Question

Match List-I with List-II and choose the correct option:

\[ \begin{array}{|l|l|} \hline \textbf{LIST-I (Function)} & \textbf{LIST-II (Expansion)} \\ \hline A. \log(1-x) & I. 1 + \frac{1}{3} + \frac{1}{6} + \frac{3}{40} + \frac{15}{336} + \dots \\ \hline B. \sin^{-1} x & II. 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \dots \\ \hline C. \log 2 & III. x + \frac{1}{2} \frac{x^3}{3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{x^5}{5} + \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{x^7}{7} + \dots, -1 < x \le 1 \\ \hline D. \frac{\pi}{2} & IV. -x - \frac{x^2}{2} - \frac{x^3}{3} - \dots, -1 \le x < 1 \\ \hline \end{array} \]

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

Hint:

Memorize the basic Maclaurin series for functions like \( e^x, \sin x, \cos x, \log(1+x), (1+x)^p \). Many other series, like \( \sin^{-1} x \), can be derived from these by differentiation or integration. For matching questions, even if one pair seems obscure, finding the other correct pairs can often lead you to the right answer.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This question requires matching functions or constants with their corresponding Taylor/Maclaurin series expansions or series representations.

Step 2: Detailed Explanation:

A. \( \log(1-x) \): The standard Maclaurin series for \( \log(1+u) \) is \( u - \frac{u^2}{2} + \frac{u^3}{3} - ... \). Substituting \( u = -x \), we get: \[ \log(1-x) = (-x) - \frac{(-x)^2}{2} + \frac{(-x)^3}{3} - ... = -x - \frac{x^2}{2} - \frac{x^3}{3} - ... \] This matches IV.
B. \( \sin^{-1} x \): The Maclaurin series for \( \sin^{-1} x \) is obtained by integrating the series for its derivative, \( (1-x^2)^{-1/2} \). The resulting series is: \[ \sin^{-1} x = x + \frac{1}{2}\frac{x^3}{3} + \frac{1 \cdot 3}{2 \cdot 4}\frac{x^5}{5} + ... \] This matches III.
C. \( \log 2 \): This is the value of the series for \( \log(1+x) \) when \( x=1 \). \[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ... \] Setting \( x=1 \), we get the alternating harmonic series: \[ \log 2 = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... \] This matches II.
D. \( \frac{\pi}{2} \): Given the other matches, \( \frac{\pi}{2} \) must correspond to series I. This is a non-standard series and likely represents a specific, less common result, or there may be a typo in the question. However, since the other three pairs A-IV, B-III, and C-II are definitive, the overall matching must be correct. For example, the series for \( \sin^{-1} x \) at \(x=1\) is \( 1 + \frac{1}{6} + \frac{3}{40} + ... = \frac{\pi}{2}\). The listed series I is different, suggesting a typo, but the matching pattern is clear.
Step 3: Final Answer:
The correct pairings are A-IV, B-III, C-II, and D-I. This corresponds to option (A).