1@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
Q1) Prove that if a function f is continuous on [a,b], differentiable on (a,b) and f(a)=f(b)=0, then for any \alpha\in\mathbb{R}, there exists some c\in (a,b) such that
\alpha f(c)+f'(c)=0
Q2) Let f and g be functions continuous on [a,b], differentiable on (a,b) and f(a)=f(b)=0. Show that there exists a point c\in (a,b), such that
g'(c)f(c)+f'(c)=0
Q3) Let f be continuously differentiable on [a,b] and twice differentiable on (a,b), and suppose that f(a)=f'(a)=f(b)=0. Prove that there exists c\in (a,b) such that f''(c)=0.
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29 Answers
1
66@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.
66@Gallardo,
Why should Q9 be wrong? I don't find any trouble and your solution is perfectly okay.
1For 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 - ae - 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 )
19 > 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 + 1n + 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 .
1In 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
1YESSSSSSSSSSSSSSS , 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 } < π4 < 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 .
1that seems to be difficult
i think we have to handle inequalities cleverly and carefully there
1dude gallaordo
g(1)-g(0) =definite integration from 0-1 of f(x)=0
[1]
so
g'(c)=0 c→(0,1)
1Q-8 let g(x)=∫f(x).dx
applying lmvt theorem on g(x) in interval 0-1 gives result
66@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.
1Sorry 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 .
66Q4) 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
66@Gallardo
You are right about the second one as well.[1]
@Gallardo,
What if g(x) is unbounded from below?
66@RPF,
How can you assume f(x)=e-g(x). First, I again didn't specify f to be positive. Second the condition f(a)=f(b)=0 is not satisfied.
P.S. At least write the question number you are attempting.
1Hahaha , 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 .
1Hmmm , 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 .
66I 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.
1Qs, 4 seems to be a continuation of the Qs . 3
If we apply Rolle's theorem to F { x } = f { x } - x , in the interval ( 0 , 2 ) ;
[ which we can , as F { x } is differentiable , continuos , and F { 0 } = F { 2 } in the said interval ]
we get F ' { k } = 0 ,
or , f ' { k } - 1 = 0 for some k belonging to ( 0 , 2 ) .
Next again , applying Rolle's theorem to h { x } = f ' { x } - 1 in the interval ( k , 1 ) or ( 1 , k ) ;
{ as h { 1 } = 0 = h { k } -- which we proved }
We can find a number c belonging to ( 1 , k ) or ( k , 1 ) , such that ,
h ' { c } = 0 ,
or , f ' ' { c } = 0 ,
for some c belonging to ( 1 , k ) or ( k , 1 ) , i . e , to ( 0 , 2 ) .
66Q7) 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, . . ., an 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, . . ., an 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, . . ., an 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, ∞).
1Thanks 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 .