Domain of the function f ( x ) = lo g e ∣ s i n x − 1∣ 1 is

∣ sin x − 1∣  = 0 .................(i) sin x  = 1 ∣ sin x − 1∣ ≤ 1 − 1 ≤ sin x − 1 ≤ 1 0 ≤ sin x ≤ 2 From (i) and (ii), sin x ∈ [ 0 , 1 ) . sin x ∈ [ 0 , 1 ) sin x = 0 → x = nπ sin x = 1 → x = nπ + ( − 1 ) n 2 π ⇒ Domain of f ( x ) is x = nπ (n any integer).

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MathematicsFunctionsDomain-RangeMedium2 minQB

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Domain of the function $f (x) = lo g_{e} \frac{1}{∣ s i n x - 1∣}$ is

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Solution:

$∣ sin x - 1∣ \neq = 0$ .................(i)

$sin x \neq = 1 ∣ sin x - 1∣ \leq 1 - 1 \leq sin x - 1 \leq 1 0 \leq sin x \leq 2$

From (i) and (ii), $sin x \in [0, 1)$ .

$sin x \in [0, 1) sin x = 0 \to x = nπ sin x = 1 \to x = nπ + (- 1)^{n} \frac{π}{2}$

$\Rightarrow$ Domain of $f (x)$ is $x = nπ$ (n any integer).

Stream:JEESubject:MathematicsTopic:FunctionsSubtopic:Domain-Range

⏱ 2mℹ️ Source: QB

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