Exams Joint Entrance Examination Mathematics Application of Derivatives Question 15 Question 15 of 212 Medium MCQ
Let ( 2 , 3 ) (2,3) ( 2 , 3 ) be the largest open interval in which the function f ( x ) = 2 log e ( x − 2 ) − x 2 + a x + 1 f(x)=2{\log }_{e}(x-2)-{x}^{2}+ax+1 f ( x ) = 2 log 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 g(x)=(x-1{)}^{3}(x+2-a{)}^{2} g ( x ) = ( x − 1 ) 3 ( x + 2 − a ) 2 is strictly decreasing. Then 100 ( a + b − c ) 100(a+b-c) 100 ( a + b − c ) is equal to :
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f ′ ( x ) = 2 x − 2 − 2 x + a ≥ 0 f ′ ′ ( x ) = − 2 ( x − 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 ∈ ( 8 5 , 2 ) 100 ( a + b − c ) = 100 ( 4 + 8 5 − 2 ) = 360 \begin{gathered} {f}^{'}(x)=\frac{2}{x-2}-2x+a\ge 0 \\ {f}^{''}(x)=\frac{-2}{(x-2{)}^{2}}-2<0 \\ {f}^{'}(x)\downarrow \\ {f}^{'}(3)\ge 0 \\ 2-6+a\ge 0 \\ a\ge 4 \\ {a}_{min}=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)(2x-2+3x-6) \\ =(x-1{)}^{2}(x-2)(5x-8)<0 \\ x\in (\frac{8}{5},2) \\ 100(a+b-c)=100(4+\frac{8}{5}-2)=360 \end{gathered} 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 min = 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 Answer the question to unlock