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Add Maths Bloopers

Add Maths Bloopers

Concentrate on the little black squares.

Believe it or not they are of the same size! This is a common observation. It is called an illusion.Maths problem can be an illusion !!

The solutions or part of solutions given below contain one or more mistakes. Circle the part that is wrong and show the correct solutions on the corresponding column on the right.

Question & SolutionsCorrections

1. Solve

5 + x = 6 ( x = 1

10 + x = 6

x = 4

2. Find the quadratic equation with roots 3 and 6.

[Sum] = 3 + 6 = 9 [Product] = (3)(6) = 18

Answer : x2 + 9x + 18

x2 9x + 18 = 0

3. Express f(x) = 3x2 6x 12 in the form of

a(x + p)2 + q. Hence, state the minimum point.

: f(x) = 3 x2+ 12x 9

= x2 + 4x 3

= (x + 2)2 5

thus, minimum point ( 2, 5) f(x) = 3 x2 +12x 9

= 3 [ x2 + 4x 3 ]

= 3 []

= 3 [ (x + 2)2 7] = 3(x + 2)2 21 maka, titik minimum ( 2, 21)

4. f(x) = x2 + (k 2)x k+ 2 has equal roots. Find the values of k.

b2 4 a c = 0

(k 2)2 4(1)(k + 2) = 0

k2 + 4 8 = 0

k2 = 4 k = 2 b2 4 a c = 0

(k 2)2 4(1)(k + 2) = 0

k2 + 4 8 = 0

k = ( 2

5. Solve x2 3x > 10

x2 3x - 10 > 0 (x - 5)(x + 2) > 0

x > 5, x > -2 (x - 5)(x + 2) > 0

x < 2 x > 5

6.Solve x2 + 5x = 6

x(x + 5) = 6

x = 6 , x + 5 = 6

x = 1 x2 + 5x 6 = 0

(x -1)(x + 6) = 0

x = 1 , x = -6

Question and Solution Correction

7.Solve simultaneously 3x + 2y = 12

2y2 x2 = xy

Solutions:

From 1 : substitute into 2

2 ( )2 x2 = x ( )

2 ( )2 x2 = x ( )

144 9x2 x2 = -6x + 3x2

13x2 + 6x + 144 = 0 2 ()x2 = x ( )

() x2 =

144 72x + 9x2 2x2 = 12x + 3x2

4x2 60x + 144 = 0

x2 15x + 36 = 0 x = 12, y = 12 ; x = 3, y = 3/2

8. 2 log2 3y + log2 y = 4

log2 3y2 + log2 y = 4

3y2 + y = 4

3y2 + y 4 = 0

(3y + 4 )( y 1) = 0 log2 (3y)2(y) = 4

9y3 = 24 = 16

y3 = 16/9

y = 1.211

9. Solve log3 ( y + 4) = log3 2y + 3

log3 y + log3 4 = log3 2y + 3

4y = 2y + 3

y = ( log3 (y + 4) log 2y = 3

y + 4 = 8 y + 4 = 16y 2y y = 4/15

10.Solve 2 x 1 + 2x = 12.

( 2 x 1 + x = 12

2x 1 = 6

x = 7/2 2x . 21 + 2x = 12

2x ( + 1 ) = 12

2x = 8

x = 3

11. Given that 4x . 4 2y = 1, find y in terms of x.

x + 2y = 1

x = 1 2y

4 x + 2y = 4 0

x + 2y = 0

y =

12. Given that y = (x 2)2 + 3. Find

(a) equation of the axis of symmetry.

(b) the y-intercept

Answer: (a)

(b) .(a) x = 2(b) y-intercept = 7 ( intersection pt

(0, 7)

(a) x = 2

(b) 7

Question & SolutionCorrection

13. Sector AOB has radius 4.5 cm and arc length 7.2 cm. Find ( , in radians.

solutions:

s = r (

7.2 = 4.5 (

( = 1.6 = 0.02793 rad incorrect when using formula s = r(,

( is already in radian form.

14. A series is given by 8, 4, 2, 1 Find the twelfth term. .

solution:

d = 4 8 = 4

T12 = 8 + (12 1) 4

= 8 + 11 4

= 15

Find out whether AP or GP

84, 42, 21, 1 = not an AP

8( , 4 ( , 2( , 1 = GP

r = ( T12 = 8

= 0.003906 //

or =

15. Solve 2 sin y cos y = 0

2 sin y (1 sin y) = 0

3 sin y = 1

sin y = 1/ 3

y = 19o 28 2sin y = cos y tan y = (positive 1 & 3 quadrant) y = 26 34 , 206 34

16. Solve cos 2A = 0.3420 for 0( ( A ( 360(

Solutions:

2A = ( 70() reference angle

2A = 110(, 250( (2 & 3 quadrant)

A = 55(, 125(

new range 0 ( 2A ( 720

2A = 110(, 250(, 470(, 510(

A = 55(, 125(, 235(, 255(

17. differentiate:

= = 2 x -2

18.Find the rate of change in the area of a circle, A if the radius, x decreases at the rate of 5 cm s-1 when x = 3 cm.

solutuins:

dA/dt = dx/dt x dA/dx A = x2

= 5 x 6 dA/dx = 2x

= 30 = 2 (3) = 6Rate of radius decreases, dx/dt = 5

dA/dt = 30


18.Given that f(x) = x2 4x + 5. Find the equation of tangent at point (1, 3)

solutions:

dy/dx = 2x 4 = 0

x = 2

( equation of tangent , y 3 = 2(x 1)

y = 2x + 1

dy/dx = 2x 4 at x = 1

= 2(1) 4 = 2

mT = 2

eq of tangent, y 3 = 2(x 1)

y = 2x + 5

19.Find the equation of a curve with gradient function 2x 1 and passes through the point (4, 5)

Solutions:

gardient function = 2x 1

thus, gradient ,

equation of a curve ( y 5 = 2 (x 4)

y = 2x 3

gradient function, dy/dx = 2x 1

Eq. of curve, y =

thus, y = x2 x + c at (4, 5) 5 = (4)2 4 + c ( c = 7

y = x2 x 7

20. Find

=

= + c

= + c

21. Given , evaluate .

=

=

=

=

=

=

22. Evaluate

= =

=

= 10


23. Find the value of ( 2i 5j 6i + 2j (

(

= ( 4 i 3j (

(

= 4i + 3 j

= (-4)2 + (-3)2 = 5

24. Find (A. Use Sine Rule

sin A = 0.6435 A = 40( ( nearest degree)if A = 40( then (C = 110( ( cannot be)Since,

BC is the longest side ( A must be the biggest

angle

A = 180 40 = 140

(Ambigous case)

25. Find x.

x2 = 52 + 72 2(5)(7) cos 50

= 74 70 kos50

= 4 (0.6428)

= 2.571

= (2.571

= 1.603 x2 = 52 + 72 2(5)(7) cos 50

= 74 70 cos50

= 29.005

x = 5.386

26. A normal distribution has mean 50 and its variance 16. Find score Z when X = 55.

solutions:

Z =

27.Solve 5 ( 3 x + 2) = 4

: 15 x + 2 = 4

(x +2) log 15 = log 4

(x + 2) = log 4 log 15

x + 2 = 0.57403

x = 2.574

(3 x + 2) = = 0.8

(x + 2)log 3 = log 0.8

(x + 2) = log 0.8 log 3

x + 2 = 0.09691 0.47712

x = 0.203114 2

x = 2.203114

THE END

After 1 hours, sitting for add maths paper :

..

Notes

My Strength

TopicSub-topic & Comment

My weaknesses

TopicSub-topic & Comment

EMBED Equation.3

(0, 7)

12 - 3x 2

12 - 3x 2

6

x

50(

5

7

12 - 3x 2

C

30(

7.77

10

2

12 - 3x 2

12 - 3x 2

4

11 33 = 4x

11 9 = 4x

2 = 4x

x = 2

yx

x2

given that xy = ax2 + EMBED Equation.3 .

Find the value of r.

solutions:

EMBED Equation.3 = 4

r = 2

Given y = EMBED Equation.3 . Find f( (x)

y(3x + 1) = x

3xy + y = x

x = EMBED Equation.3

thus , f - 1 = EMBED Equation.3

EMBED Equation.3

= EMBED Equation.3

(3x 1)2 = 9x2 + 1

dy/dx = ( x2 + 3) 1 when x = 2,

= ( 22 + 3) 1

= ( 4 + 3) 1

= 7 1

= 7

COMON MISTAKES

M

(

(

(

(

(

(

-2 5

2.69

A

B

EMBED Equation.3

B L A N K !!!!!!

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