2021/006) Let a,b,c be real with $a,b,c > 1$ and $\frac{1}{a^2-1} + \frac{1}{b^2-1} + \frac{1}{c^2-1} = 1$ Show that $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} <= 1$

We have $\frac{1}{a^2-1} + \frac{1}{b^2-1} + \frac{1}{c^2-1} = 1\cdots(1)$

Now 

$\frac{2}{a^2-1} = \frac{1}{a+1} + \frac{1}{a-1}$

As $a > 1$ so we have $a-1 > 0$ and hence

$\frac{1}{a+1} <  \frac{1}{a-1}$

adding  $\frac{1}{a+1}$ on both sides

$\frac{2}{a+1} <  \frac{1}{a-1} + \frac{1}{a+1}$

Or $\frac{2}{a+1} >  \frac{2}{a^2-1}$

Or $\frac{1}{a+1} <  \frac{1}{a^2-1}\cdots(2)$

Similarly

Or $\frac{1}{b+1} <  \frac{1}{a^2-1}\cdots(3)$

Or $\frac{1}{c+1} <  \frac{1}{a^2-1}\cdots(4)$ 

Adding above 3 we get

$\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} <  \frac{1}{a^2-1} + \frac{1}{b^2-1} + \frac{1}{c^2-1}$

or $\frac{1}{a+1} + \frac{1}{b+1} + \frac{1}{c+1} <  1$ 

2021/005) Solve $ |z - i | = |z + i |$

Method 1

Let $z = x + iy$
Then we have $ |x + iy - i | = |x + iy + i |$
Or $ |x + i(y -1) i | = |x + i(y + 1) | $
Knowing that $|x + iy| = \sqrt{x^2+y^2} $ we get
$\sqrt{x^2 + (y-1)^2} = x^2 + (y+1)^2$
Or $x^2 + (y-1)^2 = x^2 + (y+1)^2$
Or $(y-1)^2 = (y+1)^2$
or $y^2 - 2y + 1 = y^2 + 2y +1$ or $4y=0$ or $y = 0$
so the solution set is x + 0i

Another method that is Method 2 

We have z equidistant from the points (0,1), (0, -1) in the complex plane
So the point z lies on the perpendicular bisector of the segment joining (0,1) and (0,-1) .
The midpoint of the segment is (0,0) and as this is the y axis the line that is perpendicular
to this is the x axis or any point (x,0) equivalent to the number x+0i is the solution

2021/004) Given $(\sqrt{x+\sqrt{x^2+1}})(\sqrt{y+\sqrt{y^2+1}}) =1 $ find $(x+y)^2$

Let $x=\tan\theta$ $\theta$ in the range $0 ..\frac{\pi}{2}$ so we have $sec\theta$ positive

Then we have $\sqrt{x+\sqrt{x^2+1}} = \sqrt{\tan\theta + \sec\theta} $

From the given condition we have

$\sqrt{y+\sqrt{y^2+1}} = \frac{1}{\sqrt{x+\sqrt{x^2+1}}}$

Or  

$\sqrt{y+\sqrt{y^2+1}} = \frac{1}{\tan \theta + \sec\theta}$

$=\frac{\sec^\theta - tan^2\theta }{\tan \theta + \sec\theta}$

or $y+\sqrt{y^2+1} =\sec\theta - \tan \theta\cdots(1)$

if we take $y=\tan\alpha$ $\alpha$ in the range $-\frac{\pi}{2} ..\frac{\pi}{2}$ so we have $sec,\theta$ positive

then so $y + \sqrt{y^2+1}= \sec\alpha \pm \tan \alpha\cdots(2)$

comparing (1) and (2) we have $y =-\tan \theta$

so $x+y=0$ or $(x+y)^2 = 0$

 

2021/003) Which of the following numbers is a perfect square? (A) 98! · 99! (B) 98! · 100! (C) 99! · 100! (D) 99! · 101! (E) 100! · 101!

The above are of the form $(n!)^2 * (n+1)$ or  $(n!)^2 * (n+1)(n+2)$

so we need to check if (n+1) or (n+1)(n+2) is a perfect square and based ion that n = 100 is a perfect square. So ans is (C).

 

2021/002) It is given that the ratio of angles $A,\,B$ and $C$ is $1:2:4$ in a $\triangle ABC$, prove that $(a^2-b^2)(b^2-c^2)(c^2-a^2)=(abc)^2$.

We see that angles $A,\,B$ and $C$ are $\frac{\pi}{7}$, $\frac{2\pi}{7}$ and $\frac{4\pi}{7}$
and using laws of sines we need to prove

$(\sin ^2 A -\sin ^2 B)(\sin ^2 B - \sin ^2 C)(\sin ^2 C - \sin ^2 A)= (\sin\,A \sin\,B \sin\, C)^2$.
\where $A= \frac{\pi}{7}$ $B= \frac{2\pi}{7}$ and $C= \frac{4\pi}{7}$


to avoid fraction let $x =\frac{\pi}{7}$ so $A= x$ $B= 2x$ and $C= 4x$


so $\sin\,x = \sin 6x$ and $\sin 3x = \sin 4x$
we shall be using the following


$\sin\, X + \sin \,Y = 2 \sin \frac{X+Y}{2} \cos \frac{X-Y}{2}\cdots(1)$
$\sin\, X - \sin \,Y = 2 \sin \frac{X-Y}{2} \cos \frac{X+Y}{2}\cdots(2)$


So


$\sin ^2 X - \sin ^2 Y$
= $(\sin\, X + \sin \,Y)(\sin\, X - \sin \,Y)$
= $ (2 \sin \frac{X+Y}{2} \cos \frac{X+Y}{2})(2 \sin \frac{X-Y}{2} \cos \frac{X-Y}{2})$
= $ \sin (X+Y)\sin(X-Y)$


Now
LHS
= $(\sin ^2 A -\sin ^2 B)(\sin ^2 B - \sin ^2 C)(\sin ^2 C - \sin ^2 A)$
= $(\sin ^2 x -\sin ^2 2x)(\sin ^2 2x - \sin ^2 4x)(\sin ^2 4x - \sin ^2 x)$
= $(\sin ^2 2x -\sin ^2 x)(\sin ^2 4x - \sin ^2 2x)(\sin ^2 4x - \sin ^2 x)$ multiplying 1st and 2nd terms by -1
= $\sin 3x \sin \,x \sin 6x \sin 2x \sin 3x \sin \,x $
= $\sin 4x \sin \,x \sin x \sin 2x \sin 3x \sin \,x $ as $\sin 3x = \sin 4x$ and $\sin 6x =\sin \,x $
= $(\sin\,x \sin 2x \sin 4x)^2$
= $ (\sin\,A \sin\,B \sin\, C)^2$
= RHS

2021/001)Show that for any positive integers a and b $(36a+b)(a+36b)$ cannot be a power of 2

 Without loss of generality let us assume $a\ge b$. so $(36a+b) \ge (a+36b)$

Let $(36a+b)(a+36b)$ be a power of 2
Then we must have both $36a+b $ and $a + 36b$ powers of 2

So let $36a+b = 2^m $ and $a+36b = 2^n$ where $m \ge n$
So $36a+b + 3a + 36 b = 37(a+b) = 2^n(2^{m-n} + 1)$
So $37$ has to be $2^n$ or $2^{(m-n)} + 1$ and neither is possible as neither 37 not 36 is a power of 2
Hence there is contradiction
Hence it is not possible