2024/036) Show that $6∣2n^3+3n^2+n,n\in N$

 we have

$2n^3+3n^2 + n = n(2n^2+3n +1) = n(2n+1)(n+1)$

$= n((n-1) + (n+2))(n+1) = (n-1)n(n+1) + n(n+1)(n+2)$

$(n-1)n(n+1)$ is product of 3 consecutive integers and hence divisible by 6 and so is $n(n+1)(n+2)$ and hence the sum is divisible by 6

2024/035) What are the maximum and minimum values of 3x+4y on the circle $x^2+y^2=1$

As $x^2 + y^ 2 = 1$ we can chose $x = \sin\, t , y = \cos\, t$

$3x + 4 y= 3 ]sin\, t + 4 \cos\, t$

To convert $3x + 4 y= 3 ]sin\, t + 4 \cos\, t$ to the form $A \sin (x+ t)$

$A \sin (x+t) = A \sin\,t \cos\, x + A \cos\, t \sin\, x$

We can choose $3 = 5 \cos\, x$ and $4 = 5\ sin\, x$ (as $3^2 + 4^2 = 25 = 5^2$)

$= 5 \cos\, x \sin\, t + 5 \cos\, t \sin\, x = 5 \sin (x-t)$

It is maximum when $\sin (x-t) = 1$ and maximum value = $5$

minimum when $\sin (x-t) = -1$ and minimum value = $-5$

2024/034) For how many positive n less than 2025 $n^2+3n+2$ is divisible by 6

We have $n^2+ 3n + 2= (n+1)(n+2)$

This is divisible by 2 for any n

For this to be divisible by 6 either n+1 or n+ 2 is divisible by 3 that is n should not be divisible by 3

The number of numbers that is divisible by 3 is $\lfloor \frac{2024}{3} \rfloor  = 674$

So there are 664 numbers for which it is not divisible by 6 or there are 2024-674=-1350 numbers that are divisible by 6 

2024/033) If z is complex number and imaginary part of z is non zero and $\frac{2+3z-4z^2}{2 - 3z+ 4z^2} \in R $ then find $|z|^2$

 We have

 $\frac{2+3z-4z^2}{2 - 3z+ 4z^2} \in R $

So  $\frac{2+3z-4z^2}{2 - 3z+ 4z^2} -1  \in R $

Or  $\frac{6z}{2 - 3z+ 4z^2}  \in R $

Or  $\frac{2 - 3z+ 4z^2}{6z}  \in R $

Or $\frac{2 - 3z+ 4z^2}{z}  \in R $

 Or $\frac{2}{z} -3 + 4z \in R$

  Or $\frac{2}{z} + 4z \in R$

Let $z = x + iy$ and $y\ne 0$

So we have $\frac{2}{x+iy} + 4(x+iy) \in R$

Or  $\frac{2(x-iy)}{x^2+y^2} + 4(x+iy) \in R$

Or   $\frac{-2y}{x^2+y^2} + 4y = 0$ as imaginary part has to be zero

So as we have y is not zero dividing by y we get $x^2+y^2 = \frac{1}{2}$ or $|z|^2=\frac{1}{2}$