Mathematics Dictionary

Assume T(1) = 1 and T(2) = 3
Sequence : 1, 3, 7, 13, 21, ...
1st Diff : 2, 4, 6, 8, ........
2nd Diff : 2, 2, 2, 2, ........
Answer
- 1. Next number is 6th number = 21 + 10 = 31
- 2. Find T(n)
  - Since 2nd diff is common difference = 2
  - Hence T(n) = n^2 + b*n + c
  - T(1) = 1 we have 1 = 1^2 + b + c or b + c = 0
  - T(2) = 3 we have 3 = 2^2 + 2*b + c or 2*b + c = -1
  - Solve above equations, we have b = -1 and c = 1
  - Hence T(n) = n^2 - n + 1
- 3. Find S(n)
  - S(n) = Sum[n^2] - Sum[n] + Sum[1]
  - S(n) = n*(n+1)*(2*n+1)/6 - n(n+1)/2 + n
  - S(n) = n*((n+1)*(2*n+1) - 3*(n+1) + 6)/6
  - S(n) = n*(2*(n^2) + 4)/6
- Verify Question 1 : T(6) = 6^2 - 6 + 1 = 31
- Verify Question 3
  - S(3) = 1 + 3 + 7 = 11
  - S(3) = 3*(2*(3^2) + 4)/6 = 66/6 = 11

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Q02. Describe the curve y = (x^2)/((x^2 - 1)^2)

Asymptotes
- x = -1 y goes to infinite
- x = +1 y goes to infinite
y-intercept : x = 0 y = 0
Signs of y
- x LT -1, y is negative
- x EQ -1, y is infinite, Hence curve from -0 to -infinite
- x GE -1 and x LT 0, y is negative, curve from -infinte to 0 (x = 0)
- x GT +0 and x LT 1, y is positive, curve from 0 to infinite
- x GT 1, y is positive, curve from infinite to +0
Concavity
- x LT -1, it must be concave downword
- x between -1 and 0, it must be concave downword
- x between 0 and 1, it must be concave upword
- x GT 1, it must be concave upword

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Q03. a^2, b^2, c^2 in AP and then 1/(a+b), 1/(b+c), 1/(c+a) also in AP
Keyword

If x,y,z are in AP, then 2*y = (x + z)

Proof

1/(a + b) + 1/(b + c) = (a + 2*b + c)/((a + b)*(b + c))
= (a + 2*b + c)/(a*b + a*c + b^2 + b*c)
= (a + 2*b + c)/(a*b + a*c + (a^2 + b^2)/2 + b*c)
= 2*(a + 2*b + c)/(2*a*b + 2*a*c + (a^2 + b^2) + 2*b*c)
= 2*(a + 2*b + c)/((c + a)*(a + 2*b + c)
= 2/(c + a)

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Q04. Sum[t(i)]=A and i = 1 to a, Sum[t(i)]=B and i = 1 to b, ....
Question

Sum[t(i)]=A and i = 1 to a, Sum[t(i)]=B and i = 1 to b
Sum[t(i)]=C and i = 1 to c
t(i) is in AP, find A*a*b*c*(b-c) + B*a*b*c*(c-a) + C*a*b*c*(a-b)

Solution

Sum formula : S = n(t1 + tn)/2
- a*(t1 + ta)/2 = A or a*t1 + a*ta = 2*A ..... (1)
- b*(t1 + tb)/2 = B or b*t1 + b*tb = 2*B ..... (2)
- c*(t1 + tc)/2 = C or c*t1 + c*ta = 2*C ..... (3)
- Eliminate t1
  - b*(1) - a*(2) : a*b*(ta - tb) = 2*A*b - 2*B*a ... (4)
  - c*(2) - b*(3) : b*c*(tb - tc) = 2*B*c - 2*C*b ... (5)
  - a*(3) - c*(1) : c*a*(tc - ta) = 2*C*a - 2*A*c ... (6)
Last term : T(n) = T(1) + (n-1)*d
- ta = t1 + (a-1)*d ... (7)
- tb = t1 + (b-1)*d ... (8)
- tc = t1 + (c-1)*d ... (9)
- Eliminate t1
  - (7) - (8) : ta - tb = (a - b)*d ... (10)
  - (8) - (9) : tb - tc = (b - c)*d ... (11)
  - (9) - (7) : tc - ta = (c - a)*d ... (12)
Elliminate (ta - tb)
- (4) and (10) : a*b*(a - b)*d = 2*A*b - 2*B*a ... (13)
- (5) and (11) : b*c*(b - c)*d = 2*B*c - 2*C*b ... (14)
- (6) and (12) : c*a*(c - a)*d = 2*C*a - 2*A*c ... (15)
Find Left side of (13), (14), (15)
- A*a*(14) : A*a*b*c*(b - c)*d = 2*A*B*a*c - 2*A*C*a*b
- B*b*(15) : B*a*b*c*(c - a)*d = 2*B*C*b*a - 2*B*A*b*c
- C*c*(13) : C*a*b*c*(a - b)*d = 2*C*A*a*b - 2*C*B*c*a
Expression = 2*(A*B*a*c - A*C*a*b + B*C*b*a - B*A*b*c + C*A*a*b - C*B*c*a)

Verify

a*b*c*(A*(b - c) + B(c - a) + C*(a - b)) = ?
- Assume
- a = 3 and A = 1 + 2 + 3 = 6
- b = 4 and B = 1 + 2 + 3 + 4 = 10
- c = 5 and C = 1 + 2 + 3 + 4 + 5 = 15
- Find
- A*a*b*c*(b - c) = 06*3*4*5*(4 - 5) = -360
- B*a*b*c*(c - a) = 10*3*4*5*(5 - 3) = +1200
- C*a*b*c*(a - b) = 15*3*4*5*(3 - 4) = -900
- a*b*c*(A*(b - c) + B(c - a) + C*(a - b) = -360 + 1200 - 900 = -60
Find left side of (13), (14), (15)
- 2*c*C*(A*b - B*a) = 150*(24 - 30) = -900
- 2*a*A*(B*c - C*b) = 036*(50 - 60) = -360
- 2*b*B*(C*a - A*c) = 080*(45 - 30) = 1200
- Total RHS = -900 - 360 + 1200 = -60

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Q05. Describe y = (x^2 + x - 12)/(x^2 - 4) with sketch
Properties

y-intercept : x = 0 and y = 3
zeros of y : at x = -4 and x = 3
Asymptotes
- Vertical asymptotes : x = -2 and x = 2
- Horizontal asymptote : y = 1 as x goes to infinite
Signs of y = ((x+4)*(x-3))/((x+2)*(x-2))
- x LT -4 : y = (-)*(-)/(-)*(-) = (+)
- x EQ -4 : y = 0
- x GT -4 and x LT -2 : y = (+)*(-)/(-)*(-1) = (-)
- x EQ -2 : y = infinite (asymptote)
- x GT -2 and x LT -0 : y = (+)*(-)/(+)*(-) = (+)
- x EQ 0 : y = 3
- x GT 0 and x LT 2 : y = (+)*(-)/*+)*(-) = (+)
- x EQ 2 : y = infinite (asymptote)
- x GT 2 and x LT 3 : y = (+)*(-)/(+)*(+) = (-)
- x GT 3 : y = (+)*(+)/(+)*(+) = (+)
Curve range : decreasing or increasing
- x from -infinite to x = -4 : y from 1 to 0 (decrasing)
- x from -4 to -2 : y from 0 to -inifinte (decreasing)
- x from -2 to -0 : y from +infinite to 3 (decreasing)
- x from +0 to +2 : y from 3 to +infinite to 3 (increasing)
- x from +2 to +3 : y from -infinite to 0 (increasing)
- x from +3 to +infinite : y from 0 to +1 (increasing)
Concavity
- Below y = 1 and x = -2 : concave downword
- Between x = -2 and x = 2 : concave upward
- x GT 2 : concave downward

Diagram

Grphic calculator Section 3 : rational function
Method
- Open the program and enter GC
- Click Start and get menu
- Select 03 in upper box
- Select 08 in lower box
- Give coefficient : 0, 1, 1, -12, 0, 1, 0, -4

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Q06. Sketch y = x + 1/x

Method 1 : Use domain x and signs of y and asymptotes

1. Find asymptotes
- x = 0 and y = infinite, hence x = 0 is vertical asymptote
- x = infinite and y = x, hence y = x is line asymptote
- Sketch lines : x = 0 and y = x
2. Find signs of y when x LT 0
- x LE -1, y from -infinite to -2 (Increasing)
- 0 LT 0 and x GE -1, y from -2 to -infinite at x = 0 (decreasing)
- The curve is between x = 0 and y = x
- The curve is in 3rd quadrant
3. Find signs of y when x GT 0
- 0 GT x and LE 1, y from +infinite to +2 (decreasing)
- x GE 1, y from 2 to +infinite (increasing)
- The curve is between x = 0 and y = x
- The curve is in 1st quadrant
4. Extreme points at (-1,-2) and (1,2)

Method 2 : Use asymptotes and signs of y, y' and y"

1. Find y' and y"
- y = x + 1/x
- y' = 1 - 1/(x^2)
- y" = +2/x^3
2. Draw asymptotes as method 1
3. x LT -1
- y' GT 0 and y is increasing from -infinite to -2
- y" LT 0 and curve is concave downward
3. -1 GT x and x LT 0
- y' LT 0 and y is decreasing from -2 to -infinite
- y" LT 0 and curve is concave downward
4. x GT 0 and x LT 1
- y' LT 0 and y is decreasing from +infinite to 2
- y" GT 0 and curve is concave upward
5. x LG 1
- y' GT 0 and y is increasing from 2 to +infinite
- y" GT 0 and curve is concave upward
6. Extreme points at (-1,-2) and (1,2)

Diagram

Grphic of ratioanl functions Section 21 03 Rational
Open application program
Section 03 48 : y = x + 1/x

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Q07. Sketch y = x^2 + 1/x

Method 1 : Use domain x and signs of y and asymptotes

1. Find asymptotes
- x = 0 and y = infinite, hence x = 0 is vertical asymptote
- x = infinite and y = x^2, hence y = x^2 is parabola asymptote
- Sketch lines x = 0 and parabola y = x^2
2. Find signs of y when x LT 0
- x LE -1, y = 0 and y from +infinite to 0 (decreasing)
- 0 LT 0 and x GE -1, y from 0 to -infinite at x = 0 (decreasing)
- The curve is at left side of x = 0
- The curve is concave upward if x LT -1 and downward if x between -1 and 0
- The curve is inside y = x^2 when x LT -5
3. Find signs of y when x GT 0
- x GT 0, y from +infinite go to minimum and then go +infinte
- The curve is between x = 0 and y = x^2 when x is large
- The curve is decreasing and concave upward between x = 0 and minimumu point
- The curve is increasing and concave upward when x GT minimumu point

Method 2 : Use asymptotes and signs of y, y' and y"

1. Find y' and y"
- y = x^2 + 1/x
- y' = 2*x - 1/(x^2)
- y" = 2 + 2/(x^3)
2. Draw asymptotes as method 1
3. x LT -1
- y' LT 0 and y is decreasing from -infinite to 0
- y" GT 0 and curve is concave upward
- y" EQ 0 if x = -1
3. -1 GT x and x LT 0
- y' LT 0 and y is decreasing from 0 to -infinite
- y" LT 0 and curve is concave downward
4. x GT 0 and x LT minimum point
- y' LT 0 and y is decreasing from +infinite to minimum
- y" GT 0 and curve is concave upward
5. x LG 1
- y' GT 0 and y is increasing from minimum to +infinite
- y" GT 0 and curve is concave upward
6. Point of inflecion at (-1,0) and minimum at ?

Sketch using y = F(x)/G(x)

Grphic calculator Graphic Calculator (GC)
Open application program
Section 03 08 : y = x^2 + 1/x
Method
- Open the program and enter GC
- Click Start and get menu
- Select 03 in upper box
- Select 08 in lower box
- Give coefficient : 1, 0, 0, 1, 0, 0, 1, 0

Demo Diagram

Grphic of ratioanl functions 21 03 Rational
Open application program
Section 03 49 : y = x^2 + 1/x

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Q08. Sketch y = x + 4/(x^2)
Solutions

Similar as Q06

Diagram

Grphic of ratioanl functions Section 21 03 : Rational
Open application program
Enter section 03 47 : y = x + 4/(x^2)

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Q09. Use three points to sketch quadratic functions

Case 1 : b^2 - 4*a*c GT 0

Example : y = x^2 + x - 12
- Find zeros by observation or quadtic formula
- (x^2 + x - 12) = (x - 3)*(x + 4) = 0. Hence x = 3 and x = -4
- Find y-intercept : x = 0 and y = -12
Draw oxy coordinate on graphic paper
Draw three points : (0, -12), (3, 0) and (-4, 0)
Join three points using a smooth curve

Case 2 : b^2 - 4*a*c EQ 0

Example : y = x^2 + 6*x + 9
- Find zeros by observation or quadtic formula
- (x^2 + 6*x + 9) = (x + 3)^2. Hence x = -3 and x = -3. Onlu one point
- Find y-intercept : x = 0 and y = 9
- Find third point : It is (0, 9) symmetrical to the principal axis
- The principal axis : y = -b/2 = -3. Third point at x = -6
Draw oxy coordinate on graphic paper
Draw three points : (0, 9), (-3, 0) and (-6, 9)
Join three points using a smooth curve

Case 2 : b^2 - 4*a*c LT 0

Example : y = x^2 + 4*x + 9
- Find zeros by observation or quadtic formula
- No zeros
- Find y-intercept : x = 0 and y = 9
- Find vertex point : xv = -b/2 = -2 and yv = 5
- Find the symmetrical point of (0, 9) to the principal axis
- The principal axis : y = -b/2 = -2. Third point at x = -4
Draw oxy coordinate on graphic paper
Draw three points : (0, 9), (-2, 5) and (-4, 9)
Join three points using a smooth curve

Compare the diagrams for accuracy

Graphic of polynomial functions Section 21 02 : Polynomial
Plot quadratic function
- Open the application program
- Click menu
- Click 01 in upper box
- Click 02 in lower box
- Give data for y = x^ +x - 12 : 1, 1, -12
- Change y-scale : click ymax = 20 at left side list
- Click Re-plot
- Click menu to plot more

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Q10. Study y = x^4 + x^3 - 4*x^2 + x + 1
Use diagram to answer the following questions

Find y-untercept
Express it in factor form
Find minimum points and maximum point (at y' = 0)
Find domain if y GT 0
Find domain if y LT 0
Find domain if y' GT 0 (Curve is decreasing)
Find domain if y' LT 0 (Curve is decreasing)
Find domain if y" GT 0 (Curve is concave upward)
Find domain if y" LT 0 (Curve is concave downward)

Exercises : Use diagram to estimate the solutions

x^4 + x^3 - 4*x^2 + x + 1 = 0
x^4 + x^3 - 4*x^2 + x + 1 > 0
x^4 + x^3 - 4*x^2 + x + 1 < 0

Diagram

Graphic of polynomial functions Section 21 01 : Polynomial
Procedures
- Open application program (Run at current location without download)
- Click menu
- Click 01 in upper box
- Click Demo command
- Click 04 in lower box
- Chabge scale : Click xmax=5 and ymax=10 at left side box

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Q11. Study y = x^5 + 3*x^4 - 5*x^3 + -15*x^2 + 4*x + 12
Use diagram to answer the following questions

Find y-untercept
Express it in factor form
Find minimum points and maximum point
Find domain if y GT 0
Find domain if y LT 0
Find domain if y' GT 0 (Curve is increasing)
Find domain if y' LT 0 (Curve is decreasing)
Find domain if y" GT 0 (Curve is concave upward)
Find domain if y" LT 0 (Curve is concave downward)

Exercises : Use diagram to estimate the solutions

x^5 + 3*x^4 - 5*x^3 + -15*x^2 + 4*x + 12
x^5 + 3*x^4 - 5*x^3 + -15*x^2 + 4*x + 12
x^5 + 3*x^4 - 5*x^3 + -15*x^2 + 4*x + 12

Diagram

Graphic of polynomial functions Section 21 01 : Polynomial
Procedures
- Open application program (Run at current location without download)
- Click menu
- Click 01 in upper box
- Click Demo command
- Click 05 in lower box
- Chabge scale : Click xmax=5 and ymax=10 at left side box

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Q12. Study y = x^3 + 1/x

Solution

Asymptotes : x = 0 and y = x^3
Sketch asymptotes
It has no zeros
Sign of y
- It has no zeros
- When x is less than 0, y is negative. Hence curve is in 3rd quadrant
- When x is greater than 0, y is positive. Hence curve is in 1st quadrant
In 3rd quadrant
- The curve is between x = 0 and y = x^3
- Since y = x^3 + 1/x is less then y = x^3 at given domain
- Hence the curve is from -infinite to maximum point
- Hence the curve is from maximum point to -infinite (Since no zeros)
In 1st quadrant
- The curve is between x = 0 and y = x^3
- Since y = x^3 + 1/x is greater then y = x^3 at given domain
- Hence the curve is from infinite to minimum point
- Hence the curve is from minimum point to infinite

Diagram

Graphic of polynomial functions Section 21 03 : Rational
Procedures
- Open application program (Run at current location without download)
- Click menu
- Click 03 in upper box
- Click Demo command
- Click 10 in lower box
- Chabge scale : Click xmax=5 and ymax=10 at left side box

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Q13. Study y = x^3 - 1/x

Solution

Asymptotes : x = 0 and y = x^3
Sketch asymptotes
It has zeros at x = -1 and x = 1
Sign of y
- When x is less than -1, y is negative. Hence curve is in 3rd quadrant
- When x is between -1 and 0, y is positive. Hence curve is in 2nd quadrant
- When x is between +1 and 0, y is negative. Hence curve is in 4th quadrant
- When x is greater +1, y is positive. Hence curve is in 1st quadrant
In 3rd quadrant
- Since y = x^3 - 1/x is greater then y = x^3 at given domain
- The curve is at left side of x = 0 and y = x^3
- Hence the curve is from y = -infinite to y = 0 at x = -1
- The curve is concave downward
In 2nd quadrant
- Since y = x^3 - 1/x is greater then y = x^3 at given domain
- The curve is at left side of x = 0 and y = x^3
- Hence the curve is from y = 0 to y = infinite at x = 0
- The curve is concave upward
In 4th quadrant
- Since y = x^3 - 1/x is less then y = x^3 at given domain
- The curve is at right side of x = 0 and y = x^3
- Hence the curve is from y = -infinite at x = 0 and to y = 0 at x = 1
- The curve is concave downward
In 1st quadrant
- Since y = x^3 - 1/x is less then y = x^3 at given domain
- The curve is at right side of y = x^3
- Hence the curve is from y = 0 at x = 1 to y = x^3 as x is large
- The curve is concave upward

Diagram

Graphic of polynomial functions Section 21 03 : Rational
Procedures
- Open application program (Run at current location without download)
- Click menu
- Click 03 in upper box
- Click Demo command
- Click 14 in lower box
- Chabge scale : Click xmax=5 and ymax=10 at left side box

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Q14. Compare the graph of y = x^3 + 1/x and y = x^3 - 1/x

Solution

Based on signs of y and the asymptotes are given in AL 27 12 and Al 27 13
The diagrams are also mentioned in AL 27 12 and AL 27 13

Solution based on y' and y"

y = x^3 + 1/x : It has no zero value
- y' = 3*x^2 - 1/x^2
- y' = 0 when 3*x^4 - 1 = 0 or x = -0.76 or x = +0.76
- y' GT 0 when x LT -0.76. Hence the curve is increasing
- y' LT 0 when x = -0.76. Hence the curve is decreasing
- Hence the curve is concave downward in 4th quadrant
- Hence the curve has maximum at x = -0.76
- Similarly, the curve has minimum at x = + 0.76
- Hence the curve is concave upward in 1st quadrant
- y" = 6*x + 2/x^3.
- Hence y" LT zero when x LT zero (Concave downward)
- Hence y" GT zero when x GT zero (Concave upward)
y = x^3 - 1/x : It has zero values at x = -1 or x = 1
- y' = 3*x^2 + 1/x^2. Hence y' is not equal zero
- Hence the curve has no maximum or minimum
- y' GT 0 when x LT -1. Hence the curve is increasing
- y' GT 0 when x between -1 and 0. Hence the curve is decreasing
- y' GT 0 when x between +1 and 0. Hence the curve is decreasing
- y' GT 0 when x GT +1. Hence the curve is increasing
- y" = 6*x - 2/x^3.
- Hence y" LT zero when x LT -1 (Concave downward)
- Hence y" GT zero when x between 0 and -1 (Concave upward)
- Hence y" LT zero when x between 0 and +1 (Concave downward)
- Hence y" GT zero when x GT +1 (Concave upward)

Solution based on graphs (Alge.exe 03 10 and 03 14)

We can distingush them easily based on graphs in Alge.exe 03 10 and 03 14
Estimate the domain when y = 0, y' = 0 and y" = 0
Estimate the domain when y' is greater than zero
Estimate the domain when y' is less than zero
Estimate the domain when y" is greater then zero
Estimate the domain when y" is less than zero
Estimate the domain if there are extreme points
Estimate the domain if there are inflection points
Estimate the equations of the asymptotes

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