Question

If $\alpha$ is the maximum value and $\beta$ is the minimum value of $\cos^2 \frac{x}{4} + \sin \frac{x}{4}$, $x \in \mathbb{R}$, then $\alpha-\beta=$

• A. \( \frac{1}{4} \)
• B. \( \frac{9}{4} \)
• C. \( 2 \)
• D. \( 3 \)

Hint:

To find the maximum and minimum values of trigonometric expressions that can be reduced to a quadratic form, follow these steps: % Option (A) Use trigonometric identities (like $\sin^2\theta + \cos^2\theta = 1$) to express the function in terms of a single trigonometric ratio (e.g., $\sin x$ or $\cos x$). % Option (B) Substitute the trigonometric ratio with a variable (e.g., $y = \sin x$). % Option (C) Determine the valid range for this new variable (e.g., $y \in [-1, 1]$ for $\sin x$ or $\cos x$). % Option (D) The expression becomes a quadratic function in the new variable. Find the vertex of the parabola ($y = -\frac{b}{2a}$). % Option (E) Evaluate the quadratic function at the vertex (if it lies within the valid range) and at the endpoints of the range to find the maximum and minimum values.

Answer 1

The Correct Option is B

Let the given expression be $f(x) = \cos^2 \frac{x}{4} + \sin \frac{x}{4}$. To find the maximum and minimum values, we can express the function in terms of a single trigonometric ratio. Using the trigonometric identity $\cos^2 \theta = 1 - \sin^2 \theta$, let $\theta = \frac{x}{4}$. Then, $f(x) = 1 - \sin^2 \frac{x}{4} + \sin \frac{x}{4}$. Let $y = \sin \frac{x}{4}$. Since $x \in \mathbb{R}$, the angle $\frac{x}{4}$ can take any real value. Therefore, the range of $\sin \frac{x}{4}$ is $[-1, 1]$. So, $y \in [-1, 1]$. Now, the expression becomes a quadratic function of $y$: $g(y) = 1 - y^2 + y = -y^2 + y + 1$. This is a quadratic function in $y$ representing a parabola opening downwards (since the coefficient of $y^2$ is negative). The vertex of the parabola $ay^2+by+c$ occurs at $y = -\frac{b}{2a}$. For $g(y) = -y^2 + y + 1$, we have $a=-1$ and $b=1$. The vertex occurs at $y = -\frac{1}{2(-1)} = \frac{1}{2}$. Since $y = \frac{1}{2}$ lies within the interval $[-1, 1]$, the maximum value of $g(y)$ (which is $\alpha$) occurs at this vertex. $\alpha = g\left(\frac{1}{2}\right) = -\left(\frac{1}{2}\right)^2 + \frac{1}{2} + 1$ $\alpha = -\frac{1}{4} + \frac{2}{4} + \frac{4}{4}$ $\alpha = \frac{-1+2+4}{4} = \frac{5}{4}$. The minimum value of $g(y)$ (which is $\beta$) must occur at one of the endpoints of the interval $[-1, 1]$ because the parabola opens downwards and its vertex is within the interval. Evaluate $g(y)$ at $y=-1$: $g(-1) = -(-1)^2 + (-1) + 1 = -1 - 1 + 1 = -1$. Evaluate $g(y)$ at $y=1$: $g(1) = -(1)^2 + (1) + 1 = -1 + 1 + 1 = 1$. Comparing the values at the endpoints, the minimum value is $\beta = -1$. Finally, we need to find $\alpha - \beta$: $\alpha - \beta = \frac{5}{4} - (-1)$ $\alpha - \beta = \frac{5}{4} + 1$ $\alpha - \beta = \frac{5}{4} + \frac{4}{4} = \frac{9}{4}$.