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:
The domain of \( f(x) = \sqrt{4 - x^2} \) is \( -2 \leq x \leq 2 \), and the domain of \( g(x) = \sqrt{x^2 - 1} \) is \( x \leq -1 \) or \( x \geq 1 \). For the function \( h(x) = f(x) + g(x) \) to be defined, the domain must satisfy both conditions: Thus, the domain of \( h(x) \) is \( [-2, 1] \cup [1, 2] \).
Thus, the correct answer is (D).
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}. |