Question

If f(x) = sgn(sin²x - sinx - t) has exactly four points of discontinuity for x ∈ (0, nπ), n ∈ N, then

• A. minimum value of n is 5
• B. maximum value of n is 5
• C. there are exactly two possible value of n
• D. none of these

Answer 1

Answer: (D)

none of these

Answer 2

Let $f(x) = \text{sgn}(\sin^2x - \sin x - t)$. The function $f(x)$ is discontinuous when the argument of the signum function, $u = \sin^2x - \sin x - t$, changes sign. This occurs when $u=0$.

The points of discontinuity are the solutions to the equation $\sin^2x - \sin x - t = 0$.

Let $y = \sin x$. The equation becomes $y^2 - y - t = 0$.

Let $g(y) = y^2 - y - t$. The roots of $g(y) = 0$ are $y = \frac{1 \pm \sqrt{1+4t}}{2}$. For real roots, we need $1+4t \ge 0$, so $t \ge -1/4$.

The range of $\sin x$ for $x \in (0, n\pi)$ is $[-1, 1]$. The points of discontinuity occur when $\sin x = y_i$ for some root $y_i$ of $y^2 - y - t = 0$ such that $y_i \in [-1, 1]$.

We need to find the number of solutions to $\sin x = y_i$ in $x \in (0, n\pi)$ for $y_i \in [-1, 1]$.

The number of solutions to $\sin x = y$ in $(0, n\pi)$ depends on $y$ and $n$.

• If $y \in (0, 1)$, there are 2 solutions in $(k\pi, (k+1)\pi)$ for even $k$, and 0 solutions for odd $k$. The total number of solutions in $(0, n\pi)$ is $2 \times \lfloor n/2 \rfloor$ if $n$ is even, and $2 \times \lfloor n/2 \rfloor + 2 = 2 \times (n-1)/2 + 2 = n-1+2 = n+1$ if $n$ is odd. This can be summarized as $n$ if $n$ is even, and $n+1$ if $n$ is odd.
• If $y = 1$, there is 1 solution in $(k\pi, (k+1)\pi)$ for even $k > 0$ (at $x = k\pi + \pi/2$), and 0 solutions for odd $k$. Plus 1 solution at $x = \pi/2$ in $(0, \pi)$. The number of solutions in $(0, n\pi)$ is $\lfloor n/2 \rfloor$ if $n$ is even, and $\lfloor n/2 \rfloor + 1 = (n-1)/2 + 1 = (n+1)/2$ if $n$ is odd.
• If $y \in (-1, 0)$, there are 2 solutions in $(k\pi, (k+1)\pi)$ for odd $k$, and 0 solutions for even $k$. The number of solutions in $(0, n\pi)$ is $2 \times \lfloor n/2 \rfloor$ if $n$ is even, and $2 \times \lfloor (n-1)/2 \rfloor = 2 \times (n-1)/2 = n-1$ if $n$ is odd. This can be summarized as $n$ if $n$ is even, and $n-1$ if $n$ is odd.
• If $y = -1$, there is 1 solution in $(k\pi, (k+1)\pi)$ for odd $k > 0$ (at $x = k\pi + 3\pi/2$), and 0 solutions for even $k$. The number of solutions in $(0, n\pi)$ is $\lfloor (n-1)/2 \rfloor$ if $n$ is even, and $\lfloor (n-1)/2 \rfloor + 1 = (n-1)/2 + 1 = (n+1)/2$ if $n$ is odd.
• If $y=0$, $\sin x = 0$. Solutions are $x = k\pi$. For $x \in (0, n\pi)$, the solutions are $k\pi$ for $k=1, 2, \dots, n-1$. There are $n-1$ solutions.
• If $y \notin [-1, 1]$, there are no solutions.

Let the roots of $y^2 - y - t = 0$ be $y_1, y_2$.

The vertex of $g(y) = y^2 - y - t$ is at $y=1/2$. $g(1/2) = 1/4 - 1/2 - t = -1/4 - t$.

Case 1: $t = -1/4$. $g(y) = y^2 - y + 1/4 = (y-1/2)^2 = 0$. Root is $y = 1/2$. We need solutions to $\sin x = 1/2$ in $(0, n\pi)$. $y=1/2 \in (0, 1)$. Number of solutions is $n$ if $n$ is even, and $n+1$ if $n$ is odd. We want this number to be 4. If $n$ is even, $n=4$. If $n$ is odd, $n+1=4 \implies n=3$. So $n=3$ or $n=4$ are possible values.

Case 2: $-1/4 < t \le 0$. Roots are $y_{1,2} = \frac{1 \pm \sqrt{1+4t}}{2}$. Since $-1/4 < t \le 0$, $0 < 1+4t \le 1$, so $0 < \sqrt{1+4t} \le 1$. $y_1 = \frac{1 - \sqrt{1+4t}}{2}$. $0 \le 1-\sqrt{1+4t} < 1$, so $0 \le y_1 < 1/2$. If $t=0$, $y_1=0$. If $-1/4 < t < 0$, $0 < y_1 < 1/2$. $y_2 = \frac{1 + \sqrt{1+4t}}{2}$. $1 < 1+\sqrt{1+4t} \le 2$, so $1/2 < y_2 \le 1$. If $t=0$, $y_2=1$. If $-1/4 < t < 0$, $1/2 < y_2 < 1$.

Subcase 2a: $-1/4 < t < 0$. We have two roots $y_1 \in (0, 1/2)$ and $y_2 \in (1/2, 1)$. Both are in $(0, 1)$. The points of discontinuity are solutions to $\sin x = y_1$ or $\sin x = y_2$. Number of solutions for $\sin x = y_1$ is $N_1$, which is $n$ if $n$ is even, $n+1$ if $n$ is odd. Number of solutions for $\sin x = y_2$ is $N_2$, which is $n$ if $n$ is even, $n+1$ if $n$ is odd. Since $y_1 \ne y_2$, the solutions for $\sin x = y_1$ and $\sin x = y_2$ are distinct. Total number of discontinuities is $N_1 + N_2$. If $n$ is even, $N_1+N_2 = n+n = 2n$. We need $2n=4$, so $n=2$. If $n$ is odd, $N_1+N_2 = (n+1)+(n+1) = 2n+2$. We need $2n+2=4$, so $2n=2$, $n=1$. So $n=1$ or $n=2$ are possible values.

Subcase 2b: $t=0$. The roots are $y=0$ and $y=1$. We need solutions to $\sin x = 0$ or $\sin x = 1$ in $(0, n\pi)$. Number of solutions for $\sin x = 0$ in $(0, n\pi)$ is $n-1$. Number of solutions for $\sin x = 1$ in $(0, n\pi)$ is $\lfloor n/2 \rfloor$ if $n$ is even, $(n+1)/2$ if $n$ is odd. Total number of discontinuities is $(n-1) + \lfloor n/2 \rfloor$ if $n$ is even, $(n-1) + (n+1)/2$ if $n$ is odd. If $n$ is even, $n=2m$, total is $(2m-1) + m = 3m-1$. We need $3m-1=4$, $3m=5$, no integer solution for $m$. If $n$ is odd, $n=2m+1$, total is $(2m+1-1) + (2m+1+1)/2 = 2m + (2m+2)/2 = 2m + m+1 = 3m+1$. We need $3m+1=4$, $3m=3$, $m=1$. So $n=2(1)+1=3$. So $n=3$ is a possible value.

Case 3: $0 < t \le 2$. Roots are $y_{1,2} = \frac{1 \pm \sqrt{1+4t}}{2}$. Since $0 < t \le 2$, $1 < 1+4t \le 9$, so $1 < \sqrt{1+4t} \le 3$. $y_1 = \frac{1 - \sqrt{1+4t}}{2}$. $1-3 \le 1-\sqrt{1+4t} < 1-1$, so $-2 \le 1-\sqrt{1+4t} < 0$. Thus $-1 \le y_1 < 0$. If $t=2$, $y_1=-1$. If $0 < t < 2$, $-1 < y_1 < 0$. $y_2 = \frac{1 + \sqrt{1+4t}}{2}$. $1+1 < 1+\sqrt{1+4t} \le 1+3$, so $2 < 1+\sqrt{1+4t} \le 4$. Thus $1 < y_2 \le 2$. Since $y_2 > 1$, $\sin x = y_2$ has no solution. The points of discontinuity are solutions to $\sin x = y_1$ where $y_1 \in [-1, 0)$.

Subcase 3a: $0 < t < 2$. We have one root $y_1 \in (-1, 0)$. Number of solutions for $\sin x = y_1$ is $n$ if $n$ is even, $n-1$ if $n$ is odd. We need this number to be 4. If $n$ is even, $n=4$. If $n$ is odd, $n-1=4 \implies n=5$. So $n=4$ or $n=5$ are possible values.

Subcase 3b: $t=2$. We have one root $y_1 = -1$. Number of solutions for $\sin x = -1$ in $(0, n\pi)$ is $\lfloor (n-1)/2 \rfloor$ if $n$ is even, $(n+1)/2$ if $n$ is odd. We need this number to be 4. If $n$ is even, $n=2m$, $\lfloor (2m-1)/2 \rfloor = m-1$. We need $m-1=4$, $m=5$, so $n=10$. If $n$ is odd, $n=2m+1$, $(2m+1+1)/2 = m+1$. We need $m+1=4$, $m=3$, so $n=7$. So $n=7$ or $n=10$ are possible values.

Case 4: $t > 2$. $y_1 < -1$ and $y_2 > 1$. No roots in $[-1, 1]$. No discontinuities.

Combining all possible values of $n$: From $t=-1/4$: $n=3, 4$. From $-1/4 < t < 0$: $n=1, 2$. From $t=0$: $n=3$. From $0 < t < 2$: $n=4, 5$. From $t=2$: $n=7, 10$.

Possible values of $n$ are $\{1, 2, 3, 4, 5, 7, 10\}$. The minimum value of $n$ is 1. The maximum value of $n$ is 10. There are more than two possible values of $n$.

The set of possible values for $n$ is $\{1, 2, 3, 4, 5, 7, 10\}$. Minimum value of $n$ is 1. Maximum value of $n$ is 10.

The solution seems consistent.