Dear student, fx = -x3+x2-x+cos-1xdifferrentiate f'x = -3x2+2x-1-11-x2f'x = -3x2-2x+1-11-x2Discriminant of 3x2-2x+1D =b2-4ac =4-12=-8 0 eq1 for all x using eq1 f'x<0then function is decreasingfor x∈0,1fmax = f0 = 0+0+0+cos-10=π2fmin =f1 =-1+1-1+cos-11 =-1+0=-1range ∈-1,π2 Hope this information will clear your doubts If you have any more doubts just ask here on the forum and our experts will try to help out as soon as possible Regards

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Solve this: CTY1618.PH4—JEEM-PCM- This part contains 30 multiple choice questions. Each question has four choices (f which ONLY ONE is correct. The range of f(x) = —x3 + x2 —x+ cos-l x, x € 10,11 is 1. 2 The extreme values of the function f(x) = ,n€N, are denotes G.I.F.) +[cosnx-1Jx€ — 2'2

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f (x) = - x^{3} + x^{2} - x + \cos^{- 1} x d i f f e r r e n t i a t e f' (x) = - 3 x^{2} + 2 x - 1 - \frac{1}{\sqrt{1 - x^{2}}} f' (x) = - (3 x^{2} - 2 x + 1) - \frac{1}{\sqrt{1 - x^{2}}} D i s c r i m i n a n t o f (3 x^{2} - 2 x + 1) D = b^{2} - 4 a c = 4 - 12 = - 8 < 0 \Rightarrow (3 x^{2} - 2 x + 1) > 0 . . . e q 1 f o r a l l x u \sin g e q 1 f' (x) < 0 t h e n f u n c t i o n i s d e c r e a \sin g f o r x \in [0, 1] f_{m a x} = f (0) = 0 + 0 + 0 + \cos^{- 1} 0 = \frac{π}{2} f_{m i n} = f (1) = - 1 + 1 - 1 + \cos^{- 1} 1 = - 1 + 0 = - 1 r a n g e \in [- 1, \frac{π}{2}]

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