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Question

If the third term in the expansion of $$\displaystyle (\frac{1}{x}+x^{\log_{10}x})^{5} $$ is $$1,000$$, then $$x$$-equals


A
102
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B
103
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C
10
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D
None of these
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Solution

The correct option is A 10$$\displaystyle ^{2}$$
$$T_{2+1}$$
$$=T_{3}$$
$$=\:^{5}C_{2}x^{-3}x^{2log_{10}(x)}$$
$$=10^{3}$$
Therefore
$$x^{-3}x^{2log_{10}(x)}=100$$
$$x^{2log_{10}(x)-3}=100$$
Now x has to be in powers of 10.
Therefore
Let $$x=10^k$$
Above equation reduces to
$$10^{k(2k-3)}=10^{2}$$
$$2k^{2}-3k-2=0$$
$$(2k+1)(k-2)=0$$
$$k=2$$ and $$k=\frac{-1}{2}$$
Hence $$x=10^{2}$$ and $$x=10^{-\frac{1}{2}}$$

Mathematics

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