Let f(x) = 2cos⁻¹ x + 4cot⁻¹ x - 3x² - 2x + 10, x ∈ [-1, 1]. If [a, b] is the range of the function

Q: Wait, why didn't we set $f'(x) = 0$ to find critical points?

Great question! We only set $f'(x) = 0$ if we expect the function to turn around (have a local maximum or minimum). Here, we proved $f'(x)$ is *always* negative, meaning the function never turns. It just slides down from start to finish, so there are no critical points inside the interval.

Q: How do I know if the range is $[f(1), f(-1)]$ or $[f(-1), f(1)]$?

Always remember: for a strictly decreasing function, the *highest* input gives the *lowest* output. Since $1 > -1$, $f(1)$ is the minimum ($a$) and $f(-1)$ is the maximum ($b$). So the range is $[f(1), f(-1)]$.

Q: Is it possible to make a mistake with the inverse trig values?

Absolutely. Always recall the principal branches: $\cos^{-1}(-1) = \pi$ and $\cot^{-1}(-1) = \frac{3\pi}{4}$. Don't let the negative signs trip you up! Take your time with the arithmetic.

Question

Let $f(x) = 2\cos^{-1} x + 4\cot^{-1} x - 3x^2 - 2x + 10, x \in [-1, 1]$. If [a, b] is the range of the function then 4a - b is equal to:

Options:

(A) 11

(B) $11 - \pi$

(C) $11 + \pi$

(D) $15 - \pi$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on a mountain path that only goes downhill. You do not need to know every twist and turn of the path to know where you started and where you ended. This is the essence of finding the range of a monotonic function.

We are given the function $f(x) = 2\cos^{-1} x + 4\cot^{-1} x - 3x^2 - 2x + 10$ on the interval $x \in [-1, 1]$. At first glance, this looks like a chaotic mix of inverse trigonometric functions and a quadratic polynomial.

Our goal is to understand the behavior of this function. To determine if it is climbing or falling, we turn to the most powerful tool in our calculus arsenal: the derivative.

### The Monotonicity Proof

We differentiate the function with respect to $x$. Recall that the standard derivatives are:

$$\frac{d}{dx}(\cos^{-1} x) = \frac{-1}{\sqrt{1-x^2}} \quad 	ext{and} \quad \frac{d}{dx}(\cot^{-1} x) = \frac{-1}{1+x^2}$$

Applying these rules, we find the derivative:

$$f'(x) = 2\left(\frac{-1}{\sqrt{1-x^2}}ight) + 4\left(\frac{-1}{1+x^2}ight) - 6x - 2$$

This looks messy, but let us factor out a $-2$ to see the structure clearly:

$$f'(x) = -2 \left[ \frac{1}{\sqrt{1-x^2}} + \frac{2}{1+x^2} + 3x + 1 ight]$$

Now, look at the bracketed expression. For $x \in (-1, 1)$, the term $\frac{1}{\sqrt{1-x^2}}$ is always $\ge 1$, and $\frac{2}{1+x^2}$ is also $\ge 1$.

The linear part $3x+1$ ranges from $-2$ to $4$. Even when $3x+1$ is at its minimum of $-2$, the sum of the first two terms (which is at least $2$) ensures the entire bracket remains positive. Since the bracket is positive and multiplied by $-2$, $f'(x)$ is strictly negative. The function is **strictly decreasing**.

### The Final Execution

Because our function is strictly decreasing, we do not need to hunt for critical points. The maximum value $b$ must occur at the leftmost point of our interval, $x = -1$, and the minimum value $a$ must occur at the rightmost point, $x = 1$.

Let us calculate $a = f(1)$:

$$a = 2\cos^{-1}(1) + 4\cot^{-1}(1) - 3(1)^2 - 2(1) + 10$$

Since $\cos^{-1}(1) = 0$ and $\cot^{-1}(1) = \frac{\pi}{4}$, we get:

$$a = 0 + 4\left(\frac{\pi}{4}ight) - 3 - 2 + 10 = \pi + 5$$

Now for $b = f(-1)$:

$$b = 2\cos^{-1}(-1) + 4\cot^{-1}(-1) - 3(-1)^2 - 2(-1) + 10$$

With $\cos^{-1}(-1) = \pi$ and $\cot^{-1}(-1) = \frac{3\pi}{4}$, we get:

$$b = 2(\pi) + 4\left(\frac{3\pi}{4}ight) - 3 + 2 + 10 = 2\pi + 3\pi + 9 = 5\pi + 9$$

Finally, we compute $4a - b$:

$$4a - b = 4(\pi + 5) - (5\pi + 9) = 4\pi + 20 - 5\pi - 9 = 11 - \pi$$

We have conquered the problem by understanding the soul of the function rather than just blindly calculating. The final answer is **$11 - \pi$**.

Animated Solution for Mathematics - Differentiation: Let $f (x) = 2 cos^{- 1} x + 4 cot^{- 1} x - 3 x^{2} - 2 x + 10, x \in [- 1, 1]$ . If [a, b] is the range of the function then 4a - b is equal to:

Select Answer:

Visualized Solution

Define $f (x)$ and Domain

Derivative of Inverse Trig Terms

Find the Derivative $f^{'} (x)$

Simplify $f^{'} (x)$

Analyze the Sign of $f^{'} (x)$

Establish Monotonicity

Range of a Decreasing Function

Calculate Minimum Value $a$

Calculate Maximum Value $b$

Final Calculation: $4 a - b$

Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Monotonicity Proof

The Final Execution

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let f(x)=2cos−1x+4cot−1x−3x2−2x+10,x∈[−1,1]. If [a, b] is the range of the function then 4a - b is equal to:

Select Answer:

Visualized Solution

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f(x) and Domain

Derivative of Inverse Trig Terms

Find the Derivative .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)

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

Analyze the Sign of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(x)