1hey a nice approach!!!
Q>The equation f(x)=0 has 8 distinct real solutions & f(4+x)=f(4-x) then sum of all 8 solutions of f(x)=0 are-
a>12 b>32 c>16 d>15
Please describe the solution.
-
UP 0 DOWN 0 0 2
2 Answers
1Suppose that , any four solutions are of the form ,
" 4 + x1 , 4 + x2 , 4 + x3 , 4 + x4 " ;
where " x1 , x2 , x3 , x4 " are constants .
Then , Given " f ( 4 + x ) = f ( 4 - x ) " ;
Hence , if " 4 + x1 " is a solution , so is " 4 - x1 " .
So the other four solutions must be , " 4 - x1 . 4 - x2 , 4 - x3 , 4 - x4 " .
So , sum of the roots = 4 x 8 = 32
So the ans. is - ( a ) , ( c ) , ( d ) .