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

2016/013) Prove that $\cos(\tan^{-1}(\sin(\cot^{-1} x)))= \sqrt{\frac{x^2+1}{x^2+2}}$

Let $\cot^{-1} x = y$
so $\cot y = x$
so $\csc^2 y = x^2+1$
so $\sin  y = \frac{1}{\sqrt{x^2+1}}$
so $\sin\cot^{-1} x = \frac{1}{\sqrt{x^2+1}}$
similarly  $\cos \tan ^{-1}x = \sqrt{\frac{1}{x^2+1}}$
Hence $\cos(\tan^{-1}(\sin(\cot^{-1} x)))$
$=\cos(\tan^{-1}\frac{1}{\sqrt{x^2+1}})$
$=\frac{1}{1+\frac{1}{x^2+1}}$
$=\frac{1}{\sqrt{1+\frac{1}{x^2+1}}}$
$=\frac{1}{\sqrt{\frac{x^2+2}{x^2+1}}}$
$=\sqrt{\frac{x^2+1}{x^2+2}}$

2016/012) Suppose that $f(x+3)=3x^2+7x+4\cdots(1)$ and $f(x)=ax^2+bx+c\cdots(2)$ What is $a+b+c$

we have from 2nd equation $f(1) = a + b + c\cdots(3)$
Now we need to evaluate f(1)
as $f(x+3) = 3x^2 + 7x + 4$
so $f(x) = 3(x-3)^2 + 7(x-3)  + 4\cdots(4)$
putting x = 1 we get $f(1) = 3 * (-2)^2 + 7 * (-2) + 4 = 2$
so $a + b+ c = 2$

2016/011) Simplify $a^2(b+c)+b^2(c+a)+c^2(a+b)$ , if a,b,c are in Arithmetic Progression

$a^2(b+c)+b^2(c+a)+c^2(a+b)$
$= a^2(b+c+a) + b^2(c+a+b) + c^2(a+b+c) - (a^3+b^3+c^3)$
$= (a+b+c) (a^2+b^2+c^2) - ((a+b+c)(a^2+b^2 + c^2-ab-bc-ca) + 3abc)$
(because $a^3+b^3+c^3 - 3abc = (a+b+c) ( a^2+ b^2+ c^2 - ab- bc-ca))$
$= (a+b+c)( ab + bc + ca) - 3abc$
a b c are in ap so $a+c = 2b$ so $a+b+c = 3b$
now $ab+bc+ ca = b(a+c) + ca = 2b^2 + ca$
so sum = $3b(2b^2+ ca) - 3abc = 6b^3$

2016/010) Solve $|e^{it} - 1| = 2$ for $-\pi<\theta<=\pi$

$e^ {it} = \cos t + i \sin t$
so $e^{it} - 1 = (\cos t-1) + i \sin t$
take mod and square
$(\cos t-1)^2 + \sin ^2 t = 4$
or $\cos^2 t - 2 \cos t + 1 + \sin ^2 t = 2$
or $2 - 2\cos t = 4$ or $cos t = - 1$ and hence $t = \pi$

2016/009) What is the value of c such that a straight line exists which intersects $f(x)=x^4+9x^3+cx2+9x+4$ at 4 points

$f(x)=x^4+9x^3+cx^2+9x+4$
for a straight line to intersect at 4 distinct points
$f^{''}(x)$ must have 2 roots
The reason
if it has no root $f'(x)$ is either positive or -ve and so f(x) is monotonically
increasing or decreasing. as it is cubic polynomial with leading coefficient positive
it is increasing. hence no line can intersect at more than 2 points
if it has one zero then it does not have any point of inflection
now $f^{''}(x)=12x^2+54x+2c=12(x+\frac{9}{4})^2+2(c-\frac{243}{8})$
it has a double root when $c<\frac{243}{8}$
hence $c<\frac{243}{8}$

2016/008) Simplify $\frac{x^4}{(x-y)(x-z)}+\frac{y^4}{(y-z)(y-x)}+\frac{z^4}{(z-x)(z-y)}$

$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$
=  - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$
= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$
now
$x^4(y-z) + y^4(z-x) + z^4(x-y)$
= $x^4(y-z) + yz(y^3-z^3) - x (y^4-z^4)$
= $x^4(y-z) + yz(y-z)(y^2+yz+z^2) - x(y-z)(y^3 + y^2z + yz^2 + z^3)$
= $(y-z)(x^4 + yz(y^2 +yz+z^2) - xy(y^2 + yz + z^2) - xz^3)$
= $(y-z)(x^4 + (y^2+yz+z^2)(yz-xy) - xz^3)$
= $(y-z)(x(x^3-z^3) + y(z-x)(y^2 + yz + z^2)$
=$(y-z)(z-x)(y(y^2 + yz + z^2) - x(x^2 + zx + z^2)$
= $(y-z)(z-x)(y^3 + y (yz+ z^2) - x^3 - x(zx + z^2)$
= $(y-z)(z-x)(y^3-x^3 + (y^2z + yz^2 - zx^2 - z^2 x)$
= $(y-z)(z-x)((y-x) (x^2 + xy + y^2) + (z(y^2 - x^2) +z^2(y-x))$
= $(y-z)(z-x)((y-x)(x^2 + xy + y^2 + z(y+x) + z^2)$
= $(-(x-y)(y-z)(z-x)(x^2 + y^2 + z^2 + xy+yz+zx)$
So the given expression
= $x^2 + y^2 +z^2 + xy + yz+ xz$
hence $\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}= \dfrac{1}{2}((x+y)^2 + (y+z)^2 + (x+z)^2$

2016/007) How many integer values of x and y are there such that $4x+7y=3$ while $|x| < 500$ and $|y| < 500$

1st let us find one solution
this can be found by any method but as we see that $7- 4 = 3$
so $(-1,1)$ is a solution
as coeffcient of y is larger so we need to restrict x between $- 500$ to $500$
general solution is $x = -1+ 7t$ and $y = 1 + 4t$
Now $- 500 < x < 500$ or -$500 < 1 + 7t < 500$ or
$-501 < 7t < 499$ or $-71 <= t <= 71$ so 143 values