The number of solutions of the equation $x^{2} + [x] - 4 x + 3 = 0$ is (where $[]$ denotes G.I.F.)

$1$

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$0$

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$2$

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$3$

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Solution

The correct option is B
$0$

Given equation can be written as $(x^{2} - 3 x + 3) - f = 0 w h e r e f = x - [x] a n d 0 \leq f < 1$

$∴ 0 \leq x^{2} - 3 x + 3 < 1$

Now, solving $x^{2} - 3 x + 3 = 0$ we get imaginary roots.

$∴ x^{2} - 3 x + 3 \geq 0 \forall x \in R$

Solving $x^{2} - 3 x + 3 < 1$ $\Rightarrow 1 < x < 2$

If $1 < x < 2; [x] = 1$ putting $[x] = 1$ in the given equation and solving we get $x = 2$ . But $1 < x < 2$
$∴$ The given equation has no solution.

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Similar questions

The number of solutions of the equation $x^{2} + [x] - 4 x + 3 = 0$ is (where $[]$ denotes G.I.F.)

The number of solutions of the equations of the equation $x^{2} + [x] - 4 x + 3 = 0$ is Where [ ] denotes G.I.F.

x^{2} - 4 - [x] = 0

[.]

x = 3 [\sqrt{x}] + 1

x \in [0, 10]

x

x^{2} + 4 x + [x] + 6 = 0

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