Domain of definition of the function fx-sin-12x-6 for real valued x- is

Here, f(x)=√(sin^ 1(2x)+π/6), to find domain we must have, sin^ 1(2x)+ π/6≥ 0 [ but π/2≤ sin^ 1θ≤π/2 ] π/6≤ sin^ 1 (2x)≤π/2 sin ( π/6 )≤ 2n ≤ sin π/2 ...

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Standard XI

Mathematics

Domain

Domain of def...

Question

Domain of definition of the function $f (x) = \sqrt{s i n^{- 1} (2 x) + \frac{π}{6}}$ for real valued x, is

[- \frac{1}{4}, \frac{1}{2}]

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[- \frac{1}{2}, \frac{1}{2}]

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(- \frac{1}{2}, \frac{1}{9})

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[- \frac{1}{4}, \frac{1}{4}]

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Solution

The correct option is A $[- \frac{1}{4}, \frac{1}{2}]$
Here, $f (x) = \sqrt{s i n^{- 1} (2 x) + \frac{π}{6}}$ , to find domain we must have,
$s i n^{- 1} (2 x) + \frac{π}{6} \geq 0 [b u t - \frac{π}{2} \leq s i n^{- 1} θ \leq \frac{π}{2}] - \frac{π}{6} \leq s i n^{- 1} (2 x) \leq \frac{π}{2} s i n (- \frac{π}{6}) \leq 2 n \leq s i n \frac{π}{2} \frac{- 1}{2} \leq 2 x \leq 1 \frac{- 1}{4} \leq x \leq \frac{1}{2} ∵ x ϵ [- \frac{1}{4}, \frac{1}{2}]$

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Domain of definition of the function f(x)=√sin−1(2x)+π6 for real valued x, is

The correct option is A [−14, 12]Here, f(x)=√sin−1(2x)+π6, to find domain we must have, sin−1(2x)+π6≥0 [but−π2≤sin−1θ≤π2]−π6≤sin−1(2x)≤π2sin(−π6)≤2n≤sinπ2−12≤2x≤1−14≤x≤12∵ xϵ[−14,12]

Domain of definition of the function $f (x) = \sqrt{s i n^{- 1} (2 x) + \frac{π}{6}}$ for real valued x, is