Question:

Step 1: Identify the boundary points where the function definition changes. We need to check continuity at the points where the piecewise function changes its definition. These boundary points are $x = \sqrt{10}$, $x = 5$, $x = 6$, and $x = 7$ (within or at the boundary of the domain $[1, 7]$). Step 2: Check continuity at $x = \sqrt{10}$. We need to verify that the left-hand limit equals the right-hand limit and equals the function value at this point. Left-hand limit (using $f(x) = \sqrt{x^2-1}$ for $x \leq \sqrt{10}$): $$\lim_{x \to \sqrt{10}^-} f(x) = \sqrt{(\sqrt{10})^2 - 1} = \sqrt{10 - 1} = \sqrt{9} = 3$$ Right-hand limit (using $f(x) = \sqrt{10}x - 7$ for $\sqrt{10} < x < 5$): $$\lim_{x \to \sqrt{10}^+} f(x) = \sqrt{10} \cdot \sqrt{10} - 7 = 10 - 7 = 3$$ Since both limits equal 3, the function is **continuous at $x = \sqrt{10}$**. Step 3: Check continuity at $x = 5$. Left-hand limit (using $f(x) = \sqrt{10}x - 7$ for $\sqrt{10} < x < 5$): $$\lim_{x \to 5^-} f(x) = \sqrt{10}(5) - 7 = 5\sqrt{10} - 7$$ Since $\sqrt{10} \approx 3.162$, we have $5\sqrt{10} - 7 \approx 15.81 - 7 = 8.81$. Right-hand limit (using $f(x) = \sin(\pi x)$ for $5 \leq x < 6$): $$\lim_{x \to 5^+} f(x) = \sin(5\pi) = 0$$ Since $8.81 \neq 0$, the function is **discontinuous at $x = 5$**. Step 4: Check continuity at $x = 6$. Left-hand limit (using $f(x) = \sin(\pi x)$ for $5 \leq x < 6$): $$\lim_{x \to 6^-} f(x) = \sin(6\pi) = 0$$ Right-hand limit (using $f(x) = \{x\}$ for $6 \leq x \leq 7$): $$\lim_{x \to 6^+} f(x) = \{6\} = 0$$ Since both limits equal 0, the function is **continuous at $x = 6$**. Step 5: Check continuity at $x = 7$ (the right endpoint). Left-hand limit (using $f(x) = \{x\}$ for $6 \leq x \leq 7$): $$\lim_{x \to 7^-} f(x) = \lim_{x \to 7^-} \{x\} = 1$$ (The fractional part approaches 1 as $x$ approaches 7 from the left.) Function value at $x = 7$: $$f(7) = \{7\} = 0$$ (The fractional part of 7 is 0.) Since $1 \neq 0$, the function is **discontinuous at $x = 7$**. Step 6: Count the total number of discontinuities. The function is discontinuous at exactly two points: $x = 5$ and $x = 7$. **Final Answer: The number of points where $f(x)$ is discontinuous in $[1, 7]$ is 2.** The correct option is **(c) 2**.