The domain of the function fx-cos-12x-4 is

F(x)=√(cos^ 12x+π4) For f(x) to be defined, cos^ 12x+π4≥0 ⇒cos^ 12x ≥ π4 cos^ 12x ≥ π4 ∀ x as cos^ 12x∈ [0,π] ⇒ 0 ≤cos^ 12x ≤π ⇒ 1 ≤ 2x ≤ 1 [∵cos^ 1is dec ...

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Standard XII

Mathematics

Conditions on the Parameters of Logarithm Function

The domain of...

Question

The domain of the function $f (x) = \sqrt{{cos}^{- 1} 2 x + \frac{π}{4}} is$

\frac{- 1}{2} \leq x \leq \frac{1}{2 \sqrt{2}}

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\frac{- 1}{2 \sqrt{2}} \leq x \leq \frac{1}{2 \sqrt{2}}

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- 1 \leq x \leq 1

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\frac{- 1}{2} \leq x \leq \frac{1}{2}

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Solution

The correct option is D $\frac{- 1}{2} \leq x \leq \frac{1}{2}$

$f (x) = \sqrt{{cos}^{- 1} 2 x + \frac{π}{4}}$
For $f (x)$ to be defined,
${cos}^{- 1} 2 x + \frac{π}{4} \geq 0 \Rightarrow {cos}^{- 1} 2 x \geq \frac{- π}{4} {cos}^{- 1} 2 x \geq \frac{- π}{4} \forall x as {cos}^{- 1} 2 x \in [0, π] \Rightarrow 0 \leq {cos}^{- 1} 2 x \leq π \Rightarrow - 1 \leq 2 x \leq 1 [∵ {cos}^{- 1} is decreasing function]$
$\Rightarrow \frac{- 1}{2} \leq x \leq \frac{1}{2}$
$∴ Domain of f(x) is \frac{- 1}{2} \leq x \leq \frac{1}{2}$

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The domain of the function f(x)=√cos−12x+π4 is

The correct option is D −12≤x≤12 f(x)=√cos−12x+π4 For f(x) to be defined, cos−12x+π4≥0⇒cos−12x≥−π4cos−12x≥−π4 ∀ x as cos−12x∈[0,π]⇒0≤cos−12x≤π⇒−1≤2x≤1 [∵cos−1is decreasing function] ⇒−12≤x≤12 ∴Domain of f(x) is −12≤x≤12

The domain of the function $f (x) = \sqrt{{cos}^{- 1} 2 x + \frac{π}{4}} is$