How many integers a are there such that f ( x ) = ( x 2 + a x + 2 a ) e x is a strictly increasing function in ( − ∞ , ∞ ) ?
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@sauresh bhowmick , I have done up the LaTex for you. Hope that you can learn up fast.
Could u pls explain me how u did the squaring part and then dividing it by 2
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That was completing the square technique. There's a wiki page on it.
A better explanation would be why this needed to be done...
f ( x ) f ′ ( x ) = ( x 2 + a x + 2 a ) e x = ( x 2 + a x + 2 a ) e x + ( 2 x + a ) e x = ( x 2 + ( a + 2 ) x + 3 a ) e x
For f ( x ) to be increasing, f ′ ( x ) ≥ 0 . Since e x > 0 for all x , we need g ( x ) = x 2 + ( a + 2 ) x + 3 a ≥ 0 . That is when g ( x ) has no real roots or
( a + 2 ) 2 − 1 2 a a 2 + 4 a + 4 − 1 2 a a 2 − 8 a + 4 ( a − 4 + 2 3 ) ( a − 4 − 2 3 ) ≤ 0 ≤ 0 ≤ 0 ≤ 0
⟹ 4 − 2 3 ≤ a ≤ 4 + 2 3 ≈ 0 . 5 3 4 ≤ a ≤ 7 . 4 6 4 or for integers 1 ≤ a ≤ 7 , a total of 7 integral a .
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f ( x ) = e x ( x 2 + a x + 2 a ) , so, f ′ ( x ) = e x ( x 2 + ( a + 2 ) x + 3 a . For f ( x ) to be increasing f ′ ( x ) ≥ 0 . That is e x ( x 2 + ( a + 2 ) x + 3 a ) ≥ 0 ⟹ x 2 + ( a + 2 ) x + 3 a ) ≥ 0 as e x ≥ 0 , and ( x + ( a + 2 ) / 2 ) 2 + ( 3 a − ( ( a + 2 ) / 2 ) 2 ) ≥ 0 , which leads us to 1 2 a ≥ ( a + 2 ) 2 ⟹ a 2 − 8 a + 4 ≤ 0 ⟹ 0 . 5 3 ≤ a ≤ 7 . 4 6 , Since a is integer, so number of possible values of a is 7 .