Calculus

### **Detailed Step-by-Step Solution (GATE AE 2019 Q.41 / AE Q.31)** **1. Intuitive Physical Setup & Given Parameters**: - Two curves: $y_1(x) = x$ and $y_2(x) = \cos x$ for real $x \in \mathbb{R}$. - Target: Determine the number of real roots of $g(x) = x - \cos x = 0$. **2. Step-by-Step Mathematical Derivation**: 1. **Range and Domain Inspection**: - The cosine function is bounded: $-1 \le \cos x \le 1$ for all real $x$. - For an intersection to occur, $y_1 = y_2 \implies x = \cos x \implies x \in [-1, 1]$. - Outside $[-1, 1]$, $x > 1 > \cos x$ (no roots for $x > 1$) and $x < -1 < \cos x$ (no roots for $x < -1$). 2. **Derivative and Monotonicity**: Define $g(x) = x - \cos x$. $g'(x) = 1 - (-\sin x) = 1 + \sin x$ Since $-1 \le \sin x \le 1$ for all real $x$: $g'(x) = 1 + \sin x \ge 0 \quad \forall x \in \mathbb{R}$ The derivative $g'(x) = 0$ only at isolated points $x = -\pi/2 + 2k\pi$, meaning $g(x)$ is strictly monotonically increasing on $\mathbb{R}$. 3. **Existence and Uniqueness**: - At $x = 0$: $g(0) = 0 - \cos(0) = -1 < 0$. - At $x = \pi/2$: $g(\pi/2) = \frac{\pi}{2} - \cos(\pi/2) = \frac{\pi}{2} - 0 \approx 1.57 > 0$. - By the Intermediate Value Theorem, there exists at least one root in $(0, \pi/2)$. - By strict monotonicity ($g'(x) > 0$ almost everywhere), the root is strictly unique. Therefore, the two curves intersect at exactly **1 point** (the Dottie number, $x \approx 0.739085$). **3. Comprehensive Options & Distractor Analysis**: - **Question Type**: Numerical Answer Type (NAT). - **Exact Evaluation**: **1.0** (Acceptable Range: [1, 1]). **4. Official Master Key & Exam Takeaway**: - **Official Answer**: **1.0** - **Exam Rule**: If $g'(x) \ge 0$ everywhere and strictly positive almost everywhere, a continuous function can cross zero at most once.