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4-2.Quadratic Equations and Inequations
normal
Sum of the solutions of the equation $\left[ {{x^2}} \right] - 2x + 1 = 0$ is (where $[.]$ denotes greatest integer function)
A
$\frac {1}{2}$
B
$2$
C
$3$
D
$\frac {3}{2}$
Solution
$ x^{2}-1<\left[x^{2}\right] \leq x^{2} $
$\left[x^{2}\right] =2 x-1 $
$ \Rightarrow {x^2} < 2x \Rightarrow 0 < x < 2$
$0 < x < 1$
$\Rightarrow x=1/2$
$1 \le x < \sqrt 2 $
$x = 1$
$sum=3$
$\sqrt 2 \le x < 2$
$x=3/2$
Standard 11
Mathematics