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

To solve the problem, we analyze the function \( h(x) \) under the given conditions and match each case in List-I to the correct properties in List-II.

1. Understanding the Functions \( f(x) \) and \( g(x) \):
The function \( f(x) \) is defined as:

\( f(x) = \begin{cases} x|x|\sin\left(\frac{1}{x}\right), & x \neq 0, \\ 0, & x = 0. \end{cases} \)

The function \( g(x) \) is defined as:

\( g(x) = \begin{cases} 1 - 2x, & 0 \leq x \leq \frac{1}{2}, \\ 0, & \text{otherwise}. \end{cases} \)

2. Analyzing \( h(x) \) for Each Case:
The function \( h(x) \) is given by:

\( h(x) = a f(x) + b \left(g(x) + g\left(\frac{1}{2} - x\right)\right) + c (x - g(x)) + d g(x) \).

Case (P): \( a = 0, b = 1, c = 0, d = 0 \)
Here, \( h(x) = g(x) + g\left(\frac{1}{2} - x\right) \).

- For \( 0 \leq x \leq \frac{1}{2} \), \( g(x) = 1 - 2x \) and \( g\left(\frac{1}{2} - x\right) = 1 - 2\left(\frac{1}{2} - x\right) = 2x \). Thus, \( h(x) = 1 \).
- For \( x < 0 \) or \( x > \frac{1}{2} \), \( g(x) = 0 \) and \( g\left(\frac{1}{2} - x\right) = 0 \). Thus, \( h(x) = 0 \).

The range of \( h(x) \) is \( \{0, 1\} \), and \( h(x) \) is not one-one, not onto, and not differentiable everywhere (due to discontinuities at \( x = 0 \) and \( x = \frac{1}{2} \)).

Match: (P) → (5).

Case (Q): \( a = 1, b = 0, c = 0, d = 0 \)
Here, \( h(x) = f(x) \).

- The function \( f(x) \) is differentiable everywhere, including at \( x = 0 \) (as \( \lim_{x \to 0} x|x|\sin\left(\frac{1}{x}\right) = 0 \)).
- The range of \( f(x) \) is \( \mathbb{R} \) because \( x|x|\sin\left(\frac{1}{x}\right) \) oscillates between \( -x|x| \) and \( x|x| \), covering all real values as \( x \) varies.

Thus, \( h(x) \) is differentiable on \( \mathbb{R} \) and onto.

Match: (Q) → (3).

Case (R): \( a = 0, b = 0, c = 1, d = 0 \)
Here, \( h(x) = x - g(x) \).

- For \( 0 \leq x \leq \frac{1}{2} \), \( h(x) = x - (1 - 2x) = 3x - 1 \).
- For \( x < 0 \) or \( x > \frac{1}{2} \), \( h(x) = x \).

The function \( h(x) \) is one-one because it is strictly increasing (its derivative is positive where defined). It is also onto because as \( x \to \infty \), \( h(x) \to \infty \), and as \( x \to -\infty \), \( h(x) \to -\infty \).

Match: (R) → (2).

Case (S): \( a = 0, b = 0, c = 0, d = 1 \)
Here, \( h(x) = g(x) \).

- The range of \( g(x) \) is \( [0, 1] \) because for \( 0 \leq x \leq \frac{1}{2} \), \( g(x) \) decreases linearly from 1 to 0, and \( g(x) = 0 \) otherwise.
- The function \( g(x) \) is not one-one, not onto \( \mathbb{R} \), and not differentiable at \( x = 0 \) and \( x = \frac{1}{2} \).

Match: (S) → (4).

Final Answer:
The correct matching is:
\(\boxed{(P) \rightarrow (5); (Q) \rightarrow (3); (R) \rightarrow (2); (S) \rightarrow (4)}\).

Answer 2

To solve the matching problem, we need to analyze the function \(h(x)=af(x)+b(g(x)+g(\frac{1}{2}-x))+c(x-g(x))+d\ g(x)\) for each set of parameters given in List-I.

Analysis:

• (P) If \(a = 0, b = 1, c = 0, d = 0\), then \(h(x) = g(x) + g(\frac{1}{2} - x)\). 
  Analyzing \(g(x)\) and \(g(\frac{1}{2} - x)\):
  • For \(0 \leq x \leq \frac{1}{2}\), \(g(x) = 1 - 2x\).
  • \(g(\frac{1}{2} - x)\) will be nonzero only f\)
  • Hence, \(h(x) = 1 - 2x + x = 1\) for \(0 \leq x \leq \frac{1}{2}\) and \(0\) otherwise.
• (Q) If \(a = 1, b = 0, c = 0, d = 0\), then \(h(x) = f(x)\):
  • For \(x \neq 0\), \(f(x) = x|x|\sin\left(\frac{1}{x}\right)\), which is continuous but not onto as its range depends intricately on \(\sin\left(\frac{1}{x}\right)\).
• (R) If \(a = 0, b = 0, c = 1, d = 0\), then \(h(x) = x - g(x)\):
  • For \(0 \leq x \leq \frac{1}{2}\), \(h(x) = x - (1 - 2x) = 3x - 1\), \(h(x) = x\) otherwise.
• (S) If \(a = 0, b = 0, c = 0, d = 1\), then \(h(x) = g(x)\):
  • This function is defined as in the expression for \(g\), so the range is [0, 1] when considering \(g(x)\) over its constraints.\)

Final Matches:

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