2016/109) Let $-1 <=p <=1$. Show that the equation $4x^3-3x-p=0$ has a unique root in the interval $[\frac{1}{2}, 1]$ find it.

because p is between -1 and 1 so $cos^{-1}p$ is defined. let $x = \cos\,t$ so we get
$\cos\, 3t = p$
and hence $ 3t = \arccos p $ and so $x = \frac{1}{3}\cos^{-}p$
if $x > 1$ we get $4x^3-3x$ > 1 and if $x < \frac{1}{2}$ we get $4x^3 - 3x < -1$ and out side the
limit so x is between $\frac{1}{2}$ and 1

2016/108) If $S_n$ is sum of n terms of a GP show that $S_n(S_{3n} - S_{2n}) = (S_{2n} - S_{n})^2$

we have let 1st term be a and ratio be t
so $S_p = a\frac{t^{p}- t}{t-1}$
we have LHS
$= (a\frac{t^{n}- 1}{t-1})(\frac{a(t^{3n}- 1) -  a(t^{2n}- 1)}{t-1}$
$= \frac{(at^n-1)(a(t^{3n} - t^{2n}}{(t-1)}^2$
$= \frac{a^2t^2(t^n-1)^2}{(t-1)^2}$
$=(\frac{at^n(t^n-1)}{t-1})^2$
RHS = $(S_{2n} - S_{n})^2 = (\frac{at^{2n-1} - 1 - at^n + 1}{t-1})^2 =    (\frac{at^{2n}-at^n}{t-1})^2=LHS$

2016/107) Solve the system of equations

$xy+3y^2-x + 4y-7 = 0$

$2xy+y^2 -2x - 2y +1 = 0$

Solution
multiply the 1st one by 2 and subtract 2nd one to get
$5y^2 + 10y -15 = 0$
or $y^2+2y -3 = 0$ or $(y+3)(y-1)= 0$
y = -3 or y = 1
put y = -3 in 1st equation to get
$-3x + 27  - x - 12 -7 = 0$ or $ x= 2$
now check the 2nd one to get $ -12 + 9 -4 + 6 +1 = 0$ so 2nd one is satisfied and somution = $(x,y) = (2, -3)$
put y =1 in 1st one to get
$ -x + 3 -x + 4 -7$ or $x = 0$
put x= 0 and y = 1 in 2nd one and it is satisfied so solution $(x,y) = (0,1)$

2016/106) solve for x $(x^2+2)^2 + 8x^2 = 6x(x^2+2)$

taking the term on RHS to LHS
 $(x^2+2)(x^2 - 6x + 2) + 8(x^2)  = 0$
$(x^2-3x+2 + 3x )(x^2 - 3x + 2 - 3x) + 8x^2 = 0$
or $(x^2-3x+2)^2  - 9x^2 + 8x^2 = 0$
or $(x^2+ 3x +2)^2 - x^2 = 0$
or $(x^2+2x+ 2)(x^2+4x+2) = 0$
$x^2+2x+2=0$ gives $x= -1\pm i$ and $x^2+4x+2=0$ give $2\pm \sqrt{2}$

2016/105) Solve $x^4+4x^3+3x^2-14x+26=0$

As it is found that this does not have a rational factor and as we know that complex roots appear as conjugate pairs
so it can be factored to quadratic with real coefficient
so Let $x^4+4x^3+3x^2-14x+26$
$ = (x^2+px + q) (x^2 + sx + t)$ as coeffiefint of $x^4$  = 1
$= x^4 + x^3(s + p) + x^2(ps + q + t) + x(pt + qs) + qt$
Comparing coefficients
$s+ p = 4$
$ps +q + t = 3$
$pt+qs = - 14$
$qt = 26 => ( 1,26),(-1,-26),(2,13), (-2, - 13)$
it can be easily solved if we have rational coefficient (above 4 combinations)
by taking $q = 1, t = 26$ or $q = -1 t = -26$ you shall not find any solution(I have tried) but
if we take $q = 2, t = 13$ we get
$s+ p = 4 \cdots(1)$
$sp + 15 = 3$ OR  $ps = -12 \cdots(2)$
$13p + 2s = - 14 \cdots(3)$
From (1) and (3) you get p = - 2 and s = 6 and it satisfies (2)
So we get quadratic factor as $(x^2-2x+2)(x^2+6x+13) = 0$
So we are lucky to get factors with rational coefficients
$(x^2-2x+2) = 0$ gives 2 roots $1\pm i$
And $(x^2+6x+13) = 0$  gives 2 roots $-3\pm 2i$

2016/104) Find all real numbers a such that 3 $< a < 4$ and $a(a - 3\{a\})$ is an integer. where $\{x\}$ is fractional part

let a  = 3 + x
so we have (3+x) (3+x-3x) = (3+x)(3-2x) = n an integer
$9 - 3x -  2x^2 = $ integer
$2x^2 + 3x -k = 0$
the values can be 1 or 2 or 3 or 4
giving the value of x = $\frac{\sqrt{9+8k}-3}{4}$
putting k = 1 to 4 we get a = $3 + \frac{\sqrt{17}-3}{4}$,$ 3 + \frac{1}{2}$,$3 + \frac{\sqrt{33}-3}{4}$, $3 + \frac{\sqrt{41}-3}{4}$

2016/103) if p is natural number show that $p^{n+1} + (p+1)^{2n-1}$ is divisible by $p^2+p+1$

let $f(n) = p^{n+1} + (p+1)^{2n-1}$
we shall prove it principle of mathemetical induction
for n = 1 we have  $f(1) = p^2 + p+ 1 $ and obviously it is divsible by $p^2+p+1$
so we have shown it for base step
let it be dvisible for n = k so $p(k)$ is divisible by $p^2+p+1$
or $p^{k+1} + (p+1)^{2k-1} =  q(p^2+p+1)$
Now we need to show that it is true for k+ 1
$p^{k+2} + (p+1)^{2k+1} = p^(k+2) + (p+1)^2 (p+1)^{2k-1}$
$ = p^{k+2}+ (p^2 + 2p + 1) (p+1)^{2k-1}$
$= p^{k+2} + p((p+1)^{2k-1}) + (p^2 + 2p + 1)((p+1)^{2k-1})$
$=  p(p^{k+1} + (p+1)^{2k-1} + (p^2 + 2p + 1)((p+1)^{2k-1})$
$ = p(q(p^2+p+1)) + p^2 + 2p + 1)((p+1)^{2k-1})$
which is divisible by $p^2+p+1$ hence induction step is proved.
hence proved

2016/102) if${n \choose r-1} = 36$, ${n \choose r} = 84$, ${n \choose r+1} = 126$ find the values of n and r

we have
$\frac{n!}{(r-1)!(n-r+1)!} = 36$, ,
$\frac{n!}{(r!(n-r)!} = 84$
$\frac{n!}{(r+1)!(n-r-1)!} = 126$
by dividing we get
$\frac{r}{n-r+1} = \frac{3}{7}$ or $7r = 3n - 3r + 3 $ or $3n- 10r + 3 = 0$
and  $\frac{r+1}{n-r} = \frac{2}{3}$ or $3(r+1) = 2(n -r)$ or $5r -2n +3=0$
multiplying 2nd by 2 and adding gto 1st we get 9-n = 0 so n = 9 and putting in (1) we get r = 3

2016/101) The circles $x^2+y^2-10x + 16 = 0$ and $x^2+y^2 = r^2$ intersect each other in distinct points

1) $r < 2$ 2) $r > 8$ 3) $2 < r < 8$ 4) $2 <=r <= 8$

Solution
The circle $x^2+y^2 = r^2$ has centre at (0,0) and radius r
the circle $x^2+y^2 - 10x + 16 = 0$
$=>(x^2-10x + 25) + y^2 = 9$
$=>(x-5)^2 + y^2 = 3^2$
this has center at (5,0) and radius 3.
the distance between points = 5
one has radius 3 and another shall intersect if radius be between 5-3 and 5+ 3 or $2 < r < 8$ and hence(3)

2016/100) find $lim_{ x-> inf} (\frac{x+6}{x+1})^{x+4}$

we have $lim_{ x-> inf}(\frac{x+6}{x+1})^{x+4}$
= $lim_{ x-> inf}(1+ \frac{5}{x+1})^{x+4}$
= $lim_{ x-> inf}(1+ \frac{5}{x+1})^{x+1} (1+ \frac{5}{x+1})^3$
= $e^5 * 1 = e^5$

2016/099) Let $a\,b$ and $c$ be the sides of a triangle. Prove that $\frac{a}{b+c-a}+\frac{b}{a+c-b}+\frac{c}{a+b-c}\ge 3$

if we put $x = b+c-a, y = a+c-b, z = a + b -c$ we get
given expression
$= \frac{1}{2}(\frac{y+z}{x} +\frac{z+x}{y} + \frac{x+y}{z})$
$= \frac{1}{2}((\frac{y}{x}+ \frac{x}{y})+( \frac{z}{y} + \frac{y}{z}) + ( \frac{z}{x} + \frac{x}{z}))$
$= \frac{1}{2}(((\sqrt{\frac{y}{x}}- \sqrt{\frac{x}{y}})^2+ 2) +((\sqrt{\frac{y}{z}}- \sqrt{\frac{z}{y}})^2+ 2) + ((\sqrt{\frac{z}{x}}- \sqrt{\frac{x}{z}})^2+ 2)$
$>= \frac{1}{2}(2+2+2)\,or\,3$