\[ f(x) = \begin{cases} x + 1, & x < 0 \\ 1 - x, & 0 \leq x < 1 \\ x - 1, & x \geq 1 \end{cases} \]
\[ g(x) = \begin{cases} x + 1, & x < 0 \\ 1, & x \geq 0 \end{cases} \]
For \( x < 0 \), \( f(x) = x + 1 \). Substituting \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g(x + 1) = (x + 1) + 1 = x + 2. \]
For \( 0 \leq x < 1 \), \( f(x) = 1 - x \). Substituting \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g(1 - x). \]
Since \( 1 - x \geq 0 \) for \( 0 \leq x < 1 \), \( g(1 - x) = 1 \).
For \( x \geq 1 \), \( f(x) = x - 1 \). Substituting \( f(x) \) into \( g(x) \):
\[ g(f(x)) = g(x - 1). \]
Since \( x - 1 \geq 0 \) for \( x \geq 1 \), \( g(x - 1) = 1 \).
Combining all cases:
\[ g(f(x)) = \begin{cases} x + 2, & x < 0 \\ 1, & x \geq 0 \end{cases} \]
The function \( g(f(x)) \) is continuous everywhere because there are no jumps or breaks in its definition.
List - I | List - II | ||
(P) | If a = 0, b = 1, c = 0 and d = 0, then | (1) | h is one-one. |
(Q) | If a = 1, b = 0, c = 0 and d = 0, then | (2) | h is onto. |
(R) | If a = 0, b = 0, c = 1 and d = 0, then | (3) | h is differentiable on \(\R\) |
(S) | If a = 0, b = 0, c = 0 and d = 1, then | (4) | the range of h is [0, 1]. |
(5) | the range of h is {0, 1}. |
Let \( f(x) = \sqrt{4 - x^2} \), \( g(x) = \sqrt{x^2 - 1} \). Then the domain of the function \( h(x) = f(x) + g(x) \) is equal to: