2015/030) Factor $x^4 – 2ax^2+x+a^2-a$

It is not easy to solve a quartic equation but we see that it is quadratic in a

= $a^2 – a(2x^2+ 1) + x^4 + x$

now we try to factor $x^4 + x$ such that sum is $2x^2 + 1$

we have $x^4+ x = x(x^3+1) = x(x+1)(x^2- x + 1) = (x^2 + x)(x^2-x + 1)$

so the 2 factors are $x^2 + x$ and $x^2-x + 1$ as we see that sum is $2x^2+ 1$

so the original expression factors to $(a-x^2-x)(a-x^2+x-1)$ or $(x^2+x-a)(x^2-x+1-a)$

2015/029) Show that $(a\ cos\theta + b\sin\theta)^2 <= a^2 + b^2$

This follows directly from

 $(a\ cos\theta + b\sin\theta)^2 +(a\ sin\theta - b\cos\theta)^2= a^2+ b^2$

2015/028) Solve in positive integers $3x + 5y = 2xy – 1$

Let is get y in terms of x
we have $3x + 1 = 2xy – 5y = y(2x-5)$
so $y =\dfrac{3x+1}{2x-5} = 1 + \dfrac{x+6}{2x-5}$

now  $\dfrac{x+6}{2x-5}$ should be positive

 

so $(x+6)(2x-5) \gt 0$ and $x \gt 0$

so we get $x \lt - 6$ or $x > 2.5$ but as $x \gt 0$ we have $x \gt 3.5$

and $(x + 6) \ge 2x – 5$ or $x \le 11$ else we shall have y non positive

checking for x between 3 and 11 we get solutions are (11,2) and (3,10)

2015/027) (2x +3y) is divisible by 17, then for what value of k ; (9x +ky) is divisible by 17?

options :
a)3
b)5
c)7
d)9

Solution 2015/027)

(2x+ 3y) is divisible by 17

so 2x + 20y is divisible by 17
so x + 10y is divisible by 17
so 9x + 90y is divisible by 17
so k = 90 mod 17 = 5
I have solved differently at https://in.answers.yahoo.com/question/index?qid=20131025093911AAFGa34

2015/026) Given the 2 polynomials

$x^2 + ax + 1 = 0$
$x^2 + x + a = 0$

Find the value of a for which these have at least one root

solution

if both have a common root it shall satisfy the difference of the 2 polynomials

or ax – x + 1 – a = 0 or (x-1)(a-1) =0
so a = 1 or x = 1 giving a = -2

2015/025) Show that $7^{2n} - 48n - 1$ is divisible by 2304

proof:

$7^{2n} - 48n - 1 = 49^n - 48 n - 1 = (1+ 48)^n - 48n - 1$
= $1+ 48n + 48^2{n \choose 2}+ 48^2{n \choose 3}+ \cdots + 48^n - 48n - 1$
=  $48^2{n \choose 2}+ 48^2{n \choose 3}+ \cdots + 48^n$
and as each term is divisible by 48^2 so the value is dvisible by 48^2 or 2304

2015/024) find the sum $\sum_{n=6}^{+\infty}\dfrac{2}{n(n+2)}$

this is telescopic sum
we have
$\dfrac{2}{n(n+2)}= \dfrac{1}{n}- \dfrac{1}{n+2}$

as n goes to inifinite the term  $\dfrac{1}{n}$ converges to zero and when we add we are left with

$\dfrac{1}{6}+\dfrac{1}{7}= \dfrac{13}{42}$

2015/023) show that $\tan\, 9^{\circ} - \tan\, 27^{\circ}- \tan\, 63^{\circ} + \tan\, 81^{\circ}= 4$

we have $\tan\, x + \cot\, x = \tan\, x + \dfrac{1}{\tan\, x}$

 = $\dfrac{tan ^2 x + 1}{\ tan\, x} = \dfrac{sec^2 x}{\tan,x}= \dfrac{1}{\cos\,x\sin\,x}=\dfrac{2}{2\cos\,x\sin\,x} = \dfrac{2}{\sin 2x} = 2 \csc 2x$

 so

 $\tan\, 9^{\circ} + \tan\, 81^{\circ} = 2 \csc\, 18^{\circ} $

$\tan\, 27^{\circ} + \tan\, 63^{\circ} = 2 \csc\,  54^{\circ} $

so given value
= $\tan\, 9^{\circ} - \tan\, 27^{\circ}- \tan\, 63^{\circ} + \tan\, 81^{\circ}=2( \csc\, 18^{\circ} - \csc\, 54^{\circ})$ 

= $2(\dfrac{1}{\sin\,18^{\circ}} - \dfrac{1}{\sin\, 54^{\circ}})$

= $2 \dfrac{2 \cos\, 36^{\circ}  \sin\, 18^{\circ}}{ \sin \,18^{\circ}  \sin \,54^{\circ} }$
=  4

refer to https://in.answers.yahoo.com/question/index?qid=20111213202340AABYrgC

2015/022) If ab + bc + bc +cd = 16, what will at least $a^2 + b^2 + c^2 + d^2$ be?

we have

$(a-b)^2 = a^2 + b^2 - 2ab$
$(b-c)^2 = b^2 + c^2 - 2bc$
$(c-d)^2 = c^2 + d^2 - 2cd$
$(d-a) ^2 = d^2+a^2 - 2ad$

adding $(a-b)^2 + (b-c)^2 + (c-d)^2 - (d-a)^2 = 2(a^2 +b^2 +c^2 + d^2) - 2(ab+bc+cd+da)$

or $2(a^2 +b^2 +c^2 + d^2) = (a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2 + 2(ab+bc+cd+da)$

or $(a^2 +b^2 +c^2 + d^2) = \dfrac{(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2)}{2}+ (ab+bc+cd+da)$
= $\dfrac{(a-b)^2 + (b-c)^2 + (c-d)^2 + (d-a)^2)}{2}+ 16$

 

 clearly is lowest when a= b=c=d and value= 16

refer to https://in.answers.yahoo.com/question/index?qid=20120108032725AAASGBv

2015/021) If $x^3 + y^3 + z^3 = 81$ ,then prove that $x + y +z\le 9$

Applying C00auchy schwarz inequality to (x,y,z) and (1,1,1) we have
$(x+y+z)^2 \le (1+1+1)(x^2+y^2+z^2)$
or $(x+y+z)^2 \le 3(x^2+y^2+z^2)$
hence $(x+y+z)^4 \le 9(x^2+y^2+z^2)^2$
applying same inequality to $\sqrt{x},\sqrt{y},\sqrt{z}$ and  $\sqrt{x^3},\sqrt{y^3},\sqrt{z^3}$

 we get :
$(x^2+y^2+z^2)^2\le (x+y+z)(x^3+y^3+z^3)$
or $(x^2+y^2+z^2)^2 \le 81(x+y+z) \cdots (2)$

 From (1) and (2)...
$(x+y+z)^4 \le 9 * 81 (x+y+z)$

or $(x+y+z)^3\le 9^3$

or $(x+y+z)\le 9$
refer to https://in.answers.yahoo.com/question/index?qid=20120510235300AA1XWbg