Question

If the equation sin (πχ²) - sin(πχ² + 2πχ) = 0 is solved for positive roots, then in the increasing sequence of positive root

• A. first term is $\frac{-1+\sqrt{7}}{2}$
• B. first term is $\frac{-1+\sqrt{3}}{2}$
• C. third term is 1
• D. third term is $\frac{-1+\sqrt{11}}{2}$

Answer 1

Answer: (B), (C)

first term is $\frac{-1+\sqrt{3}}{2}$ and third term is 1

Answer 2

The given equation is $\sin(\pi x^2) - \sin(\pi x^2 + 2\pi x) = 0$. This can be rewritten as $\sin(\pi x^2) = \sin(\pi x^2 + 2\pi x)$.

The general solution for $\sin A = \sin B$ is $A = n\pi + (-1)^n B$, where $n$ is an integer. Let $A = \pi x^2$ and $B = \pi x^2 + 2\pi x$. So, $\pi x^2 = n\pi + (-1)^n (\pi x^2 + 2\pi x)$. Dividing by $\pi$ (since $\pi \neq 0$), we get: $x^2 = n + (-1)^n (x^2 + 2x)$.

We need to consider two cases based on the parity of $n$.

Case 1: $n$ is an even integer. Let $n = 2k$ for some integer $k \in \mathbb{Z}$. Substituting $n=2k$ into the equation: $x^2 = 2k + (-1)^{2k} (x^2 + 2x)$ $x^2 = 2k + 1 \cdot (x^2 + 2x)$ $x^2 = 2k + x^2 + 2x$ $0 = 2k + 2x$ $2x = -2k$ $x = -k$

We are looking for positive roots, so $x > 0$. Thus, $-k > 0 \implies k < 0$. Since $k$ is an integer, possible values for $k$ are $-1, -2, -3, \ldots$. For $k=-1$, $x = -(-1) = 1$. For $k=-2$, $x = -(-2) = 2$. For $k=-3$, $x = -(-3) = 3$. So, the positive roots from this case are $1, 2, 3, \ldots$.

Case 2: $n$ is an odd integer. Let $n = 2k+1$ for some integer $k \in \mathbb{Z}$. Substituting $n=2k+1$ into the equation: $x^2 = (2k+1) + (-1)^{2k+1} (x^2 + 2x)$ $x^2 = (2k+1) - (x^2 + 2x)$ $x^2 = 2k+1 - x^2 - 2x$ Rearranging the terms to form a quadratic equation in $x$: $2x^2 + 2x - (2k+1) = 0$

Using the quadratic formula $x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$: $x = \frac{-2 \pm \sqrt{2^2 - 4(2)(-(2k+1))}}{2(2)}$ $x = \frac{-2 \pm \sqrt{4 + 8(2k+1)}}{4}$ $x = \frac{-2 \pm \sqrt{4 + 16k + 8}}{4}$ $x = \frac{-2 \pm \sqrt{16k + 12}}{4}$ $x = \frac{-2 \pm \sqrt{4(4k + 3)}}{4}$ $x = \frac{-2 \pm 2\sqrt{4k + 3}}{4}$ $x = \frac{-1 \pm \sqrt{4k + 3}}{2}$

For $x$ to be a real number, the term under the square root must be non-negative: $4k+3 \ge 0 \implies 4k \ge -3 \implies k \ge -3/4$. Since $k$ is an integer, possible values for $k$ are $0, 1, 2, 3, \ldots$.

Now, we need to find positive roots ($x > 0$). Consider $x = \frac{-1 + \sqrt{4k + 3}}{2}$. For $x > 0$, we need $-1 + \sqrt{4k+3} > 0 \implies \sqrt{4k+3} > 1$. Squaring both sides (both are positive), $4k+3 > 1 \implies 4k > -2 \implies k > -1/2$. So, for $k = 0, 1, 2, \ldots$, we get positive roots. For $k=0$, $x = \frac{-1 + \sqrt{4(0)+3}}{2} = \frac{-1 + \sqrt{3}}{2}$. For $k=1$, $x = \frac{-1 + \sqrt{4(1)+3}}{2} = \frac{-1 + \sqrt{7}}{2}$. For $k=2$, $x = \frac{-1 + \sqrt{4(2)+3}}{2} = \frac{-1 + \sqrt{11}}{2}$. And so on.

Consider $x = \frac{-1 - \sqrt{4k + 3}}{2}$. For $x$ to be positive, we would need $-1 - \sqrt{4k+3} > 0 \implies -1 > \sqrt{4k+3}$. This is impossible because $\sqrt{4k+3}$ is always non-negative. So, this branch yields no positive roots.

Combining all positive roots from both cases: Set of roots $S = \{1, 2, 3, \ldots\} \cup \{\frac{-1+\sqrt{3}}{2}, \frac{-1+\sqrt{7}}{2}, \frac{-1+\sqrt{11}}{2}, \frac{-1+\sqrt{15}}{2}, \ldots\}$.

To form the increasing sequence, let's approximate the values: $\frac{-1+\sqrt{3}}{2} \approx \frac{-1+1.732}{2} = \frac{0.732}{2} = 0.366$ $\frac{-1+\sqrt{7}}{2} \approx \frac{-1+2.646}{2} = \frac{1.646}{2} = 0.823$ $1$ $\frac{-1+\sqrt{11}}{2} \approx \frac{-1+3.317}{2} = \frac{2.317}{2} = 1.1585$

Arranging the positive roots in increasing order: 1st term: $\frac{-1+\sqrt{3}}{2}$ 2nd term: $\frac{-1+\sqrt{7}}{2}$ 3rd term: $1$

Thus, options (B) and (C) are correct.