Applications Of Derivatives

Find intersection points in the first quadrant ( $x > 0, y > 0$ ): $∣ x^{2} - 1∣ = 1 ⟹ x^{2} - 1 = 1$ or $x^{2} - 1 = - 1$ . $x^{2} = 2 ⟹ x = 2$ . (Note: $x^{2} = 0$ gives intersection at $x = 0$ , but 'first quadrant' implies $x > 0$ , so we exclude $x = 0$ . The actual intersections are $(0, 1)$ and $(2, 1)$ ).
Find slopes at $x = 2$ : Curve 1: $y = x^{2} - 1$ (since $x^{2} > 1$ near $2$ ). $d y / d x = 2 x$ . At $x = 2, m_{1} = 22$ . Curve 2: $y = 1$ . Constant function. $d y / d x = 0$ . So $m_{2} = 0$ .
Calculate angle: $tan θ = ∣ \frac{m _{1} - m _{2}}{1 + m _{1} m _{2}} ∣ = ∣ \frac{2 2 - 0}{1 + 0} ∣ = 22$ . Thus, $θ = tan^{- 1} (22)$ .

This medium difficulty mathematics question is from the chapter limit, continuity and differentiability, covering the topic of applications of derivatives. It appeared in the 2025 exam.

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Applications Of Derivatives

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