Step-by-Step Solution & Explanation
### **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.