2016/020)Solve $\cos^{-1}\frac{x^2-1}{x^2+1} +\frac{1}{2} \tan^{-1}\frac{-2x}{1-x^2}=\frac{2\pi}{3}$

we have $\tan (2y) = \frac{2\tan y }{1-\tan^2 y}$
hence   $\tan (- 2y) = \frac{-2\tan y }{1-\tan^2 y}$
hence if $x = \tan y$
 $-2y = \tan ^-1 \frac{-2x}{1-x^2}$
or  $\tan ^{-1} \frac{-2x}{1-x^2}= -2 \tan ^{-1} x\cdots(1)$
further $\cos(2y) = \sin ^2 y - \cos^2 y = \frac{\sin ^2 y - \cos^2 y}{\sin ^2 y + \cos^2 y}$
or $\cos(2y) = \frac{\tan  ^2 y - 1}{\tan  ^2 y + 1}$
or $\cos(\pi - 2y) = \frac{1 - \tan  ^2 y}{\tan  ^2 y + 1}$
or $\pi - 2y = \cos ^{-1} \frac{1 - \tan  ^2 y}{\tan  ^2 y + 1}$
putting $\tan y = x$ we get  $\pi - 2 \tan ^{-1} x = \cos ^{-1} \frac{1 - x^2}{1 + x^2}\cdots(2)$
putting values from (1) and (2) in given equation we get
$\pi - 2\tan^{-1} x - (1/2)*2\tan^{-1} x = \frac{2\pi}{3}$
or  $3\tan^{-1}(x) = \frac{\pi}{3}$
or  $\tan^{-1} (x) = \frac{\pi}{9}$
Thus $x = \tan\frac{\pi}{9}$

2016/019) If $\theta = \frac{\pi}{2^{n+1}}$ then show that $2^n cos\theta \ cos 2 \theta \cos 2^2\theta \cdots cos 2^n \theta = 1$

$2^n cos\theta \cos 2 \theta \cos 2^2\theta \cdots cos 2^n \theta$
$= (2\cos\theta) (2\cos 2 \theta)(2 \cos 2^2\theta) \cdots (2cos 2^n \theta)  = 1$
$= (\frac{\sin 2\theta}{\sin \theta}) (\frac{\sin 4\theta}{\sin 2\theta})
          \cdots (\frac{\sin 2^n\theta}{\sin 2^{n-1}\theta})  = 1$
$=\frac{\sin 2^n \theta }{\sin \theta}$
$==\frac{\sin 2^n\frac{\pi}{2^n+1})}{\sin \frac{\pi}{2^n+1}}$
$==\frac{\sin (\frac{\pi}{2^n}-\frac{\pi}{2^n+1}) }{\sin \frac{\pi}{2^n+1}}$
$==\frac{\sin (\frac{\pi}{2^n+1}) }{\sin \frac{\pi}{2^n+1}}=1 $

2016/018) Show that function $f(x) = | x+2 | $ is continuous at $x = - 2$ but not differentiable at x = - 2.

We have $f(x) = | x+2 | $
or $f(x) = x + 2$ for $x >=-2$ and $f(x) = -x-2$ for $x < -2$
at x = -2 the right hand limit is 0 and the left hand limit is 0
so it is continuous at x = -2
differentiating from left we get $f'(x) = -1$ and differentiating from right $f'(x) = 1$
as left hand derivative and right hand derivative are not same of it is
not differenctiable

2016/017) In a quadrilateral ABCD ab is the smallest and CD is the largest side. Prove that angles -- (1) $A > C$ and $B > D$

Join AC. In triangle ABC $AB < BC$ so $\angle BAC > \angle BCA$
in triangle ADC $CD > AD$ so $\angle CAD > \angle DCA$
adding above 2 we get the result. Similarly the 2nd part

2016/016) Show that $4 * (29!) + 5! \equiv 0 (\,mod\, 31) $

because 31 is prime we have as per wilson's theorem
$30!  \equiv -1 (\,mod\, 31)\cdots(1) $
and also $30 * (-1) = -30  \equiv 1 (\,mod\, 31) =>30^{-1} = \equiv 1 (\,mod\, 31)\cdots(2)$
from (1) and (2)
$29!  \equiv 1 (\,mod\, 31)$
or $ 4 * 29!  \equiv 4 (\,mod\, 31) $
or $ 4 * 29! + 5!   \equiv 4 + 120 (\,mod\, 31) \equiv  124 (\,mod\, 31) \equiv 0 (\,mod\, 31)$

2016/015) Solve for x $(5+2\sqrt 6)^{x^2-3} + (5-2\sqrt 6)^{x^2-3} = 10$

We have $(5+ 2 \sqrt6)(5-2\sqrt6) = 25 - 24 =1$
So if $t= 5 + 2 \sqrt 6$ then $\dfrac{1}{t} = 5 - 2\sqrt 6$
So we get
$t^{x^2-3} + \dfrac{1}{t^{x^2-3}} = 10$
let $t^{x^3-3} = p\cdots(1) $
so we get
$ p + \dfrac{1}{p} = 10$
or $p^2 - 10p +1 = 0$
or $p = 5 + 2 \sqrt 6 $ or $p= 5 - 2\sqrt 6= (5+ 2 \sqrt 6)^{-1}$
Hence from (1) and above $x^2-3 =1  => x = \pm 2$
or $x^2 -3 = -1 => x = \pm \sqrt 2$

2016/014) if $x,y,z$ are in H.P then show that $log(x+z)+ log (x+z-2y) = 2log(x-z)$

We have $x,y,z$ are in HP
so $\dfrac{1}{x} + \dfrac{1}{z} = 2\dfrac{1}{y}$
or $ y(x+z) = 2xz$
Now $(x+z)(x+z-2y) = (x+z) ( x + z - \dfrac{4xz}{x+z})$
$= (x+z)^2 - 4xz$
$= x^2 + z^2 + 2xz -4xz$
$= x^2 + z^2 -2xz = (x-z)^2$
taking log of both sides we get the result