If the maximum value of a, for which the function f_a(x) = tan⁻¹ 2x - 3ax + 7 is non-decreasing in (

Q: Wait, why do we need the minimum value of the curve? Shouldn't we look at the maximum?

That is a fantastic question! Think of the inequality $3a \le g(x)$. If $3a$ must be less than or equal to $g(x)$ for *every* single point in the interval, it must be less than or equal to the *lowest* point of $g(x)$. If $3a$ were larger than the minimum, there would be some points where the curve dips below the line, violating our condition. We are essentially finding the 'floor' of the function.

Q: Why is the minimum of $g(x) = \frac{2}{1 + 4x^2}$ at the boundaries?

Look at the structure of $g(x)$. The numerator is constant, but the denominator $1 + 4x^2$ grows as $|x|$ increases. As the denominator gets larger, the value of the fraction gets smaller. Therefore, to make $g(x)$ as small as possible, we need the denominator to be as large as possible, which happens at the largest possible values of $|x|$—the boundaries of our interval.

Q: I got the expression $7 + \tan^{-1}(\frac{\pi}{4}) - \frac{9\pi}{4(9 + \pi^2)}$. Why is this not matching any of the options?

This is the classic JEE 'trap' moment! Many students instinctively assume $\tan^{-1}(\frac{\pi}{4}) = 1$ because they are used to seeing $\tan^{-1}(1) = \frac{\pi}{4}$. But $\frac{\pi}{4}$ is not $1$. Since $\tan^{-1}(\frac{\pi}{4})$ cannot be simplified to a rational number or a simple integer, the expression remains as it is. Trust your derivation—if it doesn't match the options, 'None of these' is a perfectly valid and often correct answer.

Question

If the maximum value of $a$, for which the function $f_a(x) = 	an^{-1} 2x - 3ax + 7$ is non-decreasing in $(-\frac{\pi}{6}, \frac{\pi}{6})$, is $\bar{a}$, then $f_{\bar{a}}(\frac{\pi}{8})$ is equal to

Options:

(A) $8 - \frac{9\pi}{4(9+\pi^2)}$

(B) $8 - \frac{4\pi}{9(4+\pi^2)}$

(C) $8 \frac{1+\pi^2}{9+\pi^2}$

(D) None of these

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Welcome, my dear student. Today, we are not just solving a problem; we are exploring the soul of a function. When we talk about a function being 'non-decreasing,' we are talking about its trajectory.

Imagine you are walking along the graph of $f_a(x) = 	an^{-1} 2x - 3ax + 7$. To be non-decreasing, you must never take a step downward. You can stand still, or you can climb, but you can never descend.

Mathematically, this translates to a simple, elegant requirement: the slope of the path, the derivative $f_a'(x)$, must never be negative. It must be $f_a'(x) \ge 0$ for every point in our domain, $x \in (-\frac{\pi}{6}, \frac{\pi}{6})$.

### The Derivative as a Gatekeeper

Let us begin by finding the slope. We differentiate $f_a(x)$ with respect to $x$. The derivative of $	an^{-1} 2x$ is $\frac{2}{1 + (2x)^2}$, and the derivative of $-3ax$ is simply $-3a$. The constant $7$ vanishes into the ether.

Thus, our slope function is:

$$f_a'(x) = \frac{2}{1 + 4x^2} - 3a$$

For the function to be non-decreasing, we demand that $\frac{2}{1 + 4x^2} - 3a \ge 0$. Rearranging this, we get the fundamental inequality:

$$3a \le \frac{2}{1 + 4x^2}$$

### The Horizontal Line Visualization

This is where the magic happens. Visualize the right-hand side, $g(x) = \frac{2}{1 + 4x^2}$, as a curve on a graph. The left-hand side, $3a$, is a horizontal line.

The condition $3a \le g(x)$ means that this horizontal line must stay below the curve $g(x)$ for the entire duration of the interval $(-\frac{\pi}{6}, \frac{\pi}{6})$.

If the line $3a$ were to rise even a tiny bit above the lowest point of the curve, it would violate our condition. Therefore, to find the **maximum** possible value of $a$, we must pin the line $3a$ exactly at the minimum value of $g(x)$. This is the 'floor' of our function.

### Finding the Floor

We need the minimum of $g(x) = \frac{2}{1 + 4x^2}$ on the interval $(-\frac{\pi}{6}, \frac{\pi}{6})$. Observe the denominator: $1 + 4x^2$. As $|x|$ increases, $x^2$ increases, and thus the denominator increases.

When the denominator of a fraction increases, the value of the fraction decreases. Therefore, $g(x)$ is at its smallest when $|x|$ is at its largest. The maximum value of $|x|$ in our interval is $\frac{\pi}{6}$.

Let us calculate $g(\frac{\pi}{6})$:

$$g\left(\frac{\pi}{6}ight) = \frac{2}{1 + 4\left(\frac{\pi}{6}ight)^2} = \frac{2}{1 + 4\left(\frac{\pi^2}{36}ight)} = \frac{2}{1 + \frac{\pi^2}{9}}$$

Simplifying this, we multiply the numerator and denominator by $9$:

$$g\left(\frac{\pi}{6}ight) = \frac{18}{9 + \pi^2}$$

This is our minimum value. To satisfy the inequality $3a \le g(x)$, we must have $3a \le \frac{18}{9 + \pi^2}$. Dividing by $3$, we find the maximum value of $a$, which we call $\bar{a}$:

$$\bar{a} = \frac{6}{9 + \pi^2}$$

### The Final Evaluation and the Trap

Now, we calculate $f_{\bar{a}}(\frac{\pi}{8})$. We substitute $\bar{a} = \frac{6}{9 + \pi^2}$ and $x = \frac{\pi}{8}$ into our original function:

$$f_{\bar{a}}\left(\frac{\pi}{8}ight) = 	an^{-1}\left(2 \cdot \frac{\pi}{8}ight) - 3\left(\frac{6}{9 + \pi^2}ight)\left(\frac{\pi}{8}ight) + 7$$

Simplifying the terms:

$$f_{\bar{a}}\left(\frac{\pi}{8}ight) = 	an^{-1}\left(\frac{\pi}{4}ight) - \frac{18\pi}{8(9 + \pi^2)} + 7$$

$$f_{\bar{a}}\left(\frac{\pi}{8}ight) = 7 + 	an^{-1}\left(\frac{\pi}{4}ight) - \frac{9\pi}{4(9 + \pi^2)}$$

Here is where the JEE examiner tests your confidence. You look at the options. You see expressions that look similar, but they contain $1$ instead of $	an^{-1}(\frac{\pi}{4})$.

Do not be tempted! $	an^{-1}(\frac{\pi}{4})$ is a transcendental value; it is not $1$. The expression we derived is the exact, correct answer. Since it does not match any of the provided options, we confidently choose **'None of these.'**

Animated Solution for Mathematics - Differentiation: If the maximum value of $a$ , for which the function $f_{a} (x) = tan^{- 1} 2 x - 3 a x + 7$ is non-decreasing in $(- \frac{π}{6}, \frac{π}{6})$ , is $\overset{a}{ˉ}$ , then $f_{\overset{a}{ˉ}} (\frac{π}{8})$ is equal to

Select Answer:

Visualized Solution

Understanding Monotonicity

Differentiating the Function

Setting up the Inequality

Analyzing the Bounds

Finding the Minimum Value

Solving for $\overset{a}{ˉ}$

Calculating $f_{\overset{a}{ˉ}} (\frac{π}{8})$

Final Conclusion

The Sigma Insight: Monotonicity

Analyzing the Setup

The Derivative as a Gatekeeper

The Horizontal Line Visualization

Finding the Floor

The Final Evaluation and the Trap

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