Question:

Step 1: Apply Taylor series expansion for cosine near $x = 0$. We use the Taylor expansion $\cos(kx) \approx 1 - \dfrac{(kx)^2}{2}$ as $x \to 0$. This gives us: $$\sum_{k=1}^{10} \cos(kx) \approx \sum_{k=1}^{10}\left(1 - \frac{k^2 x^2}{2}\right)$$ Step 2: Separate the sum into constant and variable terms. Expanding the sum: $$\sum_{k=1}^{10}\left(1 - \frac{k^2 x^2}{2}\right) = \sum_{k=1}^{10}1 - \frac{x^2}{2}\sum_{k=1}^{10} k^2 = 10 - \frac{x^2}{2}\sum_{k=1}^{10} k^2$$ Step 3: Find the numerator of the limit expression. Subtracting from 10: $$10 - \sum_{k=1}^{10}\cos(kx) \approx 10 - \left(10 - \frac{x^2}{2}\sum_{k=1}^{10} k^2\right) = \frac{x^2}{2}\sum_{k=1}^{10} k^2$$ Step 4: Evaluate the limit by dividing by $x^2$. $$\lim_{x \to 0} \frac{10 - \sum_{k=1}^{10}\cos(kx)}{x^2} = \lim_{x \to 0} \frac{\dfrac{x^2}{2}\sum_{k=1}^{10} k^2}{x^2} = \frac{1}{2}\sum_{k=1}^{10} k^2$$ Step 5: Calculate the sum of squares using the formula. Using the formula $\sum_{k=1}^{n} k^2 = \dfrac{n(n+1)(2n+1)}{6}$, with $n = 10$: $$\sum_{k=1}^{10} k^2 = \frac{10 \cdot 11 \cdot 21}{6} = \frac{2310}{6} = 385$$ Step 6: Compute the limit value. $$\lim_{x \to 0} \frac{10 - \sum_{k=1}^{10}\cos(kx)}{x^2} = \frac{1}{2} \cdot 385 = \frac{385}{2}$$ Step 7: Verify that $a$ and $b$ are coprime and find their sum. From $\dfrac{385}{2}$, we have $a = 385$ and $b = 2$. Checking: $\gcd(385, 2) = \gcd(5 \cdot 7 \cdot 11, 2) = 1$, so they are coprime. Therefore: $$a + b = 385 + 2 = 387$$ The answer is **Option 4: 387**.