2023학년도 2학년 1학기 중간고사 수 학 । 10. $12^{\frac{m+2}{3}} \times 3^{\frac{3 m-1}{2}}$ 이 자연수가 되도록 하는 100 이하의 자연수 $m$ 의 최댓값은? [4.2점] (1) 91 (2) 93 (3) 95 (4) 97 (5) 99 11. 함수 $f(x)$ 가 다음 세 조건을 만족시킨다. (가) 모든 실수 $x$ 에 대하여 $f(x+\pi)=f(x)$ 이다. 13. (나) $0 \leq x<\frac{\pi}{2}$ 일 때, $f(x)=\tan \frac{x}{2}$ (다) $\frac{\pi}{2} \leq x<\pi$ 일 때, $f(x)=\tan \left(-\frac{x}{2}+\frac{\pi}{2}\right)$ 이때 함수 $f(x)$ 의 그래프와 직선 $y=\frac{x}{2 \pi}$ 가 만나는 점의 개수는? [4.2점] (1) 2 (2) 3 (3) 4 (4) 5 (5) 6

Step 1 ：First, we need to find the value of $m$ such that $12^\frac{m+2}{3} \times 3^\frac{3m-1}{2}$ is a natural number.

Step 2 ：We can rewrite the expression as $\left(2^2 \times 3\right)^\frac{m+2}{3} \times 3^\frac{3m-1}{2}$

Step 3 ：Now, we can distribute the exponent to get $2^\frac{2(m+2)}{3} \times 3^\frac{m+2}{3} \times 3^\frac{3m-1}{2}$

Step 4 ：Combining the terms with the same base, we get $2^\frac{2(m+2)}{3} \times 3^{\frac{m+2}{3} + \frac{3m-1}{2}}$

Step 5 ：For the expression to be a natural number, both exponents must be integers.

Step 6 ：Let's first focus on the exponent of 2: $\frac{2(m+2)}{3}$. For this to be an integer, $m+2$ must be a multiple of 3.

Step 7 ：Now, let's focus on the exponent of 3: $\frac{m+2}{3} + \frac{3m-1}{2}$. For this to be an integer, the numerator must be a multiple of 6.

Step 8 ：We can rewrite the exponent of 3 as $\frac{2(m+2) + 3(3m-1)}{6}$

Step 9 ：Simplifying the numerator, we get $\frac{11m+1}{6}$

Step 10 ：For this to be an integer, $11m+1$ must be a multiple of 6.

Step 11 ：Now, we need to find the largest value of $m$ that satisfies both conditions and is less than or equal to 100.

Step 12 ：By testing values of $m$, we find that the largest value that satisfies both conditions is $m=\boxed{91}$ (1)