Differential Calculus-1
Differential Calculus-1
Allen Star Batch
Grade 12
Question:
Define $f: [0, \pi] \to \mathbb{R}$ by $f(x) = \begin{cases} \tan^2 x + \sqrt{2\sin^2 x + 3\sin x + 4 - \sqrt{\sin^2 x + 6\sin x + 2}} & x \neq \pi/2 \\ k & x = \pi/2 \end{cases}$ is continuous at $x = \pi/2$, then $k$ is equal to:
Step-by-Step Solution
Key Concept: To find the value of k that makes f continuous at x = π/2, evaluate lim(x→π/2⁻) f(x) by substituting t = sin x → 1 and rationalizing the nested radicals in the expression tan²x + √(2sin²x + 3sin x + 4 - √(sin²x + 6sin x + 2)).
Let $\sin x = t$, then we evaluate $\lim_{t \to 1^-} \frac{t^2}{\sqrt{2t^2 + 3t - 4} - \sqrt{t^2 + 6t + 2}}$ by rationalizing the denominator. Multiply by $\frac{\sqrt{2t^2 + 3t - 4} + \sqrt{t^2 + 6t + 2}}{\sqrt{2t^2 + 3t - 4} + \sqrt{t^2 + 6t + 2}}$ to get $\frac{t^2(\sqrt{2t^2 + 3t - 4} + \sqrt{t^2 + 6t + 2})}{t^2 - 6} \to \frac{(2 + 2)}{1 - 6} = -\frac{4}{5}$.
Correct Answer: 1