2015/040) Factorize $x^6+5x^3+8$

we checking rational solutions find that there is no rational root and hence there is no term of the form (x+a)

now $x^6 = (x^2)^3$ and $8= 2^3$ if we can express $5x^3$ so that one is a cube and second one is product of $x^2$,$-x$ and the 3rd term then we get into the form $a^3+b^3+c^3-3abc$

$x^6+5x^3+8 = x^6-x^3 + 2^3 -6x^3$
= $x^6 - x^3 + 2^3 - 3(x^2)(x)$
= $a^3+b^3+c^3 - 3abc$ where $a = x^2,b = -x,c = +2$

hence it factors as $(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$
 = $(x^2 -x +2)(x^4+x^2+4 + x^3 -2x - 2x^2)$
= $(x^2-x+2)(x^4+x^3-x^2+2x+4)$

2015/039) If p, q, r are in A.P., show that pth, qth, rth terms of any G.P. are themselves in G.P

$p, q, r$ are in AP

so $2q = p + r$

let for the gp 1st term is a and common factor is t

the pth term = $T_p = at^{p-1}$
qth terrm = $T_q = at^{q-1}$
r th term = $T_r = at^{r-1}$

pth term * rth term = $T_p * T_r = a^2t^{p+r-2} = a^2t^{2q-2} = (at^{q-1})^2 = T_q^2$ so they are in GP

2015/038) Find sum of n terms

 $1 + ( 1 + x ) + ( 1 + x + x^2 ) + ( 1 + x + x^2 + x^3 ) + \cdots$

let $y = 1 + (1+ x) + (1+ x + x^2) + \cdots(1+x+x^2 + \cdots x^{n-1})$
so  $y(1-x)=(1-x)+ 1(1-x^2)+ \cdots+(1-x^n) $
=  $n-(x+x^2+ \cdots+ x^n) $
=  $n-x(1+x+ \cdots+ x^{n-1}) $
= $n-x\dfrac{1-x^n}{1-x}$
or $y = \dfrac{n}{1-x} - x \dfrac{1-x^n}{1-x^2}$

the above ans is correct if x is not 1

if x = 1 then we have $1 + 2 + 3 + \cdots+n = \dfrac{n(n+1)}{2}$

2105/037) If one AM A and two GMs p and q are inserted between two given numbers, show that $\dfrac{p^2}{q} + \dfrac{q^2}{p}= 2A$

because A is AM so 

$2A = (a+b)\cdots(1)$

and p q are 2 GMS

$a,p, q, b$ are in GP

ratio = t so 
$p = at\cdots(2)$
$q = at^2\cdots(3)$
$b = at^3\cdots(4)$

now 
$(\dfrac{p^2}{q} + \dfrac{q^2}{p})$
= $\dfrac{a^2t^2}{at^2} + \dfrac{a^2t^4}{at}$ (from 2 and 3)
= $a + at^3$
= $a + b$ (from (4))
= 2A (from 1)

2015/036) Prove that $\sin(1^\circ)$ is not a rational number

we have

$\cos 2 t= 1 - 2 sin ^2 t$

so if $\sin\, 1^\circ$ is rational $\cos\, 2\circ$ is rational

now $\cos\ 0^\circ =1$ is rational

$\cos (n-2^\circ ) + \cos (n+2^\circ) = 2 \cos\, n \cos 2^\circ$

so $\cos (n+2^\circ) = - \cos (n-2^\circ) + 2 \cos\, n \cos\, 2^\circ$

if $cos\, n$ and $cos (n- 2^\circ)$ are rational then by strong induction $cos (n+2^\circ)$ is rational

hence proceeding we get $cos\,30^\circ = \dfrac{\sqrt3}{2}$ is rational which is contradiction

hence $\cos\,2^\circ$ and then $\sin\,1^\circ$ are not rational

2015/035) If $5\csc(x) - 2\cot(x)= 5$ then find the value of $5\cot(x) - 2\csc(x)$.

Let $5 \csc\, x - 2 \cot\, x = 5 \cdots (1)$

Then, squaring both sides

$25 \csc^2 x + 4 \cot^2 x - 20 \csc\, x \cot\, x = 25$

or $25 ( \csc^2x - 1 ) + 4 ( \csc^2 x - 1 ) - 20 \csc\, x \cot\, x = 0$

or $25 \cot^2 x + 4 \csc^2 x - 20 \csc\, x \cot\, x = 4$

or $(5\cot\, x - 2 \csc\, x )^2 = 4$

or $5\cot\, x - 2 \csc\, x = \pm 2$

2015/034) Given $f(x-1) + f(x+1) = \sqrt{3} * f(x)$. What is the period of f(x)

We
we need to have 2 term $f(x)$ and $f(y)$ where $y = x + t$

now $f(x-1) + f(x+1) = \sqrt{3} f(x)\cdots (1)$

putting $x-1 = y$

$f(y) + f(y+2) = \sqrt{3} f(y+1)\cdots (2)$ 


hence $f(y+2) + f(y+4) = \sqrt{3} f(y+3)\cdots (3)$  ( from 1 putting y = x - 3)

add (2) and 3 to get


$f(y ) + 2 f(y+2) + f(y+4) = \sqrt{3}(f(y+1) + f(y+3) = 3 f(y+2)$

so $f(y) + f(y+4) = f(y+2)$

putting y = x and y = x +3 we get 2 equations

$f(x) + f(x+4) = f(x+2)$
$f(x+4) + f(x+8) = f(x+6)$

add to get $f(x) + 2 f(x+4) + f(x+8) = f(x+2) + f(x+6) = f(x+4)$

so $f(x) + f(x+4) + f(x+8) = 0 \cdots(4)$

changing x to x + 4 we get

$f(x+ 4) + f(x+8) + f(x+12) = 0$ 0 and subtacting (4) from above

$f(x+12) = f(x) = 0$

so $f(x+12) = f(x)$

so period is 12 or a factor of 12

it can be checked that period is not 2 4 3 or 6.

hence it is 12

refer to https://in.answers.yahoo.com/question/index?qid=20111105051143AAcISWZ

2015/033) Find coefficient of $x^n$ in the expansion of $\sqrt{1 + 2x + 3x² + 4x³ +\cdots inf. }$

let $y = ( 1 + 2x + 3x^2 + 4x^3 + \cdots. inf.) \cdots(1)$
y converges for |x| < 1
now $xy = x + 2x^2 + 3x^3 \cdots(2)$
subtract (2) from (1)
$y - xy = 1 + x + x^2\cdots = \dfrac{1}{1-x}$
so $y(1-x) = \dfrac{1}{1-x}$
or $y = \dfrac{1}{(1-x)^2}$
so $\sqrt{1 + 2x + 3x² + 4x³ +\cdots inf. }=\dfrac{1}{1-x} = 1 + x + x^2\cdots$.
so coefficient of $x^n$  1 for all n

2015/032) If x is real,show that $\dfrac{x}{x^2-5x+9}$ always lie in the interval $[\frac{-1}{11},1]$

let $y = \dfrac{x}{x^2-5x+9}$
so $yx^2 - 5yx + 9y = x $

or $yx^2 - (5y+ 1) x + 9y = 0$

for it to have real root x discriminant > = 0 or
$(5y+1) ^2 - 36y^2 \ge 0$
=> $(5y + 1 + 6y)(5y+1 - 6y) \ge 0$

or $(11y + 1)(1-y) \ge 0$ or $y \ge \dfrac{-1}{11}$ and $y \le 1$

so $\dfrac{x}{x^2-5x+9}$ always lie in the interval $[\frac{-1}{11},1]$

I have solved it at https://in.answers.yahoo.com/question/index?qid=20130502090041AAY63AN where you can find some more solution

2015/031) if $a^2 + b^2 = c^2$ then show that $\dfrac{(c-a)(c-b)}{2}$ is a perfect square

We have
$2(c-a)(c-b) = 2 c^2- 2(a+b) c + 2ab$

= $(c^2+2ab) + (c^2- 2(a+b)c)$

= $(a^2+b^2 + 2ab) – 2(a+b)c + c^2$

= $(a+b)^2 – 2(a+b) c + c^2$

= $(a+b-c)^2$

so $\dfrac{(c-a)(c-b)}{2} = (\dfrac{(a+b-c)}{2})^2$

 

proved