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
B
103
C
10
D
None of these

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.ThereforeLet $$x=10^k$$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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