First slide

Theory of expressions

Question

$If [x^{2} - 2 x + a] = 0, where [\cdot] represents greatest integer function,has no solution then,$

Moderate

A

$- \infty < a < 1$
B

$2 \leq a < \infty$
C

$1 < a < 2$
D

$a \in R$

Solution

$\begin{array}{l} [x^{2} - 2 x + a] = 0 \\ \Rightarrow x^{2} - 2 x + a \in [0, 1) \end{array}$
For no solution
$\begin{array}{l} case (i): \\ x^{2} - 2 x + a < 0, \forall x \in R \end{array}$
which is not possible ,since $x^{2}$ coefficient is 1 is positive and expression is negative.
$case (ii):$
$\begin{array}{l} x^{2} - 2 x + a \geq 1, \forall x \in R \\ \Rightarrow x^{2} - 2 x + a - 1 \geq 0, \forall x \in R \\ \Rightarrow Disc. \leq 0 \\ \Rightarrow 4 - 4 (a - 1) \leq 0 \Rightarrow 2 \leq a \end{array}$

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