Question

lf $x$ is real and $2x^2-4x+5>0$, then $x$ lies in the interval

• A. $\varnothing$
• B. $(-\infty ,\infty )$
• C. $(0,\infty )$
• D. $(-\infty ,0)$

Answer 1

Answer: (B)

$(-\infty ,\infty )$

$2x^2-4x+5>0$, $\forall xϵR$
SInce '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 $D\ge 0$, then $2x^2-4x+5=0$ for some values of 'x')
In that case, the graph will be always above x-axis as shown in figure.
For the given expression $2x^2-4x+5=0$, $D=16-40=-24<0$
Hence , the value is always greater than zero for all values of $x$.
$\Rightarrow x\in (-\infty ,\infty )$