Problem 58 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Let the function \(f(x)\) be defined as \(f(x)=\left\\{\begin{array}{lc}\tan ^{-1} \alpha-3 x^{2}, 0

Step-by-Step Solution

Answer

(D) None of these

Step 1: Evaluate f(x) at and near x = 1

For \(x \geq 1\): \(f(1) = -6(1) = -6\).
For \(x < 1\) approaching 1: \(\lim_{x \to 1^-} f(x) = \tan^{-1}(\alpha) - 3(1)^2 = \tan^{-1}(\alpha) - 3\).

Step 2: Determine the condition for a maximum at x = 1

For \(f\) to have a maximum at \(x = 1\), we need \(f(1) \geq f(x)\) for all \(x\) near 1.

From the left: \(f(1) \geq \lim_{x \to 1^-} f(x)\), which gives \(-6 \geq \tan^{-1}(\alpha) - 3\), i.e., \(\tan^{-1}(\alpha) \leq -3\).

From the right: For \(x > 1\), \(f(x) = -6x < -6 = f(1)\), so this condition is always satisfied.

Step 3: Check feasibility of the left-side condition

The range of \(\tan^{-1}(\alpha)\) is \(\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), where \(\frac{\pi}{2} \approx 1.571\).

Therefore, \(\tan^{-1}(\alpha) < \frac{\pi}{2} \approx 1.571 < 3\), so \(\tan^{-1}(\alpha) \leq -3\) is impossible for any real \(\alpha\).

Step 4: Conclusion

Since the condition \(\tan^{-1}(\alpha) \leq -3\) can never be satisfied, \(f(x)\) cannot have a maximum at \(x = 1\) for any value of \(\alpha\).

The answer is \(\boxed{(D)\ \text{None of these}}\).

Key Concepts

Function Evaluation

Function evaluation is an important concept, especially when dealing with piecewise functions. In mathematics, evaluating a function means finding the value that the function outputs for a specific input (or 'x' value). This is particularly relevant for piecewise functions, where different rules apply to different intervals of the input domain.
For piecewise functions like in our exercise, each 'piece' of the function has its own specific formula. This means we must carefully consider which piece to use based on the value of 'x'. In the example, we have two formulas: one applying to the interval \(0 < x < 1\) and another for \(x \geq 1\).
Breaking down the evaluation process:

• For values of \(x\) between 0 and 1, including fractional values like 0.5, we use the first expression, \(\tan^{-1} \alpha - 3x^2\).
• When \(x = 1\), we re-evaluate using the second formula: \(-6x\). Thus, at \(x = 1\), \(f(1) = -6\).

Evaluating functions correctly is crucial to finding key characteristics such as maximum or minimum values.

Maximum Value Problems

Maximum value problems involve finding the largest output that a function can produce. This requires analyzing changes in the function over given intervals. For piecewise functions, the maximum may occur at a boundary point or within an interval.
In our exercise, we are tasked with deciding whether the maximum for \(f(x)\) could occur at \(x = 1\) under specific conditions. To solve such problems:

• First, identify the function's formula in each relevant interval.
• Next, evaluate the boundary points, in this case, at \(x = 1\). This is because these points could potentially hold maximum values due to the change in formula.
• Compare the values obtained to determine the highest one, which will be the maximum.
• In our problem, we focus on knowing how the parameter \(\alpha\) affects the maximum. \(\alpha\) values primarily influence the output for \(0 < x < 1\), but not for \(x \geq 1\), where the formula is simply \(-6x\).

By carefully evaluating potential maximums at each relevant point and considering the parameter \(\alpha\), we better understand how changes in this parameter shift the maximum point.

Arctan Function

The arctan function, or inverse tangent function, is a part of trigonometry dealing with angles and circles. It is typically used to find the angle whose tangent is a given number. In the piecewise function provided in the problem, the arctan function influences the first segment of the piecewise function for \(0 < x < 1\).
Learn some key properties of the arctan function:

• It ranges from \(-\frac{\pi}{2}\) to \(\frac{\pi}{2}\), covering all possible angle values that the function might output.
• Unlike tangent, it is a continuous and smooth function.
• Its graph gradually increases as inputs approach infinity or negative infinity.

In the context of this problem, the term \(\tan^{-1} \alpha\) implies that the maximum possible output from this section is bounded by the arctan's range. Any change in \(\alpha\) directly affects this part of the piecewise function, impacting whether it holds the maximum value compared to the other section.

Boundary Points

Boundary points are crucial in piecewise functions and often dictate where changes in formula occur. They are the endpoints of intervals in which each piece of the function applies. In our exercise, \(x = 1\) is a boundary point that marks the transition between two function pieces.
Here's how to handle them:

• Always evaluate the function at and around these points to ensure you're using the correct formula for any given value of \(x\).
• Boundary points like \(x = 1\) can potentially be where maximum or minimum values occur. This is because they form the endpoints of the intervals.
• To check the behavior at a boundary point, determine how the function behaves just before and just after this point. This is often done by calculating limits or directly evaluating points like \(x = 0.9999\) and \(x = 1.0001\), though not always necessary.

In the situation of the given problem, evaluating \(f(x)\) at \(x = 1\) gives us \(-6\), which gives insight into the possible maximum value compared to other segments depending on \(\alpha\). Managing boundary points effectively is essential in maximizing and minimizing functions within piecewise constructs.