Post mathematic related questions here

[rst]

Re: mathematic related questions

remainder is 1
-----------------------------------------
3^2002 +7^2002 + 2002



3^2002 =(1+2)^2001 x 3
=[2001C0 + 2001C1 x 2 + 2001C2 x 2^2 ..... +2001 C 2001 x 2^2001 ] x3
=[1 + 2001C1 x 2 + 2001C2 x 2^2 ..... +2^2001 ] x3

when divided by 29
remainder = (1 +0 +0 +.............+2) x3
=3 x 3
=9

7^2002 =(1+6)^2001 x 7
=[2001C0 + 2001C1 x 6+ 2001C2 x 6^2 ..... +2001 C 2001 x 6^2001 ] x7
=[1 + 2001C1 x 6 + 2001C2 x 6^2 ..... +6^2001 ] x7


when divided by 29
remainder = (1 +0 +0 +.............+6) x7
=7 x 7
=49
or remainder =20

2002 give remainder 1 when divided by 29

total remainder = 9+20+1
=30
which give remainder 1

 

[Amit Digga]

Re: mathematic related questions

=[1 + 2001C1 x 6 + 2001C2 x 6^2 ..... +2^2001 x 3^2001 ] x7

remainder = (1 +0 +0 +.............+2 x3) x7

Can u please explain that ...2x3 in (1 +0 +0 +.............+2 x3) x7

 

[mastercool8695]

Re: mathematic related questions

I just used Python to solve it.
I don't know the explanation.

this is cheating...
anyways redirect me to one of the download links ..

remainder is 1
-----------------------------------------
3^2002 +7^2002 + 2002



3^2002 =(1+2)^2001 x 3
=[2001C0 + 2001C1 x 2 + 2001C2 x 2^2 ..... +2001 C 2001 x 2^2001 ] x3
=[1 + 2001C1 x 2 + 2001C2 x 2^2 ..... +2^2001 ] x3

when divided by 29
remainder = (1 +0 +0 +.............+2) x3
=3 x 3
=9

7^2002 =(1+6)^2001 x 7
=[2001C0 + 2001C1 x 6+ 2001C2 x 6^2 ..... +2001 C 2001 x 6^2001 ] x7
=[1 + 2001C1 x 6 + 2001C2 x 6^2 ..... +6^2001 ] x7
=[1 + 2001C1 x 6 + 2001C2 x 6^2 ..... +2^2001 x 3^2001 ] x7

when divided by 29
remainder = (1 +0 +0 +.............+2 x3) x7
=7 x 7
=49
or remainder =20

2002 give remainder 1 when divided by 29

total remainder = 9+20+1
=30
which give remainder 1

couldn't get this.. is this any theory/theorem??

 

[rst]

Re: mathematic related questions

couldn't get this.. is this any theory/theorem??

When we divide 1 by 29 then remainder is 1
2001C1 x 2 = 2001 x 2 ,this will give remainder 0 when divisible by 29
2001C2 x 2^2 = [(2001 x 2000)/2] x 4 ,this will give remainder 0 when divisible by 29
.
.
.
2^2001 =(1+1)^2001
=[2001C0 + 2001C1 + 2001C2 ..... +2001 C 2001 ]
=[1 + 2001C1 + 2001C2 ..... +1 ]
When divisible by 29
Then remainder = (1+0+0……+1)
=2

Can u please explain that ...2x3 in (1 +0 +0 +.............+2 x3) x7

Edited

 

[mastercool8695]

Re: mathematic related questions

^^ thanks. got it now.

 

[©mß]

Re: mathematic related questions

anyways redirect me to one of the download links ..

Python is used for programming,it's not a calculator.(Think you know that.)

 

[mastercool8695]

yup..
buddy..
i know that

ok leave it, i found the link

*www.python.org/download/

found this too.. *www.learnpython.org

will learn prog lingis after june 2

 

[rst]

----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img62.imageshack.us/img62/1227/scooterd.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------[/QUOTE]

 

[Niilesh]

can someone give me the solution to the following question?

Find range of (2-x)^½ + (1+x)^½.

 

[rst]

Let f(x)=(2-x)^½ + (1+x)^½
it is defined for 2-x≥ 0 and 1+x ≥ 0
2≥ x and x ≥ -1
-1 ≤x≤ 2
domain =[-1,2]

Put values in function and find range

 

[Niilesh]

Let f(x)=(2-x)^½ + (1+x)^½
it is defined for 2-x≥ 0 and 1+x ≥ 0
2≥ x and x ≥ -1
-1 ≤x≤ 2
domain =[-1,2]

Put values in function and find range

I have already tried it, couldn't get the answer
answer is [3^½, 6^½] , Can you justify it?

Finaly got it, i just made a rough graph from sign of f'(x)

 

[rst]

let f(x)=y
y=(2-x)^½ + (1+x)^½
squaring both side we ,get
y^2 = (2-x) +(1+x) + 2(2-x)^½(1+x)^½
y^2 = 3+ 2(2-x)^½(1+x)^½

Now 2(2-x)^½(1+x)^½ ≥ 0 for all x ∈ [-1,2]
3+ 2(2-x)^½(1+x)^½ ≥ 3
y^2 ≥ 3
y ≥ 3^½

So minimum value of y is 3^½

y has maximum value if (2-x)(1+x) is maximum
i.e if -x^2+x+2 is maximum
Let z= -x^2+x+2
Differentiating we get
z' =-2x+1

For max/min z'=0
-2x+1=0
x=1/2


At x= 1/2
z"=-2 (negative)
Hence z is maximum at x= 1/2


max value of -x^2+x+2 = 9/4

max value of y^2 =3+ 2(2-x)^½(1+x)^½

y^2 =3+ 2 (9/4)^1/2
=3+2 (3/2)
y^2=6
y= 6 ^(1/2)

Hence range =[3^½,6 ^1/2 ]

 

[rst]

----------------------------------------------------------------------------------------------------------------------
NEW OBJECTIVE QUESTION

*img62.imageshack.us/img62/1227/scooterd.png

Plz explain your answer

(Also you can ask your mathematics related questions or queries
we will try to solve them)
----------------------------------------------------------------------------------------------------------------------

Ans is 5400
-------------------------------------------------
down payment =88800
Money in 60 instalment =60 x 7520
=451200

Total money =451200 +88800
=540000

So money in each 100 instalment =540000/100
=5400

 

[Niilesh]

*i.imgur.com/Tlmth3m.jpg
please explain your answer

 

[rst]

View attachment 10434
please explain your answer

√ [1- f(x)] is defined for 1- f(x) ≥ 0
1 ≥ f(x)
cubing both sides
1 ≥ f ³ (x)
1 ≥ 1 - x³ -3x
0 ≥ - x³ -3x
x³ +3x ≥0
x(x ² +3) ≥0
x ≥ 0 [As (x ² +3) is always positive]

Hence domain = [0, ∞ )
0r Domain = R + U {0}

 

[Niilesh]

√ [1- f(x)] is defined for 1- f(x) ≥ 0
1 ≥ f(x)
cubing both sides
1 ≥ f ³ (x)
1 ≥ 1 - x³ -3x
0 ≥ - x³ -3x
x³ +3x ≥0
x(x ² +3) ≥0
x ≥ 0 [As (x ² +3) is always positive]

Hence domain = [0, ∞ )
0r Domain = R + U {0}

you wrote the expression for f^3(x) wrong, it also contains f(x) at the last

 

[rst]

Anyway, what is the answer (A,B,C,or D) of this question

 

[Niilesh]

Its C.
BTW i tried hit and trial and saw that it was satisfied for x=0, 1, 2 and it was not satisfied at x = -1 so by elimination i got the answer but cannot find a proper method

 

[rst]

Then I think you misunderstood the question
f(x) is not part of f ³ (x)

 

[Niilesh]

Maybe, but there must have been a reason that it is mentioned in the question that f(1) ≠ -1