1see proving LMVT using Rolle's and vice versa can be done by just rotating the axis......yes to prove rolle's u need to rotate the axis so that the x-axis is parallel to the f|(c) .....
Q 1) Prove that f(x)=\frac{x^7}{7}-\frac{x^6}{6}+\frac{x^5}{5}-\frac{x^4}{4}+\frac{x^3}{3}-\frac{x^2}{2}+x-1 has exactly one real root
Q 2) In 1 hour a snail travels 60 meters. Prove that there was an Interval of 10 minutes where it traveled exactly 10 meters. (Not exactly Rolle's Theorem)
Q 3) Prove that f(x)=x3 - 3x + c never has both its roots in [0,1]
Q 4) if f(x)= ax3+bx2+cx+d, a≠0, Prove that f(x) cannot have more than one real roots, if b2<3ac
Q 5) Prove that there exists c in (a,b) such that f(c)f'(c)=c, given f is differentiable on (a,b) and f2(a)-f2(b)=a2-b2
Q 6) Let f is a continuous function on [a,b]
Prove that exists c in (a,b) such that \int_a^cf(x)\ \mathrm{dx}=(b-c)f(c)
Q 7) Prove LMVT using Rolle's Theorem
Q 8) Give an example to show that continuity in [a,b] and not just (a,b) is a necessary condition for both Rolle's and LMVT
Q 9) f and g are real functions continuous on [a,b] and differentiable on(a,b).
Show that there exists c in (a,b) such that f'(c)(b-c)+g'(c)(c-a)=(f(c)-f(a))+(g(b)-g(c))
Q 10) Let f , g continuous functions on [a , b] with f'(x)≠0 in (a , b).
Prove that there exists c in (a,b) such that : \frac{f'(c)}{f(a)-f(c)}+\frac{g'(c)}{g(b)-g(c)}= 1
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39 Answers
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13. (another method..plz confirm whether it's correct or not)..
f(x) = x3 -3x + c
f'(x) = 3x2 - 3
for f'(x) = 0, x= ±1
that means there is exactly one root in (-1,1)..
62Rohit i will have to kill you for not answering Q 2 correctly.. i had solved this one for you.. your argument is flawed...
62I dont know if you are able to understand what you have written :P
62if the different displacements are covered in different times...so in atleast 1 the net displacement will be zero
How did you conclude this?!
1actualyuth ething is that dx/dt will be zero sumwhere..not the interval thing is told..now is it right???the velocity in sum interval will be >1 and in sum interval <1...so in sum interval =0
12.llet us define the distance travvled in a interval of 10 minutes as x=position at (t+10)-positon at t ,
now the distance if covered in each interval is the same....then problem is alreay solved,so let us omit that
if the different displacements are covered in different times...so in atleast 1 the net displacement will be zero suppose [a,b]...then sumwhere in a,b dx/dt=0 and hence in that interval average v=10m/s and hence the distance covered is 10 m in 10 mins...hence proved
1see this....http://targetiit.com/iit-jee-forum/posts/calculus-basic-2569.html
39prophet sir here goes watever u said in #30
9Q)
a one liner , so have been waiting for others to try out, no one ever turned out
define a function h(x)=(f(x)-f(a))(x-b) -(g(x)-g(b))(x-a)
now apply rolle`s theorem for h(x)
3910Q) again this one also guys u just need to construct a function where u can apply rolle.
have a good look at what u are given... its the best hint..
3911Q)
x
take g(x)= e^(-2005x) ∫ f(u) du
a
g(x) is derivable and g(a)=g(b)=0, so now use rolle theorem
12Q) yes it can be any number other than 2005 also
3910Q)
as i said before its all about seeing the function
h(x)=e^x (f(a)-f(x))(g(b)-g(x))
now use rolle theorem for h(x)
15 > Let h { x } = f 2 { x } - x 2
Clearly , h { x } is continuos as well as differentiable in the interval ( a , b ) ,
Also , from the given condition , h { a } = h { b }
Hence , by Rolle's Theorem , for come c belonging to ( a , b ) ;
h ' { c } = 0
Or , 2 f { c } f ' { c } = 2 c
Or , f { c } f ' { c } = c .
1Nishant sir , can you give more of these type of sums ? They are really interesting !!!!!!!
112) My proof is slightly different.
f(x)=x-10, where x is the dist. covered in some 10 min. interval.
Now there exists x such that f(x)<0 which means there must exist x such that f(x)>0.
So there exists x such that f(x)=0.
Not sure whether correct or not.
3q4)f'(x) =3ax2+2bx+c which doesnt have roots if
b2<3ac.so we don have all roots of f(x) is real.
11Ans3) f(x) = x 3 - 3x + c
f(x) is continous on [0,1] as it is a polynomial
f(x) is differentiable on (0,1)
f(0) = c
f(1) = c-2
Since f(0) ≠f(1)
So Rolle's theorm is not applicable.There is no x0 exists such that f ' (x0) = 0 and 0 < x0 < 1
397)
Rolle's Theorem:
Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). If f(a) = f(b), then there is at least one point c in (a, b) where f '(c) = 0.
Mean Value Theorem
Let f be continuous on a closed interval [a, b] and differentiable on the open interval (a, b). Then there is at least one point c in (a, b) where
f '(c) = (f(b) - f(a)) / (b - a).
now, proof of LMVT:
let us assume that rolle`s theorem is applicable as stated above...
The equation of a straight line through points (a, f(a)) and (b, f(b)) is given by
g(x) = f(a) + [(f(b) - f(a)) / (b - a)](x - a)
we can see that in above equation,
g(a) = f(a) and g(b) = f(b).
this mean f-g satisfies rolle`s theorem..
(f - g)(a) = f(a) - g(a) = 0 = f(b) - g(b) = (f - g)(b) .....(1)
we can also say (f - g)'(x) = f '(x) - g'(x) = f '(x) - (f(b) - f(a)) / (b - a)....(2)
but from (1), we have a point c in (a, b) such that (f - g)'(c) = 0
so we can say f '(c) = (f(b) - f(a)) / (b - a)
hence proved
394) here we use that fact that if a function has 3 disticnt roots , then its derivative has 2 distinct roots. ( its easy to prove)
so if b2
<3ac then f`(x) cannot have 2 real roots . so f cannot have 3 real roots..
and f obviously has one root by intermediate value theorem.
3q1)f'(x)=(1+x7)/1+x which doesnt have a root,so by mean value theorem we have only one real root for f(x).
3q2)since the particle is moving continously,so the graph of x and t is continous,by legrange's theorem we will have some c for which f'(c)=average velocity.
3q7)if rolle's theorem is proved.lagrange theorem can also be proved bcos the proof of the lagrange's theorem can be made by rotating the axis and shifting the origin in order to have the conditions for rolle's theorem is satisfied.So in our new coordinates it is rolles theorem,and in our old coordinates it is lagrange's theorem.