Question:hard

The number of solutions of the equation $|\sin(\pi x)| = \frac{1}{50}(x^2 + 1)$ in $\mathbb{R}$ is

Show Hint

Plotting a rough graph of the parabolic curve intersecting the absolute sine humps makes this counting problem very visual.
For each full period of the sine function that lies below the maximum bound ($y \le 1$), there are exactly two intersection points.
Updated On: Jun 16, 2026
  • 28
  • 26
  • 14
  • 13
Show Solution

The Correct Option is A

Solution and Explanation

Step 1: Treat it as two curves meeting.
Solutions of $|\sin(\pi x)| = \frac{1}{50}(x^2 + 1)$ are the points where the wavy curve $y = |\sin(\pi x)|$ meets the upward parabola $y = \frac{1}{50}(x^2 + 1)$.

Step 2: Use the symmetry.
Both sides are even functions (unchanged when $x$ becomes $-x$), so the picture is mirror-symmetric about the $y$-axis. We count solutions for $x \gt 0$, handle $x = 0$ separately, then double the positive count.

Step 3: Find where the parabola exceeds 1.
Since $|\sin(\pi x)| \le 1$, no solution exists once $\frac{1}{50}(x^2 + 1) \gt 1$, that is $x^2 \gt 49$, so $x \gt 7$. All solutions lie within $-7 \le x \le 7$.

Step 4: Check $x = 0$.
At $x = 0$: left side $|\sin 0| = 0$, right side $\frac{1}{50}(1) = 0.02$. Not equal, so $x = 0$ is not a solution.

Step 5: Count crossings on each sine arch for $0 \lt x \le 7$.
Over each unit interval the curve $|\sin(\pi x)|$ makes one arch from 0 up to 1 and back to 0. The small parabola value (staying well below 1 until near $x = 7$) cuts each such arch twice. There are 7 arches over $(0, 7]$, giving about $2 \times 7 = 14$, but the final arch near $x = 7$ where the parabola reaches 1 yields a single tangential-type meeting, trimming the count to 13 on the positive side.

Step 6: Double for symmetry.
With 13 solutions for $x \gt 0$ and the same 13 for $x \lt 0$, and none at $x = 0$, the total is $13 + 13 = 26$.
\[ \boxed{26} \]
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