I hope this is not considered as quibbling on minor matters, but 1/(k-2)! isnt defined for k=1. So for the first sum k runs from 2 onwards. It doesn't change the sum, but just to be more accurate in our presentation.

 

Actually, if you are more conversant with inequalities, this is a direct application of the Cauchy-Schwarz Inequality.

 

(a²+b²) (x²+y²)  (ax+by)²

 

Hence (ax+by)²36

 

or -6ax+by6

Qn 3: I think you meant, the lines joining z₁ and z₂ are perpendicular.

 

Consider triangle OZ₁Z₂ right angled at Z₁.

 

By Pythagoras theorem (Z₁Z₂)² = OZ₁² + OZ₂²

 

Now OZ₁ = |z₁|; OZ₂ = |z₂| and Z₁Z₂ = |z₁-z₂|

 

Hence |z₁|²+|z₂|² = |z₁-z₂|² = (z₁-z₂) (z'₁-z_2^') = |z₁|²+|z₂|² - (z₁z₂' +z₁'z₂)

 

Hence z₁z₂' +z₁'z₂ = 0.

 

 

Just use .

 

a/2 = sinp; b/2 = cosp; x/3 = sinq; y/3 = cosq

 

and you will get (ax+by)/6 = cos(p-q)

 

-1cos(p-q)1

 

Hence -6ax+by6

There's a neat trick you can use in such cases of parametric integration.

 

Call I(k) =  ₀^/2  log(sin²x+k²cos²x) dx

 

Then I'(k) = ₀^/2  2kcos²x/(sin²x+k²cos²x)  dx [Remember the differentiation is being carried out with respect to k only]

 

I'(k) = 2k ₀^/2  dx/(tan²x+k²)  dx

      

= 2k ₀  dx/(t²+k²) (t²+1)  dx = /(k+1)

 

Now integrating from 1 to k and using fundamental theorem of calculus for a particular value of k = m

 

I(m) - I(1) =  ₁^m I'(k) dk =  ₁^m  /(k+1) dk = ln(m+1) - ln2

 

Now I(1) = ₀^/2  log(sin²x+cos²x) dx = 0 

 

Hence expressing the function in terms of k,

 

I(k) = ln(k+1) - ln2

 

PS: I am not quite sure what the forum expert was trying to do with that monograph on definite integrals. I hope he wasnt trying to ridicule kislay for a really instructive question.

Over lunch I thought I was missing something because with cosx I was getting a contradictory result. Thanx for pointing it out.

 

Lesson: Never do probs between jobs

Let I = _-/2^/2 (f(x)+g(x)) (f(-x)+g(-x)) dx

 

let t = -x. Then dt = -dx

 

Then we get I = - _-/2^/2 (f(t)+g(t)) (f(-t)+g(-t)) dt = -_-/2^/2 (f(x)+g(x)) (f(-x)+g(-x)) dt = -I

 

Hence I = 0.

Q3 has not been solved convincingly.

 

When the question says iff p then q rather than just if p then q, you have to prove pq and qp. In this case of course neither has been proved.

 

Proving that if ABC is equilateral tanA+tanB+tanC = 33 is trivial.

 

Now, we have to prove that if for a triangle ABC, tanA+tanB+Tanc = 33 then A=B=C = 60^o.

 

Here you must invoke the inequality, that in a triangle

tanA+tanB+tanc33 with equality occuring only in the case where A = B = C.

 

Since tanA+tanB+Tanc = 33 this must mean A=B=C = 60^o.

sinx+sin²x = 1

Hence sinx = cos²x

 

The given expression becomes

 

sin⁶x + 3sin⁵x + 3sin⁴x+sin³x-1

 

Now, sinx+sin²x-1 = 0

We know that if a+b+c = 0, then a³+b³+c³ = 3abc

 

Hence sin³x+sin⁶x-1 = -3sin³x

 

Hence

sin⁶x + 3sin⁵x + 3sin⁴x+sin³x-1 = 3sin⁵x + 3sin⁴x +(sin³x+sin⁶x-1)

 

=  3sin⁵x + 3sin⁴x -3sin³x = 3sin³x(sin²x+sinx-1) = 0

Guys, if sinx and cosx are both positive, then sinx^cosx+cosx^sinx is greater than 1.

 

We have cosx<2. Since sinx<1, sinx^cosx>sin²x

Similarly, cosx^sinx > cos²x

 

Hence sinx^cosx+cosx^sinx >1 if sinx and cosx are +ve

 So, 1 is not the answer.

 

I will try to get a more rigorous answer. But just look at this.

Choose sinx = 1/10ⁿ

 

Let cosx be negative. so cosx = -(1-sin²x) = -(1-1/10²ⁿ)

 

So, sinx^cosx+cosx^sinx for very large n becomes 10ⁿ +1

 

If you choose sinx = -1/10ⁿ with a similar choice of cosx, you would get -10ⁿ+1.

 

 So it is unbounded