Question

Consider a function $f: R -\{(2n-1)\frac{\pi}{2}, n \in I\} \rightarrow R$ defined as $f(x)=\int \tan^4 x dx, f(0)= 0.$ Then which of the following option(s) is/are correct?

• A. f(x) is increasing function $\forall x \in R-\{(2n-1)\frac{\pi}{2}, n \in I\}$
• B. Range of f(x) is $(-\infty, \infty)$
• C. f(x) is bijective function
• D. The sum of solutions of the equation f(x) = x, belonging to $[0, 2r \pi], r \in N$, is $r(6r+1)$

Answer 1

Answer: (A), (B), (C)

A, B, C

Answer 2

The given function is $f(x)=\int \tan^4 x dx$, with $f(0)=0$. The domain is $R -\{(2n-1)\frac{\pi}{2}, n \in I\}$.

First, let's find the function $f(x)$: We know that $\tan^2 x = \sec^2 x - 1$. $f(x) = \int \tan^4 x dx = \int \tan^2 x \cdot \tan^2 x dx$ $f(x) = \int \tan^2 x (\sec^2 x - 1) dx$ $f(x) = \int (\tan^2 x \sec^2 x - \tan^2 x) dx$ $f(x) = \int \tan^2 x \sec^2 x dx - \int (\sec^2 x - 1) dx$ For the first integral, let $u = \tan x$, then $du = \sec^2 x dx$. So $\int u^2 du = \frac{u^3}{3} = \frac{\tan^3 x}{3}$. For the second integral, $\int (\sec^2 x - 1) dx = \tan x - x$. So, $f(x) = \frac{\tan^3 x}{3} - (\tan x - x) + C$ $f(x) = \frac{\tan^3 x}{3} - \tan x + x + C$.

Given $f(0)=0$: $f(0) = \frac{\tan^3 0}{3} - \tan 0 + 0 + C = 0 - 0 + 0 + C = 0$. Thus, $C=0$. So, $f(x) = \frac{\tan^3 x}{3} - \tan x + x$.

Now let's analyze each option:

A. f(x) is increasing function $\forall x \in R-\{(2n-1)\frac{\pi}{2}, n \in I\}$ To check if $f(x)$ is increasing, we examine its derivative $f'(x)$. $f'(x) = \frac{d}{dx} \left( \frac{\tan^3 x}{3} - \tan x + x \right)$ $f'(x) = \frac{1}{3} (3\tan^2 x \sec^2 x) - \sec^2 x + 1$ $f'(x) = \tan^2 x \sec^2 x - \sec^2 x + 1$ $f'(x) = \sec^2 x (\tan^2 x - 1) + 1$ Using $\sec^2 x = \tan^2 x + 1$: $f'(x) = (\tan^2 x + 1)(\tan^2 x - 1) + 1$ $f'(x) = (\tan^4 x - 1) + 1$ $f'(x) = \tan^4 x$. Since $\tan^4 x = (\tan x)^4$, it is always non-negative for all real $x$ where $\tan x$ is defined. $f'(x) = \tan^4 x \ge 0$. $f'(x)=0$ only when $\tan x = 0$, which occurs at $x=n\pi$ for $n \in I$. These are isolated points. A function is increasing if its derivative is non-negative. Since $f'(x) \ge 0$ everywhere in its domain, $f(x)$ is an increasing function. Thus, option A is correct.

B. Range of f(x) is $(-\infty, \infty)$ The domain of $f(x)$ is a union of disjoint open intervals of the form $(k\pi - \frac{\pi}{2}, k\pi + \frac{\pi}{2})$ for $k \in I$. Let's consider the behavior of $f(x)$ at the boundaries of these intervals. As $x \to (k\pi + \frac{\pi}{2})^-$, $\tan x \to \infty$. $f(x) = \frac{\tan^3 x}{3} - \tan x + x$. The term $\frac{\tan^3 x}{3}$ dominates. So, $\lim_{x \to (k\pi + \frac{\pi}{2})^-} f(x) = \infty$. As $x \to (k\pi - \frac{\pi}{2})^+$, $\tan x \to -\infty$. The term $\frac{\tan^3 x}{3}$ dominates. So, $\lim_{x \to (k\pi - \frac{\pi}{2})^+} f(x) = -\infty$. Since $f(x)$ is continuous and increasing on each interval $(k\pi - \frac{\pi}{2}, k\pi + \frac{\pi}{2})$, its range on each such interval is $(-\infty, \infty)$. The union of these ranges over all $k \in I$ is $(-\infty, \infty)$. Thus, option B is correct.

C. f(x) is bijective function A function is bijective if it is both injective (one-to-one) and surjective (onto).

• Injectivity: Since $f(x)$ is an increasing function ($f'(x) \ge 0$ and $f'(x)=0$ only at isolated points), it is strictly increasing. A strictly increasing function is always injective.
• Surjectivity: From option B, the range of $f(x)$ is $(-\infty, \infty)$, which is the entire codomain $R$. Therefore, $f(x)$ is surjective. Since $f(x)$ is both injective and surjective, it is bijective. Thus, option C is correct.

D. The sum of solutions of the equation f(x) = x, belonging to $[0, 2r \pi], r \in N$, is $r(6r+1)$ The equation is $f(x) = x$: $\frac{\tan^3 x}{3} - \tan x + x = x$ $\frac{\tan^3 x}{3} - \tan x = 0$ $\tan x \left( \frac{\tan^2 x}{3} - 1 \right) = 0$ This implies either $\tan x = 0$ or $\frac{\tan^2 x}{3} - 1 = 0$.

Case 1: $\tan x = 0$ The solutions are $x = n\pi$ for $n \in I$. In the interval $[0, 2r\pi]$, the solutions are $x = 0, \pi, 2\pi, \dots, 2r\pi$. The sum of these solutions is $S_1 = \sum_{n=0}^{2r} n\pi = \pi \sum_{n=0}^{2r} n = \pi \frac{2r(2r+1)}{2} = r(2r+1)\pi$.

Case 2: $\frac{\tan^2 x}{3} - 1 = 0 \implies \tan^2 x = 3 \implies \tan x = \pm \sqrt{3}$. Subcase 2a: $\tan x = \sqrt{3}$ The solutions are $x = n\pi + \frac{\pi}{3}$ for $n \in I$. In the interval $[0, 2r\pi]$, we need $0 \le n\pi + \frac{\pi}{3} \le 2r\pi$. $-\frac{\pi}{3} \le n\pi \le 2r\pi - \frac{\pi}{3}$ $-\frac{1}{3} \le n \le 2r - \frac{1}{3}$. So, $n$ can be $0, 1, 2, \dots, 2r-1$. The solutions are $\frac{\pi}{3}, \pi+\frac{\pi}{3}, 2\pi+\frac{\pi}{3}, \dots, (2r-1)\pi+\frac{\pi}{3}$. The sum of these solutions is $S_2 = \sum_{n=0}^{2r-1} (n\pi + \frac{\pi}{3}) = \pi \sum_{n=0}^{2r-1} n + \sum_{n=0}^{2r-1} \frac{\pi}{3}$ $S_2 = \pi \frac{(2r-1)(2r)}{2} + 2r \cdot \frac{\pi}{3} = r\pi(2r-1) + \frac{2r\pi}{3} = r\pi \left( 2r-1 + \frac{2}{3} \right) = r\pi \left( \frac{6r-3+2}{3} \right) = \frac{r\pi(6r-1)}{3}$.

Subcase 2b: $\tan x = -\sqrt{3}$ The solutions are $x = n\pi - \frac{\pi}{3}$ for $n \in I$. In the interval $[0, 2r\pi]$, we need $0 \le n\pi - \frac{\pi}{3} \le 2r\pi$. $\frac{\pi}{3} \le n\pi \le 2r\pi + \frac{\pi}{3}$ $\frac{1}{3} \le n \le 2r + \frac{1}{3}$. So, $n$ can be $1, 2, \dots, 2r$. The solutions are $\pi-\frac{\pi}{3}, 2\pi-\frac{\pi}{3}, \dots, 2r\pi-\frac{\pi}{3}$. The sum of these solutions is $S_3 = \sum_{n=1}^{2r} (n\pi - \frac{\pi}{3}) = \pi \sum_{n=1}^{2r} n - \sum_{n=1}^{2r} \frac{\pi}{3}$ $S_3 = \pi \frac{2r(2r+1)}{2} - 2r \cdot \frac{\pi}{3} = r\pi(2r+1) - \frac{2r\pi}{3} = r\pi \left( 2r+1 - \frac{2}{3} \right) = r\pi \left( \frac{6r+3-2}{3} \right) = \frac{r\pi(6r+1)}{3}$.

The total sum of solutions is $S = S_1 + S_2 + S_3$: $S = r(2r+1)\pi + \frac{r\pi(6r-1)}{3} + \frac{r\pi(6r+1)}{3}$ $S = r\pi \left[ (2r+1) + \frac{6r-1}{3} + \frac{6r+1}{3} \right]$ $S = r\pi \left[ \frac{3(2r+1) + (6r-1) + (6r+1)}{3} \right]$ $S = r\pi \left[ \frac{6r+3 + 6r-1 + 6r+1}{3} \right]$ $S = r\pi \left[ \frac{18r+3}{3} \right]$ $S = r\pi (6r+1)$. Option D states the sum is $r(6r+1)$, which is missing the factor of $\pi$. Therefore, option D is incorrect.

Final check: Options A, B, C are correct.

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

Explanation of the solution:

1. Find f(x): Integrate $\tan^4 x$ by rewriting it as $\tan^2 x (\sec^2 x - 1)$ and then as $(\tan^2 x \sec^2 x - \sec^2 x + 1)$. The integral of $\tan^2 x \sec^2 x$ is $\frac{\tan^3 x}{3}$ (by substitution $u=\tan x$). The integral of $\sec^2 x - 1$ is $\tan x - x$. Use $f(0)=0$ to find the constant of integration, which turns out to be 0. Thus, $f(x) = \frac{\tan^3 x}{3} - \tan x + x$.
2. Check Option A (Increasing function): Calculate $f'(x)$. $f'(x) = \tan^4 x$. Since $\tan^4 x \ge 0$ for all $x$ in the domain and equals 0 only at isolated points ($x=n\pi$), $f(x)$ is an increasing function.
3. Check Option B (Range): Analyze the limits of $f(x)$ as $x$ approaches the boundaries of its domain intervals, i.e., $x \to (k\pi \pm \frac{\pi}{2})$. As $x \to (k\pi + \frac{\pi}{2})^-$, $\tan x \to \infty$, so $f(x) \to \infty$. As $x \to (k\pi - \frac{\pi}{2})^+$, $\tan x \to -\infty$, so $f(x) \to -\infty$. Since $f(x)$ is continuous and increasing on each interval, its range on each interval is $(-\infty, \infty)$. Therefore, the overall range is $(-\infty, \infty)$.
4. Check Option C (Bijective function): A function is bijective if it's both injective and surjective. Since $f(x)$ is strictly increasing (from A), it is injective. Since the range of $f(x)$ is $R$ (from B), and the codomain is given as $R$, $f(x)$ is surjective. Thus, $f(x)$ is bijective.
5. Check Option D (Sum of solutions): Set $f(x)=x$ to get $\frac{\tan^3 x}{3} - \tan x = 0$. Factor out $\tan x$ to get $\tan x (\frac{\tan^2 x}{3} - 1) = 0$. This leads to $\tan x = 0$ or $\tan x = \pm \sqrt{3}$. Find all solutions in the interval $[0, 2r\pi]$ for each case and sum them up. The sum is $r\pi(6r+1)$, which does not match $r(6r+1)$. Therefore, option D is incorrect.

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

Subject: Mathematics Chapter: Calculus Topic: Indefinite Integrals, Properties of Functions (Monotonicity, Range, Bijectivity) Difficulty Level: Medium Question Type: Multiple Choice