Home › 2020 › Page 74

Yearly Archives: 2020

Composite Function (Comparison Method) Example 3

Posted on by

Example 3:
Given f: x → hx + k and f2 : x → 4x + 15.
(a)  Find the values of h and of k.
(b)  Take h > 0, find the values of x for which f (x2) = 7x

Solution:
(a)
Step 1:
Find f2 (x)
Given f (x) = hx + k
f2 (x) = ff (x) = f (hx + k)
= h (hx + k) + k
= h2x + hk + k

Step 2:
Compare with given f2 (x)
f2 (x) = 4x + 15
h2x + hk+ k = 4x + 15
h2 = 4
h = ± 2
When, h = 2
hk + k = 15
2k + k = 15
k = 5

When, h = –2
hk + k = 15
–2k + k = 15
k = –15

(b)
h > 0, h = 2, k = 5
Given f (x) = hx + k
f (x) = 2x + 5

f (x2) = 7x
2 (x2) + 5 = 7x
2x2 – 7x+ 5 = 0
(2x – 5)(x–1) = 0
2x – 5 = 0   or  x –1= 0
x = 5/2
 or
x = 1

Example 1

Posted on by

Example 1
Given the function f : x → 6 x + 1 . Find the value of p if f ( 4 ) = 4 p + 5 .

Answer:
f : x → 6 x + 1 f ( x ) = 6 x + 1 f ( 4 ) = 6 ( 4 ) + 1 f ( 4 ) = 25

f ( 4 ) = 4 p +   5 25 = 4 p +   5 4 p = 25 − 5 = 20 p = 20 4 = 5

4.1 Simultaneous Equations

Posted on by

4.1 Simultaneous Equations

(A) Steps in solving simultaneous equations:
  1. For the linear equation, arrange so that one of the unknown becomes the subject of the equation.
  2. Substitute the linear equation into the non-linear equation.
  3. Simplify and expressed the equation in the general form of quadratic equation a x 2 + b x + c = 0
  4. Solve the quadratic equation. 
  5. Find the value of the second unknown by substituting the value obtained into the linear equation.

Example:
Solve the following simultaneous equations.
y + x = 9 x y = 20

Solution:
For the linear equation, arrange so that one of the unknown becomes the subject of the equation.
y + x = 9 y = 9 − x

Substitute the linear equation into the non-linear equation.
x y = 20 x ( 9 − x ) = 20 9 x − x 2 = 20

Simplify and expressed the equation in the general form of quadratic equation a x 2 + b x + c = 0
9 x − x 2 = 20 x 2 − 9 x + 20 = 0

Solve the quadratic equation. 
x 2 − 9 x + 20 = 0 ( x − 4 ) ( x − 5 ) = 0 x = 4  or  x = 5

Find the value of the second unknown by substituting the value obtained into the linear equation.
When  x = 4 , y = 9 − x = 9 − 4 = 5 When  x = 5 , y = 9 − x = 9 − 5 = 4

Long Questions (Question 7 & 8)

Posted on by

4.2.4 Simultaneous Equations, Long Questions 
Question 7:
Solve the following simultaneous equations.
5y – 6x= 2
4 y x − 3 x y = 4.

Solution:
5y – 6x= 2 ----- (1)
4 y x − 3 x y = 4  ------- (2) From (1), y = 2 + 6 x 5

Substitute (3) into (2),
4 ( 2 + 6 x 5 ) x − 3 x ( 2 + 6 x 5 ) = 4 8 + 24 x 5 x − 15 x 2 + 6 x = 4 ( 8 + 24 x ) ( 2 + 6 x ) − ( 15 x ) ( 5 x ) 5 x ( 2 + 6 x ) = 4

16 + 48x + 48x + 144x2 – 75x2 = 20x (2 + 6x)
69x2 + 96x + 16 = 40x + 120x2
51x2 – 56x – 16 = 0
(3x – 4)(17x + 4) = 0
3x – 4 = 0 or 17x + 4 = 0
x = 4 3   or    x = − 4 17
  

Substitute  x = 4 3  into (3), y = 2 + 6 ( 4 3 ) 5 = 2

Substitute  x = − 4 17  into (3), y = 2 + 6 ( − 4 17 ) 5 = 2 17

∴ The solutions are  ( 4 3 , 2 )  and  ( − 4 17 , 2 17 ) .




Question 8:
Solve the following simultaneous equations.
x+2y=1 2 x 2 + y 2 +xy=5
Give your answer correct to three decimal places.

Solution:
x+2y=1..........( 1 ) 2 x 2 + y 2 +xy=5..........( 2 ) x=1−2y..........( 3 ) Substitute ( 3 ) into ( 2 ), 2 ( 1−2y ) 2 + y 2 +( 1−2y )y=5 2( 1−4y+4 y 2 )+ y 2 +y−2 y 2 =5 2−8y+8 y 2 − y 2 +y−5=0 7 y 2 −7y−3=0 a=7,b=−7,c=−3 x= −b± b 2 −4ac 2a y= −( −7 )± ( −7 ) 2 −4( 7 )( −3 ) 2( 7 ) y= 7± 133 14 y=1.324  or  −0.324 From x=1−2y When y=1.324,  x=1−2( 1.324 )=1.648 When y=−0.324,  x=1−2( −0.324 )=1.648 ∴The solutions are ( 1.648,1.324 ) and ( 1.648,−0.324 ).

Long Questions (Question 5 & 6)

Posted on by

Question 5:

Lisa has a rectangular plot of land. She plants orchid and rears fish in the areas as shown on diagram above. The area used for planting orchid is 460 m2 and the perimeter of the rectangular fish pond is 48 m. Find the value of x and y.


Solution:
Area used for planting orchid = 460 m2
10 (30 – y) + xy = 460
300 – 10y + xy = 460
xy – 10y = 160
y (x – 10) = 160

y = 160 x − 10  ------- (1)

Perimeter of the rectangular fish pond = 48 m
2 (x – 10) + 2 (30 – y) = 48
2x – 20 + 60 – 2y = 48
2x – 2y= 8
x – y = 4
x = 4 + y ------ (2)


Substitute (2) into (1):
y = 160 x − 10 y = 160 4 + y − 10 y = 160 y − 6 y 2 − 6 y − 160 = 0 ( y − 16 ) ( y + 10 ) = 0 y = 16   or    y = − 10  (not accepted)

From (2),
When y = 16
x = 4 + 16 = 20



Question 6:

In the diagram above, PQRS is a rectangular piece of paper with an area of 112 cm2. A semicircle STR is cut from this piece of paper. The perimeter of the remaining piece of paper is 52 cm. Using  π = 22 7 , find the integer value of x and y.

Solution:
Area of the rectangle PQRS = 112 cm2
Therefore, (14x)(2y) = 112
 28xy= 112
 xy = 4 ------ (1)

Perimeter of PSTRQ = 52 cm
PS + QR + PQ + Length of arc STR = 52
2y + 2y + 14x + ½ (2πr) = 52
4y + 14x +   ( 22 7 ) ( 7 x ) = 52
4y + 14x + 22x = 52
4y + 36x = 52
y + 9x = 13 ------ (2)

From equation (2) : y = 13 – 9x ------ (3)

Substitute (3) into (1) :
x (13 – 9x) = 4
13x – 9x2 = 4
9x2 – 13x + 4 = 0
(x – 1)(9x – 4) = 0
x = 1   or   4 9  (not an interger, therefore not accepted)

From (3) :
When x = 1,
y = 13 – 9(1) = 4. 

Long Questions (Question 3 & 4)

Posted on by

Question 3:
Solve the following simultaneous equations.
3y – 2x= – 4
y2 + 4x2 = 2

Solution:
3y – 2x= – 4 -----(1)
y2 + 4x2= 2 -----(2)

From (1), y = 2 x − 4 3  ------- ( 3 ) Substitute (3) into (2), ( 2 x − 4 3 ) 2 + 4 x 2 = 2 ( 4 x 2 − 16 x + 16 9 ) + 4 x 2 = 2 4 x 2 − 16 x + 16 + 36 x 2 = 18     ( × 9 ) 40 x 2 − 16 x − 2 = 0 20 x 2 − 8 x − 1 = 0 ( 10 x + 1 ) ( 2 x − 1 ) = 0 x = − 1 10   or   x = 1 2 Substitute the values of  x  into (3), When  x = − 1 10 , y = 2 ( − 1 10 ) − 4 3 = − 1 2 5 When  x = 1 2 , y = 2 ( 1 2 ) − 4 3 = − 3 3 = 1 The solutions are  x = − 1 10 ,   y = − 1 2 5  and  x = 1 2 ,   y = 1.





Question 4:
Solve the simultaneous equations x – 3y = –1 and y + yx – 2x = 0.
Give your answers correct to three decimal places.

Solution:
x – 3y = –1 -----(1)
y + yx – 2x = 0 -----(2)
From (1),
x = 3y – 1 -----(3)
Substitute (3) into (2),
y + y (3y – 1) – 2(3y – 1)  = 0
y + 3y2 – y – 6y+ 2 = 0
3y2 – 6y + 2 = 0

a = 3 ,   b = − 6 ,  c = 2 y = − b ± b 2 − 4 a c 2 a y = − ( − 6 ) ± ( − 6 ) 2 − 4 ( 3 ) ( 2 ) 2 ( 3 ) y = 6 ± 12 6 y = 1.577  or 0 .423

Substitute the values of y into (3).
When y = 1.577,
x = 3 (1.577) – 1 = 3.731 (correct to 3 decimal places)

When y = 0.423,
x = 3 (0.423) – 1 = 0.269 (correct to 3 decimal places)

The solutions are x = 3.731, y = 1.577 and x = 0.269, y = 0.423.


Long Questions (Question 1 & 2)

Posted on by

Question 1:
Solve the following simultaneous equations.
y + 2 x = 2 2 x + 1 y = 5

 Solution:
y+2x=2−−−−(1) 2 x + 1 y =5−−−−(2) y=2−2x−−−−(3) substitute (3) into (2), 2 x + 1 2−2x =5 2( 2−2x )+x x( 2−2x ) =5 4−4x+x=5x( 2−2x ) 4−3x=10x−10 x 2 10 x 2 −13x+4=0 ( 5x−4 )( 2x−1 )=0 5x−4=0     or     2x−1=0 x= 4 5            or     x= 1 2 Substitute values of x into (3), When x= 4 5  ,  y=2−2( 4 5 )= 2 5 When x= 1 2 y=2−2( 1 2 )=1 The solutions are x= 4 5 , y= 2 5  and x= 1 2 , y=1




Question 2:
Solve the following simultaneous equations.
x – 3y + 5 = 3y + 5y2– 6 – x = 0

Solution:
x – 3y + 5 = 0
x = 3y – 5 -----(1)
3y + 5y2 – 6 – x = 0 -----(2)

Substitute (1) into (2),
3y + 5y2 – 6 – (3y – 5) = 0
3y + 5y2 – 6 – 3y + 5 = 0
5y2 – 1 = 0
5y2  = 1
y = ±0.447

Substitute the values of y into (1),
When y = 0.447
x = 3 (0.447) – 5
x = –3.659

When y = – 0.447
x = 3 (–0.447) – 5
x = –6.341

The solutions are x = –3.659, y = 0.447 and x = –6.341, y = – 0.447.

Simultaneous Equations (Example 1 & 2)

Posted on by

Example 1:
Solve the simultaneous equations.
x + 1 4 y = 1  and  y 2 − 8 = 4 x .

Solution:
x + 1 4 y = 1 → ( 1 ) y 2 − 8 = 4 x → ( 2 ) x = 1 − 1 4 y → ( 3 )

Substitute (3) into (2),
y 2 − 8 = 4 ( 1 − 1 4 y ) y 2 − 8 = 4 − 4 4 y
y2 + y – 12 = 0
(y + 4)(y – 3) = 0
y = –4 or y = 3 

Substitute the values of y into (3),
when y=−4,  x=1− 1 4 (−4)=2 when y=3,  x=1− 1 4 (3)= 1 4 The solutions are x=2, y=−4 and x= 1 4 , y=3


Example 2:
Solve the simultaneous equations 2x + y = 1 and 2x2+ y2 + xy = 5.
Correct your answer to three decimal places.

Solution:
2x + y = 1-----(1)
2x2 + y2+ xy = 5-----(2)

From (1),
y = 1 – 2x-----(3)

Substitute (3) into (2).
2x2 + (1 – 2x)2 + x(1 – 2x) = 5
2x2 + (1 – 2x)(1 – 2x) + x – 2x2 = 5
1 – 2x – 2x + 4x2 + x – 5 = 0
4x2 – 3x – 4 = 0

From  x = − b ± b 2 − 4 a c 2 a a = 4 ,   b = − 3 ,  c = − 4 x = − ( − 3 ) ± ( − 3 ) 2 − 4 ( 4 ) ( − 4 ) 2 ( 4 ) x = 3 ± 73 8 x = − 0.693  or  1.443

Substitute the values of x into (3).
When x = –0.693,
y = 1 – 2 (–0.693) = 2.386 (correct to 3 decimal places)

When x = 1.443,
y = 1 – 2 (1.443) = –1.886 (correct to 3 decimal places)

The solutions are x = –0.693, y  = 2.386 and x = 1.443, y = –1.886.

Quadratic Functions, SPM Practice (Long Questions)

Posted on by

Question 5:


Diagram above shows the graphs of the curves y = x2 + x – kx + 5 and y = 2(x – 3) – 4h that intersect the x-axis at two points. Find
(a) the value of k and of h,
(b) the minimum value of each curve.


Solution:
(a)
y= x 2 +x−kx+5 = x 2 +( 1−k )x+5 = [ x+ ( 1−k ) 2 ] 2 − ( 1−k 2 ) 2 +5 axis of symmetry of the graph is x=− ( 1−k ) 2

y=2 ( x−3 ) 2 −4h axis of symmetry of the graph is x=3. ∴ − 1−k 2 =3 −1+k=6 k=7

Substitute k=7 into equation y= x 2 +x−7x+5   = x 2 −6x+5 At x-axis,y=0; x 2 −6x+5=0 ( x−1 )( x−5 )=0 x=1,5

At point ( 1,0 ) Substitute x=1,y=0 into the graph: y=2 ( x−3 ) 2 −4h 0=2 ( 1−3 ) 2 −4h 4h=2( 4 ) 4h=8 h=2

(b)
For y= x 2 −6x+5 = ( x−3 ) 2 −9+5 = ( x−3 ) 2 −4 ∴ Minimum value is −4. For y=2 ( x−3 ) 2 −8, minimum value is−8.

Quadratic Functions, SPM Practice (Long Questions)

Posted on by


3.9.1 Quadratic Functions, SPM Practice (Long Questions)

Question 1:
Without drawing graph or using method of differentiation, find the maximum or minimum value of the function y = 2 + 4x – 3x2. Hence, find the equation of the axis of symmetry of the graph.

Solution:
By completing the square for the function in the form of y = a(x + p)2+ q to find the maximum or minimum value of the function.

y = 2 + 4x – 3x2
y = – 3x2 + 4x + 2 ← (in general form)
y = − 3 [ x 2 − 4 3 x − 2 3 ] y = − 3 [ x 2 − 4 3 x + ( − 4 3 × 1 2 ) 2 − ( − 4 3 × 1 2 ) 2 − 2 3 ] y = − 3 [ ( x − 2 3 ) 2 − ( − 2 3 ) 2 − 2 3 ]  

y = − 3 [ ( x − 2 3 ) 2 − 4 9 − 6 9 ] y = − 3 [ ( x − 2 3 ) 2 − 10 9 ] y = − 3 ( x − 2 3 ) 2 + 10 3 ← in the form of  a ( x + p ) 2 + q

Since a = –3 < 0,
Therefore, the function y has a maximum value of 10 3 . 
x − 2 3 = 0 x = 2 3
Equation of the axis of symmetry of the graph is x = 2 3 .   



Question 2:

The diagram above shows the graph of a quadratic function y = f(x). The straight line y = –4 is  tangent to the curve y = f(x).
(a) Write the equation of the axis of symmetry of the function f(x).
(b)   Express f(x) in the form of (x + p)2 + q , where p and q are constant.
(c) Find the range of values of x so that
(i) f(x) < 0, (ii) f(x) ≥ 0.

Solution:
(a)
x-coordinate of the minimum point is the midpoint of (–2, 0) and (6, 0)
= = − 2 + 6 2 = 2  
Therefore, equation of the axis of symmetry of the function f(x) is x = 2.



(b)
Substitute x = 2 into x + p = 0,
2 + p = 0
p = –2
and q = –4 (the smallest value of f(x))
Therefore, f(x) = (x + p)2 + q
f(x) = (x – 2)2 – 4

(c)(i) From the graph, for f(x) < 0, range of values of x are –2 < x < 6 ← (below x-axis).

(c)(ii) From the graph, for f(x) ≥ 0, range of values of x are x ≤ –2 or x ≥ 6 ← (above x-axis).