If x - R -3- then the possible values of the expression - x 2-6 x -9-3- x -2 can beA- 0B- 1C- 2D- 3

The correct options are B 1 D 3√(x^2 6x+9)3 x+2 =√((x 3)^2)3 x+2 =|x 3|3 x+2 We know that |x 3| = ± (x 3) So, √(x^2 6x+9)3 x+2 = ± 1+2 = 1, 3 nbsp; ...

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Standard XII

Mathematics

Inequalities Involving Modulus Function

If x ∈ R -3, ...

Question

If $x \in R - {3},$ then the possible value(s) of the expression $\frac{\sqrt{x^{2} - 6 x + 9}}{3 - x} + 2$ can be

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Solution

The correct options are
B $1$
D $3$
$\frac{\sqrt{x^{2} - 6 x + 9}}{3 - x} + 2$
$= \frac{\sqrt{(x - 3)^{2}}}{3 - x} + 2$
$= \frac{| x - 3 |}{3 - x} + 2$
We know that $| x - 3 | = \pm (x - 3)$
So,
$\frac{\sqrt{x^{2} - 6 x + 9}}{3 - x} + 2 = \pm 1 + 2 = 1, 3$

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If x∈R−{3}, then the possible value(s) of the expression √x2−6x+93−x+2 can be

The correct options are B 1 D 3√x2−6x+93−x+2 =√(x−3)23−x+2 =|x−3|3−x+2 We know that |x−3|=±(x−3) So, √x2−6x+93−x+2=±1+2=1,3

If $x \in R - {3},$ then the possible value(s) of the expression $\frac{\sqrt{x^{2} - 6 x + 9}}{3 - x} + 2$ can be

The correct options are
B $1$
D $3$
$\frac{\sqrt{x^{2} - 6 x + 9}}{3 - x} + 2$
$= \frac{\sqrt{(x - 3)^{2}}}{3 - x} + 2$
$= \frac{| x - 3 |}{3 - x} + 2$
We know that $| x - 3 | = \pm (x - 3)$
So,
$\frac{\sqrt{x^{2} - 6 x + 9}}{3 - x} + 2 = \pm 1 + 2 = 1, 3$