1.

The solution of 4x2 + 4x + 1 > 0 is(a) All real numbers except \(-\frac12\)(b) All real numbers (c) (– ∞, + ∞) (d) None of these

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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