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Saturday, May 18, 2024

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

Saturday, May 11, 2024

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 

Sunday, May 5, 2024

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}