Question

The sum of all the integral values of p such that the equation $3\sin^2x + 12\cos x - 3 = p, x \in \mathbb{R}$, has at least one solution, is:

• A. -54
• B. -60
• C. -75
• D. -84

Answer 1

The Correct Option is C

Step 1: Understanding the Question:
The question asks for the sum of all possible integer values of $p$ for which the given trigonometric equation has at least one real solution for $x$. This is equivalent to finding the range of the function $f(x) = 3\sin^2x + 12\cos x - 3$ and then summing all the integers within that range.
Step 2: Key Formula or Approach:
1. Express the function $f(x)$ in terms of a single trigonometric function, preferably $\cos x$. We can use the identity $\sin^2x = 1 - \cos^2x$.
2. Let $t = \cos x$. The range of $\cos x$ for $x \in \mathbb{R}$ is $[-1, 1]$. So, we need to find the range of the resulting quadratic function in $t$ over the interval $t \in [-1, 1]$.
3. Once the range $[min, max]$ of the function is found, the integer values of $p$ are all integers from $min$ to $max$. We then sum these integers.
Step 3: Detailed Explanation:
Let the function be $f(x) = 3\sin^2x + 12\cos x - 3$.
Using $\sin^2x = 1 - \cos^2x$:
$f(x) = 3(1 - \cos^2x) + 12\cos x - 3$
$f(x) = 3 - 3\cos^2x + 12\cos x - 3$
$f(x) = -3\cos^2x + 12\cos x$
Let $t = \cos x$. Since $x \in \mathbb{R}$, the values of $t$ are in the interval $[-1, 1]$.
Let $g(t) = -3t^2 + 12t$, for $t \in [-1, 1]$.
We need to find the range of this quadratic function $g(t)$ on the given interval.
$g(t)$ represents a downward-opening parabola. The vertex of the parabola $y = at^2+bt+c$ is at $t = -b/(2a)$.
For $g(t)$, the vertex is at $t = -\frac{12}{2(-3)} = -\frac{12}{-6} = 2$.
The vertex is at $t=2$, which is outside the interval $[-1, 1]$.
To find the range on the interval $[-1, 1]$, we only need to check the values of the function at the endpoints of the interval.
Since the vertex is at $t=2$ and the parabola opens downwards, the function $g(t)$ is strictly increasing over the interval $(-\infty, 2]$. Therefore, it is strictly increasing on $[-1, 1]$.
The minimum value will occur at the left endpoint, $t=-1$.
$g_{min} = g(-1) = -3(-1)^2 + 12(-1) = -3 - 12 = -15$.
The maximum value will occur at the right endpoint, $t=1$.
$g_{max} = g(1) = -3(1)^2 + 12(1) = -3 + 12 = 9$.
So, the range of $f(x)$ is $[-15, 9]$. The equation has a solution if and only if $p$ is in this range.
We need to find the sum of all integral values of $p$ in the interval $[-15, 9]$.
The integers are $-15, -14, \dots, -1, 0, 1, \dots, 8, 9$.
Sum = $(-15) + (-14) + \dots + (-10) + (-9) + \dots + (-1) + 0 + 1 + \dots + 9$.
The sum of integers from $-9$ to $9$ is 0.
So, the total sum is the sum of the remaining integers:
Sum = $(-15) + (-14) + (-13) + (-12) + (-11) + (-10)$.
Sum = $-(15 + 14 + 13 + 12 + 11 + 10)$.
This is an arithmetic progression with 6 terms. Sum = $\frac{n}{2}(a+l) = \frac{6}{2}(10+15) = 3(25) = 75$.
So, the sum is $-75$.
Step 4: Final Answer:
The sum of all integral values of p is -75.