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The solution set of inequality, \( \frac{2^{x-1}-1}{2^{x+1}+1}<2 \), contains (A) all real numbers(B) finite number of integers(C) no negative number(D) no real number |
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Answer» Correct option is (A) All real numbers \(\frac{2^{x -1} -1}{2^{x + 1} + 1} < 2\) ⇒ \(\frac{2^{x -1} -1}{2^{x + 1} + 1} -2 < 0\) ⇒ \(\frac{2^{x -1} - 1 -2.2^{x + 1} -2}{2^{x +1} + 1}< 0\) ⇒ \(2^{x-1}-2^{x +2}- 3< 0\) (∵ \(2^{x+1}+1 > 0\) (always)) ⇒ \(\frac{2^x}2 - 4.2^x - 3< 0\) ⇒ \(\frac{-7}2.2^x - 3<0\) ⇒ \(\frac 72 . 2^x > -3\) ⇒ \(2^x > \frac{-6}7\) which is true for all real values of x. |
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