calculus ..for juniors and jee aspirants

consider 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

well 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]

another 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

i request nishant sir to make a separate sub forum for us engineering students , as we need some interface to communicate and share our ideas

bindaas 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 ?

Q2. 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

the 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