Let f ( x ) = x 5 − 5 x + a , a ∈ R , Then which of the following statements are correct 1 . I f a < − 4 t h e n 0 r e a l r o o t s . 2 . I f a < − 4 t h e n 1 r e a l r o o t . 3 . I f a < − 4 t h e n 2 r e a l r o o t s . 4 . I f a < − 4 t h e n 3 r e a l r o o t s . 5 . I f a < − 4 t h e n 4 r e a l r o o t s . 6 . I f a < − 4 t h e n 5 r e a l r o o t s . 7 . I f 4 > a > − 4 t h e n 0 r e a l r o o t s . 8 . I f 4 > a > − 4 t h e n 1 r e a l r o o t . 9 . I f 4 > a > − 4 t h e n 2 r e a l r o o t s . 1 0 . I f 4 > a > − 4 t h e n 3 r e a l r o o t s . 1 1 . I f 4 > a > − 4 t h e n 4 r e a l r o o t s . 1 2 . I f 4 > a > − 4 t h e n 5 r e a l r o o t s . 1 3 . I f a > 4 t h e n 0 r e a l r o o t s . 1 4 . I f a > 4 t h e n 1 r e a l r o o t . 1 5 . I f a > 4 t h e n 2 r e a l r o o t s . 1 6 . I f a > 4 t h e n 3 r e a l r o o t s . 1 7 . I f a > 4 t h e n 4 r e a l r o o t s . 1 8 . I f a > 4 t h e n 5 r e a l r o o t s . S u b m i t t h e a n s w e r a s p r o d u c t o f t h e c o r r e c t o p t i o n s .
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Let a ( x ) = x 5 − 5 x a n d b ( x ) = − a . Draw the graph of a(x) to find out that for a<4, 1 solution, a>-4,1 solution and |a|<4, 3 solutions (By shifting the y axis).
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Let f ( x ) = x 5 − 5 x + a
By descartes' rule of signs , we note that f ( x ) has atmost 3 distinct real roots for any value of a .
Denote the points of isolated extremum as α 1 and α 2 . Note that f ′ ( α 1 ) = f ′ ( α 2 ) = 0 .
Now, f has exactly 3 real distinct roots if and only if f ( α 1 ) f ( α 2 ) < 0
And f has exactly one real (non repeated) root if and only if f ( α 1 ) f ( α 2 ) > 0
Now, f ( α 1 ) f ( α 2 ) = ( a − 4 ) ( a + 4 )
Hence the statements 2 , 1 0 and 1 4 are right.
So, the answer is 2 8 0