A) \[\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{n(n+1)}\]
B) (4)
C) \[\frac{1}{n(n+1)}\]
D) none of these
Correct Answer: A
Solution :
\[{{x}^{{{\log }_{x}}{{(1-x)}^{2}}}}=9\] \[\Rightarrow \] \[{{\log }_{x}}9={{\log }_{x}}{{(1-x)}^{2}}\] \[[\because {{a}^{x}}=N\Rightarrow {{\log }_{a}}N=x]\] \[\Rightarrow \] \[9={{(1-x)}^{2}}\] \[\Rightarrow \] \[1+{{x}^{2}}-2x-9=0\] \[\Rightarrow \] \[{{x}^{2}}-2x-8=0\] \[\Rightarrow \] \[(x+2)(x-4)=0\] \[\Rightarrow \] \[(x=-2,4)\]
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