Since ,
f ' ( x ) ≤ 1 for all " x " ,
So , f ( x ) ≤ x for all " x " .
Hence , 0∫1 f 3 ( x ) dx ≤ 0∫1 x 3 dx ≤ 14
Obviously , the numerator is simply " f ( 1 ) " , which cannot be negative since " f ' ( x ) > 0 " and " f ( 0 ) = 0 " .
Hence , the entire expression ≥ 4 f ( 1 )
But maximum value of " f ( 1 ) " is clearly 1 , as " f ( x ) ≤ x " .
Since the expression holds for any " f ( 1 ) " , so it must hold true for " f ( 1 ) = 1 " also .
Hence , it is ≥ 4 .
So , we easily see that the answer is - ( d ) .