Mean Value Theorems: For practice.

@Gallardo,
Why should Q9 be wrong? I don't find any trouble and your solution is perfectly okay.

9 > let h { x } =a _n ln ^{n + 1} xn + 1 + a _n ln ^{n - 1} xn + .......... a ₀ ln x1

Clearly , h { 1 } = 0 ,

And h { e ² } = a _n 2 ^{n + 1}n + 1 + a _{n - 1} 2 ⁿ n + ........... a ₀1 = 0 , as provided .

hence by rolle's theorem h ' { x } must have a root in ( 1 , e ² ) .

h ' { x } = { n + 1 } a _n ln ⁿ { x } / x + ............. + a ₀ / x = 0

Clearly , x ≠0 .

So a _n ln ⁿ { x } + ......... = 0 for some x belonging to ( 1 , e ² ) .

Kaymant sir , I think the question is wrong .

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

YESSSSSSSSSSSSSSS , I GOT NUMBER 5 !!!!!!!!!!

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

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

Hence , applying Lagrange's theorem ,

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

f ' { x } 1 + f ² { x } ≤ h { b } - h { a }4 , as min. value of { b - a } gives the maximum value of f ' { x }1 + f ² { x }

Now , h { b } - h { a } = tan ^{- 1} [ f { b } ] - tan ^{- 1} [ f { a } ] < π / 2 - { - π / 2 } = π

So f ' { x }1 + f ² { x } < Ï€4 < 1

Hence , f ' { x } < 1 + f ² { x }

Here , I used the fact that difference between two angles in tan inverse form can never exceed π .

dude gallaordo

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

so

g'(c)=0 c→(0,1)

Sorry sir , I didn't actually put much emphasis on the continuity and the differentiabilty of g { x } , that's why I thought maybe it could attain minus infinity as well in ( a , b ) , which is a bad mistake .

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 .

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, ∞).