2021/116) If the product of two numbers is 2,400 and their LCM is 96, then what is their HCF?

The question is ill formed because

Product = 2400

LCM = 96

So HCF = $\frac{product}{LCM} = \frac{2400}{96}=25$

But LCM has to be a multiple of HCF . ths is so because LCM is a multiple of each of the numbers and each number is multiple of HCF

As 25 does not divide 96. so the problem is incorrect, or this case is not possible 

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

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

2021/108) The polynomial: $P(x) = 1 + a_1x +a_2x^2+...+a_{n-1}x^{n-1}+x^n$ with non-negative integer coefficients has $n$ real roots. Prove, that $P(2) \ge 3^{n}$

Because all coefficients are positive so all n roots are -ve and hence

$P(x) = \prod_{k=1}^{n} (x+ a_k)$ where all $a_k$ are positive

Further $\prod_{k=1}^{n} (a_k) = 1$

So $P(2) = \prod_{k=1}^{n}(2+a_k)\cdots(1)$

Now taking AM GM between 1,1 $a_k$ we get $(2+a_k) >= 3\sqrt[3]{(a_k)}\cdots(2)$

So from (1) and (2)

$P(2) >= 3^n \sqrt[3]{\prod_{k=1}^{n} (a_k)}  = 3^n$ and hence $P(2) >= 3^n$

2021/107) Show that $\tan^{-1}(k) = \sum_{n=0}^{k-1}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right )$ - and deduce that $\sum_{n=0}^{\infty}\tan^{-1} \left ( \frac{1}{n^2+n+1} \right ) = \frac{\pi}{2}$

We have $n^2+n+1= 1+n(n+1) = \frac{1+n(n+1)}{(n+1) - n}$

Or $\frac{1}{n^2+n+1} = \frac{(n+1)-n}{1+(n+1)n}$

Using $\tan^{-1}\frac{a-b}{1+ab} = \tan^{-1} a - \tan^{-1}{b}$

We get  $\tan^{-1} \frac{1}{n^2+n+1} = \tan ^{-1}(n+1)-  \tan ^{-1}n$

Adding from 0 to k-1 we get as telescopic sum

Hence $\sum_{n=0}^{k-1} \tan^{-1} \frac{1}{n^2+n+1} =  \tan ^{-1}k-  \tan ^{-1}0 = \tan ^{-1}k$

Taking limit as $k = \infty$

$\sum_{n=0}^{\infty} \tan^{-1} \frac{1}{n^2+n+1}  = \tan ^{-1}\infty= \frac{\pi}{2}$

2021/106) Evaluate closed form of $1^2+2^2+3^2+\cdots+n^2$

We have

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

Or $(k+1)^3 - k^ 3 = 3k ^2 + 3k + 1$

Adding from 1 to n we get

$\sum_{k=1}^n((k+1)^3 - k^ 3) = 3\sum_{k=1}^n k^2 + 3\sum_{k=1}^nk + \sum_{k=1}^n1$

The LHS is a telespcopic sum = $(n+1)^3-1$

We know $\sum_{k=1}^nk = \frac{n(n+1)}{2}$

so we get $(n+1)^3 - 1 = 3\sum_{k=1}^n k^2 + 3\frac{n(n+1)}{2} + n$

or  $3\sum_{k=1}^n k^2 = (n+1)^3 - 1 -  3\frac{n(n+1)}{2} - n$

$=n^3 + 3n^2 + 3n + 1 - 3\frac{n^2+n}{2} - n$

$= \frac{1}{2}(2n^3 + 6n^2 + 6n - 3n^2 -3n -n )$

$= \frac{1}{2}(2n^3 + 3n^2 + 2n )$

$=\frac{1}{2}n(2n^2+ 3n + 2)$

$= \frac{1}{2}n(n+2)(2n+1)$

so $\sum_{k=1}^n k^2 = \frac{1}{6}(n(n+2)(2n+1)$

2021/105) Prove that if k = mn and k is a perfect square and m and n are co-primes them m and n are perfect squares

 Now let p be a prime factor of k.

So p is a prime factor of m or n but not both because GCD(m, n) = 1

Now because k is a square p shall occur even number of times say 2m

All the 2m occurences must be factor of m (as we have mention p is factor of m) 

So any prime factor of k whcich is a factor of m shall occur even number of times in m and which is not a factor of m shall occur even number of times in n making both m and n perfect squares.