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Chapter 7: Problem 495

If middle term is \(\mathrm{kx}^{\mathrm{m}}\) in expansion of \([\mathrm{x}+(1 / \mathrm{x})]^{12}\) then \(\mathrm{m}=\) (a) \(-2\) (b) \(-1\) (c) 0 (d) 1

Short Answer

Expert verified
The middle term of the expansion \([x + (1/x)]^{12}\) is computed using the binomial theorem as the 7th term, which is \(T_7 = \frac{12!}{(6!)(6!)} \cdot x^0\). Comparing this with the given form of the middle term (\(kx^m\)), we find that \(m = 0\). Therefore, the correct answer is (c) 0.

Step by step solution

01

Recall the binomial theorem

The binomial theorem tells us that for any positive integer n, \((a+b)^n = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\), where \(\binom{n}{r} = \frac{n!}{r!(n-r)!}\). Here, we have \(n = 12\), \(a = x\), and \(b = \frac{1}{x}\).
02

Find the middle term

Since the expansion has \(12+1=13\) terms, the middle term will be the 7th term (because 13 is an odd number and the middle term is simply the term in the middle). So, let's compute the 7th term of the expansion: \(T_7 = \binom{12}{6}(x)^{(12-6)}\left(\frac{1}{x}\right)^6 = \frac{12!}{6!(12-6)!} x^6 \times x^{-6}\). Now, simplify the term: \(T_7 = \frac{12!}{(6!)(6!)} \cdot x^{6-6} = \frac{12!}{(6!)(6!)}\cdot x^0\).
03

Compare the term with the given form of the middle term and find the value of \(m\)

Now, let's compare our middle term with the given form (\(kx^m\)): \(T_7 = \frac{12!}{(6!)(6!)} \cdot x^0 = kx^m\). Since x has a power of 0 in the middle term, we can infer that: \(m = 0\). Therefore, the correct answer is: \(m =\) (c) 0.

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Most popular questions from this chapter

In the expansion of \((x-y)^{10}\), (co-efficient of \(x^{7} y^{3}\) ) \(\left(\right.\) co-efficient of \(\left.x^{3} y^{7}\right)=\) (a) \({ }^{10} \mathrm{C}_{7}\) (b) \({ }^{2.10} \mathrm{C}_{7}\) (c) \({ }^{10} \mathrm{C}_{7}+{ }^{10} \mathrm{C}_{1}\) (d) 0

\(\mathrm{s}(\mathrm{k}): 1+3+5+\ldots+(2 \mathrm{k}-1)=3+\mathrm{k}^{2}\) then which statement is true ? (a) \(\mathrm{s}(\mathrm{k}) \Rightarrow \mathrm{s}(\mathrm{k}+1)\) (b) \(\mathrm{s}(\mathrm{k}) \Rightarrow \mathrm{s}(\mathrm{k}+1)\) (c) s (1) is true (d) Result is proved by Principle of Mathematical induction

\(3^{\mathrm{rd}}\) term in expansion of \(\left[(1 / \mathrm{x})+\mathrm{x}^{(\log )}{ }_{10}(\mathrm{x})\right]^{5}\) is 1000 then \(\mathrm{x}=\) (a) 10 (b) 100 (c) 1000 (d) None

\(10^{\text {th }}\) term in expansion of \(\left[2 x^{2}+(1 / x)\right]^{12}\) is \(\ldots \ldots \ldots\) (a) \(\left[(1760) / \mathrm{x}^{2}\right]\) (b) \(\left[(1760) / \mathrm{x}^{3}\right]\) (c) \(\left[(880) / \mathrm{x}^{2}\right]\) (d) \(\left[(880) / \mathrm{x}^{3}\right]\)

\(\mathrm{R}=(\sqrt{2}+1)^{2 \mathrm{n}+1}, \mathrm{n} \in \mathrm{N}\) and \(\mathrm{f}=\mathrm{R}-[\mathrm{R}]\), Where [] is an integer part function then \(\mathrm{Rf}=\) (a) \(2^{2 \mathrm{n}+1}\) (b) \(2^{2 \mathrm{n}-1}\) (c) \(2^{2 \mathrm{n}}-1\) (d) 1
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