Question

Let \(f:\R \rightarrow \R\) and \(g:\R \rightarrow\R\) be functions defined by
\(f(x)=\left\{ \begin{array}{ll}     x|x|\sin(\frac{1}{x}), &  x\ne0 \\     0, &  x = 0, \end{array} \right.\text{and} \ g(x)=\left\{ \begin{array}{ll}     1-2x, & 0\leq x\leq \frac{1}{2}, \\     0, & \text{otherwise.} \end{array} \right.\)
Let \(a,b,c,d \in \R\). Define the function \(h:\R\rightarrow\R\) by
\(h(x)=af(x)+b(g(x)+g(\frac{1}{2}-x))+c(x-g(x))+d\ g(x), x\in\R\).
Match each entry in List-I to the correct entry in List-II.

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}.

The correct option is

• A. (P) → (4) (Q) → (3) (R) → (1) (S) → (2)
• B. (P) → (5) (Q) → (2) (R) → (4) (S) → (3)
• C. (P) → (5) (Q) → (3) (R) → (2) (S) → (4)
• D. (P) → (4) (Q) → (2) (R) → (1) (S) → (3)

Answer 1

The Correct Option is C

Function Matching

1. For (P): 

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).

2. For (Q):

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).

3. For (R):

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).

4. For (S):

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).

Final Answer:

• (P) → (5)
• (Q) → (3)
• (R) → (2)
• (S) → (4)