Solve this:

​Q14. The number of integers a , b , c for which a² + b² – 8c = 3 is

         (a) 2                            (b) infinite                      (c) 0                       (d) 4


Q15. Which of the following is NOT produced by the endoplasmic reticulum ?

          (a) Lipids                  (b) Proteins                    (c) Monosaccharides               (d) Hormones

 

Dear Student,

Please find below the solution to the asked query:

Given :  a² + b² - 8 c = 3 , So

a² + b² = 3 + 8 c                                                   --- ( 1 )

Case I : If a = Even number and b = Even number , So

Let a = 2 m  and b = 2 n , Here ' m '  and ' n '  are positive integers .

Now we substitute these values in equation 1 and get :

( 2 m )² + ( 2 n )² = 3 + 8 c

$\Rightarrow$4 m² + 4 n² = 3 + 8 c

$\Rightarrow$ 4 ( m² +  n² ) = 3 + 8 c

Here , L.H.S. = 4 ( m² +  n² ) , That is multiple of " even number = 4 " , So

L.H.S. =  Even number

And we know "  Even number + Odd number = Odd number " and in R.H.S. " 8 c " is multiple of ' 8 ' so that is an even number and ' 3 ' is an odd number . So

R.H.S. = Odd number .

Therefore,

We get no solution when ' a '  and ' b '  both are even number .

Case II : If a = Odd number and b = Odd number , So

Let a = 2 m + 1  and b = 2 n + 1, Here ' m '  and ' n '  are positive integers .

Now we substitute these values in equation 1 and get :

( 2 m + 1 )² + ( 2 n  + 1 )² = 3 + 8 c

$\Rightarrow$4 m² + 1 + 4 m + 4 n² + 1 + 4 n = 3 + 8 c              ( We know ( a + b )² = a² + b² + 2 a b )

$\Rightarrow$4 m² + 4 m + 4 n²  + 4 n + 2 = 3 + 8 c

$\Rightarrow$ 2 ( m² + m  +  n² + n + 1) = 3 + 8 c

Here , L.H.S. = 2 ( m² + m  +  n² + n + 1) , That is multiple of " even number = 2 " , So

L.H.S. =  Even number

And  R.H.S. = Odd number  , As we explained in case I .

R.H.S. = Odd number .

Therefore,

We get no solution when ' a '  and ' b '  both are odd number .

Case III : When any one of ' a ' is ' b '  is an odd number , We consider  a = Even number and b = odd number , So

Let a = 2 m + 1  and b = 2 n + 1 , Here ' m '  and ' n '  are positive integers .

Now we substitute these values in equation 1 and get :

( 2 m )² + ( 2 n + 1 )² = 3 + 8 c

$\Rightarrow$4 m² + 4 n² + 1 + 4 n = 3 + 8 c

$\Rightarrow$4 m² + 4 n² + 4 n  = 2 + 8 c

$\Rightarrow$ 4 ( m² +  n² + n ) = 2 ( 1 + 4 c )

Here , L.H.S. = 4 ( m² +  n² + n ) , That is divisible by " 4 " , But R.H.S. = 2 ( 1 + 4 c ) , That is not divisible by " 4 " ,

Therefore,

We get no solution when any one of ' a '  and ' b '  is an odd number .


From Case I , Case II and Case III we get that there is no value of  a , b and c that can satisfy given equation . So

Option ( C )                                                                                   ( Ans )


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