The domain of the function f x -2-2 x - x 2 isA- -3- -3-B- -2-2- -2-3-2-3-D- -1-3-1-3-

The correct option is D [ 1 √(3), 1 + √(3)] f(x) = √(2 2x x^2) Since, 2 2x x^2 ≥ 0, x^2+2x 2 ≤ 0 ⇒ x^2 2x 2+1 1 ≤ 0 ⇒ (x 1)^2 (√(3))^2 ≤ 0 ⇒ [x ( 1 √(3) ...

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Standard XII

Mathematics

Real Valued Functions

The domain of...

Question

The domain of the function $f (x) = \sqrt{2 - 2 x - x^{2}}$ is

$[- \sqrt{3}, \sqrt{3}]$

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$[- 1 - \sqrt{3}, - 1 + \sqrt{3}]$

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$[- 2, 2]$

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$[- 2 - \sqrt{3}, - 2 + \sqrt{3}]$

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Solution

The correct option is B
$[- 1 - \sqrt{3}, - 1 + \sqrt{3}]$

$f (x) = \sqrt{2 - 2 x - x^{2}}$
Since, $2 - 2 x - x^{2} \geq 0$ ,
$x^{2} + 2 x - 2 \leq 0 \Rightarrow x^{2} - 2 x - 2 + 1 - 1 \leq 0 \Rightarrow (x - 1)^{2} - (\sqrt{3})^{2} \leq 0 \Rightarrow [x - (- 1 - \sqrt{3})] \leq 0 [x - (- 1 - \sqrt{3})] \leq 0 \Rightarrow (- 1 - \sqrt{3}) \leq x \leq (- 1 + \sqrt{3})$
Thus, $d o m (f) = [- 1 - \sqrt{3}, - 1 + \sqrt{3}]$

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Similar questions

Q. The domain of the function

f (x) = \sqrt{2 - 2 x - x^{2}}

is

(a)

[- \sqrt{3}, \sqrt{3}]

(b)

[- 1 - \sqrt{3}, - 1 + \sqrt{3}]

[- 2 - \sqrt{3}, - 2 + \sqrt{3}]

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The domain of the function f(x)=√2−2x−x2 is

The correct option is B [−1−√3,−1+√3] f(x)=√2−2x−x2 Since, 2−2x−x2≥0, x2+2x−2≤0⇒x2−2x−2+1−1≤0⇒(x−1)2−(√3)2≤0⇒[x−(−1−√3)]≤0[x−(−1−√3)]≤0⇒(−1−√3)≤x≤(−1+√3) Thus, dom(f)=[−1−√3,−1+√3]

The domain of the function $f (x) = \sqrt{2 - 2 x - x^{2}}$ is