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}. |
If a = 0, b = 1, c = 0, d = 0, then:
\[ h(x) = g(x), \quad g(x) = \begin{cases} 1 - 2x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text{otherwise}. \end{cases} \]
The range of \( g(x) \) is {0, 1}, as it attains only these two values. Thus:
(P) matches (5).
If a = 1, b = 0, c = 0, d = 0, then:
\[ h(x) = f(x), \quad f(x) = \begin{cases} x |x| \sin(1/x), & x \neq 0, \\ 0, & x = 0. \end{cases} \]
The function \( f(x) \) is continuous, and its derivative exists everywhere on \( \mathbb{R} \), making it differentiable. Thus:
(Q) matches (3).
If a = 0, b = 0, c = 1, d = 0, then:
\[ h(x) = g\left(\frac{1}{2} - x\right), \quad g\left(\frac{1}{2} - x\right) = \begin{cases} 2x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text{otherwise}. \end{cases} \]
The function \( h(x) \) maps every \( x \in \mathbb{R} \) onto the range \([0,1]\), making it onto. Thus:
(R) matches (2).
If a = 0, b = 0, c = 0, d = 1, then:
\[ h(x) = x - g(x), \quad g(x) = \begin{cases} 1 - 2x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text{otherwise}. \end{cases} \]
The range of \( h(x) \) is \([0,1]\), as verified by substituting boundary conditions. Thus:
(S) matches (4).
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: