Question

If $f: R \rightarrow [-\frac{\pi}{4}, \frac{\pi}{2}], f(x) = \tan^{-1}(x^4 - x^2 -\frac{7}{4} + \tan^{-1}\alpha)$ is a surjective function then which of the following is/are CORRECT?

• A. $\cos^{-1}(\frac{1-\alpha^2}{1+\alpha^2}) = 2$
• B. $\alpha + \frac{1}{\alpha} = 2 \cos ec2$
• C. $\sin^{-1}(\frac{2\alpha}{1+\alpha^2}) = \pi - 2$
• D. $\tan^{-1}(\frac{2\alpha}{\alpha^2 - 1}) = 2 - \pi$

Answer 1

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

A, B, C

Answer 2

The problem requires us to find the value of $\alpha$ for which the given function $f(x)$ is surjective, and then check which of the provided options are correct.

The function is $f: R \rightarrow [-\frac{\pi}{4}, \frac{\pi}{2}]$, defined by $f(x) = \tan^{-1}(x^4 - x^2 -\frac{7}{4} + \tan^{-1}\alpha)$.

Step 1: Determine the range of the inner expression $g(x) = x^4 - x^2 -\frac{7}{4}$. Let $t = x^2$. Since $x \in R$, $t \ge 0$. The expression becomes $h(t) = t^2 - t -\frac{7}{4}$. This is a quadratic function in $t$, representing a parabola opening upwards. Its vertex occurs at $t = -\frac{(-1)}{2(1)} = \frac{1}{2}$. Since $\frac{1}{2} \ge 0$, the minimum value of $h(t)$ occurs at $t = \frac{1}{2}$. The minimum value is $h(\frac{1}{2}) = (\frac{1}{2})^2 - (\frac{1}{2}) -\frac{7}{4} = \frac{1}{4} - \frac{1}{2} -\frac{7}{4} = \frac{1-2-7}{4} = -\frac{8}{4} = -2$. As $t \rightarrow \infty$ (which happens as $x \rightarrow \infty$ or $x \rightarrow -\infty$), $h(t) \rightarrow \infty$. Therefore, the range of $g(x)$ is $[-2, \infty)$.

Step 2: Use the surjectivity condition to find $\tan^{-1}\alpha$. Let $C = \tan^{-1}\alpha$. The argument of the $\tan^{-1}$ function in $f(x)$ is $Y = g(x) + C$. The range of $Y$ is $[-2+C, \infty)$. The function $f(x) = \tan^{-1}(Y)$. The standard range of $\tan^{-1}(y)$ is $(-\frac{\pi}{2}, \frac{\pi}{2})$. For $f(x)$ to be surjective onto the given codomain $[-\frac{\pi}{4}, \frac{\pi}{2}]$: The upper limit of the range of $f(x)$ must be $\frac{\pi}{2}$. This is achieved as $Y \rightarrow \infty$, which is consistent with the range of $Y$ being $[-2+C, \infty)$. The lower limit of the range of $f(x)$ must be $-\frac{\pi}{4}$. This occurs when $Y$ takes its minimum value, $Y_{min} = -2+C$. So, we must have $\tan^{-1}(Y_{min}) = -\frac{\pi}{4}$. $\tan^{-1}(-2+C) = -\frac{\pi}{4}$ $-2+C = \tan(-\frac{\pi}{4})$ $-2+C = -1$ $C = 1$. Since $C = \tan^{-1}\alpha$, we have $\tan^{-1}\alpha = 1$.

Step 3: Find the value of $\alpha$. From $\tan^{-1}\alpha = 1$, we get $\alpha = \tan(1)$. (Note: 1 is in radians).

Step 4: Check each option using $\alpha = \tan(1)$. Let $\theta = 1$ radian. So $\alpha = \tan\theta$.

• Option A: $\cos^{-1}(\frac{1-\alpha^2}{1+\alpha^2})$ We know that $\frac{1-\tan^2\theta}{1+\tan^2\theta} = \cos(2\theta)$. So, $\cos^{-1}(\frac{1-\tan^2(1)}{1+\tan^2(1)}) = \cos^{-1}(\cos(2))$. Since $2$ radians is approximately $2 \times 57.3^\circ = 114.6^\circ$, which lies in the principal value range $[0, \pi]$ for $\cos^{-1}x$. Therefore, $\cos^{-1}(\cos(2)) = 2$. Option A is CORRECT.

• Option B: $\alpha + \frac{1}{\alpha}$ Substitute $\alpha = \tan(1)$: $\tan(1) + \frac{1}{\tan(1)} = \tan(1) + \cot(1) = \frac{\sin(1)}{\cos(1)} + \frac{\cos(1)}{\sin(1)}$ $= \frac{\sin^2(1) + \cos^2(1)}{\sin(1)\cos(1)} = \frac{1}{\sin(1)\cos(1)}$. Multiply the numerator and denominator by 2: $= \frac{2}{2\sin(1)\cos(1)} = \frac{2}{\sin(2)}$. We know that $\frac{1}{\sin(2)} = \csc(2)$ or $\operatorname{cosec}(2)$. So, $\alpha + \frac{1}{\alpha} = 2\csc(2)$. Option B is CORRECT.

• Option C: $\sin^{-1}(\frac{2\alpha}{1+\alpha^2})$ We know that $\frac{2\tan\theta}{1+\tan^2\theta} = \sin(2\theta)$. So, $\sin^{-1}(\frac{2\tan(1)}{1+\tan^2(1)}) = \sin^{-1}(\sin(2))$. Since $2$ radians is approximately $114.6^\circ$, it does not lie in the principal value range $[-\frac{\pi}{2}, \frac{\pi}{2}]$ for $\sin^{-1}x$. We use the identity $\sin x = \sin(\pi - x)$. So, $\sin^{-1}(\sin(2)) = \sin^{-1}(\sin(\pi - 2))$. Since $\pi - 2 \approx 3.14159 - 2 = 1.14159$ radians, which lies in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ (approximately $[-1.57, 1.57]$). Therefore, $\sin^{-1}(\sin(2)) = \pi - 2$. Option C is CORRECT.

• Option D: $\tan^{-1}(\frac{2\alpha}{\alpha^2 - 1})$ We know that $\frac{2\tan\theta}{1-\tan^2\theta} = \tan(2\theta)$. So, $\frac{2\tan\theta}{\tan^2\theta - 1} = -\frac{2\tan\theta}{1-\tan^2\theta} = -\tan(2\theta)$. Substitute $\alpha = \tan(1)$: $\tan^{-1}(\frac{2\tan(1)}{\tan^2(1) - 1}) = \tan^{-1}(-\tan(2))$. Using the property $\tan^{-1}(-y) = -\tan^{-1}(y)$: $= -\tan^{-1}(\tan(2))$. Since $2$ radians is approximately $114.6^\circ$, it does not lie in the principal value range $(-\frac{\pi}{2}, \frac{\pi}{2})$ for $\tan^{-1}x$. We use the identity $\tan x = \tan(x - \pi)$. So, $\tan^{-1}(\tan(2)) = \tan^{-1}(\tan(2 - \pi))$. Since $2 - \pi \approx 2 - 3.14159 = -1.14159$ radians, which lies in $(-\frac{\pi}{2}, \frac{\pi}{2})$ (approximately $(-1.57, 1.57)$). Therefore, $\tan^{-1}(\tan(2)) = 2 - \pi$. So, $\tan^{-1}(\frac{2\alpha}{\alpha^2 - 1}) = -(2 - \pi) = \pi - 2$. Option D states $2 - \pi$, which is incorrect.