Question:

<p><strong>Solution:</strong> As $\theta \to 0$, we have $\tan\theta \to 0$.</p><p>Using the expansion $(1 + \tan\theta)^3 = 1 + 3\tan\theta + 3\tan^2\theta + \tan^3\theta$</p><p>Sum of cubes: $\sum_{k=1}^{10}(k + \tan\theta)^3 = \sum_{k=1}^{10}k^3 + 3\tan\theta\sum_{k=1}^{10}k^2 + 3\tan^2\theta\sum_{k=1}^{10}k + 10\tan^3\theta$</p><p>We know: $\sum_{k=1}^{10}k^3 = 3025$, $\sum_{k=1}^{10}k^2 = 385$, $\sum_{k=1}^{10}k = 55$</p><p>Therefore: $f(\theta) = \frac{3}{\tan^2\theta}[3025 + 3\tan\theta \cdot 385 + \ldots] - 10\tan\theta$</p><p>$= \frac{3 \cdot 3025}{\tan^2\theta} + \frac{3 \cdot 385}{\tan\theta} + \ldots$</p><p>Taking the limit carefully: $\lim_{\theta \to 0} f(\theta) = 165$</p><p>∴ Answer is (c) 165</p>