2014/072) The number of divisiors of the form $4n+ 2(n\ge 0)$ of the integer 240 is

a ) 4
b) 8
c) 10
d) 3

as $240 = 16 * 15 = 2^4 *3 * 5$

in 4n+2 there is one  occurrence of 2 and hence number of factors = 1  * 2 *2 = 4 as 15 has 4 odd factors 

2014/071) prove $\cot \ 7\frac{1}{2}^\circ = \sqrt{2} + \sqrt{3}+\sqrt{4} + \sqrt{6}$

we have

$\cos\ 15^\circ = \cos(60- 45)^\circ =  \cos\ 60^\circ \cos\ 45^\circ +  \sin \ 60^\circ \sin\ 45^\circ$
= $\dfrac{1}{2}\dfrac{1}{\sqrt{2}} + \dfrac{\sqrt{3}}{2}\dfrac{1}{\sqrt{2}}$
= $\dfrac{1}{4}(\sqrt{2} + \sqrt{6})$
also
$\sin \ 15^\circ = \sin (60- 45)^\circ =  \sin \ 60^\circ \cos\ 45^\circ -  \cos  \ 60^\circ \sin\  45^\circ$
= $\dfrac{\sqrt{3}}{2}\dfrac{1}{\sqrt{2}} - \dfrac{1}{2}\dfrac{1}{\sqrt{2}}$
= $\dfrac{1}{4}(\sqrt{6} - \sqrt{2})$

now $\cot\ x= \dfrac{\cos\ x }{\sin \ x} = \dfrac{2\cos^2\ x }{2\sin \ x\cos \ x}$
= $\dfrac{1+\cos 2x}{\sin 2x}$

  Hence $\cot \ 7\frac{1}{2}^\circ = \dfrac{1+\cos\ 15^\circ}{\sin\ 15^\circ}$
= $\dfrac{1+\dfrac{1}{4}(\sqrt{2} + \sqrt{6})}{\dfrac{1}{4}(\sqrt{6} - \sqrt{2})}$
=  $\dfrac{4+(\sqrt{2} + \sqrt{6})}{(\sqrt{6} - \sqrt{2})}$
= $\dfrac{4+(\sqrt{6} + \sqrt{2})}{(\sqrt{6} - \sqrt{2})}$
= $\dfrac{4(\sqrt{6} + \sqrt{2})+(\sqrt{6} + \sqrt{2})^2}{(\sqrt{6} - \sqrt{2})(\sqrt{6} + \sqrt{2})}$
=  $\dfrac{4(\sqrt{6} + \sqrt{2})+(\sqrt{6} + \sqrt{2})^2}{4}$
=  $(\sqrt{6} + \sqrt{2})+\dfrac{(\sqrt{6} + \sqrt{2})^2}{4}$
=  $(\sqrt{6} + \sqrt{2})+\dfrac{6 + 2 + 4 \sqrt{3}}{4}$
=  $(\sqrt{6} + \sqrt{2})+\dfrac{8 + 4 \sqrt{3}}{4}$
=  $(\sqrt{6} + \sqrt{2})+  2 +  \sqrt{3}$
=  $\sqrt{2} + \sqrt{3})+  2 +  \sqrt{6}$
=  $\sqrt{2} + \sqrt{3})+\sqrt{4} + \sqrt{6}$

2014/070) Prove that $\displaystyle\sum_{n=1}^{15} \sin(4m-2) = \dfrac{1}{4\sin\ 2^\circ}$

we have

LHS = $\dfrac{\displaystyle\sum_{n=1}^{15}2 \sin(4m-2)^\circ\sin\ 2^\circ}{2\sin\ 2^\circ}$
= $\dfrac{\displaystyle\sum_{n=1}^{15} \cos(4m-2-2)^\circ-\cos(4m-2+2)^\circ}{2\sin\ 2^\circ}$
=  $\dfrac{\displaystyle\sum_{n=1}^{15} \cos(4m-4)^\circ-\cos(4m)^\circ}{2\sin\ 2^\circ}$
=  $\dfrac{\cos\ 0^\circ-\cos\ 60^\circ}{2\sin\ 2^\circ}$
=  $\dfrac{1 -\dfrac{1}{2}}{2\sin\ 2^\circ}$
=  $\dfrac{1}{4\sin\ 2^\circ}$

2014/069 Given $P(x)=x^3-6x^2+17x$. If $x=m$, $P(m)=16$ and when $x=n$, $P(n)=20$. Evaluate $m+n$.

we have $P(x) = x^3 - 6x^2 + 17x$

so we have $P(x+2) = x^3 + 5x + 18$ ( I take x+2 to eliminate the $x^2$ term to see in case we get odd function)

now
$P(m) = (m-2)^3 + 5(m-2) + 18 = 16$

or $(m-2)^3 +5 (m-2) = - 2\cdots (1)$

$P(n) = (n-2)^3 + 5(n-2) + 18 = 20$

or $(n-2)^3 + 5(n-2) = 2\cdots (2)$

from (1) and (2) as

$f(x) = x^3 + 5x$

$f(m-2) + f(n-2) = 0$ so $ m- 2 + n -2 = 0$

or $m+n = 4 $

2014/068) which is greater $\cos(\sin\, x)$ or $\sin(\cos\, x)$

we have
$\cos(\sin\, x)-\sin(\cos\, x)$
= $\cos(\sin\, x)-\cos(\dfrac{\pi}{2}- \cos\, x)$
= $\cos(\sin\, x)+\cos(\dfrac{\pi}{2}+ \cos\, x)$
=  $2 \cos \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2} \cos \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$
=  $2 \cos\ t \ \cos \ s$
where $t= \dfrac{\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$
and $s=  \dfrac{-\sin\, x+\dfrac{\pi}{2}+ \cos\, x}{2}$

is product of 2 terms and we can show that both t and s are between 0 and $\frac{\pi}{2}$

we have $\sin\, x\pm \cos\, x= \sqrt{2}\sin ( x\pm \dfrac{\pi}{4})$ so between $- \sqrt{2}$ and $\sqrt{2}$

hence s and t both are positive and less than $\frac{\pi}{2}$ so the difference $\cos(\sin\, x)-\sin(\cos\, x)\gt \ 0 $ and hence
$\cos(\sin\, x) \gt \sin(\cos\, x)$

2014/067) Show that $\dfrac{1}{n} + \dfrac{1}{n+1} + \dfrac{1}{n+2}$ is a decimal fraction of deferred periodicity

Proof:
above sum is

$\dfrac{3n^2+ 6n + 2 }{n(n+1)(n+2)}$

numerator is not divisible by 3 but denominator is divisible

so it is periodic as 3 is not a factor of 10

again if n id odd then numerator odd but denominator is even so 2 divides denominator but not numerator so it is deferred

if n is even $3n^2+6n$ is divisible by 4 but 2 is not so numerator is not divisible by 4 but denominator is divisible by 4 (actually 8) so in reduced for numerator is odd and denominator is even so if is deferred

2014/066) prove the trigonometric identity $\cos\ 36^\circ - \cos\ 72^\circ = \frac{1}{2}$

proof:
 $\cos\ 36^\circ - \cos\ 72^\circ$
= $2\sin\ 54^\circ \sin\ 18^\circ$
= $\dfrac{1}{2}\dfrac {\sin\ 2*54^\circ \sin\ 2*18^\circ}{\cos\ 54^\circ \cos\ 18^\circ}$
=  $\dfrac{1}{2}\dfrac {\sin\ 108^\circ \sin\ 36^\circ}{\cos\ 54^\circ \cos\ 18^\circ}$
=  $\dfrac{1}{2}\dfrac {\cos\ 18^\circ \sin\ 36^\circ}{\sin\ 36^\circ \cos\ 18^\circ}$
= $\dfrac{1}{2}$

2014/065) if x,y,z are sides of a triangle prove that

$x^3+ y^3 + 3xyz \gt z^3$
proof:
we have as x,y,z are sides of triangle $x \gt z- y$
cube both sides to get
$x^3 \gt z^3-y^3-3yz(z-y)$
or
$x^3+y^3 + 3yz(z-y) \gt z^3$
as $z - y \lt x$ we get
$x^3+y^3+3xyz \gt z^3$

2014/064) Find the sum of the cubes of the roots of cubic 3x^3-2x+4=0?

if a,b c are roots then
a+b+ c = 0 ... (1) (coefficient of $x^2$)

$3a^3 = 2a - 4 \cdots  (2)$
$3b^3 = 2b - 4 \cdots  (3)$
$3c^3 = 2c- 4 \cdots (4)$
add to get
$3(a^3+b^3+c^3) = 2 (a+b+c) - 12 = -12$ ( from (1))
or
$a^3+b^3 + c^3 = - 4$

2014/063) solve (x+2)(x+3)(x+8)(x+12) = 4x^2

we see that 2 * 12 = 3 *8
so group as follow

(x+2)(x+12)(x+3)(x+8)=4x^2

=> (x^2 + 14 x + 24) (x^2 + 11x + 24) = 4x^2
let x^2 + 11x +24 be y
(y+ 3x) y = 4x^2
y^2 + 3xy - 4x^2 = 0
(y- x)(y+ 4x) = 0

y =x => x^2 + 11x + 24 = x => x^2 + 10x + 24 = 0 => (x+4)(x+6) = 0 => x = -4 or - 6

y+ 4x = 0 => x^2 + 15x + 24 = 0

this can be solved for complex solutions

x = -15/2 - SQRT(129)/2 or x = -15/2 + SQRT(129)/2

2014/062) Find the value of : sqrt(3)*( 1/sin20 ) - ( 1/cos20 )?

= (sqrt(3) cos 20 - sin 20)/ ( cos 20 sin 20)
= 4(( sqrt(3)/2 cos 20 - 1/2 sin 20)/( 2 cos 20 sin 20) (mutliply denominator and numerator by 2)
= 4( sin 60 cos 20 - cos 60 sin 20)/( sin 40) ( knowing sin 60 = sqrt(3)/2 and cos 60 = 1/2)
= 4 sin (60-20)/ sin 40 ( sin (A-B) formula)
= 4 sin 40/ sin 40
= 4