Question

please solve it before 25th august.explain it also.solve in a notebook and send the pic.

Answer 1

★ SAMPLE P-RMO 2017 SOLUTIONS ★
ALGEBRA SECTION SPECIAL
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⚛ QUESTION NO. 1 )

a + b = a/b + b/a
a + b = a² + b ²/ ab
ab ( a+b) = a² + b²
a²b + b²a = a² + b²
a²b - a² + b²a - b² = 0
a² ( b-1 ) + b² ( a-1 ) = 0
Given that " a" and " b " are positive integers
Hence , there's only criteria... of holding equality ... aslike ...
∴ a²( b-1 ) =0 and b² (a-1 ) =0
∴ a = 1 and b = 1
∴ a² + b² = 1 + 1 = 2

⚛QUESTION NO. 2 )

Direct application of criteria of common roots in respectively two given quadratics ... aslike ,

It will reduce into k² + k -12 =0
where k= -4 is neglected and k = 3 will be accounted

⚛QUESTION NO.3 )

P(x) = any expansion of higher degree with constant or non zero integral coefficients ...
And given that P(x) is divisible by any 'n' for being positive integer ... Hence we can conclude that the respective expansion will be divisible for any n ... aslike its expansion will result in equality with 0 ... hence , when it is subjected to 0 it will result in 0 as remainder and completely factorised .
it's an important results from theory of equations .

⚛QUESTION NO. 4 )

k² < 2014 < (k+1)²

solving for both the inequalities

and keeping in mind that any operation with prime no. does not affect its prime factors ,

k² < 2014 and 2014 < (k+1)²

prime factor from very first equation will yield
2 and in second equation
operation on k² + 2k will not affect it's factorisation ... hence factors of 2013 when resolved had 11 as its greatest
Therefore , we had obtained two prime factors 2 and 11 in which obviously 11 is greater and largest prime factor of k .

⚛QUESTION NO. 5 )

Given that , N ∈ two digited positive integer
General format = 10x + y
so, proceeding in that manner , Given that
10x + y ( 10y + x ) = perfect square
when added after reversing the order of digits
Let it be equal to m ² as its a perfect square
∴ 11 ( x+ y ) = m ²
or , m² [ 1 / x + y ] = 11 [ 1 ]
Hence equate both the components and reject 0 and -1 as a yielded result because it can't form any two digit no. here if putted into condition and as earlier described x and y are digits so, except them ... we will be left with 8 only ... Hence 8 will be the proper configuration of numbers formed .

⚛QUESTION NO. 8 )

Required condition is aslike ...
xy ( 10x + Y )
Aquired no. of digits is as ...
10x ²y + xy²
Hence , sum of digits is actually
xy ( x + y )
and when we pair up all the possibilities about its combinations we get ...

Digits 2 , 3 , 5, and 7 aslike , given that they are digits and prime in nature ...

and only possible combnation is of 2 and 3 and the rest are rejected by any of the means like ... if we choosed 7 and 3 then it will not result in largest posible value required and if we opted 3 and 5 then its not prime in nature ...
Therefore only 2 and 3 are the digits ...

Now , putting in the equation ,

in the manner to obtain the numbee of digits in the sum ... 10(4)(3) + 2(9)
and no.of digits as 1 + 3 + 8 = 12
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