2021/087)Given: $x>0,\, n\in\mathbb{N}$ Prove: $(1+x)\times\left(1+x^2 \right)\times\cdots\times\left(1+x^n \right)\geq\left(1+x^{\large{\frac{n+1}{2}}} \right)^n$

We have $x^a + x^b >= 2x^{\frac{a+b}{2}}$ by AM GM inequality

adding $1 + x^{a+b}$ on both sides we get

$(1+x^a)(1+x^b) >= 1 + 2x^{\frac{a+b}{2}} + x^{a+b}= ( 1+ x^{\frac{a+b}{2}})^2$

Putting $b = n+ 1 -a$ we get

$(1+x^a)(1+x^(n+1- a) >= (1+ x^{\frac{n+1}{2}})^2$

For n even taking a from 1 to $\frac{n}{2}$ we get n/2 expressions and multiplying them out we get the result

For n odd we have n-1 ( running a from 1 to $\frac{n-1}{2}$ we get $\frac{n-1}{2}$ terms and as middle term is $(1+x^{\frac{n+1}{2}})$ we get thr result

2021/086) Find the limit of: $lim_{x\to +\infty}\frac{ \lfloor x\rfloor}{x}$

We have

 $\frac{ \lfloor x\rfloor}{x} = \frac{ x- y}{x}$ where y is fractional part of x and hence $0 \le y \lt 1$

Hence 

$  \frac{ \lfloor x\rfloor}{x} =   1- \frac{y}{x} \ge 1 - \frac{1}{x}$

as x goes to infinitte RHS goes to 1 so the required value is between 1 and 1 so it is 1 

2021/085) Let $P(x)=x^3−2x+1$ and $Q(x)=x^3−4x^2+4x−1$. Show that if P(r)=0, then $Q(r^2)=0$

 We have

$P(r) = r^3 - 2r + 1= 0\cdots(1)$

and $Q(r^2) = r^6 - 4r^4 + 4r^2 -1 \cdots(2)$

as $r^3 = 2r  - 1$ so $r^6 = (2r-1)^2 = 4r^2 - 4r + 1$
Putting in (1)  

$Q(r^2) = 4r^2 -4r + 1 - 4r^4 + 4r^2 - 1 = -4r^4 + 8r^2 - 4r = -4r(r^3 - 2r + 1) = 0$ using (1)

2021/084) a,b,c are positive real numbers such that $a^2+b^2= c^2 $ and $ab=c$ find the value of $\frac{(a+b+c)(a+b-c)(a-b+c)(c-a+b)}{c^2}$

 We have

$a^2+b^=c^2\cdots(1)$

And $ab = c\cdots(2)$

Now

$\frac{(a+b+c)(a+b-c)}{c} = \frac{(a+b^2) - c^2 }{c} = \frac{a^2+b^2+2ab-c^2}{c} = \frac{2ab}{c}$ (uing (1)

$ = \frac{2c}{c}$ (sing (2)

$ = 2$

Or

$\frac{(a+b+c)(a+b-c)}{c^2} = 2$

similarly

$\frac{(a-b+c)(-a+b+c}{c} = 2$

multiplying we get $\frac{(a+b+c)(a+b-c)(a-b+c)(c-a+b)}{c^2}= 4$

2021/083) show that for no positive integers a, b,c can all the 3 expresssion $a^2+b+c, b^2+a + c , c^2 + a +b$ be perfect squares

 Proof:

The samllest square above $a^2$ is $a^2+2a+1$. so we must have for $a^2+b+c$ to be a prefect square

$a^2 + b+ c \ge (a+1)^2$

Or $b+c \ge 2a + 1\cdots(1)$

Similarly $c+a \ge 2b + 1\cdots(2)$

And $a + b  \ge 2c + 1\cdots(3)$

Adding above 3 equations we must have $2(a+b+c) >= 2(a + b+ c) + 3$ or $0 \ge  3$ which is contradiction

So above is impossible  or No solution exists 

2021/082) If,for all sets A, $A \cup B = A $ then prove that $B = \emptyset$

This is true.

To prove the same we have $A \cup B = A $ iff $B \subseteq A$

Let us take 2 sets $A_1,A_2$ which are disjoint and because it is true for every set $A_1 \cup B = A_1 $ so $B \subseteq A_1$

and $A_2 \cup B = A_2 $ so $B \subseteq A_2$

So from above 2 we have

$B \subseteq A_1 \cap A_2$

Because $A_1,A_2$ are disjoint sets so we have $A_1 \cap A_2= \emptyset$

So $B = \emptyset$

 

2021/081) A 5-digit number (in base 10) has digits k, k + 1, k + 2, 3k, k + 3 in that order, from left to right. If this number is $m^2$ for some natural number m, find the sum of the digits of m.

Because there is digit 3k so k <= 3

takiing k = 1 we have last 2 digits = 34 so it cannot be a square

taking k = 2 last 2 digits are 65 so it cannot be a square

taking k = 3 we get $34596= 186^2$ or m = 186 so sum of digits = 1 + 8 + 6 = 15 

2020/080) Solve in integer x $5^x - 3^x = 16$

Working mod 3 we have $(-1)^x\equiv 1 \pmod 3$

So x is even say 2n

So we have $^{2n} - 3^{2n} = 16$

Or $(5^n+3^n)(5^n-3^n) =16$

On  LHS both are even and unequal and 1st term is larget and 16 need to be factored into 2 even numbers 8 * 2

So $5^n +3^n = 8$

$5^n - 3^n = 2$

Adding we get $ 2* 5^n = 10$ or n = 1 so x = 2

  

  

2021/079) Solve in natural numbers $(x+4)(x+24) = p^n$ when p is a prime number

We have rhs is power of a prime number so each of the factors on left that ix x+ 4 and x + 20 both are power of same prime number

As x + 4 < x + 24 so we have x+ 4 is a factor of x + 20

Or x + 4 is a factor of x+24 - (x+4) = 20

As  x is natural number so x+ 4 > 4 so we need to consider only the factors of 20 which are greater than 4 that is 5, 10, 20

x+ 4 = 5 => x = 1 x + 20 = 25 power of x + 4 and p = 5 n = 3

x+4 = 10 => x= 6 x + 24 = 30 is not power of x + 4 

Similarly x + 4 = 20 does not give a soltion

  so only solution x = 1, p = 5 , n = 3