Question

The number of values of x satisfying the equation $\tan^{-1}(x^2 + 5|x| + 20) + \cot^{-1}(4\pi + \sin^{-1}(\sin 14)) = \frac{\pi}{2}$ is

Answer 1

0

Answer 2

The given equation is $\tan^{-1}(x^2 + 5|x| + 20) + \cot^{-1}(4\pi + \sin^{-1}(\sin 14)) = \frac{\pi}{2}$.

We use the identity $\tan^{-1}(y) + \cot^{-1}(y) = \frac{\pi}{2}$ for all $y \in \mathbb{R}$. For the given equation to hold, the arguments of $\tan^{-1}$ and $\cot^{-1}$ must be equal. Let $A = x^2 + 5|x| + 20$ and $B = 4\pi + \sin^{-1}(\sin 14)$. The equation implies $A = B$.

First, let's evaluate the term $\sin^{-1}(\sin 14)$. The function $\sin^{-1}(\sin \theta) = \theta - n\pi$, where $n$ is an integer such that $\theta - n\pi \in [-\frac{\pi}{2}, \frac{\pi}{2}]$. Here $\theta = 14$. We need to find an integer $n$ such that $-\frac{\pi}{2} \le 14 - n\pi \le \frac{\pi}{2}$. Dividing by $\pi$, we get $-\frac{1}{2} \le \frac{14}{\pi} - n \le \frac{1}{2}$. This is equivalent to $n - \frac{1}{2} \le \frac{14}{\pi} \le n + \frac{1}{2}$, which means $n$ is the integer closest to $\frac{14}{\pi}$. Using the approximation $\pi \approx 3.14159$, we have $\frac{14}{\pi} \approx \frac{14}{3.14159} \approx 4.4563$. The closest integer to $4.4563$ is $n=4$. Let's check if $14 - 4\pi$ is in the range $[-\frac{\pi}{2}, \frac{\pi}{2}]$. $14 - 4\pi \approx 14 - 4(3.14159) = 14 - 12.56636 = 1.43364$. Since $\frac{\pi}{2} \approx 1.5708$, we have $-1.5708 \le 1.43364 \le 1.5708$. So, $\sin^{-1}(\sin 14) = 14 - 4\pi$.

Now substitute this value back into the expression for $B$: $B = 4\pi + \sin^{-1}(\sin 14) = 4\pi + (14 - 4\pi) = 14$.

Now, we set $A = B$: $x^2 + 5|x| + 20 = 14$. $x^2 + 5|x| + 6 = 0$.

Let $y = |x|$. Since $x^2 = |x|^2 = y^2$, the equation becomes a quadratic equation in $y$: $y^2 + 5y + 6 = 0$. Factor the quadratic equation: $(y+2)(y+3) = 0$. The solutions for $y$ are $y = -2$ and $y = -3$.

Substitute back $y = |x|$: $|x| = -2$ or $|x| = -3$. The absolute value of a real number is always non-negative. Therefore, $|x| \ge 0$ for all real $x$. Since $-2 < 0$ and $-3 < 0$, neither $|x| = -2$ nor $|x| = -3$ has any real solutions for $x$.

Thus, the equation $x^2 + 5|x| + 6 = 0$ has no real solutions. This means there are no values of $x$ that satisfy the original equation. The number of values of $x$ satisfying the equation is 0.