No. of solutions of an equation involving a defined function

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Homework Statement
The number of solutions of the equation
## f(x-1)+f(x+1)=sinA ,0<A<\pi/2 ##
where
## f(x)##={##1-|x|,|x|## less than or equal to 1
={##0,|x|>1 ##
is
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20200419_095636.jpg


My attempt-:
I defined functions f(x-1) and f(x+1) using f(x).After defining them,I substituted their values in the equation f(x-1)+f(x+1)=sinA.
For different ranges of x,I got different equations.
For 1<x<2,I got 1-x=sinA.
But now I am confused.For each different value of x in the interval (0,1],there exists an A.So there would be infinite solution.But the answer was given as 4.
Please help..
 
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Physics lover said:
For each different value of x in the interval (0,1],there exists an A.
I would think they want the number of solutions for any given fixed A in the range, though why they go through a sin function instead of just giving a range 0<…<1 I don't know.
 
haruspex said:
I would think they want the number of solutions for any given fixed A in the range, though why they go through a sin function instead of just giving a range 0<…<1 I don't know.
Yes i think the same.Maybe the question is incorrect.
 
Physics lover said:
Yes i think the same.Maybe the question is incorrect.
No, with my interpretation I get the book answer, so you might as well assume that is what is intended.
 
haruspex said:
No, with my interpretation I get the book answer, so you might as well assume that is what is intended.
So shall I put any value of A in between 0 and pi/2?
 
Physics lover said:
So shall I put any value of A in between 0 and pi/2?
No, you have to show that for every A in that range there are four solutions for x.
If you sketch f(x+1)+f(x-1) it will be obvious.
 
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haruspex said:
No, you have to show that for every A in that range there are four solutions for x.
If you sketch f(x+1)+f(x-1) it will be obvious.
Yes I got it.Thanks.
 
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