IIT JEE , AIEEE Forum - Mathematics , Algebra

rajat

hey ppl. plzz solve these questions .theres a loong list

1. integration of [(cosx)^2 / (1 + a^x) ] from -pie to pie

2. if z1 and z2 are the nth roots of unity , which subtend right angle at the origin

then n is of the form

a. 4k + 1

b.4k + 2

c. 4k + 3

d. 4k

(for this quesiton plzz. give method not just the answer,only answers will be given a boink !!)

3. which of the following is differenctiable at x = 0

a. cosIxI + IxI

b cosIxI - IxI

c. sinIxI + IxI

d. sinIxI - IxI

4 . let PQ and RS be tangents at P,R of a circle of radius r where P,R are the extremeties of a diameter . if PS and RQ interesect at a point X on the circle ,

find 2r in terms of PQ,RS

5. i know this has been asked posted before but still plzz. answer it

what is maximum value of (cosa1)(cosa2)(cosa3)........(cosan)

where 0=<a1,a2,a3........an <= pie/2 given that

(cota1)(cota2)(cota3).................... = 1

6 . find the co-efficient of x^(2^ n+1) in

1/ [ (1+x)(1+x^2)(1+x^3).........(1+x^2n) ] where IxI <1

7. if f(a+x) = f(2a) for all x and a is positive ,then prove that f(x) is a periodic function of indeterminate period (also tell me what is meant by this )

8. value of tan-1( [sin2 - 1] /cos2) ??

plzzz reply

utkarshduggal

hey rajat,
i have got an answer to ur question on complex nos.
all n roots of unity lie on a circle subtending an angle 2

/n at the center.
since z1 and z2 make

/2

2

/n =

/2
thus n=4
generalising this result we get all the roots in the form of 4k.

please rate me for my effort.

sidlol

Problem 1: Put t=-x. dt = - dx.

You will get

I=the integration of -(cos² t)/(1+a^-t) from pi to -pi

= a^t (cos²t)/(1+a^t) from -pi to pi

= a^x (cos²x)/(1+a^x) from -pi to pi (changing the variable)

Therefore 2I= (1+a^x)(cos²x)/(1+a^x) from -pi to pi

= cos²x from -pi to +pi

= 2 cos²x from 0 to pi (even function)

= x + 0.5 sin 2x with limits 0 and pi

= pi

or I = pi/2.

sidlol

Question 3: The answer is 'd'.

The RHD should be the same as the LHD for the function to be differentiable.

So, for the RHD, option d is sin x - x

Differentiating, cos x - 1

= 0 when x tends to 0 from the right hand side.

For the LHD, option d is (- sin x) + x

Differentiating, - cos x + 1

= 0 when x tends to 0 from the left hand side.

None of the other functions show this.

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