If the minimum and the maximum values of the function f : [π/4, π/2] → R, defined by f(θ) = |[-sin²

Q: Why can't I just expand the determinant directly? It seems straightforward.

Expanding a $3 \times 3$ determinant with trigonometric terms is a classic trap. You will end up with a massive algebraic expression that is prone to calculation errors. Always look for properties (like $C_2 \to C_2 - C_1$) to simplify first. It transforms a 10-minute algebraic nightmare into a 30-second elegant solution.

Q: Why does the domain change from $[\frac{\pi}{4}, \frac{\pi}{2}]$ to $[\frac{\pi}{2}, \pi]$?

This is the most common mistake! Your function is $f(\theta) = 4 \cos 2\theta$. The input to the cosine function is $2\theta$, not $\theta$. If $\theta$ is in $[\frac{\pi}{4}, \frac{\pi}{2}]$, then $2\theta$ must be in $[\frac{\pi}{2}, \pi]$. Always scale your domain before evaluating the range of a trigonometric function.

Q: How do I know if the range is $[-4, 0]$ or $[0, -4]$?

The question asks for the ordered pair $(m, M)$, where $m$ is the minimum and $M$ is the maximum. Since the function decreases from $0$ to $-4$ in the second quadrant, the minimum is $-4$ and the maximum is $0$. So, the pair is $(-4, 0)$.

Question

If the minimum and the maximum values of the function $f : [\frac{\pi}{4}, \frac{\pi}{2}] 	o R$, defined by $f(	heta) = \begin{vmatrix} -\sin^2 	heta & -1-\sin^2 	heta & 1 \ -\cos^2 	heta & -1-\cos^2 	heta & 1 \ 12 & 10 & -2 \end{vmatrix}$ are $m$ and $M$ respectively, then the ordered pair $(m, M)$ is equal to :

Options:

(A) $(0,2\sqrt{2})$

(B) $(0,4)$

(C) $(-4, 4)$

(D) $(-4, 0)$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

We are given the function:

$$f(	heta) = \begin{vmatrix} -\sin^2 	heta & -1-\sin^2 	heta & 1 \ -\cos^2 	heta & -1-\cos^2 	heta & 1 \ 12 & 10 & -2 \end{vmatrix}$$

If you try to expand this directly, you will be lost in a sea of $\sin^4 	heta$ and $\cos^4 	heta$ terms. Instead, look at the columns. Notice the second column contains $-1-\sin^2 	heta$ and $-1-\cos^2 	heta$, while the first column contains $-\sin^2 	heta$ and $-\cos^2 	heta$.

If we subtract the first column from the second ($C_2 	o C_2 - C_1$), those bulky trigonometric terms will vanish.

### The Power of Column Operations

Applying the operation $C_2 	o C_2 - C_1$ yields:

$$f(	heta) = \begin{vmatrix} -\sin^2 	heta & -1 & 1 \ -\cos^2 	heta & -1 & 1 \ 12 & -2 & -2 \end{vmatrix}$$

Look at how clean that is! We have successfully reduced the complexity of the second column to simple constants.

### Creating Zeros

Now, we have a cleaner determinant, but we can do better. In the world of determinants, zeros are your best friends. Look at the first and second rows; the second and third columns are identical ($-1, 1$).

This is a golden opportunity. Let's apply the row operation $R_1 	o R_1 - R_2$.

For the first element, we get $(-\sin^2 	heta) - (-\cos^2 	heta) = \cos^2 	heta - \sin^2 	heta$. For the other elements, $(-1) - (-1) = 0$ and $1 - 1 = 0$. Our determinant now looks like this:

$$f(	heta) = \begin{vmatrix} \cos^2 	heta - \sin^2 	heta & 0 & 0 \ -\cos^2 	heta & -1 & 1 \ 12 & -2 & -2 \end{vmatrix}$$

### The Identity Reveal

Now, expanding along the first row is trivial. We have only one non-zero term: $(\cos^2 	heta - \sin^2 	heta)$. We know from our trigonometric toolkit that $\cos^2 	heta - \sin^2 	heta = \cos 2	heta$.

The determinant simplifies to:

$$f(	heta) = (\cos 2	heta) \begin{vmatrix} -1 & 1 \ -2 & -2 \end{vmatrix}$$

Calculating the $2 	imes 2$ minor: $(-1)(-2) - (1)(-2) = 2 - (-2) = 4$. Thus, our monster determinant has been tamed into the simple expression: **$f(	heta) = 4 \cos 2	heta$**.

### The Domain Trap

This is where many students lose marks. We are given the domain $	heta \in [\frac{\pi}{4}, \frac{\pi}{2}]$. We must find the range of $2	heta$.

If $\frac{\pi}{4} \le 	heta \le \frac{\pi}{2}$, then multiplying by $2$ gives us $\frac{\pi}{2} \le 2	heta \le \pi$.

Now, visualize the cosine curve in the second quadrant (from $\frac{\pi}{2}$ to $\pi$). The cosine function is strictly decreasing here. At $2	heta = \frac{\pi}{2}$, $\cos(\frac{\pi}{2}) = 0$. At $2	heta = \pi$, $\cos(\pi) = -1$.

Therefore, the range of $\cos 2	heta$ is $[-1, 0]$. Multiplying by $4$, the range of our function $f(	heta)$ is **$[-4, 0]$**.

### Conclusion

The minimum value $m$ is $-4$, and the maximum value $M$ is $0$. The ordered pair $(m, M)$ is **$(-4, 0)$**. By using properties to simplify and being careful with the domain, we turned a daunting problem into a clear, logical victory.

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Visualized Solution

Introduction to $f (θ)$

Column Operation $C_{2} \to C_{2} - C_{1}$

Simplifying $C_{2}$

Row Operation $R_{1} \to R_{1} - R_{2}$

Simplifying $R_{1}$

Expanding Along $R_{1}$

Calculating the Minor

Analyzing the Domain

Domain of $2 θ$

Finding the Range

Minimum and Maximum Values

The Sigma Insight: Properties of Determinants

Analyzing the Setup

The Power of Column Operations

Creating Zeros

The Identity Reveal

The Domain Trap

Conclusion

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Visualized Solution

Introduction to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(θ)

Column Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​→C2​−C1​

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } C2​

Row Operation .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​→R1​−R2​

Simplifying .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​

Expanding Along .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } R1​