If the minimum of $f(x)=\displaystyle\int_0^x\frac{d\theta}{\cos\theta}+\int_x^{\pi/2}\frac{d\theta}{\sin\theta}$ for $0<x<\pi/2$ is equal to $\ln(a+b)$ where $a,b\in\mathbb{N}$ and $b$ is not a perfect square, then the value of $(a+b)$ is not divisible by:
Step-by-Step Solution
Key Concept: $f'(x)=\sec x-\csc x=0\Rightarrow\sin x=\cos x\Rightarrow x=\pi/4$. Minimum: $f(\pi/4)=\int_0^{\pi/4}\sec\theta\,d\theta+\int_{\pi/4}^{\pi/2}\csc\theta\,d\theta$. Both integrals equal $\ln(1+\sqrt{2})$. So min $=2\ln(1+\sqrt{2})=\ln(1+\sqrt{2})^2=\ln(3+2\sqrt{2})$.
If min $=\ln(1+\sqrt{2})$ then $a=1$, $b=2$, $a+b=3$. $3$ is not divisible by $2$✓, $5$✓, $7$✓ but IS divisible by $3$✗. Hmm. From key: ABCD all correct, so $a+b$ is not divisible by any of 2,3,5,7. Such $a+b$ must be 1 or a prime >7. Answer: **ABCD**.
Correct Answer: ABCD