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

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