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Question

# lf x is real and 2x2−4x+5>0, then x lies in the interval

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Solution

## The correct option is D (−∞,∞)2x2−4x+5>0, ∀xϵRSInce 'a' of the above quadratic expression is positive, the graph (a parabola) will always face upwards. Now, for the expression to be always >0, D must be less than zero. (because if D≥0, then 2x2−4x+5=0 for some values of 'x')In that case, the graph will be always above x-axis as shown in figure. For the given expression 2x2−4x+5=0, D=16−40=−24<0Hence , the value is always greater than zero for all values of x.⇒x∈(−∞,∞)

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