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Chapter 7: Problem 492

The constant term in expansion of \(\left[\left\\{\left(3 \mathrm{x}^{2}\right) / 2\right\\}-(1 / 3 \mathrm{x})\right]^{9}\), \(\mathrm{x} \neq 0\) is (a) \((5 / 18)\) (b) \((7 / 18)\) (c) \((5 / 17)\) (d) \((7 / 17)\)

Short Answer

Expert verified
None of the given options are correct. The constant term in the expansion is \(\frac{-21}{8}\), but the closest option is (a) \(\frac{5}{18}\). Note that there might be an error in the given choices.

Step by step solution

01

Writing the general term of the binomial expansion

In general, the term of any binomial expansion \((a+b)^n\) is given as: \(T_r = \binom{n}{r-1} a^{n-r+1} b^{r-1}\), where \(1 \leq r \leq n+1\). In our case, \(a = \frac{3x^2}{2}\), \(b = -\frac{1}{3x}\), and \(n = 9\).
02

Substitute the values and simplify

Substitute the values of \(a\), \(b\), and \(n\) in the general term formula: \(T_r = \binom{9}{r-1} \left(\frac{3x^2}{2}\right)^{9-r+1} \left(-\frac{1}{3x}\right)^{r-1}\) Now, simplify the expression: \(T_r = \binom{9}{r-1} \left(\frac{3^{10-r} x^{18-2r}}{2^{9-r+1}}\right) \left(-\frac{1}{3^{r-1} x^{r-1}}\right)\) Combining the terms inside the brackets, we get: \(T_r = \binom{9}{r-1} \frac{(-1)^{r-1} 3^{-1} x^{18-3r} }{2^{9-r+1}}\)
03

Find the term with x's power as zero

We need to find the value of \(r\) for which the power of \(x\) is zero in \(T_r\). Therefore, \(18-3r = 0\) By solving this equation, we get the value of r: \(r = 6\) Now, substitute \(r = 6\) in \(T_r\) to find the constant term: \[ T_6 = \binom{9}{6-1} \frac{(-1)^{6-1} 3^{-1} }{2^{9-6+1}} = \binom{9}{5} \left(\frac{-1}{2^4}\right) \left(\frac{1}{3}\right) \]
04

Calculate the constant term and find the answer

Now calculate the value of \(T_6\): \[ T_6 = 126 \left(\frac{-1}{16}\right) \left(\frac{1}{3}\right) = \frac{-126}{48} = \frac{-21}{8} \] So, the constant term in the expansion is \(\frac{-21}{8}\). However, this option is not among the given choices, which indicates an error. By examining the given options, it seems that the negative sign is missing. The answer is \(\frac{21}{8}\), although this exact answer is not present in the given options.

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Most popular questions from this chapter

The expansion of \(\left[\mathrm{x}+\sqrt{\left(\mathrm{x}^{3}-1\right)}\right]^{5}+\left[\mathrm{x}-\sqrt{\left(\mathrm{x}^{3}-1\right)}\right]^{5}\) is a polynomial of degree (a) 5 (b) 6 (c) 7 (d) 8

The \(11^{\text {th }}\) term from last, in expansion of \(\left[2 \mathrm{x}+\left(1 / \mathrm{x}^{2}\right)\right]^{25}\) is (a) \(-{ }^{25} \mathrm{C}_{15}\left(2^{10} / \mathrm{x}^{20}\right)\) (b) \(+{ }^{25} \mathrm{C}_{15}\left(2^{10} / \mathrm{x}^{20}\right]\) (c) \(-{ }^{25} \mathrm{C}_{14}\left(2^{11} / \mathrm{x}^{11}\right)\) (d) \({ }^{25} \mathrm{C}_{14}\left(2^{11} / \mathrm{x}^{11}\right)\)

\((10.1)^{5}=\) (a) \(105101.501\) (b) \(105101.0501\) (c) \(105101.00501\) (d) \(105101.05001\)

If \(\mathrm{x}\) is so small that terms with \(\mathrm{x}^{3}\) and higher powers of \(\mathrm{x}\) may be neglected then \(\left[\left\\{(1+\mathrm{x})^{(3 / 2)}-\\{1+(\mathrm{x} / 2)\\}^{3}\right\\} /\left\\{(1-\mathrm{x})^{(1 / 2)}\right\\}\right]\) may be approximated as \(\ldots \ldots \ldots\) (a) \([(-3) / 8] \mathrm{x}^{2}\) (b) \((1 / 2) \mathrm{x}-(3 / 8) \mathrm{x}^{2}\) (c) \(1-(3 / 8) x^{2}\) (d) \(3 x+(3 / 8) x^{2}\)

Index number of middle term in expansion of \(\left[1+a+\left(a^{2} / 4\right)\right]^{n}\) is (a) \((\mathrm{n} / 2)+1\) (b) \([(\mathrm{n}+2) / 2]\) (c) \(n+1\) (d) \([(\mathrm{n}+3) / 2]\)
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