If the equation x cos θ = x² + p has a real solution for every θ where 0 ≤ θ ≤ π/4, then which one of the following is correct?

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CDS - 2024

If the equation $x \cos \theta = x^2 + p$ has a real solution for every $\theta$ where $0 \le \theta \le \frac{\pi}{4}$, then which one of the following is correct?

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CDS 2 2024 Maths Question Paper (01-Sep-2024)

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$p = 1/8$

$p \le 1/8$

$p \ge 1/8$

$p \le 1/4$

The correct answer is

$p \le 1/4$

Equation Analysis for Real Solutions

We are given the equation $x \cos \theta = x^2 + p$. The goal is to find the condition on the parameter $p$ such that this equation has at least one real solution for $x$ for all values of $\theta$ within the interval $0 \le \theta \le \frac{\pi}{4}$.

Rearranging the Equation

First, we rewrite the equation to express $p$ in terms of $x$ and $\theta$:

$ p = x \cos \theta - x^2 $

This form implies that for a specific value of $\theta$, a real solution $x$ exists if $p$ is less than or equal to the maximum value that the expression $x \cos \theta - x^2$ can attain.

Finding the Maximum Value of $p$ for a Fixed $\theta$

Consider the expression $f(x, \theta) = x \cos \theta - x^2$. To find the maximum value achievable by $p$ for a fixed $\theta$, we find the maximum of $f(x, \theta)$ with respect to $x$. The function $f(x, \theta)$ is a quadratic in $x$ shaped like an inverted parabola.

The maximum value occurs at the vertex. The $x$-coordinate of the vertex is:

$ x_{vertex} = \frac{-(\cos \theta)}{2(-1)} = \frac{\cos \theta}{2} $

Substituting this $x$ value back into the expression gives the maximum value:

$ \text{Max value for fixed } \theta = f\left(\frac{\cos \theta}{2}, \theta\right) = \left(\frac{\cos \theta}{2}\right) \cos \theta - \left(\frac{\cos \theta}{2}\right)^2 $

$ = \frac{\cos^2 \theta}{2} - \frac{\cos^2 \theta}{4} $

$ = \frac{\cos^2 \theta}{4} $

Therefore, for a real solution $x$ to exist for a specific $\theta$, $p$ must satisfy $p \le \frac{\cos^2 \theta}{4}$.

Determining the Overall Condition on $p$

The problem requires that a real solution exists for every $\theta$ in the interval $0 \le \theta \le \frac{\pi}{4}$. This implies that $p$ must be compatible with the range of values that $\frac{\cos^2 \theta}{4}$ can take within this interval. Specifically, for the condition $p \le \frac{\cos^2 \theta}{4}$ to be potentially met across the interval, $p$ must not exceed the maximum value that the expression $\frac{\cos^2 \theta}{4}$ can attain.

So, we need to find the maximum value of $\frac{\cos^2 \theta}{4}$ over the interval $0 \le \theta \le \frac{\pi}{4}$:

$ p \le \max_{0 \le \theta \le \frac{\pi}{4}} \left( \frac{\cos^2 \theta}{4} \right) $

Analyzing the Maximum Value in the Interval

Let's examine the behavior of $\cos^2 \theta$ in the interval $0 \le \theta \le \frac{\pi}{4}$:

The function $\cos \theta$ is decreasing throughout this interval. It ranges from $\cos(0) = 1$ down to $\cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}}$.
Consequently, $\cos^2 \theta$ is also a decreasing function in this interval.
The maximum value of $\cos^2 \theta$ will occur at the smallest value of $\theta$, which is $\theta = 0$.

Calculating this maximum value:

$ \max_{0 \le \theta \le \frac{\pi}{4}} (\cos^2 \theta) = (\cos 0)^2 = 1^2 = 1 $

Final Condition for $p$

Now we use this maximum value to determine the condition for $p$:

$ p \le \frac{1}{4} \times (\text{maximum value of } \cos^2 \theta) $

$ p \le \frac{1}{4} \times 1 $

$ p \le \frac{1}{4} $

Therefore, the condition required for the equation $x \cos \theta = x^2 + p$ to have a real solution $x$ for $\theta$ in the range $0 \le \theta \le \frac{\pi}{4}$ is $p \le \frac{1}{4}$.

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