Mean Value Theorems: For practice.

@kaymant sir agreed i messed it up

actually i was latexifying proof

i was going as wat was done by gallardo

i forgot to delete my post

are α₁,α₂....α_n integers?

@gallardo
The condition given in the problem is correct. Your solution as in #29 has a slight mistake.
h(e²) will have 2a₀ as the last term so that 2 comes out as a common factor and we get the condition given.

For qs. 7 -

Let g { x } = e ^{- x} { f ( x ) + f ' ( x ) + f ' ' ( x ) + ...... f ⁿ ( x ) }

Claerly g { x } satisfies all the conditions of Rolle's theorem.

Proof of g { a } = g { b } ---

ln { f ( b ) + f ' ( b ) + ............+ f ⁿ ( x )f ( a ) + f ' ( a ) + ............+ f ⁿ ( a ) } = b - a ,

or , f ( b ) + f ' ( b ) + ............f ( a ) + f ' ( a ) + ............ = e ^{b - a} = e ^{- a}e ^{- b}

or , e ^{- b} { f ( b ) + f ' ( b ) + ............ } = e ^{- a} { f ( a ) + f ' ( a ) + ............ }

or , g { b } = g { a } ,

Hence applying Rolle's theorem to g { x } ,

g ' { c } = 0 for some c belonging to ( a , b ) ;

or , - e ^{- x} { f ( c ) + f ' ( c ) + ............ } + e ^{- x} { f ' ( c ) + f ' ' ( c ) + ............ } = 0

or , e ^{- x} ( f ( c ) + f ' ( c ) + ...... + f ⁿ ( x ) } = e ^{- x} ( f ' ( c ) + f ' ' ( c ) + ...... + f ^{n + 1}( c ) }

or , f ( c ) = f ^{n + 1} ( c )

In qs . 6 > Progressing in the same way as in qs. 5 ,

Take h { x } = tan ^{- 1} [ f { x } ]

Similarly using Lagrange's theorem ,

h ' { c } = h { a } - h { b }b - a , for some c belonging to ( a , b ) ,

f ' { c } 1 + f ² { c } = tan ^{- 1} [ f { b } ] - tan ^{- 1} [ f { a } ]b - a ,

Noting that f { b } = - ∞ and f { a } = ∞ ,

we get L . H . S = - π / 2 - { π / 2 } b - a = - πb - a ............ 1

Now , Given that f ' { x } + 1 + f ² { x } ≥ 0 ;

Or , f ' { c } + 1 + f ² { c } ≥ 0 .........[ as the given equation holds for all x ]

Then , f ' { c }1 + f ² { c } + 1 ≥ 0.......[ dividing by 1 + f ² { c } and noting that it cannot be zero ] ...... 2

From 1 and 2 ,

- Πb - a + 1 ≥ 0 ,

or , 1 ≥ Πb - a

or , b - a ≥ Î

that seems to be difficult

i think we have to handle inequalities cleverly and carefully there

dude gallaordo

g(1)-g(0) =definite integration from 0-1 of f(x)=0
[1]

so

g'(c)=0 c→(0,1)

5 seems to be tan inverse function coming into the picture

trying

@Gallardo,
In any case, –∞ cannot be attained by any variable. Remember that ±∞ do not belong to the set of real numbers.

Something being +∞ is just a way to say that it could be as large a positive real number as you like. The same goes for &ndash∞. In this sense, ∞ is a variable quantity.

Q4) Let f be continuous on [0,2] and twice differentiable on (0,2). Show that if f(0)=0, f(1)=1, and f(2)=2, then there is x₀ in (0,2) such that f''(x₀) = 0.

Q5) Suppose that f:[a,b]\to\mathbb{R} where b-a\ge 4, is differentiable on (a,b). Prove that there is x_0\in (a,b) such that
f'(x_0)<1+f(x_0)^2.

Q6) Prove that if f is differentiable on (a,b), and if \lim_{x\to a^+}f(x)=+\infty, \lim_{x\to b^-}f(x)=-\infty, and
f'(x)+f(x)^2+1\ge 0\quad \forall\ x\in (a,b)
then b-a\ge \pi

Hahaha , the second sum , in this case too , is the continuation of the first one .[1]

2 > Let h { x } = f { x } e ^{g { x }}

Apply Rolle's theorem in the interval ( a , b ) ,

[ keeping in mind the continuity and differentiabilty of the function h { x } and noting that

h { a } = h { b } as f { a } = f { b } = 0 ]

We get , for some c belonging to ( a , b ) ;

h ' { c } = 0 ,

or , f ' { c } e ^{g { c }} + f { c } g ' { c } e ^{g { c }} = 0

But e ^{g { x }} > 0 for all g { x } , ............[ it must be mentioned that g { x } is

not unbounded from below ]

So , cancelling out e ^{g { c }} gives ,

f ' { c } + f { c } g ' { c } = 0 .

That's correct [1].

Hmmm , seems I got the first one ,

Let h { x } = f { x } e ^{a x}

Clearly , h { a } = h { b } ............[ as f { a } = f { b } = 0 ]

Again , h { x } is diffrentiable as well as continuos in the interval ( a , b ) .

Applying Rolle's theorem , for some c belonging to ( a , b) ,

h ' { c } = 0 ,

or , f ' { c } e ^{a c} + a e ^{a c} f { c } = 0

As e ^{a c} > 0 for all c and a belonging to Real numbers ,

Hence we can cancel out e ^{a c} from L . H . S to obtain ,

a f { c } + f ' { c } = 0 .

just first thoughts

in q1
has it to do anything with

?

I have mixed up Rolle's and Lagrange's theorem because giving them separately would give you one direction of thought. But mixing them would force you to exercise your brain more.

Q7) Let f:\mathbb{R}\to\mathbb{R} be (n+1) times differentiable. Show that for each pair of real numbers a, b, (a<b), such that
\ln\left(\dfrac{\sum_{k=0}^nf^{(k)}(b)}{\sum_{k=0}^nf^{(k)}(a)}\right)=b-a
there is a number c\in(a,b) for which f^{(n+1)}(c)=f(c).

Here, f^{(k)}(x) denotes the k-th order derivative of f; the zeroth order derivative being the function itself.

Q8) Assume that a₀, a₁, . . ., a_n are real numbers such that
\dfrac{a_0}{n+1}+\dfrac{a_1}{n}+\ldots +\dfrac{a_{n-1}}{2}+a_n=0
Prove that the polynomial
P(x)=a_0x^n +a_1x^{n-1}+\ldots +a_n
has at least one root in (0,1).

Q9) For real constants a₀, a₁, . . ., a_n such that \dfrac{a_0}{1}+\dfrac{2a_1}{2}+\dfrac{2^2a_2}{3}+\ldots +\dfrac{2^{n-1}a_{n-1}}{n}+\dfrac{2^na_n}{n+1}=0
show that the function
f(x)=a_n\ln^n x+a_1\ln^{n-1}x+\ldots +a_2\ln^2x+a_1\ln x +a_0
has at least one root in (1,e²).

Q10) For non-zero a₁, a₂, . . ., a_n and for distinct \alpha_1,\ \alpha_2,\ \ldots, \ \alpha_n, prove that the equation
a_1x^{\alpha_1}+a_2x^{\alpha_2}+\ldots +a_nx^{\alpha_n}=0,\quad x\in(0,\infty)
has at most n-1 roots in (0, ∞).

Thanks to Anant sir for providing a few good qs. -

Last one seems a bit easy .

If we apply Rolle's theorem to f { x } ,

then we get that f ' { k } = 0 for some k belonging to ( a , b ) .

Now we apply Rolle's theorem for F { x } = f ' { x } in the interval ( a , k ) .

It's given that f ' { a } = 0 , and we proved that f ' { k } = 0 .

So F { a } = F { k } , also as f is twice differentiable , so F is also differentiable as well as continuos .

Hence , F ' { c } = 0 for some c belonging to ( a , k ) , i . e , belonging to ( a , b ) .

Or , f ' ' { c } = 0 .