Question:

Step 1: Define the components of N and find their sum of squares. Let $A = \sqrt{9\cos^2\theta + 16\sin^2\theta}$ and $B = \sqrt{16\cos^2\theta + 9\sin^2\theta}$. Then $N = A + B$. Computing $A^2 + B^2$: $$A^2 + B^2 = 9\cos^2\theta + 16\sin^2\theta + 16\cos^2\theta + 9\sin^2\theta$$ $$= 25\cos^2\theta + 25\sin^2\theta = 25(\cos^2\theta + \sin^2\theta) = 25$$ Step 2: Express $N^2$ in terms of $AB$. Expanding $N^2$: $$N^2 = (A + B)^2 = A^2 + B^2 + 2AB = 25 + 2AB$$ To find the maximum and minimum values of $N^2$, we need to find the maximum and minimum values of $AB$. Step 3: Set up the product $AB$ as a function of $\sin^2\theta$. Let $t = \sin^2\theta$ where $t \in [0,1]$. Then $\cos^2\theta = 1 - t$. $$A^2 = 9(1-t) + 16t = 9 + 7t$$ $$B^2 = 16(1-t) + 9t = 16 - 7t$$ Therefore: $$f(t) = A^2 \cdot B^2 = (9 + 7t)(16 - 7t)$$ Step 4: Expand and simplify $f(t)$. $$f(t) = 144 - 63t + 112t - 49t^2 = 144 + 49t - 49t^2$$ Step 5: Find the maximum of $f(t)$ by taking the derivative. $$f'(t) = 49 - 98t$$ Setting $f'(t) = 0$: $$49 - 98t = 0 \implies t = \frac{1}{2}$$ Evaluating at $t = \frac{1}{2}$: $$f\left(\frac{1}{2}\right) = 144 + 49 \cdot \frac{1}{2} - 49 \cdot \frac{1}{4} = 144 + \frac{49}{2} - \frac{49}{4} = 144 + \frac{49}{4} = \frac{625}{4}$$ Therefore: $AB_{\max} = \sqrt{\frac{625}{4}} = \frac{25}{2}$ This gives: $N^2_{\max} = 25 + 2 \cdot \frac{25}{2} = 25 + 25 = 50$ Step 6: Find the minimum of $f(t)$ at the endpoints. At $t = 0$: $$f(0) = 144$$ So $AB_{\min} = \sqrt{144} = 12$ At $t = 1$: $$f(1) = 144 + 49 - 49 = 144$$ So $AB = 12$ Therefore: $N^2_{\min} = 25 + 2(12) = 25 + 24 = 49$ Step 7: Calculate the sum of maximum and minimum values of $N^2$. $$N^2_{\max} + N^2_{\min} = 50 + 49 = 99$$ The sum of the maximum and minimum values of $N^2$ is **99**. The answer is **Option 3**.