The inequality 2cos x - -1-sin 2x - -1-sin 2x - -2 where x-0-2- holds true in the interval -a-2-b-2- Then the value of a-b is

| √(1+sin 2x) √(1 sin 2x )| nbsp;= ||cos x +sin x | |cos x sin x| | nbsp;=2 min{|cos x|, |sin x |} Now, 2cos x ≤ 2 min{|cos x|, |sin x |}≤√(2) ⇒co ...

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Standard XIII

Mathematics

General Solutions of (sin theta)^2 = (sin alpha)^2 , (cos theta)^2 = (cos alpha)^2 , (tan theta)^2 = (tan alpha)^2

The inequalit...

Question

The inequality $2 cos x \leq ∣ ∣ \sqrt{1 + sin 2 x} - \sqrt{1 - sin 2 x} ∣ ∣ \leq \sqrt{2}$ where $x \in [0, 2 π]$ holds true in the interval $[\frac{a π}{2}, \frac{b π}{2}]$ . Then the value of $a + b$ is

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Solution

$∣ ∣ \sqrt{1 + sin 2 x} - \sqrt{1 - sin 2 x} ∣ ∣ = ∣ ∣ ∣ | cos x + sin x | - | cos x - sin x | ∣ ∣ ∣ = 2 min {| cos x |, | sin x |}$

Now,
$2 cos x \leq 2 min {| cos x |, | sin x |} \leq \sqrt{2} \Rightarrow cos x \leq min {| cos x |, | sin x |} \leq \frac{1}{\sqrt{2}}$

Clearly, from the graph,
$x \in [\frac{π}{4}, \frac{7 π}{4}] a + b = \frac{1}{2} + \frac{7}{2} = 4$

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Similar questions

Q. The inequality

2 cos x \leq ∣ ∣ \sqrt{1 + sin 2 x} - \sqrt{1 - sin 2 x} ∣ ∣ \leq \sqrt{2}

where

x \in [0, 2 π]

holds true in the interval

[\frac{a π}{2}, \frac{b π}{2}]

. Then the value of

a + b

Q. Determine all the values of

x

in the interval

x ϵ [0, 2 π]

which satisfy the inequality

2 cos x \leq ∣ ∣ \sqrt{1 + sin 2 x} - \sqrt{1 - sin 2 x} ∣ ∣ \leq 2

Q. The solution of the equation

{(sin x + cos x)}^{1 + sin 2 x} = 2, - π \leq x \leq π

, is

If sin (sin^-1 x) = x, the interval in which x lies is__

Q. The range of

x

for which the formula

3 {sin}^{- 1} x = {sin}^{- 1} [x (3 - 4 x^{2})]

holds is :

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The inequality 2cosx≤∣∣√1+sin2x−√1−sin2x∣∣≤√2 where x∈[0,2π] holds true in the interval [aπ2,bπ2]. Then the value of a+b is

∣∣√1+sin2x−√1−sin2x∣∣=∣∣∣|cosx+sinx|−|cosx−sinx|∣∣∣=2min{|cosx|,|sinx|} Now, 2cosx≤2min{|cosx|,|sinx|}≤√2⇒cosx≤min{|cosx|,|sinx|}≤1√2 Clearly, from the graph, x∈[π4,7π4]a+b=12+72=4

The inequality $2 cos x \leq ∣ ∣ \sqrt{1 + sin 2 x} - \sqrt{1 - sin 2 x} ∣ ∣ \leq \sqrt{2}$ where $x \in [0, 2 π]$ holds true in the interval $[\frac{a π}{2}, \frac{b π}{2}]$ . Then the value of $a + b$ is