IIT JEE , AIEEE Forum - Mathematics , Algebra

md_2674062

q1 FACTORISE a(b^2+c^2-a^2)+b(c^2+a^2-b^2)+c(a^2+b^2-c^2)-2abc

q2

the number of solution of 3^x+4^x=5^x are?

q3

the remainder when 4^101 divided by 101?.(tenth class question)

q4 the number of integral solution of x^2-3y=2000 are?

feynmann

q2 = 1 ( only x= 2 )

q3 Apply Fermat's theorem

101 is a prime no

so 4^ 101 = 4 mod ( 101)

so remainder is 4

feynmann

q1. taking c common out of the appropriate tems form c ( a - b ) ^2

In this way absorb -2abc

rest is easy

nadeemoidu

1)
Here's a good method of factorizing.
Take the given expression as f(a) = a ( b^2 c^2 - a^2 ) + b ( c^2 + a^2 - b^2 ) + c ( a^2 + b^2 - c^2 ) - 2 abc

Find f [ -(b+c) ] . Here it is equal to 0.

So ( a - b - c ) is a factor of the given expression.

By the symmetry in a,b,c , we can see that ( - a + b - c) and ( - a - b + c) are also factors.

So the final answer should be ( -a + b +c ) ( a - b + c )( a + b - c)

In general if u get f( b) =0 , then (a-b) is a factor. if f( -b) =0 then ( a+b) is a factor and so on.

md_2674062

pl solve each qs

nadeemoidu

4)
x ² - 3y = 2000

x² = 1,0 ( mod 3)
x² - 3y= 1,0 ( mod 3)

2000 = 2 (mod 3)

So there are no integral solutions.

feynmann

In the factorization problem

plz explain why threre should not be any const coeff in tne factor.

nadeemoidu

It can easily be verified.

u can always verify the factors by multiplying . ( working with 'a' alone would do as they are symmetric in a, b,c )

hsbhatt

I think this one was missed out:

3^x+4^x = 5^x

Rewrite as (3/5)^x+(4/5)^x = 1

x=2 is a solution. To check if any other solutions

Let f(x) = (3/5)^x+(4/5)^x -1

This is a monotonically decreasing function. It is easy to see that there are no solutions (-inf,2) and also (2, inf)

Hence x=2 is the unique soln.

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