Question

If $cos^{-1}(e^x) + sin^{-1}(x^2) = \frac{\pi}{2}$, then number of solutions of this equation

• A. 1
• B. 3
• C. 2
• D. 0

Answer 1

Answer: (A)

1

Answer 2

The equation is $cos^{-1}(e^x) + sin^{-1}(x^2) = \frac{\pi}{2}$. We know that $cos^{-1}(y) + sin^{-1}(y) = \frac{\pi}{2}$ for $y \in [-1, 1]$. For the given equation to hold, we must have $e^x = x^2$. Also, the arguments must be in the domain of the respective inverse trigonometric functions:

1. For $cos^{-1}(e^x)$: $-1 \le e^x \le 1$. Since $e^x > 0$, this means $0 < e^x \le 1$, which implies $x \le 0$.
2. For $sin^{-1}(x^2)$: $-1 \le x^2 \le 1$. Since $x^2 \ge 0$, this means $0 \le x^2 \le 1$, which implies $-1 \le x \le 1$. Combining these, the valid range for $x$ is $[-1, 0]$.

Now, we need to find the number of solutions for $e^x = x^2$ in $[-1, 0]$. Let $h(x) = e^x - x^2$. We are looking for roots of $h(x)=0$ in $[-1, 0]$. $h(-1) = e^{-1} - (-1)^2 = \frac{1}{e} - 1 < 0$. $h(0) = e^0 - 0^2 = 1 > 0$. By the Intermediate Value Theorem, there is at least one root in $(-1, 0)$. Let's examine the derivative: $h'(x) = e^x - 2x$. $h''(x) = e^x - 2$. For $x \in [-1, 0]$, $e^x \in [1/e, 1]$, so $h''(x) < 0$. This means $h'(x)$ is decreasing in $[-1, 0]$. $h'(-1) = e^{-1} + 2 > 0$. $h'(0) = 1 > 0$. Since $h'(x)$ is decreasing and positive at the endpoints, $h'(x) > 0$ in $[-1, 0]$. Thus, $h(x)$ is strictly increasing in $[-1, 0]$. A strictly increasing continuous function can have at most one root. Since we found at least one root, there is exactly one root. Therefore, there is 1 solution.