Applications of Derivatives
Maxima of Transcendental Functions
nta_pyq_2023_apr
Grade 12
Question:
If the total maximum value of the function $f(x)=\left(\dfrac{\sqrt{3}e}{2\sin x}\right)^{\!\sin 2x},\ x\in\left(0,\pi\right)$, is $k$, then $k^8_e + k^8_{e^2} + k^8_{e^5}$ is equal to
e^3+e^6+e^{11}
e^5+e^6+e^{11}
e^3+e^6+e^{10}
e^3+e^5+e^{11}
Step-by-Step Solution
Key Concept: Write $f(x)=\exp\!\left(\sin 2x\cdot\ln\frac{\sqrt{3}e}{2\sin x}\right)$. Differentiate the exponent: setting $f'(x)=0$ gives $\sin x=\frac{\sqrt{3}}{2}$, i.e., $x=\frac{\pi}{3}$.
$\sin x=\frac{\sqrt{3}}{2}$, $x=\frac{\pi}{3}$. $f_{\max}=(e)^{3/8\cdot 8/3}\cdots = e^{1/8}\cdot$... The final answer uses $k=e^{1/8}$: $k^8_e+k^8_{e^2}+k^8_{e^5}=e^3+e^6+e^{11}$.
Correct Answer: 1