Question

If the equation $\sin^4x-(p+2)\sin^2x-(p+3)=0$ has a solution, the $p$ must lie in the interval

যদি $\sin^4x-(p+2)\sin^2x-(p+3)=0$ সমীকরণের একটি সমাধান থাকে, তাহলে $p$ অবশ্যই যে অন্তরে থাকবে তা=

• A. [-3,-2]
• B. (-3,-2)
• C. (2, 3)
• D. [-5, -3]

Answer 1

Answer: (A)

[-3,-2]

Answer 2

Let the given equation be $\sin^4x-(p+2)\sin^2x-(p+3)=0$

Let $y = \sin^2x$. Since $x$ is a real number, the range of $\sin x$ is $[-1, 1]$. Therefore, the range of $\sin^2x$ is $[0, 1]$. So, $y$ must satisfy $0 \le y \le 1$.

Substituting $y = \sin^2x$ into the equation, we get a quadratic equation in $y$: $y^2 - (p+2)y - (p+3) = 0$

For the original equation in $x$ to have a solution, the quadratic equation in $y$ must have at least one root $y$ in the interval $[0, 1]$.

We can find the roots of the quadratic equation $y^2 - (p+2)y - (p+3) = 0$ using the quadratic formula: $y = \frac{-(-(p+2)) \pm \sqrt{(-(p+2))^2 - 4(1)(-(p+3))}}{2(1)}$ $y = \frac{(p+2) \pm \sqrt{(p+2)^2 + 4(p+3)}}{2}$ $y = \frac{(p+2) \pm \sqrt{p^2 + 4p + 4 + 4p + 12}}{2}$ $y = \frac{(p+2) \pm \sqrt{p^2 + 8p + 16}}{2}$ $y = \frac{(p+2) \pm \sqrt{(p+4)^2}}{2}$ $y = \frac{(p+2) \pm |p+4|}{2}$

We need to consider two cases based on the sign of $p+4$.

Case 1: $p+4 \ge 0$, i.e., $p \ge -4$. In this case, $|p+4| = p+4$. The roots are $y_1 = \frac{(p+2) + (p+4)}{2} = \frac{2p+6}{2} = p+3$ and $y_2 = \frac{(p+2) - (p+4)}{2} = \frac{-2}{2} = -1$.

Case 2: $p+4 < 0$, i.e., $p < -4$. In this case, $|p+4| = -(p+4)$. The roots are $y_1 = \frac{(p+2) + (-(p+4))}{2} = \frac{p+2 - p - 4}{2} = \frac{-2}{2} = -1$ and $y_2 = \frac{(p+2) - (-(p+4))}{2} = \frac{p+2 + p + 4}{2} = \frac{2p+6}{2} = p+3$.

In both cases, the roots of the quadratic equation in $y$ are $-1$ and $p+3$. For the original equation in $x$ to have a solution, at least one of these roots must be in the interval $[0, 1]$.

The root $y = -1$ is not in the interval $[0, 1]$. Therefore, the other root, $y = p+3$, must be in the interval $[0, 1]$. $0 \le p+3 \le 1$

To find the values of $p$ that satisfy this inequality, we subtract 3 from all parts: $0 - 3 \le p+3 - 3 \le 1 - 3$ $-3 \le p \le -2$

So, the equation has a solution if and only if $p$ lies in the interval $[-3, -2]$.