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Yeah i know.I've only been posting ques on functions.But wat can i do.At the moment i am stuck wid this fordsaken chapter.But i do try to put up ood ques dat will not make u guys feel bored. So here are a few more of my doubts.(But pls solve my prev doubts dat i put up in my last post).
1)Let f(x,y)be a periodic function, satisfying the condition f(x,y)=f(2x+2y,2y-2x)for all x,yεR and let g(x)be a function defined as g(x)=f(2x,0).prove that g(x)is a periodic function and find its period.
2)Find the domains of the following functions:
(i)f(x)=1/[|x-1|]+[|7-x|]-6 (In the denominator the bracketed expressions are greatest integer of mod of that expression.Only 1 is in the numerator)
(ii)f(x)=√sin-1(log2x) + √cos(sinx) +
sin-1(1+x2/2x)
3)solve the equation
||x2-5x+4|-|2x-3||=|x2-3x+1|
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7 Answers
11let a=x2 -5x +4
b = 2x-3
then the given equationn can be written as
||a|-|b|| = |a+b|
squaring both sides we get
-|a||b| = ab
ab + |ab| =0
→ ab<0
→either a<0 or b<0
1thanx for the well explained sol.now pls solve the remaining 2 in da same manner.
12nd
denominator is not defined if
[|x-1|] + [|x-7|] = 6
make cases
case 1, x>7
then [x-1] + [x-1-6] = 6
=> 2[x-1] -6 = 6 { use the fact that [x+a] = [x] + a if a is integer}
=> [x-1] = 6
=> 7<=x<8
nomake other cases urself ( x belongs to [1,7] and less than 7)
check boundary points {1,7} separately also
13)
sin-1(1+x2/2x) ...only x = 1 and -1 from here
0<logx<1 (logx with base 2)
solve now
341f(x,y)=f(2x+2y, 2y-2x)=f(8y,-8x)=f( 16(y-x), -16(x+y) ) = f(-64x, -64y)
Hence we have f(x,y) = f(-64x, -64y)
This means f(-64x, -64y) = f(642x, 642y)
Hence f(2x,0) = f(2x+12,0)
which implies g(x) = g(x+12)