Question

If $x^4 + 3\cos(ax^2+bx+c) = 2(x^2-2)$ has two solutions with a,b,c $\in$ (2,5) then

• A. a + b + c = $\pi$
• B. a - b + c = $\pi$
• C. a + b + c = 3$\pi$
• D. a - b + c = 3$\pi$

Answer 1

Answer: (C)

C

Answer 2

The given equation is $x^4 + 3\cos(ax^2+bx+c) = 2(x^2-2)$. Rearranging the terms, we get $x^4 - 2x^2 + 4 + 3\cos(ax^2+bx+c) = 0$.

We can rewrite the term $x^4 - 2x^2 + 4$ as follows: $x^4 - 2x^2 + 1 + 3 = (x^2-1)^2 + 3$.

Substitute this back into the equation: $(x^2-1)^2 + 3 + 3\cos(ax^2+bx+c) = 0$, which can be written as $(x^2-1)^2 + 3(1 + \cos(ax^2+bx+c)) = 0$.

We know the following properties:

1. $(x^2-1)^2 \ge 0$ for any real $x$.
2. The range of $\cos(\theta)$ is $[-1, 1]$. Therefore, $1 + \cos(\theta) \ge 1 + (-1) = 0$. So, $3(1 + \cos(ax^2+bx+c)) \ge 0$.

The sum of two non-negative terms is zero if and only if both terms are individually zero. Therefore, we must have:

1. $(x^2-1)^2 = 0$
2. $3(1 + \cos(ax^2+bx+c)) = 0$

From condition (1): $(x^2-1)^2 = 0 \implies x^2-1 = 0 \implies x^2 = 1 \implies x = \pm 1$. These are the two solutions mentioned in the problem statement.

From condition (2): $1 + \cos(ax^2+bx+c) = 0 \implies \cos(ax^2+bx+c) = -1$. This condition must hold for both solutions $x=1$ and $x=-1$.

Case 1: For $x=1$

Substitute $x=1$ into the argument of cosine: $\cos(a(1)^2 + b(1) + c) = -1$, so $\cos(a + b + c) = -1$. For $\cos(\theta) = -1$, $\theta$ must be an odd multiple of $\pi$. So, $a + b + c = (2n+1)\pi$ for some integer $n$.

We are given that $a, b, c \in (2,5)$. Let's find the range of $a+b+c$: $2 < a < 5$, $2 < b < 5$, $2 < c < 5$. Adding these inequalities: $2+2+2 < a+b+c < 5+5+5 \implies 6 < a+b+c < 15$.

Now, let's check which odd multiple of $\pi$ falls into this range:

If $n=0$, $a+b+c = \pi \approx 3.14$ (not in $(6,15)$).

If $n=1$, $a+b+c = 3\pi \approx 9.42$ (is in $(6,15)$).

If $n=2$, $a+b+c = 5\pi \approx 15.70$ (not in $(6,15)$).

Thus, for $x=1$, we must have $a+b+c = 3\pi$. This matches option (C).

Case 2: For $x=-1$

Substitute $x=-1$ into the argument of cosine: $\cos(a(-1)^2 + b(-1) + c) = -1$, so $\cos(a - b + c) = -1$. Similarly, $a - b + c = (2m+1)\pi$ for some integer $m$.

Let's find the range of $a-b+c$: $2 < a < 5$, $-5 < -b < -2$ (multiplying $2 < b < 5$ by -1 and reversing inequalities), $2 < c < 5$. Adding these inequalities: $2+(-5)+2 < a-b+c < 5+(-2)+5 \implies -1 < a-b+c < 8$.

Now, let's check which odd multiple of $\pi$ falls into this range:

If $m=-1$, $a-b+c = -\pi \approx -3.14$ (not in $(-1,8)$).

If $m=0$, $a-b+c = \pi \approx 3.14$ (is in $(-1,8)$).

If $m=1$, $a-b+c = 3\pi \approx 9.42$ (not in $(-1,8)$).

Thus, for $x=-1$, we must have $a-b+c = \pi$. This matches option (B).

Both conditions, $a+b+c=3\pi$ and $a-b+c=\pi$, must be true simultaneously for the equation to hold for $x=\pm 1$ under the given constraints on $a,b,c$.

We can check for consistency:

Adding the two derived equations: $(a+b+c) + (a-b+c) = 3\pi + \pi \implies 2a+2c = 4\pi \implies a+c = 2\pi$.

Since $a,c \in (2,5)$, $a+c \in (4,10)$. $2\pi \approx 6.28$ which is indeed in $(4,10)$.

Subtracting the second equation from the first: $(a+b+c) - (a-b+c) = 3\pi - \pi \implies 2b = 2\pi \implies b = \pi$.

Since $b \in (2,5)$, $\pi \approx 3.14$ which is indeed in $(2,5)$.

All conditions are consistent.

Since both options (B) and (C) are necessarily true based on the problem statement, and assuming this is a single-choice question, there might be an issue with the question itself. However, if forced to select one, we present one of the derived correct statements.