Problem 117 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

\((1+x)^{p} \leq 1+x^{p}\), where (A) \(p>1\) (B) \(0 \leq p \leq 1\) (C) \(x>0\) (D) \(x<0\)

Step-by-Step Solution

Answer

The inequality holds when \(0 \leq p \leq 1\) for all \(x\), including negatives, but fails for \(p > 1\) with \(x > 0\).

Step 1: Understand the Problem

We are given an inequality \((1+x)^p \leq 1+x^p\) and need to determine the conditions under which it holds true. Conditions to consider include values of \(p\) and \(x\).

Step 2: Analyze the Base Case

First, consider the case when \(x = 0\). Substituting, we obtain \((1+0)^p = 1^p = 1\) and \(1+0^p = 1\), which satisfies the inequality \((1+x)^p \leq 1+x^p\).

Step 3: Evaluate with \(x > 0\)

For \(x > 0\), the function \((1+x)^p - (1 + x^p)\) should be evaluated. Using the binomial theorem, expanding \((1+x)^p\) gives greater terms for \(p > 1 \), hence \((1+x)^p \geq 1+x^p\). Thus, the inequality doesn't hold for \(p > 1\).

Step 4: Evaluate with \(x < 0\)

With \(x < 0\), the expression \((1+x)^p\) becomes smaller, and the result can vary depending on the magnitude of \(p\). If \(0 \leq p \leq 1\), then \((1+x)^p \leq 1+x^p\) for negative \(x\) since both terms involve decreasing magnitudes.

Step 5: Evaluate with \(0 \leq p \leq 1\)

Testing when \(0 \leq p \leq 1\): This case makes \((1+x)^p\) a convex function, which is closer to 1; hence, \((1+x)^p \leq 1+x^p\) holds for both \(x > 0\) and \(x < 0\). Thus, the inequality holds for non-negative \(p\).

Step 6: Conclusion

Based on the evaluation, \((1+x)^p \leq 1+x^p\) holds for non-negative \(p\) values, especially when \(0 \leq p \leq 1\), and the behavior changes depending on \(x\). For \(p > 1\) and \(x > 0\), inequality doesn't hold.

Key Concepts

Binomial Theorem

The Binomial Theorem is a powerful tool in algebra that allows us to expand expressions that are raised to a power, such as \((1+x)^p\).
The theorem states that \((1+x)^p = \sum_{k=0}^{n} \binom{n}{k} x^k\).
For small values of \(x\) and a fixed \(p\), this expansion can be very helpful in comparing or approximating different functions.

• When expanding, keep in mind that higher powers of \(x\) become significant factors especially for larger \(p\). This is why when \(p > 1\), we observe that \((1+x)^p\) typically gets larger than \(1 + x^p\), especially for positive \(x\).
• For smaller values of \(p\), particularly when \(0 \leq p \leq 1\), the higher powers influence the expression much less drastically.

This theorem aids in understanding inequalities by breaking down complex exponential expressions into manageable sums, facilitating comparisons across different bounds.

Power Functions

Power functions are expressions of the form \(x^p\) where \(p\) is any real number.
The behavior of these functions varies depending on the exponent \(p\)'s range, making them versatile yet sometimes tricky to evaluate directly.

• When \(p > 1\), power functions grow significantly as \(x\) increases, leading to the observation that \(x^p\) increases faster than \(x\), affecting comparisons against other functions.
• When \(0 \leq p \leq 1\), power functions display a slower growth.
• For such cases, the function is more linear, providing a more contained increase or decrease which allows for balance in inequalities such as \((1+x)^p \leq 1 + x^p\).

Understanding how \(x^p\) behaves according to \(p\) is essential to solving inequalities by providing insight into the nature of \(x\) for various bounds like in the original exercise.

Convex Functions

Convex functions are types of functions where the line segment between any two points on the graph of the function lies above or on the graph.
This property plays a crucial role in optimization and in proving inequalities.

• A function \(f(x)\) is convex if its second derivative \(f''(x)\) is non-negative across its domain.
• For powers where \(0 \leq p \leq 1\), the function \((1+x)^p\) is convex for \(x > 0\).
• This convexity assures us that \((1+x)^p\) doesn't abruptly increase beyond a linear addition constraint \(1 + x^p\), giving it a smooth progression which makes it approachable in inequality settings.

Utilizing convexity helps establish boundaries and provides certainty about function behavior, especially when examining inequalities in particular ranges, aiding in deeper understanding and application.