Solving for Integral Values of '$p$'
We are given the equation $3\sin^2 x + 12\cos x - 3 = p$, where $x \in \mathbb{R}$. We need to find the sum of all integral values of '$p$' for which this equation has at least one solution.
Simplifying the Trigonometric Equation
First, we simplify the equation using the identity $\sin^2 x = 1 - \cos^2 x$.
Substituting this into the equation:
$3(1 - \cos^2 x) + 12\cos x - 3 = p$ $3 - 3\cos^2 x + 12\cos x - 3 = p$ $p = -3\cos^2 x + 12\cos x$Analyzing the Function
Let $y = \cos x$. Since $x \in \mathbb{R}$, the possible values for $y$ lie in the interval $[-1, 1]$. The equation becomes:
$p = -3y^2 + 12y$We need to find the range of the function $f(y) = -3y^2 + 12y$ for $y \in [-1, 1]$. This function represents a downward-opening parabola.
Finding the Range of $f(y)$
- The vertex of the parabola $f(y) = -3y^2 + 12y$ occurs at $y = -\frac{b}{2a} = -\frac{12}{2(-3)} = 2$.
- Since the vertex ($y=2$) is outside the interval $[-1, 1]$ and the parabola opens downwards, the function is strictly increasing on the interval $[-1, 1]$.
- The minimum value of $f(y)$ occurs at $y = -1$: $f(-1) = -3(-1)^2 + 12(-1) = -3(1) - 12 = -15$
- The maximum value of $f(y)$ occurs at $y = 1$: $f(1) = -3(1)^2 + 12(1) = -3(1) + 12 = 9$
- Therefore, the range of $f(y)$ for $y \in [-1, 1]$ is $[-15, 9]$.
The equation $p = f(y)$ has at least one solution if and only if $p$ is within this range. So, $p$ must be in the interval $[-15, 9]$.
Calculating the Sum of Integral Values
We need the sum of all integers $p$ such that $-15 \le p \le 9$. These integers are $-15, -14, \dots, -1, 0, 1, \dots, 8, 9$.
The sum can be calculated as:
$ \sum_{p=-15}^{9} p = \sum_{p=-15}^{-1} p + \sum_{p=0}^{9} p $ $ = \left( \sum_{p=1}^{15} (-p) \right) + \left( \sum_{p=1}^{9} p \right) $ $ = -\frac{15(15+1)}{2} + \frac{9(9+1)}{2} $ $ = -\frac{15 \times 16}{2} + \frac{9 \times 10}{2} $ $ = -(15 \times 8) + (9 \times 5) $ $ = -120 + 45 $ $ = -75 $The sum of all integral values of $p$ is -75.
Similar Questions
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Important Questions from Trigonometry
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Let $\cos(\alpha + \beta) = -\frac{1}{10}$ and $\sin(\alpha - \beta) = \frac{3}{8}$, where $0 < \alpha < \frac{\pi}{3}$ and $0 < \beta < \frac{\pi}{4}$. If $\tan 2\alpha = \frac{3(1 - r\sqrt{5})}{\sqrt{11}(s + \sqrt{5})}, r, s \in \mathbb{N}$, then $r+s$ is equal to ________.
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- Number of solutions of $\sqrt{3}\cos 2\theta + 8\cos \theta + 3\sqrt{3} = 0, \theta \in [-3\pi, 2\pi]$ is :
