If -1-sin x-1-sin x-23- then a possible interval for x is

|1 |sin x|1+|sin x||≥23 ⇒1|1+|sin x||≥23 Since 1+|sin x|≥0 ∀ x∈ℝ ∴ nbsp;11+|sin x|≥23 ⇒ |sin x|≤12 From the graph, nbsp; |sin x|≤12 ⇒ x∈[ π6,π6] Since pe ...

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Byju's Answer

Standard XII

Mathematics

Critical Point

If |1-|sin x|...

Question

If $∣ ∣ ∣ 1 - \frac{| sin x |}{1 + | sin x |} ∣ ∣ ∣ \geq \frac{2}{3},$ then a possible interval for $x$ is

[- \frac{7 π}{6}, - \frac{5 π}{6}]

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[\frac{π}{6}, \frac{5 π}{6}]

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[\frac{7 π}{6}, \frac{11 π}{6}]

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[- \frac{5 π}{6}, - \frac{π}{6}]

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Solution

The correct option is A $[- \frac{7 π}{6}, - \frac{5 π}{6}]$
$∣ ∣ ∣ 1 - \frac{| sin x |}{1 + | sin x |} ∣ ∣ ∣ \geq \frac{2}{3} \Rightarrow \frac{1}{| 1 + | sin x | |} \geq \frac{2}{3}$
Since $1 + | sin x | \geq 0 \forall x \in R$
$∴ \frac{1}{1 + | sin x |} \geq \frac{2}{3} \Rightarrow | sin x | \leq \frac{1}{2}$

From the graph,
$| sin x | \leq \frac{1}{2} \Rightarrow x \in [- \frac{π}{6}, \frac{π}{6}]$
Since period of $y = | sin x |$ is $π,$
$∴ x \in ⋃ n \in Z [n π - \frac{π}{6}, n π + \frac{π}{6}]$

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Similar questions

Q. If

∣ ∣ ∣ 1 - \frac{| sin x |}{1 + | sin x |} ∣ ∣ ∣ \geq \frac{2}{3},

then

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Q. Assertion :If n is a positive integer then

\int_{0}^{n π} ∣ ∣ ∣ \frac{sin x}{x} ∣ ∣ ∣ d x \geq \frac{2}{π} (1 + \frac{1}{2} + \frac{1}{3} + . . . + \frac{1}{n})

Reason: In the interval

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Q. If

∣ ∣ ∣ ∣ \begin{matrix} s i n x & c o s x & c o s x c o s x & s i n x & c o s x c o s x & c o s x & s i n x \end{matrix} ∣ ∣ ∣ ∣ = 1

in the interval

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If ∣∣∣1−|sinx|1+|sinx|∣∣∣≥23, then a possible interval for x is

The correct option is A [−7π6,−5π6] ∣∣∣1−|sinx|1+|sinx|∣∣∣≥23⇒1|1+|sinx||≥23 Since 1+|sinx|≥0 ∀ x∈R ∴11+|sinx|≥23⇒|sinx|≤12 From the graph, |sinx|≤12⇒x∈[−π6,π6] Since period of y=|sinx| is π, ∴x∈⋃n∈Z[nπ−π6,nπ+π6]

If $∣ ∣ ∣ 1 - \frac{| sin x |}{1 + | sin x |} ∣ ∣ ∣ \geq \frac{2}{3},$ then a possible interval for $x$ is