2014/116) What is the value of m in the quadratic equation $x^2-mx+4=0$ to have real and equal roots.

equal roots

discriminant = 0 ($b^2-4ac, a = 1 b = -m . c = 4$)

$m^2 - 4*4 = 0$ or m= 4 or -4

m = 4 means $^2-4x+4 = 0$   oots are 2 and 2
m = -4 means $^2+4x+4 = 0$r  roots 2 and -2

roots are real as well


refer to https://in.answers.yahoo.com/question/index?qid=20111021024215AACQbhH

2014/115) how many factors of 240 are therer which are of the form 4n+ 2

240 can be factored as $2^4 * 3 * 5$

now for a factor to be of the form it should be 2 multiplied by by an odd number,
so it is 1 * (1+1) * (1+1) ( one way of choosing 2, 2 ways of choosing power of 3 ( $3^0, 3^1$) and same way 2 ways of choosing 5) or it is 4

2014/114) solve the equation $(x-3)^4 + (x-7)^4 = 24832$

Before I provide the solution I would like to mention that generally a quartic polynomial is not easy to solve but this type of equation

$(x-a)^4 + (x-b)^4= c$ can be converted to a quadratic equation by transformation of

$y= \dfrac{(x-a) + (x-b)}{2}$

as below

we shall put

$y= \dfrac{(x-3) + (x-7)}{2}$ = x- 5

so we get
$(y+2)^4 + (y-2)^4 = 24832$
or $2(y^4 + 6 y^2 (-2)^2 + 16) =  24832$
or $y^4 + 24 y^2 = 12400$
$y^4 + 24 y^2 – 12400 = 0$
 or $(y^2 – 100)(y^2 + 124)  = 0$
so $y^2 = 100$
so $y = \pm 10$

or x = -5 or 15

2014/113) find the number of solutions of $\sin\,x = \dfrac{x}{100}$

first let us look at positive x and same number of solutions shall be for -ve x

as $\sin,x$ and $x$ both are odd functions.

as $\sin\,x$ is less than 1 so x shall be less than 100 and so if we draw a sin  curve there shall be $\dfrac{100}{2\pi}$ or 15.91 ( around 16) units so there shall be 16 parts( 1/2 oscilaltions) in positive side and 16 in -ve side.

for each on the curve y = x shall intersect the sin curve 2 times so 32 in positive side including zero) so there is 1 value at 0, 31 positive values and 31 -ve values or 63 values

2014/112) if x+y is divisble by 3 show that $x^3+y^3$ is divisible by 9

we have
$x^3+y^3= (x+y)^3 - 3xy(x+y)$
if $(x+y)$ is divisible by 3 then $(x+y)^3$ is divisible by 9 ( it is divisible by 27 but 9 is required) and $3xy(x+y)$ is divisible by 9 and hence the difference.

2014/111) Show that the function below has same remainder when divided by $x(x+1)$ and $x(x+1)^2$

$f(x)=2008+2007x+2006x^2+\cdots+3x^{2005}+2x^{2006}+x^{2007}$

1st we provide the premise

this shall have same remainder when divided by $x(x+1)$ and $x(x+1)^2$

provided this shall have same remainder when divided by $(x+1)$ and $(x+1)^2$

when we divide by (x+1)^2 we shall have a linear polynomial say m(x+1) + c and when we divide by (x+1) it shall be d and both are same if m= 0 and m+c = d

so if we convert

f(x) as a polynomial of (x+1) the coefficient of x should be zero

of f(x-1) should have coefficient of x to be zero

now we provide the solution based on premise

$f(x-1)=2008+2007(x-1)+2006(x-1)^2+\cdots$
   $+3(x-1)^{2005}+2(x-1)^{2006}+(x-1)^{2007}$

should have coefficient of x to be zero

the coefficient of x

$= 2007 + 2006 * (-2) + 2005 * 3 \cdots + 2 * 2006 - 2007 = 0$

hence proved

I have solved the problem at http://mathhelpboards.com/challenge-questions-puzzles-28/division-polynomial-13760.html#post65389 where you can find some more different solutions

2014/110) Find closed form of

 $\cos\,1^\circ \cos\,2^\circ + \cos\,2^\circ \cos\,3^\circ\cdots \cos\,88^\circ \cos\,89^\circ$

we have $2 \cos(x) \cos (y) = \cos (x+y) + \cos (y-x)$

so $2 (\cos\,1^\circ \cos\,2^\circ ) = \cos\,3^\circ + \cos\,1^\circ$
 $2 (\cos\,2^\circ \cos\,3^\circ ) = \cos\,5^\circ + \cos\,1^\circ$
so on till
$2 (\cos\,44^\circ \cos\,45^\circ ) = \cos\,89^\circ + \cos\,1^\circ$

 $2 (\cos\,45^\circ\cos\,46^\circ ) = \cos\,91^\circ + \cos\,1^\circ$

or  $2 (\cos\,45^\circ \cos\,46^\circ ) = - \cos\,89^\circ + \cos\,1^\circ$


so on till

$2 (\cos\,88^\circ \cos\,89^\circ ) = - \cos\,3^\circ + \cos\,1^\circ$

on adding above for each positive term in 1st half for the first term there is a -ve term for the second half and we are left wth 88 times $\cos\,1^\circ$

so $2( \cos\,1^\circ \cos\,2^\circ + (\cos\,2^\circ \cos\,3^\circ +\cdots \cos\,88^\circ \cos\,89^\circ)  = 88 \cos\,1^\circ$

or  $( \cos\,1^\circ \cos\,2^\circ + (\cos\,2^\circ \cos\,3^\circ +\cdots \cos\,88^\circ \cos\,89^\circ)  = 44 \cos\,1^\circ$

problem 2014/109) prove that $\dfrac{\sin\,x - \cos\, x + 1}{\sin\, x + \cos\, x – 1} = \dfrac{1 + \tan \frac{x}{2}}{1- \tan\frac{x}{2}}$

Proof

convert it into $\dfrac{x}{2}$ form

numerator = $2 sin \dfrac{x}{2}\cos\dfrac{x}{2} + 2 \sin ^2\dfrac{x}{2}$ (as $\cos\, x= 1 - 2 \sin ^2 \dfrac{x}{2}$) = $2 \sin \dfrac{x}{2}( \cos \dfrac{x}{2} + \sin \dfrac{x}{2})$
denominator = $2 \sin \dfrac{x}{2} \cos \dfrac{x}{2} - 2 \cos^2 \dfrac{x}{2}$ (as $cos\, x= 2 \cos^2 \dfrac{x}{2} - 1$ = $2 \sin \dfrac{x}{2}(\cos \dfrac{x}{2}- \sin \dfrac{x}{2})$
dividing
we get
$\dfrac{\sin\,x - \cos\,x + 1}{\sin\,x + \cos\,x - 1} = \dfrac{\cos \dfrac{x}{2} + \sin \dfrac{x}{2}}{\cos\dfrac{x}{2}-\sin \dfrac{x}{2}}$
= $\dfrac{(1 + \tan \dfrac{x}{2})\cos\dfrac{x}{2}}{(1 - \tan \dfrac{x}{2})\cos\dfrac{x}{2}}$
= $\dfrac{1 + \tan \dfrac{x}{2}}{(1 - \tan \dfrac{x}{2}}$

2014/108) Find the remainder when $2^{1990}$ is divided by 1990

We have 1990 = 10*199 = 2* 5* 199

now we have

$2^4$ mod 5 = 1 as per fermat theorem
so $2^{1990}$ mod 5 = $2^2$ mod 5 = 4 mod 5

$2^{ 198} = 1$ mod 199

so $2^{1990}$ mod $199$ = $2^10$ mod 199 = 1024 mod 199 = 29 mod 199

$2^{1990}$ mod 2 = 0

so $2^{ 1990}$ mod 199 = 29
$2^{1990}$ mod 5 = 4

using Chinese Remainder Theorem you can proceed

continuing further

$2^{1990} = 0$ mod 2
$2^{1990} = 4$ mod 5

this gives $2^{1990} = 4$ mod 10

now using
$a = 4 (mod\, 10)$
$a = 29 (mod\, 199)$

Notice that 20*10+(-1)*199=1, thus 20*10≡1 (mod 199) and -199≡1 (mod 10)

let a= 29*(20•10)+4*(-199), then it is clear that
a = 29*(0)+4*(1)=4 (mod 10) and a=29•(1)+4•(0)=29 (mod 199) so this a works for what we want.

a= 5004=1024 (mod 1990)

so the remainder is 1024

Q2014/107) find coefficient of a^2 in the following expression

 $(1-a^2)+(1-a^2)^2+(1-a^2)^3+(1-a^2)^4+(1-a^2)^5+(1-a^2)^6$
       $+(1-a^2)^7+(1-a^2)^8+(1-a^2)^9+(1-a^2)^{10}+(1-a^2)^{11}+(1-a^2)^{12}$

if we put $x= (1-a^2)$
we get given expression as
$x+x^2+x^3+x^4+x^5+x^6+x^7+x^8+x^9+x^{10}+x^{11}+x^{12}$= $x\dfrac{1-x^{12}}{1-x}$
= $(1-a^2)\dfrac{1-(1-a^2)^{12}}{a^2}$
= $\dfrac{(1-a^2)-(1-a^2)^{13}}{a^2}$

now $\dfrac{1-a^2}{a^2}$ shall not contribute to $a^4$ so we need to find the coefficient of $a^6$ in $-(1-a^2)^{13}$ which is ${13\choose 3}$

that is the ans

2014/106) find sum of real roots of the following

$2x^8-9x^7+20x^6-33x^5+46x^4-66x^3+80x^2-72x+32=0$
we have
$2x^8-9x^7+20x^6-33x^5+46x^4-66x^3+80x^2-72x+32=0$
= $2(x^8+ 16) -9(x^7+8x) +20(x^6+ 4x^2) -33(x^5+2x^3) + 46x^4=0$
or deviding by $x^4$ as x = 0 is not a solution  we get
$2(x^4 + (\dfrac{2}{x})^4)-9(x^3 + (\dfrac{2}{x})^3)+20((x^2 + (\dfrac{2}{x})^2)-33(x (+\dfrac{2}{x}))+46 = 0$

now if we put  $x+\dfrac{2}{x}=t$
we get
$x^2+(\dfrac{2}{x})^2=t^2-4$
$x^3+(\dfrac{2}{x})^3=t^3-6t$
$x^4+(\dfrac{2}{x})^4=t^4-8t + 8$

so given relation reduces to
$2(t^2-8t^2 +8) -9(t^3-6t) +20(t^2-4) - 33t + 46= 0$
or $2t^4-9t^3+4t^2+21t-18=0$

now we see that t = 1 and t = 3 are solutions and hence we get
 $2t^4-9t^3+4t^2+21t-18=0$
= $2t^3(t-1) - 7t^2(t-1) -3t(t-1) + 18(t-1)=0$
or$(t-1)(2t^3-7t^2-3t+18) = 0$
gives a solution t = 1

or
$2t^3-7t^2- 3t + 18 = 0$ as 3 is a root we get
$2t^2(t-3) - t(t-3) - 6(t-3) = 0$
or $(t-3)(2t^2 - t^2-3) = 0$
so t = 3
or $2t^2 - t - 3 = 0 $
or $(2t-3)(t+1) = 0$

so t = 1 or 3 or - 1 or $-\dfrac{3}{2}$

now $t = x+ \dfrac{2}{x}$ and if x is positive then by AM GM inequality lowest value = $2\sqrt{2}$
or only possible value from above is

t = 3 (as t cannot be between  -$2\sqrt{2}$ and $2\sqrt{2}$)

t = 3 gives x = 1 or 2 and so sum of real roots = 3

I had the privilege to solve it at http://mathhelpboards.com/challenge-questions-puzzles-28/find-sum-real-roots-13522.html#post64344