FUNCTIONS G2

 

SEQUENCE & SERIES G1

 

The 1st term of an arithmetic progression is 60 and the common difference is 40. Find the nth term

 

solution

 

A1=60, d= 40, n=?

 

An =A1 + (n-1) d

 

An=60 + (n-1)40

 

An=60 + 40n -40

 

An=40n + 60-40

 

An=40n + 20

 

Hence the nth term is An=40n + 20  

 

TRY THIS………………….. 

 

The 1st term of arithmetic progression is 148 and the common difference is 170. Find the nth term

FUNCTIONS G1

 

Find the maximum value of the quadratic equation:7-6t-8t².

 

Solution

 

a=-8, b=-6, c=7

 

Maximum = 4ac-b²

                        4a

 

Maximum = (4 x -8 x 7) - (-6)²

                          4(-8)

 

Maximum = (-224)- 36

                         32

 

Maximum= -260                         since (-224)- 36= -260

                       32

 

Maximum = -130 =   -65

                       16          8

 

Hence maximum value is -65/8

 

TRY THIS………………………………..

 

NECTA  2003 QN. 10c

 

Find the maximum value of the quadratic equation:3+30t-5t².

 

SIMPLIFY G1

 Simplify (6a)² - 30a². 

 

Solution

 

= (6a)² - 30a²    

 

= 6².a² - 30a²               Since (mn)² =m² . n²

 

= 36a² - 30a²   {since the square number of 6 is 36}

 

= 6a²

 

TRY THIS………………

 

Simplify (8m)² – 80m².

PROBABILITY F1

 

A bag contains 4 red balls and 8 blue balls. Two balls are taken from the bag. What is the probability that they are both red?

 

Solution

 

(This is a problem with replacement)

 

n(R) = 4, n(B) = 8, n(S) = 12

 

P(R) = n(R)

            n(S)

 

P(R) = 4

          12

 

1st pick = ⁴/12

2nd pick = ⁴/12 as well.

 

 

P(RR) =  4      x    4

             12            12

 

P(RR) =   ¹⁶ /144 = ¹/9       

 

Therefore Probability of drawing a red ball is 1/9     

 

 

TRY THIS ..............

 

A bag contains 7 purple stones and 11 yellow stones. Two stones are taken from the bag. What is the probability that they are both purple?

ALGEBRA 3F

 The sum of four consecutive even numbers is 316. Find the 3rd number.

 

Solution

 

Let the numbers be as shown in the table below

 

1st number	2nd number	3rd number	4th number	TOTAL
n	n+2	n+4	n+6	316

 

Then, n + (n+2) + (n+4) + (n+6) = 316

 

4n + 2+4+6= 316

 

4n + 12 = 316

 

4n = 316 – 12

 

4n = 304 

 

4n = 304

 4        4

 

n = 76

 

3rd number = n +4

 

                   = 76 + 4

 

                   = 80

 

Hence the 3rd number is 80

 

TRY THIS………………….. 

 

The sum of four consecutive even numbers is 372. Find the largest number.

ALGEBRA F3

 The sum of four consecutive even numbers is 316. Find the 3rd number.

 

Solution

 

Let the numbers be as shown in the table below

 

1st number	2nd number	3rd number	4th number	TOTAL
n	n+2	n+4	n+6	316

 

Then, n + (n+2) + (n+4) + (n+6) = 316

 

4n + 2+4+6= 316

 

4n + 12 = 316

 

4n = 316 – 12

 

4n = 304 

 

4n = 304

 4        4

 

n = 76

 

3rd number = n +4

 

                   = 76 + 4

 

                   = 80

 

Hence the 3rd number is 80

 

TRY THIS………………….. 

 

The sum of four consecutive even numbers is 372. Find the largest number.

LOGARITHMS F9

 Evaluate Log1,000,000,000 -  log 0.01 + log₃243

 

Solution

 

= Log1,000,000,000 -  log 0.001 + log₃243

 

= Log10⁹ -  log 10^-3 + log₃3^-5

 

= 9Log10 -  (-3log 10) + (-5log₃3)

 

= (9x1) - (-3x1) + (-5x1)

 

= 9 - (-3) + (-5)

 

= 7

 

Hence Log1,000,000,000 +  log 0.001 + log₃243 = 7

 

TRY THIS……………..

 

 

Evaluate Log1,000,000,000  -   log 0.0001 + log₃81

VARIATIONS F1

 

x is inversely proportional to y. x=22 while y=5. Find y when x is 10.

 

Solution

 

x ⍺ k

      y

 

x = k

      y

 

22 = k

        5

 

k = 22 x 5

 

k = 110

 

,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

 

k = 110, y= ?, x = 10.

 

x = k

      y

 

10  = 110

          y

 

y x 10 = 110  x  y   (multiply by y on both sides)

               y

 

 

10y = 110

10      10

 

Hence y = 11.

 

 

TRY THIS…………….

 

 

x is inversely proportional to y. x=80 while y=4. Find y when x is 8. 

LOGARITHMS F8

 

Simplify Log₂256 - Log₃729

 

Solution

 

= Log₂256 - Log₃729

 

= Log₂2⁸ - Log₃3⁵    [Since 256=2⁸ and 729=3⁶].

 

= 8Log₂2 - 6Log₃3    [since Log_aa = 1]

 

= (8 x 1) - (6 x 1)

 

= 8 - 6

 

= 2

 

Hence Log₂256 -Log₃729 = 2

 

TRY THIS..................

 

Simplify Log₂2048 - Log₅625 

LENGTH OF AN ARC F1

 

Find the length of an arc if the radius of the circle is 70cm.

 

Solution

 

L = ∏r

      180⁰

 

L = ∏ x  70

             180⁰

 

L = 7∏ cm

       18

 

Hence length of an arc is ^7∏/18 cm

 

TRY THIS…………………………..

 

 

Find the length of an arc if the radius of the circle is 30cm.

COMPOUND INTEREST F1

Mwakajonga invested a certain amount of money in a bank which gives an interest rate of 20% compounded annually. How much did she invest at the start if she got 20,000 sh at the end of 3 years?

 

Solution

 

n=3, t=1, R=20%, A₃=8,500, P=?

A_n = P(1 + ^RT/100)ⁿ

A₃ = P(1 + ^(20x1)/100)³

A₃= P(1 + ²⁰/100)³

A₃ = P(¹⁰⁰/100 + ²⁰/100)³

A₃ = P(¹²⁰/100)³  But A₃=20,000.

20,000 = P(1.2)³  = P(1.2x1.2x1.2)  

 

20,000 = 1.728P

 

20000  =  1.728P

1.728       1.728

 

20000  =  P

1.728

 

P ≈ 11574/=

 

Hence at the start she invested Tsh 11574/=

 

TRY THIS………………..

 

 

Tunsubilege invested a certain amount of money in a bank which gives an interest rate of 20% compounded annually. How much did she invest at the start if she got 16,000 sh at the end of 5 years?

EXPAND F1

 

Expand 11w(5w + 3 - 11w)

 

Solution

 

= 11w(5w + 3 - 11w)

 

= (11w x 5w) + (11w x 3) - (11w x 11w)

 

= 55w² + 33w - 121w²

 

= [55w² - 121w²] + 33w collecting like terms

 

= - 66w² + 33w

 

= 33w - 66w² answer

 

 

TRY THIS………..

  

Expand 6a(5a+ 12 - 9a)

LOGARITHMS F7

 If log_x 2401 – log₃729 = -2; find x

 

solution

 

log_x 2401 – log₃729 = -2

 

log_x 2401 – log₃3⁶ = -2

 

log_x 2401 – 6log₃3 = -2

 

log_x 2401 – 6 x 1 = -2

 

log_x 2401 – 6 = -2

 

log_x 2401 = -2 + 6

 

log_x 2401 = 4 

 

2401 = x⁴        2401=7x7x7x7=7⁴ by prime factors

 

7⁴ = x⁴      Powers cancel out

 

x = 7

 

Hence x = 7.

 

SETS F2

 If n(A)= 55 , n(B)= 90 and n(AuB)= 136, find n(AnB).

 

Solution

 

n(AuB) = n(A) + n(B) - n(AnB)

 

136 = 55 + 90 - n(AnB)

 

136 = 145 - n(AnB)

 

136 - 145 = - n(AnB) [after transferring 145 on the left hand side]

 

-9 = -n(AnB)

 

n(AnB) = 9 [after dividing by -1 both sides]

 

Hence n(AnB) = 9 answer

MATRICES F1

 

SETS F1

 

In a certain school, 160 students take either Chemistry or biology. 140 take chemistry and 100 take both subjects. Find those who take biology only.

 

Solution

 

In most cases, OR stands for union whereas AND/BOTH, stands for intersection.

 

Let Chemistry = n(C), Biology = n(B).

 

n(C)= 140 ,

n(B)= ?

n(CuB) = 160,

n(CnB)= 100

 

 

Solution

 

n(CuB) = n(C) + n(B) - n(CnB)

 

160  =  140 + n(B)  – 100

 

160  =  n(B) + 40

 

160 - 40  =  n(B)

 

n(B)= 120

 

physics only = n(B) – n(CnB)

 

                    = 140 – 100

 

                    = 40 

 

 

Hence those taking only biology are 40.    

 

TRY THIS………….

 

In a certain school, 196 students take either Chemistry or French. 150 take chemistry and 110 take both subjects. Find those who take French only.

FUNCTIONS F3

 If F(x) = 2x + 50; Find F^-1(200)

 

Solution

 

HINT: F^-1(x) means inverse.

 

PROCEDURE:

Make x the subject and then interchange x and y variables.

 

Let y=F(x)

 

So,  y= 2x + 50

 

y – 50 = 2x

 

y – 50  = 2x    dividing by 2 on both sides

   2          2

 

y – 50   = x

   2

 

x  =   y – 50     after rearranging

           2

 

 y^-1 = x – 50     after interchanging x and y variables.

             2

 

F^-1(x) = x – 50     after interchanging x and y variables.

                2

 

Now we find F^-1(200)

 

F^-1(200) = 200– 50     

                     2

 

F^-1(200) = 150     

                   2

 

F^-1(200) = 75   

 

TRY THIS……………………………

 

If F(x) = 2x + 80; Find F^-1(20)