2019/020) Express $sin ^7x$ as $a\sin7x+b\sin5x+c\sin3x+d\sin\,x$.

We have $\sin\,x = \frac{e^{ix}-e^{-ix}}{2i}$
To avoid fraction we have
$2i\sin\,x = e^{ix}-e^{-ix}$
Take power 7
$-128i\sin^7 x = (e^{ix}-e^{-ix})^7 = {7 \choose 0} e^{7ix} -   {7 \choose 1} e^{5ix} +  {7 \choose 2} e^{3ix} -  {7 \choose 3} e^{ix}$
                                                         $ +  {7 \choose  4} e^{-ix} -  {7 \choose 5} e^{-3ix} +  {7 \choose 6} e^{-5ix}  -  {7 \choose 7} e^{-7ix}$
$= {7 \choose 0} e^{7ix} -   {7 \choose 1} e^{5ix} +  {7 \choose 2} e^{3ix} -  {7 \choose 3} e^{ix}$
   $ +  {7 \choose  3} e^{-ix} -  {7 \choose 2} e^{-3ix} +  {7 \choose 1} e^{-5ix}  -  {7 \choose 0} e^{-7ix}$
$= {7 \choose 0} (e^{7ix} - e^ {-7ix}) -   {7 \choose 1} (e^{5ix} -  e^{-5ix})+  {7 \choose 2} (e^{3ix} - e^{-3ix}) -  {7 \choose 3} (e^{ix} - e^{-ix})$

so $-64\sin^7 x =  {7 \choose 0} \frac{(e^{7ix} - e^ {-7ix})}{2i} -   {7 \choose 1} \frac{(e^{5ix}-  e^{-5ix})}{2i} +  {7 \choose 2} \frac{(e^{3ix} + e^{-3ix})}{2i} -  {7 \choose 3}\frac{ (e^{ix} + e^{-ix})}{2i}$
 $= 1 \sin\, 7x  -   7 \sin\, 5x +  21 \sin\, 3x - 35 \sin\,x$
or $-64\sin^7 x = 1 \sin\, 7x  -   7 \sin\, 5x +  21 \sin\, 3x - 35 \sin\,x$

Hence
$\sin^7 x = - \frac{1}{64} \sin\, 7x  +\frac{7}{64}\sin\, 5x -  \frac{21}{64}\sin\, 3x + \frac{35}{64} \sin\,x$

2019/019) Solve for positive x $4^x+6^x = 9^x$

we have $4=2^2$, $6= 2 * 3$, $9=3^2$
So we get
$2^{2x}+( 2 * 3)^x = 3^{2x}$
divding by $2^{2x}$ we get
$1 + (\frac{3}{2})^x =   (\frac{3}{2})^{2x}$
Putting $(\frac{3}{2})^x = y$ we get
$1 + y = y^2$
so y = golden ratio $\phi = \frac{1+\sqrt{5}}{2}$
Giving
$(\frac{3}{2})^x = \frac{1+\sqrt{5}}{2}$
taking log on both sides we get
$x= \frac{\log \frac{1+\sqrt{5}}{2}}{\log \frac{3}{2}}$

2019/018) $x^2 + bx + a$ leaves same remainder when divided by x + 2 or x - a (where a ≠ -2). Show that a + b = 2.

We have
$f(x) = x^2 + bx +a$
Same remainder when divided by x + 2 or (x-a)

So $f(-2) = 4- 2b + a = f(a) =  a^2 +ab +a$
Or 4 - 2b = a^2 + ab
Or 2b+ ab = 4 - a^2
Or b(2+a) = (2-a)(2+a)
As a is not -2 so we have b = 2-a or a + b = 2

Alternatively as same remainder when divided by x+ 2 and x -a and it is degree 2 polynomial

We have $f(x) = x^2 + bx + a = (x+2)(x-a) +c $ where  c is a constant
Or  $f(x) = x^2 + bx + a = x^2+(2-a)x + c - 2a$
Comparing coefficient of x we have b= 2- a or a+b =2 

2019/017) Each of the numbers $x_1,x_2,\cdots,x_{101}$ is $\pm1$. what is the smallest positive value of $\sum_{1 \le x_i\lt x_j\le 101 }x_ix_j$

Let the given sum be S

We have $(\sum_{1 \le  n \le 101 }x_n)^2= \sum_{1 \le  n \le 101 }x_n^2 + 2 \sum_{1 \le  x_i\lt x_j\le 101 }x_ix_j  $

so $2S = (\sum_{1 \le  n \le 101 }x_n)^2 - \sum_{1 \le  n \le 101 }x_n^2$

 as each $x_i$ is $pm1$ so $x_i^2 = 1$

so $\sum_{1 \le  n \le 101 }x_n^2 = 101$
so $2S = (\sum_{1 \le  n \le 101 }x_n)^2 - 101$
 For S to be positive we must have
$(\sum_{1 \le  n \le 101 }x_n)^2 \gt   101$ and odd
for S to be smallest we must have
$(\sum_{1 \le  n \le 101 }x_n)^2 $ smallest number greater than 101 and it is true when $(\sum_{1 \le  n \le 101 }x_n)^2 = 121$
this is possible and this give S = 10.

2019/016) Let the rational number $\frac{p}{q}$ be closest to but not equal to$\frac{22}{7}$ among all rational numbers with denominator less than 100. What is the value of $p-3q$

Solution

We have $\lvert \frac{p}{q}-\frac{22}{7}\rvert$ as close to zero
Or $\lvert \frac{7 p- 22q}{7q}\rvert$
Above value is lowest when 7q is close to the highest and 7p-22q is close to the lowest( as both may not be at the same point)
q highest is when q = 99  this gives p = 311 and 7p-22q = 1 so this is lowest when q is highest
so p = 311, q = 99 and p - 3q = 311 - 3 * 99 = 14

2019/015) Let a and b be positive real numbers such that $a+b=1$. Prove that $a^ab^b +a^bb^a <=1$

without loss of generality we can assume $a>=b$
We have
$1= a+ b = a^{a+b} + b^{a+b}$
so $1- (a^ab^b + a^b b^a)$
$=  a^{a+b} + b^{a+b} - (a^ab^b + a^b b^a)$
$= a^a(a^b-b^b) + b^a(b^b-a^b) = (a^a - b^a)(a^b - b^b)$
for a > b both the terms are non -ve so we have

$1- (a^ab^b + a^b b^a) >=0$ and hence the result

2019/014) Find the values of n such that $n^4+4$ is a prime

We have $n^4+4 = n^4+4n^2 + 4 - 4n^2 = (n^2+2)^2 -(2n)^2 = (n^2+2n+2)(n^2-2n +2)$
$n^4+4$ is a prime iff $n^2+2n+2$ is a prime and $n^2-2n+2=1$
$n^2-2n+2=1=>(n-1)^2 = 0$  or n = 1
And for n= 1 $n^2+2n+2=5$ which is a prime
we could also compute $n^4+4= 1 + 4 =5$
So 1 is the only choice  for n