2016/085)Let n be a positive integer. if $2^{nd}$ , $3^{rd}$ and $4^{th}$ term of expansion of $(1+x)^n$ are in AP find n

we are given
$ {n \choose 1} +  {n \choose 3}=  2{n \choose 2}$
or
$n + \frac{n(n-1)(n-2)}{6} = 2\frac{n(n-1)}{2}$
dividing both sides by $n$ we get
$ 1  + \frac{(n-1)(n-2)}{6} = (n-1)$
or $ (n-2)(1-\frac{n-1}{6} = 0$ giving n= 2 or 7
but $n > =3$ so n = 4

2016/084) If $\cos(x-y) = \frac{4}{5}$ and $\sin(x+y) = \frac{5}{13}$ and x,y are between 0 and $\frac{\pi}{4}$ then find $\tan (2x)$

 we have $\cos(x-y) = \frac{4}{5}=> \sin (x-y) = \sqrt{1-(\cos(x-y))^2} = \sqrt{1-\frac{16}{25}} = \frac{3}{5}$
 and  $\sin(x+y) = \frac{5}{13}=> \cos (x+y) = \sqrt{1-(\sin(x+y))^2} = \sqrt{1-\frac{144}{169}} = \frac{12}{13}$

giving $\tan (x-y) = \frac{\sin (x-y}{\cos(x-y)} = \frac{\frac{3}{5}}{\frac{4}{5}} = \frac{3}{4}$
and $\tan (x+y) = \frac{\sin (x+y}{\cos(x+ y)} = \frac{\frac{5}{13}}{\frac{12}{13}} = \frac{5}{12}$
hence
$\tan(2x) = \tan (x+y+x-y) = \frac{\tan (x+y) + \tan (x-y)}{1- \tan (x+y)\tan (x-y)} = \frac{\frac{5}{12} + \frac{3}{4}}{1- \frac{5}{12}\frac{3}{4}} = \frac{46}{33}$

2016/083) Find the smallest positive p such that $\cos(p\sin\, x) = \sin (p\cos\,x)$

we have
$\cos(p\sin\, x) = \sin (p\cos\,x)= \cos (\frac{\pi}{2} -  p\cos\,x)$
hence
$p\sin \,x  = \frac{\pi}{2} - p \cos \,x$ (other values shall given -ve / larger p)
hence
$p(\cos \,  x + \sin\, x) = \frac{\pi}{2}$
or $\sqrt{2}p( \sin \frac{\pi}{4} \cos \, x + \cos \frac{\pi}{4} \sin \, x) = \frac{\pi}{2}$
or $\sqrt{2}p( \sin ( x + \frac{\pi}{4}) = \frac{\pi}{2}$
the largest value of $\sin ( x + \frac{\pi}{4})$ is 1 hence smallest $p$ is $\frac{\pi}{2\sqrt{2}}$

2016/082) Let P(x) be a polynomial of x. Show that there exists a polynomial Q(x) such that P(x)Q(x) is a polynomial of $x^3$

We have
$P(x) = \sum_{n=0}^{\infty} a_nx^n$
this can be written as $A(x) + x B(x) + x^2C(x)$
where
$A(x) = \sum_{k=0}^{\infty} a_{3n}x^{3n}$
$B(x) = \sum_{k=0}^{\infty} a_{3n+1}x^{3n}$
$C(x) = \sum_{k=0}^{\infty} a_{3n+2}x^{3n}$
Now let
$R(x) =  A(x) + x B(x)w + x^2C(x)w^2$ where w is cube root of one
and
$S(x) = A(x) + x B(x)w^2 + x^2C(x)w$
using $(a+b+c)(a+bw^2+cw)(a+bw+cw^2) = a^3+b^3+c^3 - 3abc$
we get $P(x)R(x)S(x) = (\sum_{k=0}^{\infty} a_{3n}x^{3n})^3 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^3 x^3 + (\sum_{k=0}^{\infty} a_{3n+2}x^{3n})^3x^6$
$- 3(\sum_{k=0}^{\infty} a_{3n}x^{3n})(\sum_{k=0}^{\infty} a_{3n+1}x^{3n})(\sum_{k=0}^{\infty} a_{3n+2}x^{3n})x^3$
which is a polynomial of $x^3$
now using  $(a+bw+cw^2)( a+ bw^2 +cw) = (a^2+b^2+c^2-ab-bc-ca)$
we have
$R(x)S(x) = (A(x) + x B(x)w + x^2C(x)w^2)(A(x) + x B(x)w^2 + x^2C(x)w)$
$= (\sum_{k=0}^{\infty} a_{3n}x^{3n})^2 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^2)x^2 + (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})^2x^4$
   $- (\sum_{k=0}^{\infty} a_{3n+1}x^{3n}) (\sum_{k=0}^{\infty} a_{3n+1}x^{3n})x$
  $- (\sum_{k=0}^{\infty} a_{3n+1}x^{3n}))(\sum_{k=0}^{\infty} a_{3n+2}x^{3n})x^3$
  $-(\sum_{k=0}^{\infty} a_{3n+2}x^{3n}\sum_{k=0}^{\infty} a_{3n}x^{3n})x^3$
multiplying by the above polynomial say Q(x) we get a polynomial of $x^3$

2016/081) find the number of integer-sided isosceles obtuse angled triangles with perimeter 2008.

let the sides by x,x,y.
so 2x + y = 2008
now 2x >y so y < 1004.
y has to be even.
x cannot be the longest side becuase if x is longest we have isosceles triangle with 2 large angles so cannot be obtuse
so the 3 sides are (y= 1004-2k, x= 502+k, x= 502 + k) and $k < 502$ and k is positive
because it is obtuse
$x^2+x^2 < y^2$ or $2(502+k)^2 < (1004-2k)^2$
solving these we get $k < 502(3-2\sqrt{2})$ or  $k > 502(3+2\sqrt{2})$
because k < 502 so we heve  $k < 502(3-2\sqrt{2})$ giving $k < 86.1432$
so there are 86 triangles

2016/080)Find the coefficient $x^3$ in $1+ (1+x) + (1+x)^2 + (1+x)^3\cdots(1+x)^n$

we have $1+ (1+x) + (1+x)^2 + (1+x)^3\cdots(1+x)^n= \frac{(1+x)^{n+1} -1 }{1+x-1}$
$= \frac{(1+x)^{n+1} -1 }{x}$
so coeffient $x^3$ in $1+ (1+x) + (1+x)^2 + (1+x)^3\cdots(1+x)^n$ is coefficient of $x^4$ in $(1+x)^{n+1} -1$
or ${n+1 \choose 4}$

2016/079) The lines $x^2-3xy+2y^2=0$ are shifted parallel to themselves so that their point of intersection comes to (1,1). find the combined equation of lines in the new position.

we have $x^2-3xy+y^2=0$ represent two lines $(x-y)(x - 2y)= 0$ giving $(x=y=0)$. if we need point of
 intersection 1,1
then we need to replace x by x-1 and y by y-1 in given equation giving
$(x-1)^2 -3(x-1)(y-1) + 2(y-1)^2 = 0$
or $x^2 - 2x +1 -3xy + 3x + 3y -3 + 2y^2-4y + 2= 0$
or $x^2+2y^2 - 3xy + x - y= 0$

2016/078)if $x=cy+bz$, $y= az + cx$ and $z=bx+ay$ then show that $a^2+b^2+c^2 + 2abc = 1$

we put the equations in standard form
$x-cy - bz = 0\cdots(1)$                    
$cx -y + az = 0\cdots(2)$    
$bx + ay - z = 0\cdots(3)$                            
multiply (1) by c and subtract it from (2) to get
$(c^2-1) y + (bc+ a) z = 0\cdots(4)$
mulltiply (1) by b and subtract from (3) to get
$(a+bc) y - (1-b^2) z = 0\cdots(5)$
as (4) and (5) are consistant we get
$(c^2-1)(b^2 - 1) -   (bc+ a)^2 = 0$
or $b^2c^2 - b^2 - c^2 + 1 -  b^2c^2 - a^2 - 2bca$
or $a^2+b^2+c^2 + 2abc = 1$

2016/077) Prove that $1+\cos 2x + \cos 4x + \cos 6x$$ = 4\cos\,x \cos 2x \cos 3x$

$1+\cos 2x + \cos 4x + \cos 6x$
= $(1+\cos 2x) + (\cos 4x + \cos 6x)$
= $(2\cos^2 x) + (\cos 4x + \cos 6x)$ using $cos 2x = 2cos^2 x - 1$
= $(2\cos^2 x) + 2\cos x  \cos 5x)$ using $\cos\, A + \cos \, B = 2(\cos \frac{A+B}{2} +  \cos \frac{B- A}{2})$
= $2\cos\,x(\cos \, x + \cos 5x)$
=$2 \cos\,x(2 \cos 2x  \cos 3x)$ using $\cos\, A + \cos\, B = 2(\cos \frac{A+B}{ 2} +  \cos \frac{B- A}{ 2})$
= $4\cos \, x\cos 2x \cos 3x$

2016/076) If a,,b,c are 3 unequal terms of AP then show that $\frac{b-c}{a-b}$ is rational.

proof
let $1^{st}$ term be p and difference d. Let a be $q^{th}$,  b be $r^{th}$,  c be $s^{th}$
so $a = p+(q-1)t$
$b= p+(r-1)t$
$c= p + (s-1)t$
hence $b-c = (r-s)t$
and $a-b = (q-r)t$
or  $\frac{b-c}{a-b} =  \frac{q-r}{r-s}$
as $q,r,s$ are intgers rhs is rational and hence the result

2016/075) derive the following given $u_n$ the $n^{th}$ fibonacci number $u_{n+1}^2 - 4u_nu_{n-1} = u_{n-2}^2$

we have
$u_{n+1}^2 -  u_{n-2}^2$
$= (u_n + u_{n-1})^2 -  (u_n - u_{n-1})^2$
$ =4u_nu_{n-1}$ using $(a+b)^2 - (a-b)^2 = 4ab$
hence $u_{n+1}^2 - 4u_nu_{n-1} = u_{n-2}^2$

2016/074) you have 25 horses and you have to pick fastest 3 out of the 25. In each race only 5 horses can run at the same time as there are only 5 tracks. what is the minimum number of races to pick the 3 horses without using a stopwatch.

you make the horses into 5 groups of 5 horses each say A,B,C,D,E . have 5 races. then pick the winner of the 5 races and have
a race that is race number 6.. Now pick the 3
winners out of the 5. Say the winner is from group A, the 2nd ranked is from Group B and 3rd one is from group C.
now let the 1st 3 postions in group A be A2,A2,A3. in group B be B1,B2,B3. and in group C be C1,C2, C3.
A1 is the 1st. B3 cannot be in top 3 beacuse A1, B1, B2 are faster.
So B3 is ruled out.
C2 and C3 cannot be in top 3 as A1,B1,C1 are faster. So have a race among A2,A3,B1,B2,C1 and choose the 2 fastest. the ranks
shall be 2nd and 3rd.
So we need 7 races.

2016/073) $\frac{1}{x} + \frac{1}{y} +\frac{1}{z}=1$ what is the value of $x,y$ and $z$ if $x>y>z$ and they are integers

$\frac{1}{x} + \frac{1}{y} +\frac{1}{z} > 3\frac{1}{z} = \frac{3}{z} = 1$

hece $ z < 3$ and as z cannot be 1 so $z = 2$
hence $\frac{1}{x} + \frac{1}{y} =\frac{1}{2}$
hence $2y+ 2x = xy$
hence $xy-2y-2x = 0$
or $xy-2x-2y + 4 = 4$
or $(x-2)(y-2) = 4$
as $x> y$ so we get $x = 2 = 4, y=2 =1$ giving $x = 6, y= 3$
hence $z=2,y=3,x = 6$