Exams Joint Entrance Examination Mathematics Differentiation Question 20 Question 20 of 67 Medium NUMERICAL
Let f : R → R f:R\to R f : R → R be a thrice differentiable odd function satisfying f ′ ( x ) ≥ 0 , f ′ ′ ( x ) = f ( x ) , f ( 0 ) = 0 , f ′ ( 0 ) = 3 {f}^{'}(x)\ge 0,{f}^{''}(x)=f(x),f(0)=0,{f}^{'}(0)=3 f ′ ( x ) ≥ 0 , f ′′ ( x ) = f ( x ) , f ( 0 ) = 0 , f ′ ( 0 ) = 3 . Then 9 f ( log e 3 ) 9f({\log }_{e}3) 9 f ( log e 3 ) is equal to \_\_\_\_\_\_\_\_\_\_ .
Answer: 36
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f ′ ( x ) ≥ 0 , f ′ ′ ( x ) = f ( x ) \begin{gathered} {f}^{'}(x)\ge 0,{f}^{''}(x)=f(x) \end{gathered} f ′ ( x ) ≥ 0 , f ′′ ( x ) = f ( x )
f ( x ) = A e x + B e − x f ( 0 ) = 0 ⇒ A = − B ⇒ f ( x ) = A ( e x − e − x ) f ′ ( x ) = A e x + A e − x = A ( e x + e − x ) f ′ ( 0 ) = 3 = A ( e 0 + e − 0 ) = 2 A ⇒ A = 3 2 f ( x ) = 3 2 ( e x − e − x ) ( ln 3 ) = 27 2 ( e ln 3 − e − ln 3 ) = 27 2 ( 3 − 1 3 ) = 27 2 ⋅ 8 3 = 36 \begin{gathered} f(x)=A{e}^{x}+B{e}^{-x} \\ f(0)=0\Rightarrow A=-B \\ \Rightarrow f(x)=A({e}^{x}-{e}^{-x}) \\ {f}^{'}(x)=A{e}^{x}+A{e}^{-x}=A({e}^{x}+{e}^{-x}) \\ {f}^{'}(0)=3=A({e}^{0}+{e}^{-0})=2A\Rightarrow A=\frac{3}{2} \\ f(x)=\frac{3}{2}({e}^{x}-{e}^{-x}) \\ (\ln 3)=\frac{27}{2}({e}^{\ln 3}-{e}^{-\ln 3})=\frac{27}{2}(3-\frac{1}{3})=\frac{27}{2}\cdot \frac{8}{3} \\ =36 \end{gathered} f ( x ) = A e x + B e − x f ( 0 ) = 0 ⇒ A = − B ⇒ f ( x ) = A ( e x − e − x ) f ′ ( x ) = A e x + A e − x = A ( e x + e − x ) f ′ ( 0 ) = 3 = A ( e 0 + e − 0 ) = 2 A ⇒ A = 2 3 f ( x ) = 2 3 ( e x − e − x ) ( ln 3 ) = 2 27 ( e l n 3 − e − l n 3 ) = 2 27 ( 3 − 3 1 ) = 2 27 ⋅ 3 8 = 36 Answer the question to unlock