Problem 41 - A Complete Resource Book in Mathematics for JEE Main 2017

Question

Which of the following expansions will have term containing \(x^{3}\) ? (A) \(\left(x^{-\frac{1}{5}}+2 x^{\frac{3}{5}}\right)^{25}\) (B) \(\left(x^{\frac{3}{5}}+2 x^{-\frac{1}{5}}\right)^{24}\) (C) \(\left(x^{\frac{3}{5}}-2 x^{-\frac{1}{5}}\right)^{23}\) (D) \(\left(x^{\frac{3}{5}}+2 x^{-\frac{1}{5}}\right)^{22}\)

Step-by-Step Solution

Answer

Option (A) contains the term with \(x^3\).

Step 1: Understand the Binomial Term

The general term in the expansion of \((a + b)^n\) is given by \(T_{k+1} = \binom{n}{k}a^{n-k}b^k\). In this exercise, we identify the general term in the binomial expansion.

Step 2: Identify the Powers in Each Option

For option (A), use the general term: \(T_{k+1} = \binom{25}{k}(x^{-\frac{1}{5}})^{25-k}(2x^{\frac{3}{5}})^k\). Simplifying, the power of \(x\) will be \(-\frac{1}{5}(25-k) + \frac{3}{5}k = -\frac{25}{5} + \frac{k}{5} + \frac{3k}{5} = \frac{4k - 25}{5}\). We set this equal to 3 to find the relevant terms for the problem.

Step 3: Solve for Option A

Set \(\frac{4k - 25}{5} = 3\). Multiplying through by 5 gives \(4k - 25 = 15\). Solving for \(k\), we get \(4k = 40\) and hence \(k = 10\). The term containing \(x^3\) indeed exists in option (A).

Step 4: Solve for Option B

For option (B), the general term is \(\binom{24}{k}(x^{\frac{3}{5}})^{24-k}(2x^{-\frac{1}{5}})^k\). Simplifying, the power of \(x\) becomes \(\frac{3}{5}(24-k) - \frac{k}{5} = \frac{72}{5} - \frac{3k}{5} - \frac{k}{5} = \frac{72 - 4k}{5}\). Setting this equal to 3 yields the equation \(72 - 4k = 15\), thus, \(4k = 57\). Since \(k\) must be an integer, there is no term containing \(x^3\) in option (B).

Step 5: Solve for Option C

For option (C), the term is \(\binom{23}{k}(x^{\frac{3}{5}})^{23-k}(-2x^{-\frac{1}{5}})^k\). The power of \(x\) is \(\frac{3}{5}(23-k) - \frac{k}{5} = \frac{69}{5} - \frac{4k}{5}\). Solving \(\frac{69 - 4k}{5} = 3\) gives \(69 - 4k = 15\), and \(4k = 54\) resulting in a non-integer \(k = 13.5\). Therefore, no term with \(x^3\) exists in option (C).

Step 6: Solve for Option D

For option (D), the term is \(\binom{22}{k}(x^{\frac{3}{5}})^{22-k}(2x^{-\frac{1}{5}})^k\). The power of \(x\) becomes \(\frac{3}{5}(22-k) - \frac{k}{5} = \frac{66}{5} - \frac{4k}{5}\). Setting this equal to 3 gives \(66 - 4k = 15\), so \(4k = 51\). Since \(k\) must be an integer, there are no terms containing \(x^3\) in option (D).

Key Concepts

Binomial Theorem

The binomial theorem is a powerful tool used to expand expressions that are raised to any power. When you have an expression like \((a + b)^n\), the binomial theorem helps you find each term in the expansion without multiplying the expression over and over. In mathematical terms, it states that:

• \((a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{n-k} b^k\)

This means each term is determined by the binomial coefficient \(\binom{n}{k}\) which indicates combinations, and \(a^{n-k}b^k\) shows the distribution of powers between \(a\) and \(b\). Taking this scientific approach simplifies dealing with high powers of binomials.
Think of the binomial theorem as a shortcut to swiftly unpack the powers, which comes in handy, especially in problems like the one we have at hand. With it, finding specific terms in expansive polynomials has never been easier and more structured.

Power of x

Understanding the power of \(x\) in binomial expansions is crucial. In our exercise, the key was to determine the form of \(x\) in the expansion that gives us the desired power of \(x^3\). Each term in the binomial expansion such as \((x^{a} + bx^{b})^n\), impacts the resulting power of \(x\).Here's the breakdown for each option:

• Option A: \(T_{k+1} = \binom{25}{k}(x^{-\frac{1}{5}})^{25-k}(2x^{\frac{3}{5}})^k\) simplifies to \(\frac{4k - 25}{5}\). For power \(x^3\), this simplifies to \(k=10\).
• Option B: Results in \(\frac{72 - 4k}{5}\). Here, a suitable \(k\) does not exist.
• Options C & D: Similar reasoning shows non-integer or non-suitable values for \(k\) that achieve \(x^3\).

Thus, correctly breaking down these calculations determines our success in spotting applicable powers needed. Recognizing how different terms affect \(x\)'s power lets us efficiently find necessary components like the term with \(x^3\).

General Term in Expansion

In a binomial expansion, identifying the general term paints a clear picture of all possible products in the multiplied form. We use the expression \(T_{k+1} = \binom{n}{k}a^{n-k}b^k\), where each part has its specific function:

• The binomial coefficient \(\binom{n}{k}\) tells us how many ways we can pick \(k\) elements from a set \(n\), giving us the term's "weight."
• The powers \(a^{n-k}\) and \(b^k\) adjust accordingly, distributing the respective terms throughout the expansion.

In the provided exercise, correctly simplifying the powers continued from this general term structure guides you to identify which term contains the specific power, such as \(x^3\). Finding \(k\) such that the expression yields exactly that power is crucial in solving these types of binomial problems. Once mastered, the general term becomes an efficient tool in cutting through complicated polynomials to focus on terms of interest.