Question

Consider $f(x) = (x - 1) \tan^{-1}x - \frac{1}{2} \ln(1 + x^2), x > 1$. Identify which of the following statement(s) is(are) correct?

• A. equation $f(x) = 0$ has exactly one real root in $(1, \infty)$
• B. $f'(x)$ is increasing in $(1, \infty)$
• C. $f(x+4) - f(x) > 2\pi \ \forall \ x \in (1, \infty)$
• D. $f(x + 4) - f(x) < 2\pi \ \forall \ x \in (1, \infty)$

Answer 1

Answer: (A), (B), (D)

A, B, D

Answer 2

The problem asks us to analyze the given function $f(x) = (x - 1) \tan^{-1}x - \frac{1}{2} \ln(1 + x^2)$ for $x > 1$ and determine the correctness of the provided statements.

Step 1: Calculate the first derivative, $f'(x)$. Using the product rule for $(x-1)\tan^{-1}x$ and the chain rule for $\frac{1}{2}\ln(1+x^2)$: $\frac{d}{dx} [(x-1)\tan^{-1}x] = 1 \cdot \tan^{-1}x + (x-1) \cdot \frac{1}{1+x^2}$ $\frac{d}{dx} \left[ \frac{1}{2}\ln(1+x^2) \right] = \frac{1}{2} \cdot \frac{1}{1+x^2} \cdot 2x = \frac{x}{1+x^2}$ So, $f'(x) = \tan^{-1}x + \frac{x-1}{1+x^2} - \frac{x}{1+x^2}$ $f'(x) = \tan^{-1}x + \frac{x-1-x}{1+x^2}$ $f'(x) = \tan^{-1}x - \frac{1}{1+x^2}$

Step 2: Calculate the second derivative, $f''(x)$. $f''(x) = \frac{d}{dx} \left( \tan^{-1}x - \frac{1}{1+x^2} \right)$ $f''(x) = \frac{1}{1+x^2} - \frac{-(2x)}{(1+x^2)^2}$ $f''(x) = \frac{1}{1+x^2} + \frac{2x}{(1+x^2)^2}$ To combine these terms, find a common denominator: $f''(x) = \frac{1+x^2}{(1+x^2)^2} + \frac{2x}{(1+x^2)^2}$ $f''(x) = \frac{1+x^2+2x}{(1+x^2)^2}$ $f''(x) = \frac{(1+x)^2}{(1+x^2)^2}$

Step 3: Analyze statement (B) - $f'(x)$ is increasing in $(1, \infty)$. For $x > 1$, $(1+x)^2 > 0$ and $(1+x^2)^2 > 0$. Therefore, $f''(x) > 0$ for all $x \in (1, \infty)$. Since $f''(x) > 0$, $f'(x)$ is strictly increasing in $(1, \infty)$. Thus, statement (B) is correct.

Step 4: Analyze statement (A) - equation $f(x) = 0$ has exactly one real root in $(1, \infty)$. To determine the number of roots, we need to understand the behavior of $f(x)$. First, evaluate the limits of $f'(x)$: $\lim_{x \to 1^+} f'(x) = \tan^{-1}(1) - \frac{1}{1+1^2} = \frac{\pi}{4} - \frac{1}{2}$ Since $\pi \approx 3.14159$, $\frac{\pi}{4} \approx 0.785$, so $\frac{\pi}{4} - \frac{1}{2} \approx 0.785 - 0.5 = 0.285 > 0$. $\lim_{x \to \infty} f'(x) = \lim_{x \to \infty} \left( \tan^{-1}x - \frac{1}{1+x^2} \right) = \frac{\pi}{2} - 0 = \frac{\pi}{2}$ Since $f'(x)$ is strictly increasing and $f'(1^+) > 0$, it implies that $f'(x) > 0$ for all $x \in (1, \infty)$. This means $f(x)$ is strictly increasing in $(1, \infty)$.

Now, evaluate the limits of $f(x)$: $\lim_{x \to 1^+} f(x) = (1-1)\tan^{-1}(1) - \frac{1}{2}\ln(1+1^2) = 0 - \frac{1}{2}\ln(2) = -\frac{1}{2}\ln(2)$ Since $\ln(2) > 0$, $f(1^+) < 0$. $\lim_{x \to \infty} f(x) = \lim_{x \to \infty} \left[ (x-1)\tan^{-1}x - \frac{1}{2}\ln(1+x^2) \right]$ As $x \to \infty$, $\tan^{-1}x \to \frac{\pi}{2}$. So, the first term approaches $(x-1)\frac{\pi}{2}$. The second term is $-\frac{1}{2}\ln(1+x^2) \approx -\frac{1}{2}\ln(x^2) = -\ln x$. So, for large $x$, $f(x) \approx \frac{\pi}{2}x - \frac{\pi}{2} - \ln x$. Since the linear term $\frac{\pi}{2}x$ grows much faster than the logarithmic term $\ln x$, $\lim_{x \to \infty} f(x) = \infty$.

Since $f(x)$ is continuous, strictly increasing, starts from a negative value ($-\frac{1}{2}\ln(2)$) and goes to positive infinity, by the Intermediate Value Theorem, there must be exactly one real root for $f(x) = 0$ in $(1, \infty)$. Thus, statement (A) is correct.

Step 5: Analyze statements (C) and (D) - $f(x+4) - f(x)$ vs $2\pi$. By the Mean Value Theorem, for any $x \in (1, \infty)$, there exists a $c \in (x, x+4)$ such that: $f(x+4) - f(x) = (x+4 - x) f'(c) = 4 f'(c)$ Since $c \in (x, x+4)$ and $x > 1$, we have $c > 1$. We know that $f'(x)$ is strictly increasing in $(1, \infty)$ and its limit as $x \to \infty$ is $\frac{\pi}{2}$. This means that for any finite value of $c > 1$, $f'(c)$ will always be less than its limiting value. So, $f'(c) < \frac{\pi}{2}$ for all $c \in (1, \infty)$. Therefore, $4 f'(c) < 4 \cdot \frac{\pi}{2}$ $f(x+4) - f(x) < 2\pi$ This holds for all $x \in (1, \infty)$. Thus, statement (D) is correct. Consequently, statement (C) is incorrect.

Conclusion: Statements (A), (B), and (D) are correct.

The final answer is $\boxed{A, B, D}$

Explanation of the solution:

1. Calculate Derivatives: Find the first derivative $f'(x) = \tan^{-1}x - \frac{1}{1+x^2}$ and the second derivative $f''(x) = \frac{(1+x)^2}{(1+x^2)^2}$.
2. Analyze $f''(x)$ for Monotonicity of $f'(x)$: Since $x > 1$, $f''(x) > 0$. This implies $f'(x)$ is strictly increasing in $(1, \infty)$. So, (B) is correct.
3. Analyze $f'(x)$ for Monotonicity of $f(x)$: Calculate $\lim_{x \to 1^+} f'(x) = \frac{\pi}{4} - \frac{1}{2} > 0$. Since $f'(x)$ is increasing and starts positive, $f'(x) > 0$ for all $x \in (1, \infty)$. This means $f(x)$ is strictly increasing in $(1, \infty)$.
4. Analyze $f(x)$ for Roots: Calculate $\lim_{x \to 1^+} f(x) = -\frac{1}{2}\ln(2) < 0$. Calculate $\lim_{x \to \infty} f(x) = \infty$ (since $\frac{\pi}{2}x$ dominates $\ln x$). As $f(x)$ is continuous and strictly increasing from a negative value to positive infinity, it crosses the x-axis exactly once. So, $f(x)=0$ has exactly one real root in $(1, \infty)$. Thus, (A) is correct.
5. Analyze $f(x+4)-f(x)$ using Mean Value Theorem: Apply MVT: $f(x+4) - f(x) = 4f'(c)$ for some $c \in (x, x+4)$. Since $c > 1$ and $f'(x)$ is increasing with $\lim_{x \to \infty} f'(x) = \frac{\pi}{2}$, we have $f'(c) < \frac{\pi}{2}$. Therefore, $4f'(c) < 4 \cdot \frac{\pi}{2} = 2\pi$. So, $f(x+4) - f(x) < 2\pi$. Thus, (D) is correct and (C) is incorrect.

Answer: The correct statements are (A), (B), and (D).

The final answer is $\boxed{A, B, D}$