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 equation will have a solution for $x$ if and only if the value of $p$ falls within the range of the function on the left-hand side. The problem therefore reduces to finding the range of $f(x) = 3\sin^2x + 12\cos x - 3$, and then summing all the integers within that range.
Step 2: Key Formula or Approach:
1. Simplify the function: Use the Pythagorean identity $\sin^2x = 1 - \cos^2x$ to rewrite the function in terms of a single trigonometric ratio, $\cos x$.
2. Create a Quadratic: Substitute $t = \cos x$. Note that the domain for $t$ is the closed interval $[-1, 1]$. The function becomes a quadratic in $t$.
3. Find the Range: Determine the maximum and minimum values of this quadratic function on the interval $t \in [-1, 1]$. This will be the range of $p$.
4. Sum the Integers: Calculate the sum of all integers that lie in the determined range.
Step 3: Detailed Explanation:
Let the given function be $f(x) = 3\sin^2x + 12\cos x - 3$.
Convert $\sin^2x$ to $1 - \cos^2x$:
\[ f(x) = 3(1 - \cos^2x) + 12\cos x - 3 = 3 - 3\cos^2x + 12\cos x - 3 \] \[ f(x) = -3\cos^2x + 12\cos x \] Let $t = \cos x$. Since $x \in \mathbb{R}$, we have $t \in [-1, 1]$. The problem is now to find the range of the quadratic function $g(t) = -3t^2 + 12t$ on the domain $t \in [-1, 1]$.
This is a downward-opening parabola. Let's find its vertex. The t-coordinate of the vertex is $t = -b/(2a)$:
\[ t_{vertex} = \frac{-12}{2(-3)} = 2 \] The vertex is at $t=2$, which lies outside the domain $[-1, 1]$. Since the parabola opens downward, the function $g(t)$ is continuously increasing on the interval $(-\infty, 2]$. Therefore, on our interval $[-1, 1]$, the function is also strictly increasing.
The minimum value will occur at the leftmost point of the interval, $t=-1$.
\[ g_{min} = g(-1) = -3(-1)^2 + 12(-1) = -3 - 12 = -15 \] The maximum value will occur at the rightmost point of the interval, $t=1$.
\[ g_{max} = g(1) = -3(1)^2 + 12(1) = -3 + 12 = 9 \] So, the range of $f(x)$ is $[-15, 9]$. The possible integer values for $p$ are all integers from -15 to 9.
We need to compute the sum: $S = -15 + (-14) + \dots + 0 + \dots + 8 + 9$.
We can split the sum:
\[ S = (-15 - 14 - 13 - 12 - 11 - 10) + (-9 - 8 - \dots + 8 + 9) \] The sum of the balanced part from -9 to 9 is 0. So we are left with:
\[ S = -15 - 14 - 13 - 12 - 11 - 10 \] This is the sum of an arithmetic progression. Number of terms $n=6$, first term $a=-15$, last term $l=-10$.
\[ S = \frac{n}{2}(a+l) = \frac{6}{2}(-15 - 10) = 3(-25) = -75 \] Step 4: Final Answer:
The sum of all integral values of p is -75.