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Let f(x) and g(x) be two functions which cuts each other orthogonally. At their common point of intersction (x 1 ) , both f(x) and g(x) have equal to n, where n belongs to N, and n ≠ 1. Also if |f ' (x 1 )| = | g ' (x 1 )| at the common point of intersection. Then show that the limit (x approaches x 1 ) [f(x).g(x)] is equals to n-1 , where [.] represents greatest integral functions.
Hi, Since the two functions cut orthogonally, => f'(x1).g'(x1) = -1 Now since |f ' (x1)| = | g ' (x1)| Therefore : f ' (x1)= - g'(x1) Hence Either f'(x1) = 1 and g'(x1) = -1 or f'(x1) = -1 and g'(x1) = 1 Now, What does this statement mean? "both f(x) and g(x) have ?????? equal to n, where n belongs to N, and n ≠ 1" Shoould there be something in place of ?????? Waiting for your reply. Rajat Askiitian Expert
Hi,
Since the two functions cut orthogonally,
=>
f'(x1).g'(x1) = -1
Now since
|f ' (x1)| = | g ' (x1)|
Therefore :
f ' (x1)= - g'(x1)
Hence
Either f'(x1) = 1 and g'(x1) = -1
or
f'(x1) = -1 and g'(x1) = 1
Now, What does this statement mean?
"both f(x) and g(x) have ?????? equal to n, where n belongs to N, and n ≠ 1"
Shoould there be something in place of ??????
Waiting for your reply.
Rajat
Askiitian Expert
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