If A - x - R - -4- x -3 and fx-sin x -x- then fA-

The correct option is A [√(3)2 π3, 1√(2) π4] Which is decreasing function in [π4,π3] ∴π4≤ x ≤π3 ⇒ f(π4)≥ f(x)≥ f(π3) ⇒1√(2) π4≥ f(x)≥√ ...

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Byju's Answer

Standard XII

Mathematics

Definition of Functions

If A = x ε ...

Question

If $A = x ε R / \frac{π}{4} \leq x \leq \frac{π}{3}$ and f(x)=sin x -x, then f(A)=

[\frac{\sqrt{3}}{2} - \frac{π}{3}, \frac{1}{\sqrt{2}} - \frac{π}{4}]

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[- \frac{1}{\sqrt{2}} - \frac{π}{4}, \frac{\sqrt{3}}{2} - \frac{π}{3}]

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[- \frac{π}{3}, \frac{- π}{4}]

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[\frac{π}{4}, \frac{π}{3}]

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Solution

The correct option is A $[\frac{\sqrt{3}}{2} - \frac{π}{3}, \frac{1}{\sqrt{2}} - \frac{π}{4}]$
Which is decreasing function in $[\frac{π}{4}, \frac{π}{3}]$
$∴ \frac{π}{4} \leq x \leq \frac{π}{3}$
$\Rightarrow f (\frac{π}{4}) \geq f (x) \geq f (\frac{π}{3})$
$\Rightarrow \frac{1}{\sqrt{2}} - \frac{π}{4} \geq f (x) \geq \frac{\sqrt{3}}{2} - \frac{π}{3}$
$f (A) ε [\frac{\sqrt{3}}{2} - \frac{π}{3}, \frac{1}{\sqrt{2}} - \frac{π}{4}]$

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If A=xεR/π4≤x≤π3 and f(x)=sin x -x, then f(A)=

The correct option is A [√32−π3,1√2−π4]Which is decreasing function in [π4,π3]∴π4≤x≤π3⇒f(π4)≥f(x)≥f(π3)⇒1√2−π4≥f(x)≥√32−π3f(A)ε[√32−π3,1√2−π4]

If $A = x ε R / \frac{π}{4} \leq x \leq \frac{π}{3}$ and f(x)=sin x -x, then f(A)=