2021/115) 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 choose $x = \sin\, t$, $y = \cos\, t$

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

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

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

2021/114 solve in integers $3x^2 + 5y^2 = 345$

working in mod 3 we have $5y^2 = 0 \pmod 3$ or $y =0 \pmod 3$

so y = 3a for some a

similarly x = 5 b for som b

so ge get $75 b^2 + 45 y^2 = 345$ 

deviding by 15 we get $5b^2 + 3a^2 = 23$

we need to check for $5b^2 < 23$ or $b <=2$

putting b = 1 we get  $3a^2 = 18$ or $a^2 = 6$ not an integer

b = 2 gives $3a^3 =3$ or a = 1

so we have a= 1 , b= 2 giving x = 10 and y = 3 

2021/113) prove the following identity: $\binom{n}{k}=\binom{n-2}{k}+2\binom{n-2}{k-1}+\binom{n-2}{k-2}$

We can solve the same in 2 ways. Combinotorics way or alegraic ways

We present here to solve in combinonorics way

From n objects we can choose k objects in   $\binom{n}{k}$ ways

let us group the n objects into (n-2,1,1) ways

for choosing k objects this can be done in 3 ways

k  objects from n-2 objects that is from 1st set 0 from 2nd set and 0 from3rd set in $\binom{n-2}{k}$ ways

k-1  objects from n-2 objects that is from 1st set 1 from 2nd set or  1 from 3rd set in $2 * \binom{n-2}{k-1}$ ways

k-2  objects from n-2 objects that is from 1st set 1 from 2nd set and  1 from 3rd set in $\binom{n-2}{k-2}$ ways

as all above 3 are mutually exclusive so no of ways  =$\binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$

in  2 ways we have computed the number of choosing k objects from n obects so they must be same or

$\binom{n}{k} = \binom{n-2}{k} + 2 * \binom{n-2}{k-1} + \binom{n-2}{k-2 }$

2021/112) Let G= $\{a+bi$ in complex: $a^2 +b^2=1\}$. Is G a group under multiplication?

For it to be group folllowing must be tue.

1) it should be closed

that is if  x = a + ib and y = c + id and $(a^2+b^2) = 1$ and $c^2+d^2=1$ 

and xy = m + ni then $m^2+n^2 =1$

we have $xy = m + ni = (a+ib)(c+id) = (ac - bd) + (bc + ad)i$

we have m = ac - bd and n = bc + ad

$m^2 + n^2 = (ac-bd)^2 + (bc + ad)^2 = a^2c^2 - 2abcbd + b^2d^2 + b^2 c^2 + 2abcd + a^2d^2$

 $= a^2c^2 + b^2d^2 + b^2c^2 + a^2d^2 = (a^2+b^2)(c^2 + d^2) = 1$ 

So it is closed 

2) It should have an identity

1 or 1+0i is identity element as $(a+bi)(1+0i) = a+ bi$

3) it should have an inverse  

because $a^2+b^2=1$ so it it not zero and hence it has inverse and we need to show that if

m+in is inverse then $m^2+ n^2 =1$ that is the inverse is in this group

$m + in = \frac{1}{a+ib} = \frac{a-ib}{(a+ib)(a=ib)} = \frac{a-ib}{a^2+b^2} = a - ib$

'so m = a , n = - b and $m^2 + n^2 = a^2 + (-b)^2 = a^2 + b^2 =1$

so it has an inverse

4) assosiativity law holds as unde rcomplex number multiplication assosiativity holds 

  

2021/111) For which primes p, 7p+4 is a perfect square?

 7p + 4 is a perfect square say $m^2$

so $7p = m^2 -4 = (m+2)(m-2)$

now there are 2 cases

1. p is 2 which gives 7p + 4 = 18 which is not a perfect square

so

2. p is odd

so 7p is odd and  $m^2 -4$ is odd

so m+2 and m-2 are co-primes as they differ by 4

so m+2 = 7, m-2 = p gives p = 3 which is prime and m = 5

or m+2 = p and m-2 = 7 giving m = 9 and p =11 which is a prime

so p = 3 or 11 

2021/110) express $\cos\, 5t$ in term of power of $\cos\, t$

We have

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

$= 2 \cos\, 3t \cos\, 2t - \cos\, t$ using $\cos\, A + \cos\, B = 2 \cos \frac{A+B}{2} \cos \frac{A-B}{2}$

$= 2 * (4 \cos ^3 t - 3 \cos\, t)(2\cos ^2 t - 1) - \cos\, t$ using formula for $cos 3t$ and $cos 2t$

$= 16 \cos^5 t - 20 \cos^3 t + 6 \cos t -\cos t$

$= 16 \cos^5 t - 20 \cos^3 t + 5 \cos t$

2021/109) Find the sum of the series $\sum_{n=1}^{\infty}\frac{1}{n^2}$

We have tailor  expansion of $\sin\, x$  as  

$P(x) = \sin\,x = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots  (1) $

This is a polynomial of degree infinite with zeroes at  0 and npi so

this is  $Ax(1- \frac{x}{\pi})(1+ \frac{x}{pi})(1-\frac{x}{2pi})(1-\frac{x}{2\pi})\cdots$

or $P(x) =Ax(1- \frac{x^2}{\pi^2})(1- \frac{x^2}{2^pi}^2)(1-\frac{x^2}{3^2pi^2})\cdots$

comparing above with (1) we get A = 1

So $P(x) =x(1- \frac{x^2}{\pi^2})(1- \frac{x^2}{2^pi}^2)(1-\frac{x^2}{3^2pi^2})\cdots$

The coefficient of $x^3$ is    $- \sum_{n=1}^{\infty}\frac{1}{n^2\pi^2}$

from (1)  coefficient of $x^3$ is $-\frac{1}{6}$

as both are same so  $- \sum_{n=1}^{\infty}\frac{1}{n^2\pi^2} = -\frac{1}{6}$

multiplying both sdes by $- \pi^2$ we get

 $\sum_{n=1}^{\infty}\frac{1}{n^2} = \frac{\pi^2}{6}$