Let ( 2 , 3 ) be the largest open interval in which the function f ( x ) = 2 lo g e ( x − 2 ) − x 2 + a x + 1 is strictly increasing and (b, c) be the largest open interval, in which the function g

Question

Let ( 2 , 3 ) be the largest open interval in which the function f ( x ) = 2 lo g e ​ ( x − 2 ) − x 2 + a x + 1 is strictly increasing and (b, c) be the largest open interval, in which the function g ( x ) = ( x − 1 ) 3 ( x + 2 − a ) 2 is strictly decreasing. Then 100 ( a + b − c ) is equal to :

TestHub · Accepted Answer

​ f ′ ( x ) = x − 2 2 ​ − 2 x + a ≥ 0 f ′′ ( x ) = ( x − 2 ) 2 − 2 ​ − 2 < 0 f ′ ( x ) ↓ f ′ ( 3 ) ≥ 0 2 − 6 + a ≥ 0 a ≥ 4 a m i n ​ = 4 g ( x ) = ( x − 1 ) 3 ( x + 2 − a ) 2 g ( x ) = ( x − 1 ) 3 ( x − 2 ) 2 g ′ ( x ) = ( x − 1 ) 3 2 ( x − 2 ) + ( x − 2 ) 2 3 ( x − 1 ) 2 = ( x − 1 ) 2 ( x − 2 ) ( 2 x − 2 + 3 x − 6 ) = ( x − 1 ) 2 ( x − 2 ) ( 5 x − 8 ) < 0 x ∈ ( 5 8 ​ , 2 ) 100 ( a + b − c ) = 100 ( 4 + 5 8 ​ − 2 ) = 360 ​

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