2016/052) if $n\sin \theta =\sin (\theta+ 2a)$ then $\tan (\theta+a) = ?$

$\frac{\sin (\theta+2a)}{\sin\theta} = n$
using componendo dividendo we have
$\frac{\sin (\theta+2a) +  \sin \theta}{\sin (\theta+2a) -  \sin \theta} = \frac{n+1}{n-1}$
Or $\frac{2 \sin (\theta+a)  \cos\,  a}{2 \cos (\theta+a)  \sin\, a} = \frac{n+1}{n-1}$
Or $\frac{tan  (\theta+a)}{\tan\,   a} = \frac{n+1}{n-1}$

Or  $\tan  (\theta+a) =    \frac{n+1}{n-1} \tan\, a$

2016/051) If $1, w, w^2$ are cube roots of unity and $a+b+c=0$ then prove that $(a+bw+cw^2)^3+(a+bw^2+cw)^3=27abc$

let $X = a+bw+cw^2$
$Y = a+bw^2+cw$
$X+Y = 2a + b(w+w^2) + c(w^2+w) = 2a - b -c = 3a - (a+b+c) = 3a$
let $Z = -3a$
so $X+Y+Z=0$
so $X^3 + Y^3 + Z^3 = 3XYZ = -9a(a+bw+cw^2)(a+bw^2+cw)$
           $ = -9a(a^2 + b^2 + c^2 - ab - ac - bc)$
or $X^3 + Y^3 + (-3a)^3 = -9a((b+c)^2 + b^2 + c^2 - a(b +c) - bc)$
or $X^3 + Y^3 + (-3a)^3 = -9a((b+c)^2 + (b+c)^2 - 2bc + (b +c)^3 - bc)$
or $X^3 + Y^3 = 27a^3 -9a(3(b+c)^2 + 3 bc)$
$ = 27a(a^2 -  (b+c)^2  + bc)$
$= 27a((b+c)^2 - (b+c)^2 + bc) = 27abc$
hence  $(a+bw+cw^2)^3+(a+bw^2+cw)^3=27abc$

2016/050) Let a,b be roots of $x^2-3x + A =0$ and c,d be roots of $x^2-12x+B=0$ if $a,b,c,d$ form an increasing GP then find A and B.

because a,b,c,d are in GP let common ratio be t then we have $b=at,c=at^2,d=at^3$
a,b are roots of $x^2-3x + A =0$ so $a+at= 3$ or $a(1+t) = 3\cdots(1)$
c,d are roots of $x^2-12x+C=0$ so $c+d = 12$ or $at^2+bt^3= 12$ or $at^2(1+t) = 12\cdots(2)$
from (1) and (2) $t^2 = 4$ ot $t=2$ because t has to be positive otherwise $a,b,c,d$ cannot be increasing
from (1) $a = 1$ so $a=1,b=2,c=4,d=8$ and hence $A=ab= 2,C=cd = 32$

2016/049) Find the equation of the normal to the curve $x^2= 4y$ that passes through the pont(1,2) (IIT 1984)

equation of the curve $4y = x^2$
so slope of the tangent = $\frac{dy}{dx} = \frac{x}{2}$
so slope of the normal  = $- \frac{2}{x_1}$
the line passes through   $x_1,\frac{x_1^2}{4}$ and $(1,2)$ so slope of line
$\frac{frac{x_1^2}{4} - 2}{x_1-1} = - \frac{2}{x_1}$
$x_1^3 - 8x_1 = - 8x_1 + 8$ or $x_1 = 2$
giving slope of normal = - 1
so equation of normal $= (y-2) = - (x-1)$ or $x+y=3$

2016/048) Let $f(x) = Ax^2 + Bx +C$ where A,B,C are real numbers. prove that if (x) is integer for all integers x then 2A, A + B, C are integers. prove the converse as well.

we have $f(x) = A x^2 + Bx + C = A (x^2-x) + (A+B) x + C=2A\frac{x(x-1)}{2} + (A+B) x + C$
now x and $\frac{x(x-1)}{2}$ are integers for integer x. so if (A+B),2A and C are integers the $f(x)$ is integer for integer x
if f(x) is integer for all x then $f(0) = C$ is integer.
$f(1) = (A+B) 1 + C$ is integer so $A+B$ is integer
$f(2) = 2A + 2(A +B) + C$ is integer so 2A is integer

2016/047) if $\alpha+\beta+\gamma = \pi$ then prove that $\sin ^2\alpha + \sin ^2 \beta - \sin ^2 \gamma = 2 \sin\,\alpha\sin\,\beta \cos\,\gamma$

$\sin ^2\alpha + \sin ^2 \beta - \sin ^2 \gamma$
$= \sin ^2\alpha + \sin^2  \beta - \sin ^2 \gamma(\sin ^2 \beta + \cos  ^2 \beta)$
$= \sin ^2\alpha + \sin^2  \beta(1 - \sin ^2 \gamma) - \sin ^2 \gamma \cos  ^2 \beta$
$= \sin ^2\alpha + \sin^2  \beta\cos ^2 \gamma - \sin ^2 \gamma \cos  ^2 \beta$
$= \sin ^2\alpha + (\sin\, \beta\cos\, \gamma + \sin\, \gamma \cos\, \beta)(\sin\, \beta\cos\, \gamma - \sin\, \gamma \cos\, \beta)$
$= \sin ^2\alpha + (\sin(\beta + \gamma)\sin(\beta- \gamma)$
$=\sin\,\alpha \sin (\sin(\beta + \gamma) + \sin\,\alpha\sin(\beta - \gamma)$ as $\sin\,\alpha= \sin(\beta + \gamma)$
$=\sin\,\alpha (\sin(\beta + \gamma)+\sin(\beta - \gamma))$
$=\sin\,\alpha (2\sin\,\beta \sin\,\gamma)$
$=2 \sin\,\alpha \sin\,\beta \sin\,\gamma$

2016/046) Show that there is no combination of integers a,b,c for which $f(n) = n^3 + an^2 + bn+ c$ is perfect square for all n

If we can show that it cannot be for n = 1,2,3,4 then we are through
for n = 1 we have $f(1) = 1 + a + b+ c\cdots(1)$
for n= 2 we have $f(2) = 8 + 4a + 2b + c\cdots(2)$
$f(3) = 27 + 9a + 3b + +c\cdots(3)$
$f(4) = 64 + 16a + 4b + c\cdots(4)$
if f(1) and f(3) are perfect squares then from (1) and (3) we get $26+ 8a+ 2b \equiv 0 \pmod 4$ or $2b \equiv 2 \pmod 4\cdots(5)$
if f(2) and f(4) are perfect squares then from (2) and (4) we get $56+ 12a+ 2b \equiv 0 \pmod 4$ or $2b \equiv 0 \pmod 4\cdots(6)$
(5) and (6) cannot be satisfied at the same time so for n = 1 to 4 it cannot be a perfect square

2016/045) Show that there exists infinitely many pairs of coprime integers (a,b) such that both $x^2+ax+b=0$ and $x^2+2ax+b=0$ have integer solution

from the above b can be expressed as product of 2 numbers in 2 different ways that is at least 3 integers say uvw.
now a = uv + w
and 2a = u + vw
so 2(uv+w) = u + vw
or u(2v-1) = w (v- 2)
the above holds if we choose (49) w = 2v- 1 and u = v-2.
that gives
$a = v(v-2) + 2v-1 = v^2 - 1 = (v-1)(v+1)$
$b= v(v-2)(2v-1)$
for a b to be coprime (v-1) should be co prime to v, (v-2) and (2v-1) which is true for any v
and (v+1) should be co prime to v, (v-2) and (2v-1) which is true unless v+1 is divisible by 3.
so we can choose v not divisible by 3 and get corresponding a,b for the same.
hence there are inifinite of them. for example v = 7 giving a = 48 and b= 13 * 35 = 455
$x^2+48x + 455 = 0$ gives solution x = -33,-35
$x^2+96x + 455 = 0$ gives solution -91,- 5

2016/044) Sum to infinity $1+\frac{3}{1!}+\frac{5}{2!}+\frac{7}{3!}\cdots$

we have for n > 1$n^th$ term = $\frac{2n-1}{(n-1)!} = \frac{2n-2+1}{(n-1)!} = 2 \frac{1}{(n-2)!} +  \frac{1}{(n-1)!}$
so the sum = $1 + \sum_{n=2}^{+\infty} (2 \frac{1}{(n-2)!} +  \frac{1}{(n-1)!})$
$= 1 + \sum_{n=2}^{+\infty} (2 \frac{1}{(n-2)!} +  \frac{1}{(n-1)!})$
$= 1 + \sum_{n=2}^{+\infty} \frac{1}{(n-1)!} + 2 \sum_{n=2}^{+\infty} \frac{1}{(n-2)!}$
$= 1 + \sum_{n=1}^{+\infty} \frac{1}{n!} + 2 \sum_{n=0}^{+\infty} \frac{1}{n!}$
$= \sum_{n=0}^{+\infty} \frac{1}{n!} + 2 \sum_{n=0}^{+\infty} \frac{1}{n!}$
$= 3 \sum_{n=0}^{+\infty} \frac{1}{n!} =3e$

2016/043) Factor in real $(1-(ab+bc+ca))^2 + (a+b+c-abc)^2$

using $a^2+b^2 = (a+ib)(a-ib)$
we get  $(1-(ab+bc+ca))^2 + (a+b+c-abc)^2$
$ = ( 1 +(a+b+c) i - (ab+bc+ca) - abci))  ( 1 - (a+b+c) i - (ab+bc+ca) + abci)$
$= ( 1 +(a+b+c) i + (ab+bc+ca)i^2  + abci^3 ))$
 $ ( 1 - (a+b+c) i + (ab+bc+ca)^2 - abci^3)$
$= ( 1 +ai)(1+bi)(1+ci)(1- ai)(1-bi)(1-ci)$
$= ( 1 +ai)(1-ai)(a+bi)(1- bi)(1+ci)(1-ci)$
$= ( 1 +a^2)(1+b^2)(1+c^2)$

2016/042) If $(m_r,\frac{1}{m_r})$ for r from 1 to 4 are 4 points that lie on a circle show that $m_1m_2m_3m_4= 1$

we have general equation of a circle
$x^2+y^2+ 2gx + 2fy + c = 0$
let a point $p,\frac{1}{p}$ be on the circle
then we get
$p^2 + \frac{1}{p^2} + 2gp + \frac{2f}{p} + c = 0$
or $p^4 + 1 + 2gp^3 + 2fp + cp^2=0$
or $p^4 + 2gp^3 + cp^2 + 2fp + 1=0$
the 4 roots are $m_1,m_2,m_3,m_4$ and hence product of roots = $m_1m_2m_3m_4=1$ (the constant term)

2016/041) If $ax + y + 1= 0, x +by+1=0,$ $x+y+c=0$ are concurrent then prove that $\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = 1$

proof
We have
$a = - \frac{y+1}{x}$
or $1- a = \frac{x+y+1}{x}$
or $\frac{1}{1-a} = \frac{x}{x+y + 1} \cdots 1$
Similarly
or $x +by+1=0$
or  $x+1 = - by$
or $b = -\frac{x+1}{y}$
or $1-b= \frac{x+y+1}{y}$
or $\frac{1}{1-b} = \frac{y}{x+y+1}\cdots (2)$

and $x+y+c=0$
$=>c = x + y$
$=>1-c = x+y+1$
or $\frac{1}{1-c} = \frac{1}{x+y+1}\cdots(3)$

adding all 3 we get the
$\frac{1}{1-a} + \frac{1}{1-b} + \frac{1}{1-c} = \frac{x+ y + 1}{x+y+1} =1$

Proved