A) \[-1\]
B) \[1\]
C) \[2\]
D) \[-2\]
E) \[-\frac{1}{2}\]
Correct Answer: A
Solution :
\[\int_{a}^{0}{\frac{{{x}^{2}}-1}{1-x}dx=-\frac{1}{2}}\] \[\Rightarrow \] \[\int_{0}^{a}{\frac{(x-1)(x+1)}{(x-1)}}dx=-\frac{1}{2}\] \[\Rightarrow \] \[\int_{0}^{a}{(x+1)}dx=-\frac{1}{2}\] \[\Rightarrow \] \[[{{x}^{2}}/2+x]_{0}^{a}=-\frac{1}{2}\] \[\Rightarrow \] \[{{a}^{2}}/2+a=-\frac{1}{2}\] \[\Rightarrow \] \[{{a}^{2}}+2a+1=0\] \[{{(a+1)}^{2}}=0\] \[\Rightarrow \] \[a=-1\]
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