0

Guest

Courses New

if the function satisfies the relation f(x+y)=y|x-1|/(x-1)f(x)+f(y) with f(1)=2, then lim f`(x) where x tends to 1 is

if the function satisfies the relation f(x+y)=y|x-1|/(x-1)f(x)+f(y) with f(1)=2, then lim f`(x) where x tends to 1 is

Grade:12

1 Answers

Nirmal Singh.
askIITians Faculty 44 Points
10 years ago
f(x+y)=[y|x-1| ]/ [(x-1)f(x)+f(y)]
We take log on both sides and have
Log f(x+y) = log y + log mod x-1 –log (x-1) f(x) +f(y0
We find derivatie now
F’(x+y)(1+y’)/f(x+y) = y’/y + 1/mod x-1 -1/x-1 f(x) + f(y)
When x>1 1/x-1 absolute value = 1/x-1.
When x<1 we cannot find as x-1 becomes negative and log does not exist for negative numbers.
As here y is also there, I am unable to arrive at the exact answer for f’(x)
If it single variable easily we can find out
So what we do is we put x=1 and y =0 and try that
We get f(1) =2 which means 2
Regards,
Nirmal Singh
Askiitians Faculty

Think You Can Provide A Better Answer ?

Other Related Questions on Differential Calculus

View all Questions »

ASK QUESTION

Get your questions answered by the expert for free

Answer and earn gift vouchers

Win Gift vouchers upto Rs 500/-

View courses by askIITians

Design classes One-on-One in your own way with Top IITians/Medical Professionals

Click Here Know More

Complete Self Study Package designed by Industry Leading Experts

Click Here Know More

Live 1-1 coding classes to unleash the Creator in your Child

Click Here Know More

a Complete All-in-One Study package Fully Loaded inside a Tablet!

Click Here Know More