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 #1
avatar+128579 
+1

\(f(x)=1-\sqrt{1-x^2} \)

\(-1\le x\le 1 \)

 

 

We need to find the inverse of the  given function....for f(x), we can write, y

 

y = 1 - √  [1 - x^2]      rearrange as

 

√  [1 - x^2]   = 1 - y      square both sides

 

1 - x^2  = (1 - y)^2

 

1 - x^2   =  y^2 - 2y + 1    subtract 1 from both sides

 

-x^2  = y^2  - 2y        mutiply through  by -1

 

x^2  = 2y  - y^2        take both roots

 

x  = ±√ [ 2y - y^2 ]      "swap"  x  and  y

 

y  = ±√ [2x - x^2 ]

 

We only need the top half of this graph   ⇒   y  = √ [2x - x^2 ]

 

Here is the graph  of  both functions  along with the graphs of  y  = x   and y  = 1

https://www.desmos.com/calculator/ozvzaxlwxe

 

The area enclosed by both functions will be twice the area of the area between  y = x  and  y  = 1 - √  [1 - x^2]

 

Note that the area  bounded  by the y axis from 0 to 1 , the line y = 1 from 0 to 1   and the line  y  =  x   will from a triangle with a base  and height of 1....so...it's area =  (1/2) base * height  = (1/2)(1)(1)  = 1/2  units^2

 

And note that the area formed  y axis from 0 to 1, the line y  =1 from 0 to 1  and the function  y = 1 - √  [1 - x^2]    will be the area of a quarter circle with a radius of  1  =

(1/4) pi * (1)^2   =  pi /4  units^2

 

So....this area  less the area of the triangle will   =  1/2  of the area we are seeking

 

So....the total area  =  2 [ pi/4  -1/2] =   [ pi/2  - 1] units^2 ≈   0.571  units^2

 

 

cool cool cool

Jul 23, 2018
 #4
avatar+128579 
+1

5.Consider the sequence 1,3,4,9,10,12,13,... which consists of every positive integer that can be expressed as a sum of distinct powers of 3. What is the 75th term of this sequence?

 

Note the pattern that emerges :

3^0  = 1                                                                                                         1

3^1  = 3    3^0 + 3^1 = 4                                                                           1        1

3^2 = 9   3^2 + 3^0 = 10   3^2 + 3^3 = 12   3^2 + 3^1 + 3^0 = 13         1      2        1

Note that the  number of terms - 7-  is just the sum of the first 2 rows  of Pascal's Triangle  [ the first entry is "Row 0" ]

 

Note that the next terms are

3^3  = 27

3^3 + 3^0 = 28

3^3 + 3^1  = 30

3^3 + 3^1 + 3^0 = 31

3^3 + 3^2  = 36

3^3 + 3^2 + 3^0 = 37

3^3 + 3^2 + 3^1  = 39

3^3 + 3^2 + 3^1 + 3^0  = 40

 

There  are 8 terms  which is the sum  of the elements of the  3 row of Pascal's triangle =  1  3  3   1

And each row adds 2^n  more  terms   [ where n is the row number ]

So...after the 4th row we have 15 + 2^4  =  15 + 16  =  31 terms

After the 5th row we have  31 + 2^5  =  31 + 32  = 63 terms

 

The first term in the 6th row  will represent 64 term  ....this will be  = 3^6   = 729

Notice that the next few terms are

3^6 + 3^0  =  730

3^6 + 3^1  = 732

3^6 + 3^1 + 3^0  = 733  

3^6 + 3^2 = 738  ....

And note that the terms in each case appended after the first term just follow the pattern of the first 15 terms

So...the 75th term  will be the sum of this first term  plus the 11th term in the above  series [ 31]  

 

So....the 75th term is    729  + 31  =  760

 

 

cool cool cool

Jul 21, 2018