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digits

This can be expressed as the sum of a GP with common ratio 10.

So ,

$a_n = \frac{10^{3^n}-1}{9}$

Similarly

$a_{n-1} = \frac{10^{3^{n-1}}-1}{9}$

Now ,

$a_n = \frac{10^{3^n} -1 }{9} \\=\frac{(10^{3^{n-1}})^3-1}{9}$

Expand this using the formula a^3 - b^3 . The first term will be a_(n-1) and it can be easily shown that the second term is divisible by3

This is now equivalent to finding the coefficient of $(\frac{1}{a})^3(\frac{1}{b})^2(\frac{1}{c})$ in the expansion of $( \frac{1}{a}+\frac{1}{b}+\frac{1}{c})^6$ . This is the usual multinomial expansion .

The answer is 6! / 3! 2! = 60

It is the same as finding the coefficient of $\frac{(abc)^6}{a^3b^2c}$ in the expansion of $( \frac{1}{a}+\frac{1}{b}+\frac{1}{c})^6$

nothing fishy. whenever u square an equation , u should substitute and check in the end . if it does not satisfy the equation , then it is not a root. that's all . u should do that always when u square.

There is no problem . whenever u square an equation , u should substitute and check in the end . if it does not satisfy the equation , then it is not a root. that's all . u should do that always when u square.

D)
$Z(\theta , 2\theta ) = \cos \theta + e^{i2\theta} \sin \theta$

$= \cos \theta + (\cos 2\theta+ i \sin 2\theta ) \sin \theta$

$|Z|^2 = ( \cos \theta + \cos 2 \theta \sin \theta ) ^2 + (\sin 2 \theta \sin\theta)^2$

Now simplify ,
$|Z|^2 = 1 + \frac{\sin 4 \theta }{2}$

sometimes , inexperienced problem setters will make questions where we can get the answer through some special method . they will not check whether the question is completely correct.
e.g. sinA may become greater than 1, the triangle inequality may not be valid etc. we can still get some answer.

cant say about this question though.

2. 2pi/3
Let us denote the sides as
$a = \sin \alpha \\ b= \cos \alpha \\ c= \sqrt{1 + \sin \alpha \cos \alpha}$

Now use cosine formula to find the angles.

$\cos C = \frac{a^2 + b^2 - c^2}{2ab} \\ \\ = \frac{\sin^2\alpha + \cos^2\alpha - (1 + \sin\alpha\cos\alpha )}{2\sin\alpha\cos\alpha} \\ \\ = -1/2$

So C= $\frac{2\pi}{3}$ . This is the greatest angle.

@sboosy
It's true that if this process goes on , the final digit will be 1 . but how do we know that after 4 operations , the result will be a 1 digit no.?

I still don't understand.

Consider x³ - 2x +2

Here 4a³ + 27 b² = - 32 + 108 > 0 . But the equation has only 1 real root.

You can also verify that if it is less than 0 , the equation will have 3 roots.

u r going to learn all those when u r in 12th??????

for c++ : go here http://www.cplusplus.com/

PS: this may not be useful if u r studying computer science in cbse class 12. cbse teaches a decades-old version of the language called turboC++

Maybe if u consider ABCD in the clockwise direction , then the centroid will be at - $i\frac{z_1}{3}$

But the usual practise is to take in the anticlockwise direction.

The centroid is at $i\frac{z_1}{3}$ .

multiplying a complex no. with 'i' will rotate it by $\frac{\pi}{2}$ degrees.

so here,

A=z_1

centroid = iz1/3

@sboosy, For one real root , shouldn't it be $f(x_1)f(x_2)\geq0$ ?

Also , if a>0, the equation will always have one real root.

Check the answer now . sorry for the earlier wrong answer.