ago)
to me

Yes, thank you so much, sir.

Brain Teaser 7: Line of Symmetry, Maxima (or, Minima) Of Quadratic Function in Terms Of a, b and c

The Brain Teaser: For quadratic function y = ax² + bx + c, show that ax² + bx + c ≡ a(x – s)² + m, where x = s = - (b/2a) is its line of symmetry; m = c - (b²/4a) is its maxima (or minima, if a > 0)

Any quadratic function graph has a line of symmetry x = s and its maxima (or, minima whichever is applicable) lies at the point of intersection of the graph and its line of symmetry.

In other words, if you substitute x = s into y = ax² + bx + c, you would get the value of maxima or minima i.e. m. Hence, if you know the value of the line of symmetry of a quadratic function, you can use it to calculate its maxima or minima in lieu of completing the square method or the dy/dx method. 

But what is the value of s in terms of a, b and c? And, what is m in terms of a, b and c? This, in essence, is what this Brain Teaser 7 is all about!

In a week or so, I will post the working in getting ax² + bx + c ≡ a(x – s)² + m, where:

1)       x = s = - (b/2a)  (Do you not see this line of symmetry inside the Quadratic Root Formula?)

2)      m = c - (b²/4a) 

Test these with y = (x + 2)(x – 3) = x² – x – 6, where a = 1, b = -1 and c = -6 and

a    *   The line of symmetry x = s = - (b/2a) = - (-1/2) = ½; and

*   Minima (a = 1 > 0), m = c - (b²/4a) = -6 – (1/4x1) = -6 – ¼ = - 6.25

Have fun with maths!

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Simple proof:

If the quadratic function y = ax² + bx + c is expressed as

                                       y = a(x – s)² + m,

show that:

·        its line of symmetry is x = s = -b/2a

·        its minima (if a > 0) or maxima (if a < 0)

m = c – (b²/4a) = y when x = s = -b/2a

Solution:

            y = ax² + bx + c

               = a[x² + (b/a)x] + c

               = a[(x + b/2a)² – (b/2a)²] + c

               = a(x + b/2a)² - a(b/2a)² + c

               = a(x + b/2a)² + c – b²/4a

Thus, if y = a(x – s)² + m,

              s = -b/2a, and

              m = c – (b²/4a)

Point of maxima or minima occurs when x = s (i.e. at its line of symmetry).

At x = s = -b/2a, maximum or minimum value of y is as follows:

     y =  ax² + bx + c

        = a(-b/2a)² + b(-b/2a) + c

        = b²/4a – b²/2a + c

        = b²/4a – 2b²/4a + c

        = c – (2b²/4a - b²/4a)

        = c – (b²/4a) = m,   as in the expression y = a(x – s)² + m         …(Shown)

Trigo Question: Edexcel 4PMO, 2012 May 17 Paper 1 Question 7 – Q and A

Question 7:                              cos(A +B) = cosA cosB – sinA sinB

a)      Express cos(2x + 45^o) in the form M cos2x + N sin2x, where M and N are constants, giving the exact value of M and the exact value of N.                       (2)

b)      Solve, for 0^o ≤ x ≤ 180^o, the equation cos2x –sin2x = 1                                    (5)

The maximum value of cos2x –sin2x is k.

c)      Find the exact value of k.                                                                                  (2)

d)      Find the smallest positive value of x for which a maximum occurs.                      (3)

My Proposed Solutions

a)      Use:                 cos(A +B) = cosA cosB – sinA sinB

Therefore:         cos(2x + 45^o) = cos2x cos45^o – sin2x sin45^o

[Use:                cos45^o = 1/√2 = sin45^o (trigo of convenient angle 45^o)]

Therefore:         cos(2x + 45^o) = (1/√2) cos2x – (1/√2) sin2x

                                                = (1/√2) cos2x + (-1/√2) sin2x

                                                ≡ M cos2x + N sin2x

Where, M = 1/√2 and N = -1/√2                                             (Answer)

b)      For 0^o ≤ x ≤ 180^o, solve cos2x –sin2x = 1.

cos2x –sin2x = 1

            x 1/√2:             (1/√2) cos2x – (1/√2) sin2x = 1/√2

Therefore:         cos(2x + 45^o) = 1/√2

Imply:               (2x + 45^o) = cos^-1 (1/√2) (= Reference Angle 45^o)

[Principle Angle Domain: 0^o ≤ x ≤ 180^o implies 45^o ≤ (2x + 45^o) ≤ 360^o + 45^o]

Therefore:         (2x + 45^o) is in 1^st qdrant (qd), 4^th qd and 1^st qd 1^st co-terminal

1^st qd:               2x + 45^o = 45^o;            2x = 0^o;            x = 0^o   (Answers)

4th qd:              2x + 45^o = 360^o -  45^o; 2x = 270^o;       x = 135^o

5^th qd:               2x + 45^o = 360^o +  45^o; 2x = 360^o;       x = 180^o

c)      Max. value of cos2x – sin2x = k

x 1/√2:             (1/√2) cos2x – (1/√2) sin2x = k/√2

Imply:               cos(2x + 45^o) = k/√2

Therefore:         k = (√2) cos(2x + 45^o)

k maximum, when cos(2x + 45^o) = 1; and, max. k = (√2)(1) = √2 (Answer)

d)      For principle angle domain: 45^o ≤ (2x + 45^o) ≤ 360^o + 45^o,

the maximum value occurs when cos(2x + 45^o) = 1 = cos 360^o.

(cos0^o = 1 too but 0^o  is outside the domain)

Therefore:         (2x + 45^o) = 360^o;        2x = 315^o;        x = 157.5^o        (Answer)

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Good luck to Ms Kim and J Min!

Solve 9 sin2 Ө – 9 sin Ө = 11

Last night at Mont Kiara, the above question was forwarded to me to solve.

Anyway, you will note from my proposed solutions below that different domains for Ө will mean different answers. 

For examples:

* For 0 ≤ Ө ≤ π radians, no value of Ө can satisfy the equation: 9 sin² Ө – 9 sin Ө = 11;

* For 0 ≤ Ө ≤ 2π radians, two values Ө satisfy the equation

My Proposed Solutions

                                  9 sin² Ө – 9 sin Ө = 11

-11:                   9 sin² Ө – 9 sin Ө – 11 = 0

                        (compare: ax² + bx + c = 0)

where:                   a = 9, b = -9 and c = -11

therefore, roots:     [α = (-b + √(b² – 4ac))/2a; or β = (-b - √(b² – 4ac))/2a]

imply:          sin Ө = -(-9) - √(-9)² – 4(9)(-11))/(2x9) = - 0.713351648; or,

                    sin Ө = -(-9) + √(-9)² – 4(9)(-11))/(2x9) =  1.713351648;

For 0 ≤ Ө ≤ 2π radians:

                    sin Ө = - 0.713351648 implies Ө in 3^rd or 4^th quadrants

                          Ө = π + sin^-1 (0.713351648) or 2π - sin^-1 (0.713351648)

                          Ө = 3.935861836 (3rd Qd) or 5.488916105 (4th Qd)

                          Ө = 3.94 or 5.49 rad (both to 3 sf)

But for 0 ≤ Ө ≤ π radians (1^st and 2^nd quadrants):

                    For all values of Ө, sin Ө must be +ve.

                    Therefore, sin Ө ≠ - 0.713351648.

Also for all values of Ө in any quadrant, sin Ө ≠ 1.713351648 because 1 ≥  sin Ө ≥ -1.

So remember: If you want to solve a trigonometric equation, you must know or be given the domain because the answers depend on the domain for Ө! In fact, Q 10(e) of Edexcel Further Pure Maths (4PMO) Yr 2013 May Paper 1 amply demonstrates this!!

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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