A) \[56{{\left( \frac{1-t}{1+t} \right)}^{3}}\]
B) \[56{{\left( \frac{1+t}{1-t} \right)}^{3}}\]
C) \[70{{\left( \frac{1-t}{1+t} \right)}^{4}}\]
D) \[70{{\left( \frac{1+t}{1-t} \right)}^{4}}\]
Correct Answer: C
Solution :
| [c] \[[({{t}^{-1}}-1)x+{{({{t}^{-1}}+1)}^{-1}}{{x}^{-1}}){{]}^{8}}\] |
| \[={{\left[ \left( \frac{1}{t}-1 \right)x+{{\left( \frac{1}{t}+1 \right)}^{-1}}\frac{1}{x} \right]}^{8}}\] |
| Let \[{{T}_{r+1}}\] be the term independent of x, then |
| \[{{T}_{r+1}}={{\,}^{8}}{{C}_{r}}{{\left( \frac{1}{t}-1 \right)}^{8-r}}{{x}^{8-r}}\cdot {{\left( \frac{1}{t}+1 \right)}^{-r}}{{\left( \frac{1}{x} \right)}^{r}}\] |
| \[={{\,}^{8}}{{C}_{r}}{{\left( \frac{1}{t}-1 \right)}^{8-r}}\cdot {{\left( \frac{1}{t}+1 \right)}^{-r}}\cdot {{x}^{8-2r}}\] |
| \[\therefore 8-2r=0\Rightarrow r=0\] |
| \[\therefore \,\,\,\,{{T}_{5}}\] is the term independent of x and |
| \[{{T}_{5}}={{\,}^{8}}{{C}_{4}}{{\left( \frac{1}{t}-1 \right)}^{4}}\cdot {{\left( \frac{1}{t}+1 \right)}^{-4}}\] |
| \[={{\,}^{8}}{{C}_{4}}{{\left( \frac{1-t}{t} \right)}^{4}}\cdot {{\left( \frac{1+t}{t} \right)}^{-4}}\] |
| \[={{\,}^{8}}{{C}_{4}}{{\left( \frac{1-t}{1+t} \right)}^{4}}=\frac{8.7.6.5}{4.3.2.1}{{\left( \frac{1-t}{1+t} \right)}^{4}}\] |
| \[=\,70.{{\left( \frac{1-t}{1+t} \right)}^{4}}\] |
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