Question

Match List-I with List-II.
\[ \begin{array}{|c|c|} \hline \textbf{List-I (Function)} & \textbf{List-II (Interval in which function is increasing)} \\ \hline \frac{x}{\log_e x} & (-\infty, -2) \cup (2, \infty) \\ \hline \frac{x}{2} + \frac{2}{x}, x \neq 0 & \left(-\frac{\pi}{4}, \frac{\pi}{4}\right) \\ \hline x^x, x > 0 & \left(\frac{1}{e}, \infty\right) \\ \hline \sin x - \cos x & (e, \infty) \\ \hline \end{array} \]
Choose the correct answer from the options given below:

• A. (A-II), (B-I), (C-III), (D-IV)
• B. (A-I), (B-III), (C-IV), (D-II)
• C. (A-IV), (B-I), (C-III), (D-II)
• D. (A-III), (B-IV), (C-I), (D-II)

Answer 1

The Correct Option is C

To solve this type of matching problem, we analyze each function from List-I and determine where it is increasing. Let's go through each one: 

(A) \(\frac{x}{\log_e x}\)

To find where \(\frac{x}{\log_e x}\) is increasing, we can take its derivative and determine where it is positive. The derivative is:

\[ f'(x) = \frac{\log_e x - 1}{(\log_e x)^2} \]

Set \(f'(x) > 0\):

\[ \log_e x - 1 > 0 \Rightarrow \log_e x > 1 \Rightarrow x > e \]

The function is increasing in \((e, \infty)\).

(B) \(\frac{x}{2} + \frac{2}{x}, x \neq 0\)

Taking the derivative:

\[ f'(x) = \frac{1}{2} - \frac{2}{x^2} \]

Set \(f'(x) > 0\):

\[ \frac{1}{2} > \frac{2}{x^2} \Rightarrow x^2 > 4 \Rightarrow x > 2 \text{ or } x < -2 \]

The function is increasing in \((-\infty, -2) \cup (2, \infty)\).

(C) \(x^x, x > 0\)

Taking the derivative using the logarithm derivative rule:

\[ f(x) = x^x \Rightarrow \log y = x \log x \]

\[ f'(x) = x^x (\log x + 1) \]

Set \(f'(x) > 0\):

\[ \log x + 1 > 0 \Rightarrow \log x > -1 \Rightarrow x > \frac{1}{e} \]

The function is increasing in \(\left(\frac{1}{e}, \infty\right)\).

(D) \(\sin x - \cos x\)

Taking the derivative:

\[ f'(x) = \cos x + \sin x \]

Set \(f'(x) > 0\):

Use the range expression: \(\cos x + \sin x = \sqrt{2} \sin(x + \frac{\pi}{4})\), so

\[ \sin(x + \frac{\pi}{4}) > 0 \Rightarrow -\frac{\pi}{4} < x < \frac{\pi}{4} \]

The function is increasing in \((- \frac{\pi}{4}, \frac{\pi}{4})\).

Matching Results:

Comparing these intervals with List-II:

List-I (Function)	List-II (Interval in which function is increasing)
(A) \(\frac{x}{\log_e x}\)	(e, \infty)
(B) \(\frac{x}{2} + \frac{2}{x}, x \neq 0\)	(-\infty, -2) \cup (2, \infty)
(C) \(x^x, x > 0\)	\(\left(\frac{1}{e}, \infty\right)\)
(D) \(\sin x - \cos x\)	\((- \frac{\pi}{4}, \frac{\pi}{4})\)

Therefore, the correct answer is: (A-IV), (B-I), (C-III), (D-II)

Answer 2

(A) The function \( \frac{x}{\log_e x} \) is increasing for \( x > e \) (interval (IV)).

(B) The function \( \frac{x^2 + 1}{x - 2} \) is increasing in the intervals \( (-\infty, -2) \cup (2, \infty) \) (interval (I)).

(C) The exponential function \( e^x \) is increasing for \( x > 0 \), and the interval where the function is increasing for \( e^x \) is \( \left( \frac{1}{e}, \infty \right) \) (interval (III)).

(D) The function \( \sin x - \cos x \) is increasing in the interval \( \left( -\frac{\pi}{4}, \frac{\pi}{4} \right) \) (interval (II)).