Let A- and a function f-A- B be defined as fx-x-1-x-1 - -2-x-2-x- Then

For domain, |x| 1|x|+1≥0 ⇒ |x|≥1 ⋯ (1) and nbsp; |x|+2|x| 2≤0 ⇒ |x| lt;2 ⋯ (2) From (1) and (2), x∈( 2, 1]∪[1,2) ⇒ A={ 1,1} Now, for x=±1, we have ...

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Byju's Answer

Standard XI

Mathematics

Range

Let A⊂ℤ and a...

Question

Let $A \subset Z$ and a function $f : A \to B$ be defined as $f (x) = \sqrt{\frac{| x | - 1}{| x | + 1}} - \sqrt{\frac{2 + | x |}{2 - | x |}} .$ Then

n (A) \geq 2

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n (A) = 2

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n (B) = 1

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n (B) \geq 1

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Solution

The correct option is D $n (B) \geq 1$
For domain,
$\frac{| x | - 1}{| x | + 1} \geq 0$
$\Rightarrow | x | \geq 1 \dots (1)$
and $\frac{| x | + 2}{| x | - 2} \leq 0$
$\Rightarrow | x | < 2 \dots (2)$

From $(1)$ and $(2),$
$x \in (- 2, - 1] \cup [1, 2)$
$\Rightarrow A = {- 1, 1}$
Now, for $x = \pm 1,$ we have $f (\pm 1) = - \sqrt{3}$
So, codomain $B$ can have at least one element.

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Let A⊂Z and a function f:A→B be defined as f(x)=√|x|−1|x|+1 − √2+|x|2−|x|. Then

The correct option is D n(B)≥1For domain, |x|−1|x|+1≥0 ⇒|x|≥1 ⋯(1) and |x|+2|x|−2≤0 ⇒|x|<2 ⋯(2) From (1) and (2), x∈(−2,−1]∪[1,2) ⇒A={−1,1} Now, for x=±1, we have f(±1)=−√3 So, codomain B can have at least one element.

Let $A \subset Z$ and a function $f : A \to B$ be defined as $f (x) = \sqrt{\frac{| x | - 1}{| x | + 1}} - \sqrt{\frac{2 + | x |}{2 - | x |}} .$ Then