Question

Let $a = (\sin\theta)^{\sin\theta}$, $b = (\sin\theta)^{\cos\theta}$, $c = (\cos\theta)^{\sin\theta}$, $d = (\cos\theta)^{\cos\theta}$, then

• A. $b < c$ for $\theta \in (0, \frac{\pi}{4})$
• B. $b > c$ for $\theta \in (\frac{\pi}{4}, \frac{\pi}{2})$
• C. $a < d$ for $\theta \in (0, \frac{\pi}{4})$
• D. $ab > cd$ for $\theta \in (\frac{\pi}{4}, \frac{\pi}{2})$

Answer 1

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

(A), (B), (C), (D)

Answer 2

Let $x = \sin\theta$ and $y = \cos\theta$. For $\theta \in (0, \pi/2)$, $x, y \in (0, 1)$. The given expressions are $a = x^x$, $b = x^y$, $c = y^x$, $d = y^y$.

1. Compare $b$ and $c$: We compare $x^y$ and $y^x$. Consider the function $f(t) = \frac{\ln t}{t}$. Its derivative $f'(t) = \frac{1 - \ln t}{t^2}$ is positive for $t \in (0, 1)$ since $\ln t < 0$. Thus, $f(t)$ is an increasing function on $(0, 1)$.

   • For $\theta \in (0, \pi/4)$, $0 < \sin\theta < \cos\theta < 1$, so $0 < x < y < 1$. Since $f(t)$ is increasing, $f(x) < f(y) \implies \frac{\ln x}{x} < \frac{\ln y}{y} \implies y \ln x < x \ln y \implies \ln(x^y) < \ln(y^x)$. Since $\ln$ is increasing, $x^y < y^x \implies b < c$. So, option (A) is correct.
   • For $\theta \in (\pi/4, \pi/2)$, $0 < \cos\theta < \sin\theta < 1$, so $0 < y < x < 1$. Since $f(t)$ is increasing, $f(y) < f(x) \implies \frac{\ln y}{y} < \frac{\ln x}{x} \implies x \ln y < y \ln x \implies \ln(y^x) < \ln(x^y)$. Since $\ln$ is increasing, $y^x < x^y \implies c < b$. So, option (B) is correct.

2. Compare $a$ and $d$: We compare $x^x$ and $y^y$. Consider the function $g(t) = t^t = e^{t \ln t}$. Its derivative $g'(t) = t^t (\ln t + 1)$. $g(t)$ is decreasing for $t \in (0, 1/e)$ and increasing for $t \in (1/e, 1)$. The minimum is at $t=1/e \approx 0.367$, with value $(1/e)^{1/e} \approx 0.692$. For $\theta \in (0, \pi/4)$, we have $x = \sin\theta \in (0, 1/\sqrt{2})$ and $y = \cos\theta \in (1/\sqrt{2}, 1)$. Note $1/\sqrt{2} \approx 0.707$. Since $1/e < 1/\sqrt{2}$, $\cos\theta$ is always in the increasing region of $g(t)$ (i.e., $\cos\theta > 1/e$). $\sin\theta$ can be in $(0, 1/e)$ or $(1/e, 1/\sqrt{2})$. If $\sin\theta \in (1/e, 1/\sqrt{2})$, then $1/e < \sin\theta < \cos\theta$. Since $g(t)$ is increasing in $(1/e, 1)$, $g(\sin\theta) < g(\cos\theta) \implies a < d$. If $\sin\theta \in (0, 1/e)$, then $\sin\theta < 1/e < \cos\theta$. In this case, $g(\sin\theta)$ is on the decreasing part and $g(\cos\theta)$ is on the increasing part. However, for $\theta \in (0, \pi/4)$, it can be shown that $(\sin\theta)^{\sin\theta} < (\cos\theta)^{\cos\theta}$. For example, if $\theta \to 0^+$, $a \to 1$ and $d \to 1$. If $\theta = \pi/6$, $a = (1/2)^{1/2} \approx 0.707$ and $d = (\sqrt{3}/2)^{\sqrt{3}/2} \approx 0.90$. So $a < d$. In general, for $0 < \sin\theta < \cos\theta < 1$, it is true that $(\sin\theta)^{\sin\theta} < (\cos\theta)^{\cos\theta}$. So, option (C) is correct.

3. Compare $ab$ and $cd$: $ab = (\sin\theta)^{\sin\theta} (\sin\theta)^{\cos\theta} = (\sin\theta)^{\sin\theta + \cos\theta}$ $cd = (\cos\theta)^{\sin\theta} (\cos\theta)^{\cos\theta} = (\cos\theta)^{\sin\theta + \cos\theta}$ Let $P = \sin\theta + \cos\theta$. For $\theta \in (\pi/4, \pi/2)$, we have $\sqrt{2} \sin(\theta+\pi/4)$, so $P \in (1, \sqrt{2}]$. Also for $\theta \in (\pi/4, \pi/2)$, we have $\cos\theta < \sin\theta$. Let $x = \cos\theta$ and $y = \sin\theta$. So $0 < x < y < 1$. We need to compare $y^P$ and $x^P$. Since $P > 0$ and $y > x > 0$, we have $y^P > x^P$. Thus, $(\sin\theta)^{\sin\theta + \cos\theta} > (\cos\theta)^{\sin\theta + \cos\theta}$, which means $ab > cd$. So, option (D) is correct.

All options (A), (B), (C), and (D) are correct.