Question

$\sin^{-1}\frac{4x}{x^2+4}+2\tan^{-1}(-\frac{x}{2})$ is a constant function of x, then

Answer 1

-2 \le x \le 2

Answer 2

Let the given function be $f(x) = \sin^{-1}\frac{4x}{x^2+4}+2\tan^{-1}(-\frac{x}{2})$.

We can simplify the terms using trigonometric identities. Consider the first term $\sin^{-1}\frac{4x}{x^2+4}$. Let $x = 2\tan\theta$, where $\theta \in (-\pi/2, \pi/2)$. Then $\frac{4x}{x^2+4} = \frac{4(2\tan\theta)}{(2\tan\theta)^2+4} = \frac{8\tan\theta}{4\tan^2\theta+4} = \frac{8\tan\theta}{4(\tan^2\theta+1)} = \frac{2\tan\theta}{\sec^2\theta} = 2\sin\theta\cos\theta = \sin(2\theta)$. So, $\sin^{-1}\frac{4x}{x^2+4} = \sin^{-1}(\sin(2\theta))$.

We know that $\sin^{-1}(\sin z) = z$ if $z \in [-\pi/2, \pi/2]$. Since $\theta = \tan^{-1}(x/2)$, and $\theta \in (-\pi/2, \pi/2)$, the range of $2\theta$ is $(-\pi, \pi)$. We need to consider the cases based on the value of $2\theta$.

Case 1: $-\pi/2 \le 2\theta \le \pi/2$. This implies $-\pi/4 \le \theta \le \pi/4$. Since $\theta = \tan^{-1}(x/2)$, this means $\tan(-\pi/4) \le x/2 \le \tan(\pi/4)$, which is $-1 \le x/2 \le 1$, or $-2 \le x \le 2$. In this interval, $\sin^{-1}(\sin(2\theta)) = 2\theta = 2\tan^{-1}(x/2)$.

Case 2: $\pi/2 < 2\theta < \pi$. This implies $\pi/4 < \theta < \pi/2$. Since $\theta = \tan^{-1}(x/2)$, this means $\tan(\pi/4) < x/2 < \tan(\pi/2)$, which is $1 < x/2 < \infty$, or $x > 2$. In this interval, $\sin^{-1}(\sin(2\theta)) = \sin^{-1}(\sin(\pi - 2\theta))$. Since $\pi/2 < 2\theta < \pi$, we have $0 < \pi - 2\theta < \pi/2$. So, $\sin^{-1}(\sin(\pi - 2\theta)) = \pi - 2\theta = \pi - 2\tan^{-1}(x/2)$.

Case 3: $-\pi < 2\theta < -\pi/2$. This implies $-\pi/2 < \theta < -\pi/4$. Since $\theta = \tan^{-1}(x/2)$, this means $\tan(-\pi/2) < x/2 < \tan(-\pi/4)$, which is $-\infty < x/2 < -1$, or $x < -2$. In this interval, $\sin^{-1}(\sin(2\theta)) = \sin^{-1}(\sin(-\pi - 2\theta))$. Since $-\pi < 2\theta < -\pi/2$, we have $-\pi/2 < -\pi - 2\theta < 0$. So, $\sin^{-1}(\sin(-\pi - 2\theta)) = -\pi - 2\theta = -\pi - 2\tan^{-1}(x/2)$.

Now consider the second term $2\tan^{-1}(-\frac{x}{2})$. Using the property $\tan^{-1}(-z) = -\tan^{-1}(z)$, we have $2\tan^{-1}(-\frac{x}{2}) = -2\tan^{-1}(\frac{x}{2})$.

Now, let's write the function $f(x)$ in different intervals: For $-2 \le x \le 2$: $f(x) = 2\tan^{-1}(x/2) + (-2\tan^{-1}(x/2)) = 0$. The function is a constant (zero) in this interval.

For $x > 2$: $f(x) = (\pi - 2\tan^{-1}(x/2)) + (-2\tan^{-1}(x/2)) = \pi - 4\tan^{-1}(x/2)$. This is not a constant function of $x$ for $x > 2$.

For $x < -2$: $f(x) = (-\pi - 2\tan^{-1}(x/2)) + (-2\tan^{-1}(x/2)) = -\pi - 4\tan^{-1}(x/2)$. This is not a constant function of $x$ for $x < -2$.

The question states that the function is a constant function of $x$. This means the function must evaluate to a single value for all $x$ in its domain. From our analysis, the function is constant only when the domain of $x$ is restricted to the interval $[-2, 2]$. If the function is constant, it must be equal to the value it takes in the interval where it is constant, which is 0. The function is constant if and only if the domain of $x$ is a subset of $[-2, 2]$. If the question implies that the function is identically constant over its natural domain (which is all real numbers since $\frac{4x}{x^2+4}$ is always between -1 and 1, and $\tan^{-1}(-x/2)$ is defined for all real $x$), then this is only possible if the domain is restricted to $[-2, 2]$.

Assuming the question asks for the condition on $x$ for the function to be constant, that condition is $-2 \le x \le 2$. Since the options are not provided, we infer the intended meaning from the structure of such questions in JEE/NEET. The question implies that the given expression simplifies to a constant value under some condition on $x$. That condition is the interval where the piecewise definition of the function results in a constant value.

The function is constant when $x \in [-2, 2]$, and the constant value is 0.