Question

$\huge\tt\pink{Chapter \: :- \: Functions}$

$\huge\bold{Question \: :}$

The range of |x-2| is ?

$\huge\sf\purple{✓No \: Google$
$\huge\sf\purple{✓No \: Spamming}$​

Answer 1

We need to find range of,

$\longrightarrow f(x)=|x-2|$

Let,

$\longrightarrow y=|x-2|$

or,

$\longrightarrow y=\sqrt{(x-2)^2}$

Since $\sqrt x\in[0,\ \infty)$ for every $x$ in its domain, we get,

$\longrightarrow y\in[0,\ \infty)\quad\quad\dots(1)$

Then,

$\longrightarrow y^2=(x-2)^2$

$\longrightarrow y^2=x^2-4x+4$

$\longrightarrow x^2-4x+(4-y^2)=0$

Now we have a quadratic equation in which the discriminant should be non - negative.

$\longrightarrow (-4)^2-4(4-y^2)\geq0$

$\longrightarrow 16-16+4y^2\geq0$

$\longrightarrow4y^2\geq0$

$\longrightarrow y^2\geq0$

$\Longrightarrow y\in\mathbb{R}\quad\quad\dots(2)$

Taking $(1)\land (2),$ we get,

$\longrightarrow y\in[0,\ \infty)$

Hence the range is,

$\longrightarrow \underline{\underline{ f(x)\in[0,\ \infty)}}$

Answer 2

We need to find range of,

\longrightarrow f(x)=|x-2|⟶f(x)=∣x−2∣

Let,

\longrightarrow y=|x-2|⟶y=∣x−2∣

or,

\longrightarrow y=\sqrt{(x-2)^2}⟶y=

(x−2)

2

Since \sqrt x\in[0,\ \infty)

x

∈[0, ∞) for every xx in its domain, we get,

\longrightarrow y\in[0,\ \infty)\quad\quad\dots(1)⟶y∈[0, ∞)…(1)

Then,

\longrightarrow y^2=(x-2)^2⟶y

2

=(x−2)

2

\longrightarrow y^2=x^2-4x+4⟶y

2

=x

2

−4x+4

\longrightarrow x^2-4x+(4-y^2)=0⟶x

2

−4x+(4−y

2

)=0

Now we have a quadratic equation in which the discriminant should be non - negative.

\longrightarrow (-4)^2-4(4-y^2)\geq0⟶(−4)

2

−4(4−y

2

)≥0

\longrightarrow 16-16+4y^2\geq0⟶16−16+4y

2

≥0

\longrightarrow4y^2\geq0⟶4y

2

≥0

\longrightarrow y^2\geq0⟶y

2

≥0

\Longrightarrow y\in\mathbb{R}\quad\quad\dots(2)⟹y∈R…(2)

Taking (1)\land (2),(1)∧(2), we get,

\longrightarrow y\in[0,\ \infty)⟶y∈[0, ∞)

Hence the range is,

\longrightarrow \underline{\underline{ f(x)\in[0,\ \infty)}}⟶

f(x)∈[0, ∞)