Question

Match the LIST-I with LIST-II

LIST-I Indefinite Integral	LIST-II Solution (where \(c\) is an arbitrary constant)
A. \(\int\sqrt{16-x^2}\,dx\)	I. \(\sin^{-1}\left(\frac{x}{4}\right)+c\)
B. \(\int\frac{dx}{\sqrt{16-x^2}}\)	II. \(\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\left(\frac{x}{4}\right)+c\)
C. \(\int\frac{dx}{\sqrt{x^2-16}}\)	III. \(\frac{1}{8}\log\left|\frac{4+x}{4-x}\right|+c\)
D. \(\int\frac{dx}{16-x^2}\)	IV. \(\log\left|x+\sqrt{x^2-16}\right|+c\)

Choose the correct answer from the options given below:

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

Hint:

Use the standard formulas for \(\sqrt{a^2-x^2}\), \(\frac{1}{\sqrt{a^2-x^2}}\), \(\frac{1}{\sqrt{x^2-a^2}}\) and \(\frac{1}{a^2-x^2}\) with \(a=4\).

Answer 1

The Correct Option is D

Step 1: Approach:
We check the matches by differentiating the answers in List-II. If the derivative gives the integrand in List-I, the pair is correct.

Step 2: Differentiate II:
$\frac{d}{dx}\left[\frac{x}{2}\sqrt{16-x^2}+8\sin^{-1}\frac{x}{4}\right]=\frac12\sqrt{16-x^2}-\frac{x^2}{2\sqrt{16-x^2}}+\frac{8}{\sqrt{16-x^2}}$. Put over a common denominator: $\frac{(16-x^2)-x^2+16}{2\sqrt{16-x^2}}=\frac{32-2x^2}{2\sqrt{16-x^2}}=\sqrt{16-x^2}$. So A goes with II.

Step 3: Differentiate I:
$\frac{d}{dx}\sin^{-1}\frac{x}{4}=\frac{1/4}{\sqrt{1-x^2/16}}=\frac{1}{\sqrt{16-x^2}}$. So B goes with I.

Step 4: Differentiate III and IV:
$\frac18\frac{d}{dx}\left[\log(4+x)-\log(4-x)\right]=\frac18\left[\frac{1}{4+x}+\frac{1}{4-x}\right]=\frac18\cdot\frac{8}{16-x^2}=\frac{1}{16-x^2}$, so D goes with III. And $\frac{d}{dx}\log\left(x+\sqrt{x^2-16}\right)=\frac{1}{\sqrt{x^2-16}}$, so C goes with IV.

Step 5: Choose:
The set A-II, B-I, C-IV, D-III is option 4.

Final Answer:
\[ \boxed{\text{A-II, B-I, C-IV, D-III}} \]