We have a prime number is 2 or 3 of of the form $6n\pm 1$ for $n \gt 0$ let us compute $n^2+2$
$2^2 + 2 = 6 = 2 * 3$ not a prime
$3^2+ 2 = 11$ is a prime
$(6n\pm 1)^2+ 2 = (36n^2 \pm 12 n + 1) + 2 = (36n^2 \pm 12 n + 3) = 3(12n^2 \pm 4 n + 1)$ which is not a prime
hence 11 is the only prime
We have $n^{th}$ triangular number $t_n= \sum_{k=1}^n k = \frac{n(n+1)}{2}$
So we must have $\frac{n(n+1)}{2} \equiv -1 \pmod {11}$
Or $n(n +1) \equiv -2 \pmod {11}$
Or $n^2 + n + 2 \equiv 0 \pmod {11}$
Or $4 n^2 + 4n + 8 \equiv 0 \pmod {11}$ (the purpose of doing this is to covert to perfect square as evident from next line)
Or $(2n+1)^2 + 7 \equiv 0 \pmod {11}$
Or $(2n+1)^2 = \equiv -7 \pmod {11}$
Or $(2n+1)^2 = \equiv 4 \pmod {11}$
Let us find the square mod 11 for n = 0 to 5 we get (0,0),(1,1),(2,4),(3,9),(4,5),(5,3)$
So the numbers are 2 and 9 ( that is 11 -2)
$2n + 1 \equiv 2 \pmod {11}$ gives n = 6 and $2n + 1 \equiv 9 \pmod {11}$ gives n = 4
So we have the triangular numbers are $t_{11k+4}$ and $t_{11k+6}$ for any non negative k
We have by definition triangular number $t_k = \frac{k(k+1)}{2}$
So $t_{3k}+t_{4k+1}$
$= \frac{(3k(3k+1)}{2} + \frac{(4k+1)(4k+2)}{2}$
$= \frac{(9k^2+3k}{2} + \frac{16k^2 + 12 k + 2}{2}$
$= \frac{25k^2+15k + 2}{2}$
$= \frac{(5k+1)(5k + 2)}{2}= t_{5k+1} $
Then
a) exactly one real root is -ve
b) all real roots are positive
c) exactly one real root is positive
d) no real root is positive.
Solution
we have
$x^{2021} + x^{2020} + \cdots+x - 1=0$
or $x^{2021} + x^{2020} + \cdots+x + 1=2$
Note that 1 is not a root of this equation.
Multiplying both sides by $x-1$ we get
$x^{2022} - 1 = 2(x-1)$
or $f(x) = x^{2022} - 2x + 1 = 0$
Note that as we have multiplied by x-1 so x = 1 so there is at least one positive root that is 1
As there is change of sign two times as per Descarte rule there are two or zero positive roots root but as it has at least one positive root so there are two positive roots so original equation has one positive root.
Now $f(-x) = x^{2022} + 2x + 1 = 0$
As there is no change of sign so there is no -ve root
So the equation has has one root and it is positive.
So answer is (c)-
We have the terms 8,15,24,35
Let us compute 1st order difference 7,9,11 which is an AP
Next difference is 13 and so next term is 35 + 13 = 48
So the nth term of the given sequence is quadratic say $an^2+bn + c$
Putting n =1 ,n= 2 and n =3 we get following
$a+b + c = 8\cdots(1)$
$4a+2b + c = 15\cdots(2)$
$9a+3b + c = 24\cdots(3)$
Subtracting (1) from (2) we get
$3a+b = 7\cdots(4)$
Subtracting (2) from (3) we get
$5a+b = 9\cdots(5)$
Subtracting (4) from (3) we get $2a=2$ or $a=1$
Putting $a=1$ in (4) we get $b=4$ and putting in (1) we get $c=3$
So $n^{th}$ term = $n^2+4n + 3$
we get the same results by putting n=1,2,3,4 so on
The number can contain only the digits 1,2 besides 0. 1 * 44 = 44 and
there is no overflow( if in the number the sum is one then it becomes 8,
and if it is 2 the the sum of digits is 16) so the sum of digits is 8
times. if any digit is 3 to 9 then sum of digits less than 8 times so
this shall give a lesser sum
(just an observation and not required for the result) Again 1 may be preceded/succeeded by 0 or 1 as 11 * 44 = 484 . but 2
has to be preceded/succeeded by 0 as 21 * 44 = 924 and 12 * 44 = 538
and the sum of digits become less
So the number shall have 0 1 and 2 meeting above conditions so that sum
of digits 100 and when we multiply by 3 (that is 3n) the digits shall be
0,3,6 and there is no overflow and sum of digits 300.
Let the number of numbers be n and it starts at a
So we have the sum $= an + frac{n(n-1)}{2} = 2024$
Or $2an + n(n-1) = 4048$ and $a\ge 1$
Or $n(2a+n-1)= 4048 = 16 * 253 = 2^4 * 23 *11$
Now out of n and 2a+n-1 one is even and one is odd so n = 16 ( 2a +n -1 = 253) or 1 ( 2a +n -1 = 4048) or 11 ( 2a +n -1 = 368) or 23 ( 2a +n -1 = 176)
Clearly n = 23 is the largest giving 2a + 22 = 176 and a = 77
So largest number of consecutive positive integers is 23 and it starts with 77
LCM of 48 and 60 (that is 12 * 4 and 12 *5) = 240
Now LCM 1680 = 240 * 7
So
3rd number should have a factor 7 and it should be multiple of 12(else
GCD cannot be be 12) and can have other that is 3rd factor that is a
factor of 240/12 or 20 it could be 1,2,4,5,10,20. as GCD of 48 and 60 is 12 so taking a higher multiple shall not change GCD
So 3rd number could be 84/168/324/420/840/1680 that is any of these 6 numbers
We shall use 2 facts
$\cos 3x = 4 \cos^3 x - 3 \cos\, x\cdots(1)$
And
if $a+b+c=0$ then $a^3+b^3+c^3 = 3abc\cdots(2)$
We are given
$\cos\, A +\ cos\, B + \cos\, C = 0\cdots(3)$
Hence $\cos^3 A +\ cos^3 B + \cos^3 C = 3 \cos\, A \cos\, B \cos\, C\cdots(4)$
Now $\cos 3A + \cos 3B +\cos 3C$
$ = 4 \cos^3 A - 3 \cos\, A$ + $ 4 \cos^3 B - 3 \cos\, B $ + $ 4 \cos^3 C - 3 \cos\, C$ using(1)
$=4(\cos^3 A +\ cos^3 B + \cos^3 C - 3 (\cos\, A + \cos\, B + \cos\, C)$
$ =4 * 3 \cos\, A \cos\, B \cos\, C- 3 *0 $ using (3) and (4)
$ =12 \cos\, A \cos\, B \cos\, C$
Proved
We have a,b,c are in AP so
$2b = a+ c$
Hence $a+2b -c = a + a+c -c = 2a\cdots(1)$
Let the $26^{th}$ digit be k
so $N = 1111\cdots( 50\, times) + (k-1) * 10^{24}$
or $N = \frac{10^{50}-1}{9} + (k-1) * 10^{24}$
or $9N = 10^{50}-1 + 9 * ((k-1) * 10^{24})$
because $GCD(9,13)=1$ so if n is divisible by 13 9N is divisible by 13.
as 13 ia prime we have
$10^{12} \equiv 1 \pmod {13}$
so
$10{24} \equiv 1 \pmod {13}\cdots(1)$
and
$10{48} \equiv 1 \pmod {13}$
hence
$10{50} \equiv 100 \pmod {13}$
or
$10{50} \equiv 9 \pmod {13}$
$9N = 10^{50}-1 + 9 * ((k-1) * 10^{24}) \pmod {13}$
$= 9-1 + 9 * ((k-1) * 1) \pmod {13}$
$= 9k -1 \pmod {13}$
this is zero and because k is a digit trying from 0 to 9 we get k=3 satisfies the condition
so the $26^{th}$ digit is 3
LHS
If n = 2 we have $10^2+3 = 103$ this is divisible by 103 .
Hence $10^2 + 3 = 0 \pmod {103}\cdots(1)$
We know that 103 is a prime but we shall use this as a starting point to show that there are infinite numbers of the form $10^n+3$ is divisible by 103.
As 103 is a prime so using Fermats little theorem
$10^{102} = - 1 \pmod {103}$
Squaring both sides we get
$10^{204} = 1 \pmod {103}$
So from (1) and above we have
$10^{204n + 2} + 3= 0 \pmod {103}$
except for n = 0 when the value is 103 all other values are above 103 and have a factor 103 so not a prime
So there are infinitely many composite numbers as they are divisible by 103. We have not taken all cases and have found infinitely many composite numbers. Hence proved
We shall use Euler's theorem that is a generalization of Fermat's little theorem: For any n and any integer a coprime to n, one has
$a^{\phi(n)} \equiv -1 \pmod n$
Where $\phi(n)$ is number of numbers that is co-prime to n,
Let us compute $\phi(49)$
If $n = p_1^{k_1}p_2^{k_2}p_3^{k-3} ... $ when each $p_i$ is prime then
$\phi(n)=n(1-\frac{1}{p_1})(1-\frac{1}{p_2})(1-\frac{1}{p_3})$
As $49= 7^2$ so $\phi(49) = 49 * (1-\frac{1}{7}) = 49 * \frac{6}{7} = 42$
Hence
$6^{\phi(49)} \equiv -1 \pmod {49}n$
or $6^{(42)} \equiv -1 \pmod {49}$
Squaring both sides
$6^{(84)} \equiv 1 \pmod {49}$
Dividing by 6 we get
$6^{(83)} \equiv \frac{1}{6} \pmod {49}\cdots(1)$
Where $\frac{1}{6}} is not fraction but inverse of 6 mod 49
Similarly
$8^{(83)} \equiv \frac{1}{8} \pmod {49}\cdots(2)$
Adding (1) and(2) we get
$6^{(83)}+ 8^{(83)} \equiv \frac{1}{6} + \frac{1}{8} \pmod {49}\cdots(2)$
Now $\frac{1}{6} + \frac{1}{8} \pmod {49}$
$= \frac{8 + 6}{48} \pmod{49}$
$= \frac{14}{48} \pmod{49}$
$= \frac{14}{-1} \pmod{49}$ as $48 \equiv -1 \pmod {49}$
$= -14 \pmod{49}$
$=35$ taking positive number
Hence remainder = 35
Probability that you choose one unfair coin = $\frac{1}{1000}$
iI you choose one unfair coin the probability that all 10 heads come = 1
So Probability that you choose one unfair coin and all heads come = $\frac{1}{1000}$
Probability that you choose one fair coin = $\frac{999}{1000}$
If you choose one fair coin the probability that all 10 heads come = $\frac{1}{2^{20}} = \frac{1}{1024}$
So Probability that you choose one fair coin and all heads come = $\frac{999}{1024*1000}$
So probability that all heads come = $\frac{999}{1024*1000} + \frac{1}{1000} = \frac{2023}{1024000}$
So probability that coin is unfiar = $\frac{\frac{1}{1000}}{\frac{2023}{1024000}}=\frac{1024}{2023}$
We have $\sin \, ax + \cos \, bx = \frac{1}{2} (\sin (a+b) x + \sin (a-b) x$
Knowing that
$\frac{d}{dx} \ cos \, nx = - n \sin \, nx$
Hence $\int (\sin \, ax \cos \, bx) dx = \int (\frac{1}{2} (\sin (a+b) x + \sin (a-b) x)dx $
= $\frac{\cos(a+b)x}{a+b} - \frac{\cos(a-b)x}{a-b} + C $
Proof;
We shall use the fact that if x and y are co-prime then xy is a perfect cube if both x and y are perfect cubes.
Let the 1st number be x so the product is $x(x+1)(x+2)(x+3)$
Either x is odd or even
If x is odd then we have $GCD(x,x+2) = GCD(x+1, x+2) = GCD(x+2,x+3) = 1$
For x + 2 to be a perfect cube minimum x is 6
So $GCD(x+2,x(+1)(x+2)) = 1$
So we need to prove that $x(x+1)(x+3)$ is not a perfect cube.
$x(x+1)(x+3) = x^3 + 4x^2 + 3$
We have $x(x+3) - (x-1)^2 = x^3+3x - x^2 -2x -1= x-1 > 0$ as x is minimum 6
So $x(x+1)(x+3) >= (x+1)^3$
$x^3 + 4x^2 + 3$ is below $(x+2)^3 = x^3 + 6x^2 + 12x + 8$
As it is between 2 cubes it cannot be a cube
Similarly we can prove taking $x +1$ when $x$ is even
1,2,4,6
We know that $ x+ y | x^3+y^3 $
Hence $2^n + n | (2^3)^n + n^3$
as $2^n + n | 8^n + n $
so $2^n + n | n^3-n $
so we must have $ n^3-n =0 $ or $2^n + n < n^3-n $
$n^3-n= 0$ gives n = -1,0, 1 and out of theses only 1 is solution
we need to solve $2^n + n < n^3-n $ or $2^n < n^3-2n $
let us find an upper bound for n putting a condition
$2^n < n^3$ for $n \lt 10$
putting n from 1 to 9 we see that $n \in \{1, 2,4,6\} $ satisfy the case and there is no other solution
We are given
$x^2-y=111\cdots(1)$
$y^2-x=111\cdots(2)$
From (1) and (2)
$x^2-y = y^2 -x$
Or $x^2-y^2 + x -y = 0$
Or $(x-y)(x+y) + (x-y) = 0$
Or $(x-y)(x+y+1) = 0$
As $ x\ne y$ dividing by x-y we get $x+y+1=0\cdots(3)$
Or $y = -(x+1)$
Putting the value in (1) we get $x^2 + (x+ 1) = 111$ or $x^2 + x - 110 = 0$
Or $(x-10)(x+11) = 0 $
Or $ x= 10$ or $x = -11$
Putting in (3) if $x= 10$ then $y = -11$
If $x= -11 $ then $y = 10$
So Solution set $x=10,y= -11$ or $x=-11,y=10$
To show that 123123 is a factor of $2^{60}-1$ we need to show that each prime factor of 123123 to the highest power is a factor
Now let us factor 123123
$123123= 123 * 1001 = 3 * 41 * 7 * 11 * 13$
As each prime occurs once we need to prove each of 3,7,11,13,41 divides $2^{60}-1$
We know $x^y-1 | x^{my}-1$ for any m
We have $2^2-1 | 2^{60}-1$ and $3| 2^{2} -1$ using FLT(Fermat's Little Theorem) so $3| 2{60}^-1$
$2^6-1 | 2^{60}-1$ and $7| 2^{6} -1$ using FLT(Fermat's Little Theorem) so $7| 2{60}^-1$
$2^{10}-1 | 2^{60}-1$ and $11| 2^{10} -1$ using FLT(Fermat's Little Theorem) so $11| 2{60}^-1$
$2^{12}-1 | 2^{60}-1$ and $13| 2^{12} -1$ using FLT(Fermat's Little Theorem) so $13| 2{60}^-1$
Now we need to prove for 41
using FLT we know $41| 2^{40} -1$
So let us find $GCD(2^{(60}-1,2^{40}-1)$
$GCD(2^{(60}-1,2^{40}-1)= GCD((2^{(60}-1) - (2^{40-1},2^{40}-1)$
$=GCD((2^{60} - 2^{40}), 2^{40}-1)$
$=GCD((2^{40}( 2^{20}-1), 2^{40}-1)$
$=GCD((2^{20}-1), 2^{40}-1)$ as second number is odd so removing even factors of 1st number
$= 2^{20}-1$ as this is a factor of $ 2^{40}-1$
now $= 2^{20}-1)= (2^{10}-1)(2^{10}+1)$
$= 1023 * 1025$ as $41 | 1025$ so $41 | 2^{60}-1 $
as each of 3,7,11,13,41 divides $2^{60}-1$ so 123123 is a factor
We have $a^3-a= a(a^2-1) = a(a-1)(a+1) = (a-1)a(a+1)$
As $(a-1)a(a+1)$ is product of 3 consecutive if is s divisible by 6 say 6m for some m
So $a^3 = a + 6m\cdots(1)$
Similarly
$b^3 = b + 6p\cdots(2)$
and
$c^3 = c + 6q\cdots(3)$
Adding (1),(2) and (3) we get $a^3+b^3+c^3 = (a+b+c) +6(m+p+q)$
So if $(a+b+c)$ is divisible by 6 then $a^3+b^3+c^3$ is divisible by 6
We have $6^2=36$ that is it ends with 6.
So we have all the powers of 6 end with 6
So $6^n+1$ shall end with 7
So we need to find n such that all the digits of $6^n+1$ must have all digits 7 and let it be k 7's/
So $6^n+1 = \frac{7}{9} (10^k -1 )$
Or $9(6^n+1) = 7 (10^k -1 )$for
Or $9 * 6^n + 16 = 7 * 10^k\cdots(1)$
We have $2 | 6$ so $2^2 | 6^2 $ so if $n \ge 5$ then we have
$9 * 6^n + 16 \equiv 16 \pmod {32} $ for $n \ge 5$
So $9 * 6^n + 16 \equiv m \pmod {32} $ where $m \le 16$ and not zero
For $k \ge 5$ $ 7 * 10^5 \equiv 0 \pmod {32} $
So $k \le 4$
So we need to check for n such that $9 * 6^n + 16 \le = 70000$
Or $6^n \le 7776$
Or $6^n+1 \le 7777$
We calculate for n = 1 $6^1 + 1 = 7$ meets criteria
n = 2 $6^2 + 1 = 37$ does not meet criteria
n = 3 $6^3 + 1 = 217$ does not meet criteria
n =4 $6^4 + 1 = 1297$ does not meet criteria
n = 5 $6^5 + 1 = 7777 $ meets criteria
Other value of n takes the value out of range
So $ n \in \{1,5\}$
We first factorize 437
$437 = 19 * 23$
Now let compute mod relative to19 and 23
As 19 is prime number as as per Wilson'sTheorem we have
$18 ! \equiv -1 \pmod {19}\cdots(1) $
As 23 is prime number as as per Wilson'sTheorem we have
$22 ! \equiv -1 \pmod {23} $
Now
$22 \equiv -1 \pmod {23}\cdots(2) $
$21 \equiv -2 \pmod {23} \cdots(3)$
$20 \equiv -3 \pmod {23}\cdots(4) $
$19 \equiv -4 \pmod {23}\cdots(5) $
As $22! = 22 * 21 * 20 * 19 * 18! $
So $22 ! \equiv -1 \pmod {23} $
$\implies 22 * 21 * 20 * 19 *18 ! \equiv -1 \pmod {23} $
$\implies (-1) *(-2) * (-3) * (-4) *18 ! \equiv -1 \pmod {23} $
$\implies 24 *18 ! \equiv -1 \pmod {23} $
$\implies 24 *18 ! \equiv -1 \pmod {23} $
$\implies 1 *18 ! \equiv -1 \pmod {23} $
$\implies 18 ! \equiv -1 \pmod {23}\cdots(6) $
Using (1) and (3) we get
$18 ! \equiv -1 \pmod {437} $
Proved
We have $8 = 2 * 4 = 2 * 2^2$
so $\sqrt{8} = 2 \sqrt{2}$
now $\sqrt{9} = 3$
so $\sqrt{\sqrt{9} - \sqrt{8}}$
= $\sqrt{3 - 2\sqrt{2}}$
this is of the form $\sqrt{n - 2\sqrt{n-1}}$ when n = 3
so $\sqrt{3 - 2\sqrt{2}} = \sqrt{2 + 1 - 2 * \sqrt{2} * 1}$
$= \sqrt{(\sqrt{2})^2 + 1^2 - 2 * \sqrt{2} * 1} $
$= \sqrt{2} - 1$
We have the constraint $(a-5)^2+(b-7)^2=4$ this is set of points lying in a circle with centre (5,7) and radius 2.
We need to find the minimum of square of the distance of (a,b) from (-7,-2) and this is minimum when (a,b) lies in the line from (5,7) to (-7,-2)
Distance from (5,7) to (-7,2) = $\sqrt{(5+7)^2 + (7+2)^2}= \sqrt{12^2 + 9^2} = 15$
So distance from (a,b) to (-7, -2) is $15-2 = 13$
Minimum Value of $(a+7)^2+(b+2)^2= 13^2 =169$
Note that we cannot have $x=k$ or $x = -k$
So multiplying both sides by $x^2-k^2$
We get
$((x-k) + k(x+k) + 2k) )(| x-k | -k) = 0$
Or $(x+k)(k+1))(| x-k | -k)=0$
As x cannot be -k $(| x-k | -k)=0$
So we get 2 values of x that is 2k or 0
Hence no solution
The power of digit of 3 shall have 1 or 3 or 7 or 9 as the unit digit.
Let us see some power of 3 that is 1,3,27,81(after this the sequence in the 1st digit repeats.
So we have (20n + x) where x is 1 or 3 or 7 or 9.
When we multiply by 3 we get 60n + 3 or 60n + 9 or 60n + 21 or 60n + 27 that is 60n + 3 or 60n + 9 or 20(3n+1) + 7 or 20(3n+1) + 9
So tens digit is even
Because $m > 1$ $2m + 1\ge 4$ So $2^{2m+1}$ is divisible by 16.
$2^{2m+1} $ is double of $(2^m)^2$ and is not a square.
Now we consider 2 cases
Let us consider when n is odd
Then $n= (2k+1)$
So $n^2 = (2k+1)^2 = 4k^2 + 4k + 1 = 4k(k+1)+1$
Ss k or k+1 is even we $n^2 \equiv 1 \pmod 8$
So $2^{2m+1}$ is divisible by 8 and minimum number above $n^2$ which is divisible by 8 is $n^2 + 7$
So we have proved for the case n is odd.
Let us consider when n is even
If $n^2$ is divisible by 8 then $n^2$ is divisible by 16 and as $2^{2m+1}$ is divisible by 16 and n^2 is divisible by 16
So $2^{2m+1}-n^2 = 16k$ for some positive k
So $2^{2m+1}-n^2 \ge16$
Hence $2^{2m+1}-n^2 \gt 7$ or $2^{2m+1} \gt n^2+ 7$
If $n^2$ is even and is not divisible by 8 then n is of the form 4k+2
$n^2= 16k^2 + 16k + 4$
Or $n^2 \equiv 4 \pmod {16}$
As $2^{2m+1} = 0 \pmod {16}$
So $2^{2m+1} \ge n^2 + 12 $
Hence $2^{2m+1} \gt n^2 + 7 $
we have proved for all 3 cases hence done
We have $640= 128 * 5 = 2^7 * 5$
To make it a power of 10 we need to multiply by $5^6$ giving $10^7$
so $a=5^6$ and b = 7 are the smallest values
Here let $frac{1}{10} = x$
We have for $|x| \lt 1$
$\sum_{n=0}^{\infty} x^n= \frac{1}{1-x}$
Differentiating both sides wrt x we get
$\sum_{n=0}^{\infty} nx^{n-1}= \frac{1}{(1-x)^2}$
Multiplying both sides by x we get
$\sum_{n=0}^{\infty} nx^n= \frac{x}{(1-x)^2}$
Differenting both sides wrt x we get
$\sum_{n=0}^{\infty} n^2x^{n-1}= \frac{d}{dx} \frac{x}{(1-x)^2}$
Usimg $\frac{d} ({dx} \frac{u}{v})= \frac{v\frac{du}{dx} - u \frac{dv}{dx}}{v^2}$
We get
$\sum_{n=0}^{\infty} n^2x^{n-10} =\frac{(1-x)^2 . 1 - x . (x-2)}{(1-x)^4} = \frac{x +1}{(1-x)^3}$
Multiply by x on both sides we get $\sum_{n=0}^{\infty} n^2x^n = \frac{x^2 +x}{(1-x)^3}$
Putting value of $x = \frac{1}{10}$ we get the result
Let $a = 2^{\frac{1}{3}}\cdots(1)$
So $a^3 = 2\cdots(2)$
And $\frac{1}{a} = 2^{-\frac{1}{3}}\cdots(3)$
And
$x = a+ \frac{1}{a}\cdots(4)$
Cubing both sides $x^3 = a^3 + \frac{1}{a^3} + 3 * a * \frac{1}{a} ( a + \frac{1}{a})$
Or $x^3 = a^3 + \frac{1}{a^3} + 3 * ( a + \frac{1}{a})$
Or $x^3 = 2 + \frac{1}{2} + 3 *x$ from (2), 3) and (4)
Or $x^3 = \frac{5}{2} + 3 *x$
Or $2x^3 = 5 + 6x$
Or $2x^3-6x = 5$
Proved
Let $\frac{x-a}{x-b} = m$ and $\frac{a}{b} =n $
So we get $ m + \frac{1}{m} = n + \frac{1}{n}$
Or $m^2n + n = n^2m + m$
Or $m^2n - n^2m = m - n$
Or $mn(m -n ) = m - n$
Or m-n = 0 Or $mn = 1$
case 1
m = n gives $\frac{x-a}{x-b} = \frac{b}{a}$
Or $b(x-a) = x(a-b)$ or $bx-ab = ax-ab$ or $bx = ax$ or $x(a-b)=0$ or $x = 0$
Case 2)
mn =1 gives
$\frac{x-a}{x-b}*\frac{a}{b}=1$
or $a(x-a)=b(x-b)$
Or $ax-a^2=bx-b^2$
Or $ax-bx = a^2-b^2$
Or $x(a-b) = a2-b^2$
Or $x = a + b$
