Question

If $f(x) = \frac{x}{x^2+x+1}$, then

If $\int f(\tan x)dx = x - \frac{2}{\sqrt{3}} \tan^{-1}(g(x)) + c$, then for $x \in (-\frac{\pi}{3}, \frac{\pi}{3})$ range of $g(x)$ is -

• A. $(\frac{1}{\sqrt{3}}-2, \frac{1}{\sqrt{3}}+2)$
• B. $(-\sqrt{3}, \sqrt{3})$
• C. $(-\frac{\sqrt{3}}{4-\sqrt{3}}, \frac{\sqrt{3}}{4+\sqrt{3}})$
• D. none of these

Answer 1

Answer: (A)

$(\frac{1}{\sqrt{3}}-2, \frac{1}{\sqrt{3}}+2)$

Answer 2

The given function is $f(x) = \frac{x}{x^2+x+1}$. We are given the integral equation $\int f(\tan x)dx = x - \frac{2}{\sqrt{3}} \tan^{-1}(g(x)) + c$. Substitute $x$ with $\tan x$ in $f(x)$: $f(\tan x) = \frac{\tan x}{(\tan x)^2 + \tan x + 1} = \frac{\tan x}{\tan^2 x + \tan x + 1}$.

Differentiate the given integral equation with respect to $x$: $\frac{d}{dx} \left( \int f(\tan x)dx \right) = \frac{d}{dx} \left( x - \frac{2}{\sqrt{3}} \tan^{-1}(g(x)) + c \right)$ $f(\tan x) = 1 - \frac{2}{\sqrt{3}} \frac{1}{1+(g(x))^2} g'(x)$ $\frac{\tan x}{\tan^2 x + \tan x + 1} = 1 - \frac{2}{\sqrt{3}} \frac{g'(x)}{1+(g(x))^2}$ $\frac{2}{\sqrt{3}} \frac{g'(x)}{1+(g(x))^2} = 1 - \frac{\tan x}{\tan^2 x + \tan x + 1} = \frac{\tan^2 x + \tan x + 1 - \tan x}{\tan^2 x + \tan x + 1} = \frac{\tan^2 x + 1}{\tan^2 x + \tan x + 1} = \frac{\sec^2 x}{\tan^2 x + \tan x + 1}$. $\frac{g'(x)}{1+(g(x))^2} = \frac{\sqrt{3}}{2} \frac{\sec^2 x}{\tan^2 x + \tan x + 1}$.

We know that $\frac{d}{dx} \tan^{-1}(h(x)) = \frac{h'(x)}{1+(h(x))^2}$. We need to find $g(x)$ such that $\frac{g'(x)}{1+(g(x))^2}$ matches the expression on the right-hand side. Consider the denominator $\tan^2 x + \tan x + 1$. We can complete the square for $\tan x$: $\tan^2 x + \tan x + 1 = (\tan x + \frac{1}{2})^2 + 1 - \frac{1}{4} = (\tan x + \frac{1}{2})^2 + \frac{3}{4}$. So, $\frac{\sec^2 x}{\tan^2 x + \tan x + 1} = \frac{\sec^2 x}{(\tan x + \frac{1}{2})^2 + \frac{3}{4}}$. We want to express this in the form $\frac{h'(x)}{1+y^2}$: $(\tan x + \frac{1}{2})^2 + \frac{3}{4} = \frac{3}{4} \left[ \frac{(\tan x + \frac{1}{2})^2}{\frac{3}{4}} + 1 \right] = \frac{3}{4} \left[ \left( \frac{\tan x + \frac{1}{2}}{\frac{\sqrt{3}}{2}} \right)^2 + 1 \right] = \frac{3}{4} \left[ \left( \frac{2 \tan x + 1}{\sqrt{3}} \right)^2 + 1 \right]$. So, $\frac{\sec^2 x}{\tan^2 x + \tan x + 1} = \frac{\sec^2 x}{\frac{3}{4} \left[ \left( \frac{2 \tan x + 1}{\sqrt{3}} \right)^2 + 1 \right]} = \frac{4}{3} \frac{\sec^2 x}{1 + \left( \frac{2 \tan x + 1}{\sqrt{3}} \right)^2}$.

Now substitute this back into the equation for $\frac{g'(x)}{1+(g(x))^2}$: $\frac{g'(x)}{1+(g(x))^2} = \frac{\sqrt{3}}{2} \cdot \frac{4}{3} \frac{\sec^2 x}{1 + \left( \frac{2 \tan x + 1}{\sqrt{3}} \right)^2} = \frac{2}{\sqrt{3}} \frac{\sec^2 x}{1 + \left( \frac{2 \tan x + 1}{\sqrt{3}} \right)^2}$.

Consider the function $h(x) = \frac{2 \tan x + 1}{\sqrt{3}}$. Its derivative is $h'(x) = \frac{2 \sec^2 x}{\sqrt{3}}$. Then $\frac{h'(x)}{1+(h(x))^2} = \frac{\frac{2 \sec^2 x}{\sqrt{3}}}{1 + \left( \frac{2 \tan x + 1}{\sqrt{3}} \right)^2} = \frac{2}{\sqrt{3}} \frac{\sec^2 x}{1 + \left( \frac{2 \tan x + 1}{\sqrt{3}} \right)^2}$. This matches the expression for $\frac{g'(x)}{1+(g(x))^2}$. This implies that $g(x)$ is proportional to $h(x)$ or $g(x)$ is $h(x)$ plus a constant. However, the form of the integral suggests $g(x)$ is exactly $h(x)$. Let's verify the integral by integrating $\frac{2}{\sqrt{3}} \frac{\sec^2 x}{1 + \left( \frac{2 \tan x + 1}{\sqrt{3}} \right)^2}$. Let $u = \frac{2 \tan x + 1}{\sqrt{3}}$. Then $du = \frac{2 \sec^2 x}{\sqrt{3}} dx$. $\int \frac{2}{\sqrt{3}} \frac{\sec^2 x}{1 + \left( \frac{2 \tan x + 1}{\sqrt{3}} \right)^2} dx = \int \frac{du}{1+u^2} = \tan^{-1}(u) + C' = \tan^{-1}\left(\frac{2 \tan x + 1}{\sqrt{3}}\right) + C'$. The given integral is $\int f(\tan x)dx = x - \frac{2}{\sqrt{3}} \tan^{-1}(g(x)) + c$. We calculated $\frac{2}{\sqrt{3}} \frac{g'(x)}{1+(g(x))^2} = 1 - f(\tan x)$. Integrating this gives $\int \frac{2}{\sqrt{3}} \frac{g'(x)}{1+(g(x))^2} dx = \int (1 - f(\tan x)) dx$. $\frac{2}{\sqrt{3}} \int \frac{g'(x)}{1+(g(x))^2} dx = x - \int f(\tan x) dx$. $\frac{2}{\sqrt{3}} \tan^{-1}(g(x)) = x - \int f(\tan x) dx + \text{constant}$. $\int f(\tan x) dx = x - \frac{2}{\sqrt{3}} \tan^{-1}(g(x)) + \text{constant}$. This confirms that $g(x) = \frac{2 \tan x + 1}{\sqrt{3}}$.

Now we need to find the range of $g(x)$ for $x \in (-\frac{\pi}{3}, \frac{\pi}{3})$. For $x \in (-\frac{\pi}{3}, \frac{\pi}{3})$, the function $\tan x$ is strictly increasing. The range of $\tan x$ for $x \in (-\frac{\pi}{3}, \frac{\pi}{3})$ is $(\tan(-\frac{\pi}{3}), \tan(\frac{\pi}{3})) = (-\sqrt{3}, \sqrt{3})$. Let $y = \tan x$. The range of $y$ is $(-\sqrt{3}, \sqrt{3})$. $g(x) = \frac{2 \tan x + 1}{\sqrt{3}} = \frac{2y + 1}{\sqrt{3}}$. We need to find the range of $\frac{2y+1}{\sqrt{3}}$ for $y \in (-\sqrt{3}, \sqrt{3})$. The function $\frac{2y+1}{\sqrt{3}}$ is a linear function of $y$ with a positive slope, so it is strictly increasing. The range is $\left( \frac{2(-\sqrt{3}) + 1}{\sqrt{3}}, \frac{2(\sqrt{3}) + 1}{\sqrt{3}} \right)$. Lower bound: $\frac{-2\sqrt{3} + 1}{\sqrt{3}} = \frac{1}{\sqrt{3}} - \frac{2\sqrt{3}}{\sqrt{3}} = \frac{1}{\sqrt{3}} - 2$. Upper bound: $\frac{2\sqrt{3} + 1}{\sqrt{3}} = \frac{2\sqrt{3}}{\sqrt{3}} + \frac{1}{\sqrt{3}} = 2 + \frac{1}{\sqrt{3}}$. The range of $g(x)$ is $(\frac{1}{\sqrt{3}} - 2, \frac{1}{\sqrt{3}} + 2)$.

The range of $g(x)$ is $(\frac{1}{\sqrt{3}}-2, \frac{1}{\sqrt{3}}+2)$.