If A ={x∈R|π4≤x≤π3} and f(x)=sinx−x, then f(A)=

f(x)=sinx−xf1(x)=cosx−1< for π4 ≤x≤π3f(x) is decreasingf(A)=[f(π3), f(π4)]=[32−π3, 12−π4]

If A ={x∈R|π4≤x≤π3} and f(x)=sinx−x, then f(A)=

f(x)=sinx−xf1(x)=cosx−1< for π4 ≤x≤π3f(x) is decreasingf(A)=[f(π3), f(π4)]=[32−π3, 12−π4]

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$I f A = {x \in R | \frac{π}{4} \leq x \leq \frac{π}{3}} a n d f (x) = \sin x - x, t h e n f (A) =$

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

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

$[- \frac{π}{3}, \frac{- π}{4}]$

$[\frac{π}{4}, \frac{π}{3}]$

answer is A.

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Detailed Solution

$\begin{array}{l} f (x) = \sin x - x \\ f^{1} (x) = \cos x - 1 < f o r \frac{π}{4} \leq x \leq \frac{π}{3} \\ f (x) i s d e c r e a \sin g \\ f (A) = [f (\frac{π}{3}), f (\frac{π}{4})] \\ = [\frac{\sqrt{3}}{2} - \frac{π}{3}, \frac{1}{\sqrt{2}} - \frac{π}{4}] \end{array}$