Wednesday, 18 October 2017

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.

Monday, 9 October 2017

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 

= m70-52

= 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=?

An =A1 + (n-1) d

An=24 + (n-1)40

An=24 + 40n - 40

An=40n + 24 - 40

An=40n – 16

Hence the nth term is An=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, An = 97

Sn = n(A1 + An)
       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 1st term.