f ( x ) = { lim n → ∞ x ( x 2 n − x − 2 n ) x 2 n + x − 2 n ( x ≠ 0 ) 0 ( x = 0 ) \displaystyle f ( x ) = {\begin{cases} \lim\limits _{n \rightarrow \infty} {\frac{x \left( x ^{2 n} - x ^{- 2 n} \right)}{x ^{2 n} + x ^{- 2 n}}} & \left( x \ne 0 \right) \\ 0 & \left( x = 0 \right) \end{cases}} f ( x ) = ⎩ ⎨ ⎧ n → ∞ lim x 2 n + x − 2 n x ( x 2 n − x − 2 n ) 0 ( x = 0 ) ( x = 0 ) f ( x ) = { − x ( − 1 < x < 1 ) 0 ( x = ± 1 ) x ( x < − 1 , x < 1 ) \displaystyle f ( x ) = {\begin{cases} - x & \left( - 1 < x < 1 \right) \\ 0 & \left( x = \pm 1 \right) \\ x & \left( x < - 1 , x < 1 \right) \end{cases}} f ( x ) = ⎩ ⎨ ⎧ − x 0 x ( − 1 < x < 1 ) ( x = ± 1 ) ( x < − 1 , x < 1 )
이 때, f ( x ) = ( x − k ) 2 f ( x ) = ( x - k ) ^{2} f ( x ) = ( x − k ) 2 3 3 3 k = 0 k = 0 k = 0
ㄴ) y = − x y = - x y = − x y = ( x − k ) 2 y = \left( x - k \right) ^{2} y = ( x − k ) 2 − x = ( x − k ) 2 - x = ( x - k ) ^{2} − x = ( x − k ) 2 x 2 − ( 2 k − 1 ) x + k 2 = 0 x ^{2} - ( 2 k - 1 ) x + k ^{2} = 0 x 2 − ( 2 k − 1 ) x + k 2 = 0 D = ( 2 k − 1 ) 2 − 4 k 2 D = ( 2 k - 1 ) ^{2} - 4 k ^{2} D = ( 2 k − 1 ) 2 − 4 k 2 = − 4 k + 1 = 0 = - 4 k + 1 = 0 = − 4 k + 1 = 0 k = 1 4 \displaystyle k = \frac{1}{4} k = 4 1
ㄱ),ㄴ)의 사이 영역에서 실근 개수가 3개 성립
∴ \therefore ∴ 0 < k < 1 4 \displaystyle 0 < k < \frac{1}{4} 0 < k < 4 1 ∴ \therefore ∴ a + b = 0 + 1 4 = 1 4 \displaystyle a + b = 0 + \frac{1}{4} = \frac{1}{4} a + b = 0 + 4 1 = 4 1
