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The solution of 4x2 + 4x + 1 > 0 is(a) All real numbers except \(-\frac12\)(b) All real numbers (c) (– ∞, + ∞) (d) None of these |
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Answer» (a) all real numbers except \(-\frac12\). 4x2 + 4x + 1 > 0 ⇒ (2x + 1)2 > 0 ⇒ (2x + 1) < 0 or (2x + 1) > 0 ⇒ x > \(-\frac12\) or x < \(-\frac12\) ∴ The inequality holds for all real numbers except x = \(-\frac12\). |
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