106consider g(x) = f2(x) - x2
g(a) = g(b) (given)
=> there exists some c belongs to [a,b] st
g'(c) = 0 (by rolle's theorem)
=> 2f(c).f'(c) - 2c = 0
=> f(c)*f'(c) = c
hence proved
thx for the question .. going to be useful for my midsems
let be a real valued function satisfying f:[a,b]\rightarrow [a,b] \\ a,b \in \mathrm{R}
1)continous in [a,b]
f^2(a)-f^2(b)=a^2-b^2
prove that
f(x).f'(x)=x
for some x \in [a,b]
P.S. RICKY IS PROHIBTIED FROM ANSWERING
-
UP 0 DOWN 0 0 12
12 Answers
1061well i had cauchy's mean value theorem in my mind
i chose
h(x)=f^2(x) \ and\\ g(x)=x^2
by cauchy MVT
\frac{h'(c)}{g'(c)}=\frac{f^2(b)-f^2(a)}{b^2-a^2} \\ \frac{2f(c)f'(c)}{2c}=1 \\ f(c)f'(c)=c \\ \texttt{for some c}\in[a,b]
1another one for asish
a simpler one this time
\boxed{Q1}
\texttt{let f and g be continous , real valued functions satisfying the following conditions} \\ \texttt{i)f and g are continous in [a,b]}\\ \texttt{ii)f and g are differntiable in (a,b)}\\ \texttt{iii)f(a)=f(b)=0} \\ \texttt{then shoe that for some c } \in [a,b]\\ f(c)g'(c)+f(c)=0
\boxed{Q2}
\texttt{let f be a continous and differntiable function f(0)=0 and f'(0)=1 }\\ \texttt{then prove that } \\ \lim_{x\rightarrow \infty}\frac{1}{x}\left(f(x)+f(\frac{x}{2})+f(\frac{x}{3})\cdots+f(\frac{x}{k}) \right)=1+\frac{1}{2}+\frac{1}{3}+\cdots+\frac{1}{k}
Q2 my doubt
1i request nishant sir to make a separate sub forum for us engineering students , as we need some interface to communicate and share our ideas
23bindaas i think for 2nd one lim shud be x -> 0 ,
u can see it by L'Hospital
or since f(0)= 0, \lim_{x\rightarrow 0}\frac{f(\frac{x}{r})-f(0)}{x}=\frac{f^{!}(\frac{x}{r})}{r}=\frac{1}{r}
so \lim_{x\rightarrow 0}\sum{\frac{f(\frac{x}{r})}{x}}=\sum{\frac{1}{r}}
for 1st one u made some typing error ?
106Q2. isnt it x--> 0 instead of x--> inf,
we have similar question in our assignment but with x-->0 and the method is obviously LH rule
Q1 probably a typo in it... pl check the question
1qwerty's argument seems fine
Q1 is having a minor error which i think u guys cud have made out
\texttt{let f and g be continous , real valued functions satisfying the following conditions} \\ \texttt{i)f and g are continous in [a,b]}\\ \texttt{ii)f and g are differntiable in (a,b)}\\ \texttt{iii)f(a)=f(b)=0} \\ \texttt{then shoe that for some c } \in [a,b]\\ f(c)g'(c)+f'(c)=0
66For the above question set
h(x) = f(x) eg(x)
and the result follows from Rolle's theorem on h(x) in [a,b]. And by the way, c should be in the open interval (a,b) and not the closed one.
1needless to say kaymant sir is always right :)
wanted some juniors participation in it
106yeah but how do we think abt that ..
needless to say this has been discussed in our tutorials...
1the basic idea is very easy
try solving for g(c)
g'(c)=-\frac{f'(c)}{f(c)}
the term \frac{f'(c)}{f(c)}
says something to do with ln or e^
then its not tough to square onto that function