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$= rac{1}{2sin^2 rac{ heta}{2} + i2 sin rac{ heta}{2} cos rac{ heta}{2}}$

= rac{1}{2sin rac{ heta}{2}} rac{1} {left( sin rac{ heta}{2} + icos rac{ heta}{2} ight)}

= rac{1}{2sin rac{ heta}{2}} left( sin rac{ heta}{2} - icos rac{ heta}{2} ight)

= rac{1}{2} - rac{i}{2} cot rac{ heta}{2}

Sorry, I seem to have solved another prob!

Your answer is right, but your working is not entirely correct. Notice, you haven't used the information that TQ = 4. I am not sure how you got at the coordinates of B as (10,0) without using this. In any case, using coord-geo for this problem is overkill.

Anyway, using your working, we have BR = 10 and hence BT = BT-RT = 2. So the projection T' of T on the y-axis is at a distance 2 from the origin O.

Hence OT' = 2

You have been given TQ = 4. Now there is a theorem you would have learnt in school according to which QT.PT = RT.TS

Hence PT = 24 Rightarrow PQ = 28

This gives PT' = 14.

Hence $OP = \sqrt{OT$

= $\sqrt{200} = 10\sqrt 2$

Let $2^x = 3^y = 6^{-z} = k$

Then $2 = k^{\frac{1}{x}}; 3 = k^{\frac{1}{y}}; 6= k^{-\frac{1}{z}}$

Since 2. 3 = 6

$k^{\frac{1}{x}} k^{\frac{1}{y}} = k^{-\frac{1}{z}}$

Hence $\frac{1}{x} + \frac{1}{y} = -\frac{1}{z}$

There is an identity known as Hermite's Identity that you can use here:

[x]+[x+rac{1}{n}]+[x+rac{2}{n}]+[x+rac{3}{n}]+..+[x+rac{n-1}{n}] = [nx]

Hence,

Now it is easy to see that all points x such that x = rac{k}{3}, k in mathbb{Z}

, are points of discontinuity.

in the interval [-1,1] there are seven such points (-1, -2/3, -1/3, 0, 1/3, 2/3, 1)

Just curious: How is x! defined for numbers other than whole numbers?

As a related exercise can you use the above result to prove that

(a^2+b^2)(c^2+d^2) = (ac-bd)^2 + (ad+bc)^2

More succintly,

Hence

|(a+ib)(c+id)|^2 = |a+ib|^2 |c+id|^2 = (a^2+b^2)(c^2+d^2) = 1

3rd one: Pls see http://www.goiit.com/posts/list/algebra-target-iit-jee-2008-challenging-problems-from-maths-42418.htm#215465

This one is a well known prob so pls dont rate.

1. $1+n+n^2+n^3+...+n^{127} = rac{n^{128}-1}{n-1} = rac{(n^{64}-1)(n^{64}+1)}{n-1}$

$n^{64} -1$ is divisible by n -1

Hence $n^{64}+1 ext{divides} 1+n+n^2+n^3+...+n^{127}$

There is one more method that used complex numbers. Anybody try?

$\sin \theta + \cos \theta < \frac{\pi}{2}$

Hence, $\sin \theta < \frac{\pi}{2} - \cos \theta$

Also, since $\theta \in [0, \frac{\pi}{2}]$ , $\sin \theta , cos \theta \in [0, \frac{\pi}{2}]$

where cosx is a decreasing function.

Hence $\cos (\sin \theta) > \cos (\frac{\pi}{2} - \cos \theta) = \sin (\cos \theta)$

It is given that $f(x) > 0, x in mathbb{R^+}$

This is possible under two cases:

1. f(x) has no real roots (or zero root), and the leading coefficient m>0

Thus,

and

This gives 1 le m le 9

2. f(x) has real roots, both roots are non-positive, and the leading coefficient m>0.

So, the following has to be satisfied:

1.

2.

3. product of roots = rac{1}{m} > 0

4. Sum of roots = m-3 < 0

This gives 0<m<1

.

Hence,

yeah, he has given for 2ⁿ case actually. Anyway procedure is the same for the 2n case (nice flip-flop no?

)

He says 2n and also 2ⁿas far as I can see

f(x) = (1+x)(1+x²)(1+x⁴)...(1+x²ⁿ)

If you use product rule for differentiation

f'(x) = (1+x²)(1+x⁴)...(1+x²ⁿ)+ x g(x)

as the differentiation of every other factor produces an x term. e.g. d/dx (1+x²) = 2x.

Hence f'(0) = 1.

Ditto for 2ⁿ case

Finding the last two digits is the same as finding the remainder when the number is divided by 100.

You can write $3^{999} = (3^2)^{499} * 3 = (10-1)^{499} * 3$

If you expand $(10-1)^{499}$ using binomial theorem, all the terms will be

divisible by 100 except the terms inom{499}{1} 10 = 4990

and -1

i.e. $3^{998}$ is of the form 100k+4989 = 100k+4900+89 = 100k'+89.

So $3^{999}$ = $3^{998}*3$ is of the form (100k'+89)*3 = 100k"+47.

So 4,7 are in the tens place and units place respectively

Ok, another way, using trig:

$| rac{z_1}{z_2}| = 1$

Hence, $rac{z_1}{z_2} = cos heta + isin heta$

Now, $rac{z_1+z_2}{z_1 - z_2} = rac{ rac{z_1}{z_2}+1}{ rac{z_1}{z_2}-1}$

=

rac{cos heta + isin heta +1} {cos heta + isin heta -1}

= $rac {2 cos^2 rac{ heta}{2} + i2sin rac{ heta}{2} cos rac{ heta}{2}} {i2sin rac{ heta}{2} cos rac{ heta}{2} - 2 sin^2 rac{ heta}{2}}$

=

rac {2 cos rac{ heta}{2} (cos rac{ heta}{2} + isin rac{ heta} {2})} {2 sin rac{ heta}{2} (icos rac{ heta}{2} - sin rac{ heta}{2})}

=cot rac{ heta}{2} rac {cos rac{ heta}{2} + isin rac{ heta}{2}} {i(cos rac{ heta}{2} + isin rac{ heta}{2})}

=

which is a purely imaginary number or zero