See that F(n) has inside (1-x^2)^n which has a degree of 2n but as we take n derivatives of it we would have a polynomial of degree n. Now this means that F(n) can have atmost n roots. ------(1) Now consider F(1) and prove using LMVT Theorem that it has 1 root, ..., similarly prove that F(n) has at least n roots. -------(2) By combining statement 1 and 2 we get our answer.
Let a,c be nonnegative real numbers and let f:[a,b]→[c,d] be a bijective increasing function. Prove that there is a unique real number μ∈(a,b) such that within limits a to b ∫f(t)dt=(μ−a)c+(b−μ)d. Bhaiya... please explain the solution for this question.
By using generalized Mean Value Theorem that says if g(x) is continuous [a, b] and f(x) has derivative function which is continuous and never changes its sign in [a, b]. Then there exists some c∈[a, b] such that (limits a to b) ∫f(x)g(x)dx = f(a) (limits a to c) ∫g(x)dx + f(b) (limits c to b) ∫g(x)dx Now to prove you question simply put g(x) = 1 and c = μ. you will get your answer.
you can simply check the cases by making group and then doing the permutation of those groups. there would be 4 cases for 1 group of size 3 and 6 cases for 2 groups of size 2.
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Prove that the Legendre polynomial
F^n (x)=d^ny/dx (1−x2)^n
has n
distinct real roots between (-1,1) using LMVT theorem?
See that F(n) has inside (1-x^2)^n which has a degree of 2n but as we take n derivatives of it we would have a polynomial of degree n. Now this means that F(n) can have atmost n roots. ------(1)
Now consider F(1) and prove using LMVT Theorem that it has 1 root, ..., similarly prove that F(n) has at least n roots. -------(2)
By combining statement 1 and 2 we get our answer.
Let a,c
be nonnegative real numbers and let f:[a,b]→[c,d]
be a bijective increasing function. Prove that there is a unique real number μ∈(a,b)
such that within limits a to b
∫f(t)dt=(μ−a)c+(b−μ)d.
Bhaiya... please explain the solution for this question.
By using generalized Mean Value Theorem that says if g(x) is continuous [a, b] and f(x) has derivative function which is continuous and never changes its sign in [a, b]. Then there exists some c∈[a, b] such that
(limits a to b) ∫f(x)g(x)dx = f(a) (limits a to c) ∫g(x)dx + f(b) (limits c to b) ∫g(x)dx
Now to prove you question simply put g(x) = 1 and c = μ. you will get your answer.
In how many ways can 4 identical apples and 4 identical oranges can distributed among 6 children if each child gets at least 1 fruit?
you can simply check the cases by making group and then doing the permutation of those groups. there would be 4 cases for 1 group of size 3 and 6 cases for 2 groups of size 2.
Bhaiya aap iit se ho kya bta do 🙏
No
Bhaiya mera 22 ko hai mai nhi drunga fod ke aaunga fir adv ki tayyari mai lag jaunga...