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

The function $f(x) = \text{sgn}(\sin^2x - \sin x - t)$ is discontinuous when the argument of the sgn function is zero. So, the points of discontinuity occur when $\sin^2x - \sin x - t = 0$. Let $y = \sin x$. The equation is $y^2 - y - t = 0$. This is a quadratic equation in $y$. Let its roots be $y_1, y_2$. $y = \frac{1 \pm \sqrt{1+4t}}{2}$. For real roots to exist, we must have $1+4t \ge 0$, so $t \ge -1/4$.

Case 1: $t = -1/4$. The equation is $y^2 - y + 1/4 = 0$, which is $(y - 1/2)^2 = 0$. The only root is $y = 1/2$. We need to find the number of solutions for $\sin x = 1/2$ in the interval $x \in (0, n\pi)$. The values of $x$ in $(0, \pi)$ for which $\sin x = 1/2$ are $\pi/6$ and $5\pi/6$. In the interval $(0, n\pi)$, the graph of $\sin x$ consists of $n$ arches.

If $n$ is odd, $n=2k+1$, the intervals are $(0, \pi), (\pi, 2\pi), \dots, (2k\pi, (2k+1)\pi)$. The intervals with positive $\sin x$ are $(0, \pi), (2\pi, 3\pi), \dots, (2k\pi, (2k+1)\pi)$. There are $k+1$ such intervals. In each such interval, $\sin x = 1/2$ has two solutions. The total number of solutions is $2(k+1) = 2(\frac{n-1}{2}+1) = n-1+2 = n+1$.

If $n$ is even, $n=2k$, the intervals are $(0, \pi), (\pi, 2\pi), \dots, ((2k-1)\pi, 2k\pi)$. The intervals with positive $\sin x$ are $(0, \pi), (2\pi, 3\pi), \dots, ((2k-2)\pi, (2k-1)\pi)$. There are $k$ such intervals. In each such interval, $\sin x = 1/2$ has two solutions. The total number of solutions is $2k = n$.

We are given that there are exactly four points of discontinuity.

If $n$ is odd, $n+1 = 4 \implies n=3$. If $n$ is even, $n = 4$.

So, if $t=-1/4$, $n$ can be 3 or 4.

Case 2: $t > -1/4$. There are two distinct roots $y_1 = \frac{1 - \sqrt{1+4t}}{2}$ and $y_2 = \frac{1 + \sqrt{1+4t}}{2}$. We know $y_1 < 1/2$ and $y_2 > 1/2$. The points of discontinuity occur when $\sin x = y_1$ or $\sin x = y_2$. We need to count the number of solutions in $x \in (0, n\pi)$ for which $\sin x$ takes these values, provided $y_1, y_2$ are in the range of $\sin x$ for $x \in (0, n\pi)$.

The range of $\sin x$ for $x \in (0, n\pi)$ is $(0, 1]$ if $n=1$, and $[-1, 1]$ if $n \ge 2$.

Subcase 2a: $n=1$. $x \in (0, \pi)$. $\sin x \in (0, 1]$. We need $y_1, y_2 \in (0, 1]$.

$y_2 = \frac{1 + \sqrt{1+4t}}{2}$. $y_2 \le 1 \implies 1 + \sqrt{1+4t} \le 2 \implies \sqrt{1+4t} \le 1$. Since $t > -1/4$, $1+4t > 0$. Squaring gives $1+4t \le 1 \implies 4t \le 0 \implies t \le 0$. So, $-1/4 < t \le 0$.

If $t=0$, $y^2-y=0 \implies y(y-1)=0 \implies y=0, 1$. $\sin x = 0$ has $n-1=0$ solution in $(0, \pi)$. $\sin x = 1$ has 1 solution in $(0, \pi)$. Total 1 solution. Not 4.

If $-1/4 < t < 0$, $y_2 < 1$. $y_1 = \frac{1 - \sqrt{1+4t}}{2}$. Since $-1/4 < t < 0$, $0 < 1+4t < 1$, so $0 < \sqrt{1+4t} < 1$. $y_1 = \frac{1 - \sqrt{1+4t}}{2} \in (\frac{1-1}{2}, \frac{1-0}{2}) = (0, 1/2)$. $y_2 = \frac{1 + \sqrt{1+4t}}{2} \in (\frac{1+0}{2}, \frac{1+1}{2}) = (1/2, 1)$. Both $y_1, y_2 \in (0, 1)$. For $x \in (0, \pi)$, $\sin x = y_1$ has two solutions and $\sin x = y_2$ has two solutions. Total $2+2=4$ solutions. So, if $-1/4 < t < 0$, $n=1$ is a possible value.

Subcase 2b: $n \ge 2$. $x \in (0, n\pi)$. $\sin x \in [-1, 1]$. We need $y_1, y_2 \in [-1, 1]$.

$y_2 = \frac{1 + \sqrt{1+4t}}{2} \le 1 \implies t \le 0$. $y_1 = \frac{1 - \sqrt{1+4t}}{2} \ge -1 \implies 1 - \sqrt{1+4t} \ge -2 \implies 3 \ge \sqrt{1+4t}$. Since $t > -1/4$, $1+4t > 0$. Squaring gives $9 \ge 1+4t \implies 8 \ge 4t \implies t \le 2$. So, for $n \ge 2$, we need $-1/4 < t \le 2$.

Let's analyze the number of solutions for $\sin x = y_0$ in $(0, n\pi)$ for $y_0 \in [-1, 1]$. If $y_0 \in (-1, 0) \cup (0, 1)$, there are 2 solutions in each full period of $2\pi$ where $\sin x$ reaches that value. In $(0, n\pi)$, there are $\lfloor n/2 \rfloor$ full periods $(0, 2\pi), (2\pi, 4\pi), \dots$. If $n$ is even, $n=2k$, the interval is $(0, 2k\pi)$. $\sin x$ completes $k$ full cycles. If $y_0 \in (-1, 0) \cup (0, 1)$, there are $2k = n$ solutions. If $y_0 = 1$, there are $k = n/2$ solutions. If $y_0 = -1$, there are $k-1 = n/2-1$ solutions for $k \ge 1$. If $y_0 = 0$, there are $n-1$ solutions ($\pi, 2\pi, \dots, (n-1)\pi$).

If $n$ is odd, $n=2k+1$, the interval is $(0, (2k+1)\pi)$. This is $k$ full cycles plus an interval of length $\pi$. If $y_0 \in (0, 1)$, there are $2k$ solutions in $(0, 2k\pi)$ and 2 solutions in $(2k\pi, (2k+1)\pi)$ if $y_0 < 1$. Total $2k+2 = n+1$ solutions if $y_0 \in (0, 1)$. If $y_0=1$, $k+1$ solutions. If $y_0 \in (-1, 0)$, there are $2k$ solutions in $(0, 2k\pi)$. Total $2k = n-1$ solutions. If $y_0 = 0$, there are $n-1$ solutions. If $y_0 = -1$, there are $k = (n-1)/2$ solutions.

Let's consider the ranges of $y_1, y_2$ for $-1/4 < t \le 2$. If $-1/4 < t < 0$, $y_1 \in (0, 1/2)$, $y_2 \in (1/2, 1)$. Both are in $(0, 1)$. If $n \ge 2$ is even ($n=2k$), number of solutions for $\sin x = y_1$ is $2k=n$. Number of solutions for $\sin x = y_2$ is $2k=n$. Total $2n$. We need $2n=4 \implies n=2$. If $n=2$, total solutions $2 \times 2 = 4$. So $n=2$ is possible if $-1/4 < t < 0$. If $n \ge 2$ is odd ($n=2k+1$), number of solutions for $\sin x = y_1$ is $n+1$. Number of solutions for $\sin x = y_2$ is $n+1$. Total $2(n+1)$. We need $2(n+1)=4 \implies n+1=2 \implies n=1$. But we assumed $n \ge 2$. No odd $n \ge 2$ is possible in this range of $t$.

If $t=0$, $y=0, 1$. For $n \ge 2$. If $n$ is even ($n=2k$), $\sin x = 0$ has $n-1$ solutions. $\sin x = 1$ has $n/2$ solutions. Total $n-1+n/2 = 3n/2 - 1$. We need $3n/2 - 1 = 4 \implies 3n/2 = 5 \implies n=10/3$. Not an integer. If $n$ is odd ($n=2k+1$), $\sin x = 0$ has $n-1$ solutions. $\sin x = 1$ has $(n+1)/2$ solutions. Total $n-1+(n+1)/2 = (2n-2+n+1)/2 = (3n-1)/2$. We need $(3n-1)/2 = 4 \implies 3n-1=8 \implies 3n=9 \implies n=3$. So $n=3$ is possible if $t=0$.

If $0 < t \le 2$. $y_1 = \frac{1 - \sqrt{1+4t}}{2}$. $0 < t \le 2 \implies 1 < 1+4t \le 9 \implies 1 < \sqrt{1+4t} \le 3$. $y_1 \in (\frac{1-3}{2}, \frac{1-1}{2}] = [-1, 0)$. $y_2 = \frac{1 + \sqrt{1+4t}}{2} \in (\frac{1+1}{2}, \frac{1+3}{2}] = (1, 2]$. Since the range of $\sin x$ is $[-1, 1]$ for $n \ge 2$, we need $y_2 \le 1$. This requires $t \le 0$. So there are no solutions for $y_2$ in the range of $\sin x$ when $0 < t \le 2$ and $n \ge 2$, unless $y_2=1$, which requires $t=0$.

Let's recheck the range of $y_1, y_2$ for $t > -1/4$. $y^2 - y - t = 0$. Roots are $y_1, y_2$. Vertex at $y=1/2$. $g(y) = y^2-y-t$. $g(-1) = 1 - (-1) - t = 2-t$. $g(1) = 1 - 1 - t = -t$. We need roots in $[-1, 1]$. One root in $[-1, 1]$ and the other outside. $y_1 \in [-1, 1]$ and $y_2 \notin [-1, 1]$ (since $y_2 \ge 1/2$, this means $y_2 > 1$). Requires $g(1) < 0$ and $g(-1) \ge 0$. $-t < 0 \implies t > 0$. $2-t \ge 0 \implies t \le 2$. So, if $0 < t \le 2$, one root $y_1 \in [-1, 1]$ and the other root $y_2 > 1$. $y_1 = \frac{1 - \sqrt{1+4t}}{2}$. For $0 < t \le 2$, $1 < 1+4t \le 9$, $1 < \sqrt{1+4t} \le 3$. $y_1 \in [\frac{1-3}{2}, \frac{1-1}{2}) = [-1, 0)$. So for $0 < t \le 2$, we have one root $y_1 \in [-1, 0)$ and the other root $y_2 > 1$. The only possible values for $\sin x$ are in $[-1, 1]$, so only $\sin x = y_1$ can have solutions. $y_1 \in [-1, 0)$.

If $n \ge 2$ is even ($n=2k$), $y_1 \in (-1, 0)$. Number of solutions for $\sin x = y_1$ is $2k=n$. If $y_1=-1$, number of solutions is $k-1 = n/2-1$ (for $k \ge 1$, i.e. $n \ge 2$). We need $n=4$ or $n/2-1=4 \implies n/2=5 \implies n=10$. So if $n$ is even, $n=4$ or $n=10$ are possible if $y_1 \in (-1, 0]$ and $y_2 > 1$. $y_1 \in (-1, 0]$ means $-1 < y_1 \le 0$. $y_1 \le 0 \implies \frac{1 - \sqrt{1+4t}}{2} \le 0 \implies 1 \le \sqrt{1+4t} \implies 1 \le 1+4t \implies 0 \le 4t \implies t \ge 0$. $y_1 > -1 \implies \frac{1 - \sqrt{1+4t}}{2} > -1 \implies 1 - \sqrt{1+4t} > -2 \implies 3 > \sqrt{1+4t} \implies 9 > 1+4t \implies 8 > 4t \implies t < 2$. So for $0 \le t < 2$, $y_1 \in (-1, 0]$. If $t=0$, $y_1=0$. $n-1=4 \implies n=5$. But we assumed $n$ is even. If $0 < t < 2$, $y_1 \in (-1, 0)$. Number of solutions is $n$. We need $n=4$. So $n=4$ is possible if $0 < t < 2$. If $t=2$, $y_1 = -1$. Number of solutions is $n/2-1$. We need $n/2-1=4 \implies n/2=5 \implies n=10$. So $n=10$ is possible if $t=2$.

If $n \ge 2$ is odd ($n=2k+1$), $y_1 \in [-1, 0)$. If $y_1 \in (-1, 0)$, number of solutions is $2k = n-1$. We need $n-1=4 \implies n=5$. So $n=5$ is possible if $0 < t < 2$. If $y_1 = -1$, number of solutions is $k = (n-1)/2$. We need $(n-1)/2 = 4 \implies n-1=8 \implies n=9$. So $n=9$ is possible if $t=2$.

Both roots outside $[-1, 1]$. $y_1 < -1$ and $y_2 > 1$. $y_1 < -1 \implies t > 2$. $y_2 > 1 \implies t > 0$. So if $t > 2$, $y_1 < -1$ and $y_2 > 1$. Neither root is in $[-1, 1]$. So there are no solutions for $\sin x = y_1$ or $\sin x = y_2$. Number of discontinuities is 0. Not 4.

Summary of possible values of $n$: If $t=-1/4$, $n=3, 4$. If $-1/4 < t < 0$, $n=1, 2$. If $t=0$, $n=3$. If $0 < t < 2$, $n=4, 5$. If $t=2$, $n=9, 10$.

The possible values of $n$ are $1, 2, 3, 4, 5, 9, 10$. We are looking for statements about $n \in N$. Minimum value of $n$ is 1. The statement "minimum value of n is 5" is false. Maximum value of $n$ is 10. The statement "maximum value of n is 5" is false. There are exactly two possible values of $n$. The possible values are $\{1, 2, 3, 4, 5, 9, 10\}$, which has 7 values. The statement "there are exactly two possible value of n" is false. Therefore, none of the given statements are true.

The final answer is $\boxed{none of these}$.