2016/013) Prove that $\cos(\tan^{-1}(\sin(\cot^{-1} x)))= \sqrt{\frac{x^2+1}{x^2+2}}$

Let $\cot^{-1} x = y$
so $\cot y = x$
so $\csc^2 y = x^2+1$
so $\sin  y = \frac{1}{\sqrt{x^2+1}}$
so $\sin\cot^{-1} x = \frac{1}{\sqrt{x^2+1}}$
similarly  $\cos \tan ^{-1}x = \sqrt{\frac{1}{x^2+1}}$
Hence $\cos(\tan^{-1}(\sin(\cot^{-1} x)))$
$=\cos(\tan^{-1}\frac{1}{\sqrt{x^2+1}})$
$=\frac{1}{1+\frac{1}{x^2+1}}$
$=\frac{1}{\sqrt{1+\frac{1}{x^2+1}}}$
$=\frac{1}{\sqrt{\frac{x^2+2}{x^2+1}}}$
$=\sqrt{\frac{x^2+1}{x^2+2}}$

2016/012) Suppose that $f(x+3)=3x^2+7x+4\cdots(1) $ and $f(x)=ax^2+bx+c\cdots(2)$ What is $a+b+c$

we have from 2nd equation $f(1) = a + b + c\cdots(3)$
Now we need to evaluate f(1)
as $f(x+3) = 3x^2 + 7x + 4$
so $f(x) = 3(x-3)^2 + 7(x-3)  + 4\cdots(4)$
putting x = 1 we get $f(1) = 3 * (-2)^2 + 7 * (-2) + 4 = 2$
so $a + b+ c = 2$

2016/011) Simplify $a^2(b+c)+b^2(c+a)+c^2(a+b)$ , if a,b,c are in Arithmetic Progression

$a^2(b+c)+b^2(c+a)+c^2(a+b)$
$= a^2(b+c+a) + b^2(c+a+b) + c^2(a+b+c) - (a^3+b^3+c^3)$
$= (a+b+c) (a^2+b^2+c^2) - ((a+b+c)(a^2+b^2 + c^2-ab-bc-ca) + 3abc)$
(because $a^3+b^3+c^3 - 3abc = (a+b+c) ( a^2+ b^2+ c^2 - ab- bc-ca))$
$= (a+b+c)( ab + bc + ca) - 3abc$
a b c are in ap so $a+c = 2b$ so $a+b+c = 3b$
now $ab+bc+ ca = b(a+c) + ca = 2b^2 + ca$
so sum = $3b(2b^2+ ca) - 3abc = 6b^3$

2016/010) Solve $|e^{it} - 1| = 2$ for $-\pi<\theta<=\pi$

$e^ {it} = \cos t + i \sin t$
so $e^{it} - 1 = (\cos t-1) + i \sin t$
take mod and square
$(\cos t-1)^2 + \sin ^2 t = 4 $
or $\cos^2 t - 2 \cos t + 1 + \sin ^2 t = 2 $
or $2 - 2\cos t = 4$ or $cos t = - 1$ and hence $t = \pi$

2016/009) What is the value of c such that a straight line exists which intersects $f(x)=x^4+9x^3+cx2+9x+4$ at 4 points

$f(x)=x^4+9x^3+cx^2+9x+4$
for a straight line to intersect at 4 distinct points
$f^{''}(x)$ must have 2 roots
The reason
if it has no root $f'(x)$ is either positive or -ve and so f(x) is monotonically
increasing or decreasing. as it is cubic polynomial with leading coefficient positive
it is increasing. hence no line can intersect at more than 2 points
if it has one zero then it does not have any point of inflection
now $f^{''}(x)=12x^2+54x+2c=12(x+\frac{9}{4})^2+2(c-\frac{243}{8})$
it has a double root when $c<\frac{243}{8}$
hence $c<\frac{243}{8}$

2016/008) Simplify $\frac{x^4}{(x-y)(x-z)}+\frac{y^4}{(y-z)(y-x)}+\frac{z^4}{(z-x)(z-y)}$

$\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}$
=  - ($\dfrac{x^4}{(x-y)(z-x)}+\dfrac{y^4}{(y-z)(x-y)}+\dfrac{z^4}{(z-x)(y-z)})$
= - $(\dfrac{x^4(y-z) + y^4(z-x) + z^4(x-y)}{(x-y)(y-z)(z-x)})$
now
$x^4(y-z) + y^4(z-x) + z^4(x-y)$
= $x^4(y-z) + yz(y^3-z^3) - x (y^4-z^4)$
= $x^4(y-z) + yz(y-z)(y^2+yz+z^2) - x(y-z)(y^3 + y^2z + yz^2 + z^3)$
= $(y-z)(x^4 + yz(y^2 +yz+z^2) - xy(y^2 + yz + z^2) - xz^3)$
= $(y-z)(x^4 + (y^2+yz+z^2)(yz-xy) - xz^3)$
= $(y-z)(x(x^3-z^3) + y(z-x)(y^2 + yz + z^2)$
=$(y-z)(z-x)(y(y^2 + yz + z^2) - x(x^2 + zx + z^2)$
= $(y-z)(z-x)(y^3 + y (yz+ z^2) - x^3 - x(zx + z^2)$
= $(y-z)(z-x)(y^3-x^3 + (y^2z + yz^2 - zx^2 - z^2 x)$
= $(y-z)(z-x)((y-x) (x^2 + xy + y^2) + (z(y^2 - x^2) +z^2(y-x))$
= $(y-z)(z-x)((y-x)(x^2 + xy + y^2 + z(y+x) + z^2)$
= $(-(x-y)(y-z)(z-x)(x^2 + y^2 + z^2 + xy+yz+zx)$
So the given expression
= $x^2 + y^2 +z^2 + xy + yz+ xz$
hence $\dfrac{x^4}{(x-y)(x-z)}+\dfrac{y^4}{(y-z)(y-x)}+\dfrac{z^4}{(z-x)(z-y)}= \dfrac{1}{2}((x+y)^2 + (y+z)^2 + (x+z)^2$

2016/007) How many integer values of x and y are there such that $4x+7y=3$ while $ |x| < 500$ and $|y| < 500$

1st let us find one solution
this can be found by any method but as we see that $7- 4 = 3$
so $(-1,1)$ is a solution
as coeffcient of y is larger so we need to restrict x between $- 500$ to $500$
general solution is $x = -1+ 7t$ and $y = 1 + 4t$
Now $- 500 < x < 500$ or -$ 500 < 1 + 7t < 500$ or
$-501 < 7t < 499$ or $-71 <= t <= 71$ so 143 values