FUNCTIONS 3

If f(x) = x⁴ - kx² - 6x + 7 has a remainder of 22 when divided by x+2; find k.

solution

f(x) = x⁴ - kx² - 6x + 7

x + 2 = 0

x = -2

f(x) = (-2)⁴ - k(-2)² - 6(-2) + 7 = 22

16 - 4k + 12 + 7 = 22

16 - 4k + 19 = 22

16 + 19 = 22 + 4k

16 + 19 - 22=4k

35 - 22=4k

13 = 4k

k = ¹³/4    after dividing by 4 both sides.

hence k=¹³/4

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

If f(x) = x⁴ - kx² - 5x - 31 has a remainder of 18 when divided by x-2; find k.

PERCENTAGE PROFIT1

A man got a profit of 7400/= after selling an item. Find the buying price if the percentage profit was 5%.

Solution

%’ge profit = Profit   X  100    where B. P. represents Buying Price.

                        B.P

5% = 7400   X  100   

           B.P.

B.P. x 5% = 740,000   x B.P.  [multiplying by B.P. both sides]

                       B.P.

B.P. x 5 = 740,000   

B.P. x 5¹ =   740,000  ¹⁴⁸⁰⁰⁰ 

  ₁5                   5₁

B.P. = 148000

                      

Hence Buying Price was 148000/=

  

TRY THIS………………

A man got a profit of 27,500/= after selling an item. Find the buying price if the percentage profit was 20%.

VARIATION 2

x is directly proportional to y. x=24 while y=4. Find y when x is 60.

Solution

x ⍺ y

x = ky

24 = k x 4

24 = 4k

4      4

k = 6

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

k = 6, y= ? x = 60.

x = ky

60 = 6 x y

60 = 6y

6      6

y = 10.

TRY THIS…………….

x is directly proportional to y. x=48 while y=12. Find y when x is 70.

LOGARITHMS 3

If log 2= 0.3010; find the value of log 400,000 without using tables.

solution

= log 400,000

=log(100,000 x 4)

=log100,000 + log4

=log10⁵ + log2²

=5log10 + 2log2

=(5x1) + 2(0.3010)

=5– 2(0.3010)

=5 – 0.6020

=4.398

Hence log2500=3.398

TRY THIS……………

If log 2= 0.3010; find the value of log 80,000 without using tables.

EXPONENTIALS 3

Simplify the following by writing in power form:

(m⁵⁵ x m¹⁵)/(m⁴⁶ x m⁶)

  

Solution

= (m⁵⁵ x m¹⁵)/(m⁴⁶ x m⁶)

= m⁵⁵⁺¹⁵

   m⁴⁶⁺⁶ 

= m⁷⁰

    m⁵² 

= m^70-52

= m¹⁸

TRY THIS……………….

Simplify the following by writing in power form:

(a⁶⁰ x a²⁰)/(a⁴⁶ x a¹⁴)

A. P. 1

The 1st term of arithmetic progression is 24 and the common difference is 40. Find the nth term

solution

A₁=24, d= 40, n=?

A_n =A₁ + (n-1) d

A_n=24 + (n-1)40

A_n=24 + 40n - 40

A_n=40n + 24 - 40

A_n=40n – 16

Hence the nth term is A_n=40n – 16  

TRY THIS………………….. 

The 1st term of arithmetic progression is 60 and the common difference is 122. Find the nth term.

AREAS 1

Area of a rectangle is 84 cm². Find its perimeter if its width is 7cm.

Solution

First we find length

A =  l  x w

84 =  l  x 7

84 = 7l

7       7

12 =  l

Now length is 12cm.

We can find perimeter since we have both length and width.     

Let l = length and w = width;

P = 2(l + w)

   = 2(12 + 7)

   = 2 x 19

   = 38

Hence perimeter is 38 cm. 

TRY THIS………………….. 

Area of a rectangle is 100 cm². Find its perimeter if its width is 4cm.

FUNCTIONS 2

If F(x) = 4x + 16; Find F^-1(x).

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= 4x + 16

y – 16 = 4x

y – 16  = 4x

   4          4

y – 16   = x

   4

x  =   y – 16     after rearranging

           4

F^-1(x) = x – 16    after interchanging x and y variables.

                4

Hence, F^-1(x) = x – 16     

                              4

TRY THIS………………   

If F(x) = 11x + 30; Find F^-1(x).

BINARIES 1

If p*k = 4pk - p – 2k; find 3*2.

Solution

p*k = (4 x p x k) - p – (2xk)   [rewriting the given expression more clearly]

3*2 = (4 x 3 x 2) - 7 – (2x2)   [substituting 7 for p and 2 for k]

3*2 = 24 - 7 – 4

3*2 = 24 – 11   [since( -7-4) = -11]

3*2 = 13.

Hence   (3*2) = 13.

 TRY THIS........

If p*k = 5p + k – 5pk; find (-2*7).

SUM OF AP 2

The first term of an AP is 21 and the last term is 97. If the arithmetic progression consists of 20 terms, calculate the sum of all the terms. 

Solution

A₁ = 21, n=20, A_n = 97

S_n = n(A₁ + A_n)

       2

S₂₀ = 20(21 + 97)

         2

S₂₀ = 10 x 118

S₂₀ = 1180

Hence the sum of all 20 terms is 1180.

TRYTHIS………………….

The sum of first 20 terms of an AP is 1380, and the last term is 97. Find the 1^st term. 

EXPONENTIALS 2

Simplify the following by writing in power form:

w⁶³

w⁴⁰

Solution

= w^{63 - 40} 

=w²³

TRY THIS……………….

Simplify the following by writing in power form:

m⁵⁵

m²¹ 

VARIATION 1

x is inversely proportional to y. x=30 while y=5. Find y when x is 15.

Solution

x ⍺ k

      y

x = k

      y

30 = k

        5

k = 30 x 5

k = 150

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

k = 150, y= ?, x = 15.

x = k

      y

15 = 150

         y

y x 15 = 150  x  y   (multiply by y on both sides)

                y

15y = 150 ¹⁰        dividing by 15 both sides

15      15

Hence y = 10  .

TRY THIS…………….

x is inversely proportional to y. x=80 while y=4. Find y when x is 40.

SUM OF AP 1

The sum of first 20 terms of an AP is 1380, and the last term is 97. Find the 1^st term. 

Solution

A₁ = ?, n=20, A_n = 97, S_n =1380

S_n = n(A₁ + A_n)

       2

1380 = 20(A₁ + 97)

             2

1380 = 10(A₁ + 97)

1380 = 10A₁ + 970

1380 - 970= 10A₁

410= 10A₁

410= 10A₁

10     10

41= A₁

Hence 1^st term is 41.

TRYTHIS……………

The sum of first 30 terms of an AP is 2475, and the last term is 105. Find the 1^st term. 

EXPONENTIALS 1

Simplify the following by writing in power form:

(m⁵⁵ x m⁵)/ m⁴⁶ 

   

Solution

= (m⁵⁵ x m⁵)/ m⁴⁶ 

= m⁵⁵⁺⁵

    m²¹ 

= m⁶⁰

    m²¹ 

= m^60-21

= m³⁹

TRY THIS……………….

Simplify the following by writing in power form:

 (y⁶⁵ x y⁵¹)/ y⁹⁴ 

LOGARITHMS 2

If Log_ax = 0.9, find Log_a(x⁴)

Solution

= Log_a(x⁴)

= 4Log_ax

= 4 x Log_ax

= 4 x 0.9

= 3.6

TRY THIS………..

If Log_aw= 3.6, find Log_a(w³)

LOGARITHMS 1

Evaluate Log1000000 +  log 0.001 -  log 0.00000001

Solution

= Log1000000 +  log 0.001 -  log 0.00000001

= Log10⁶ +  log 10^-3 – log 10^-8

= 6Log10 + (-3 log 10) – (-8log 10)

= (6x1) + (-3x1) - (-8x1)

=6 + (-3) – (-8)

= 6 – 3 + 8

=11

Hence Log1000000 +  log 0.001-  log 0.00000001 = 11

POLYGONS 1

A regular polygon has 66 sides. Find the total degrees of that polygon.

Solution

n = 66

Total degrees = (n - 2)180⁰

                       = (66 – 2)180⁰

                       = 64 x 180⁰

                       = 11520⁰

Total degrees = 11520⁰

TRY THIS……………………….

A regular polygon has 33 sides. Find the total angles of that polygon.

FUNCTIONS 1

Find the maximum value of the quadratic equation:5-3t-8t².

Solution

a=-8, b=-3, c=5

Maximum = 4ac-b²

                        4a

Maximum = (4 x -8 x 5) - (-3)²

                          4(-8)

Maximum = (-160)- 9

                        -32

Maximum= -169                         since (-160)- 9= -169

                    -32

Maximum = 169

                     32       

Hence maximum value is 169/32

TRY THIS………………………………..

Find the maximum value of the quadratic equation:6+10t-4t².