It is given that 0<a<12That means our x will lie in the interval 0, 12Let x=tan y, where y∈0, π6Note: In the given interval of y, x lies in tan 0, tan π6 i e 0, 13, which contains interval 0, 12 also 3y∈0, π2, so we can directly write tan-1tan 3y=3y, if it occursfx=cot-13x-x31-3x2=cot-13tan y-tan3y1-3tan2y=cot-1tan 3y=π2-tan-1tan 3y=π2-3ygx=cos-11-x21+x2=cos-11-tan2y1+tan2y=cos-1cos 2ySince 2y∈0, π3 which is in range of cos inverse functionThereforegx=2yNowlimx→afx-fagx-gaIts zero by zero form apply L hopsital rulelimx→afx-fagx-ga=limx→ad dxfx-fad dxgx-ga=limx→ad dxfx-0d dxgx-0=limx→ad dxfxd dxgx=limx→ad dxπ2-3yd dx2y=limx→a-3d ydx2d ydx=limx→a-3d ydx2d ydx=limx→a-32=-32

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I t i s g i v e n t h a t 0 < a < \frac{1}{2} T h a t m e a n s o u r x w i l l l i e i n t h e i n t e r v a l [0, \frac{1}{2}] L e t x = \tan y, w h e r e y \in [0, \frac{π}{6}] N o t e : I n t h e g i v e n i n t e r v a l o f y, x l i e s i n [\tan 0, \tan \frac{π}{6}] i . e . [0, \frac{1}{\sqrt{3}}], w h i c h c o n t a i n s i n t e r v a l [0, \frac{1}{2}] a l s o 3 y \in [0, \frac{π}{2}], s o w e c a n d i r e c t l y w r i t e \tan^{- 1} (\tan 3 y) = 3 y, i f i t o c c u r s f (x) = c o t^{- 1} (\frac{3 x - x^{3}}{1 - 3 x^{2}}) = c o t^{- 1} (\frac{3 \tan y - \tan^{3} y}{1 - 3 \tan^{2} y}) = c o t^{- 1} (\tan 3 y) = \frac{π}{2} - \tan^{- 1} (\tan 3 y) = \frac{π}{2} - 3 y g (x) = \cos^{- 1} (\frac{1 - x^{2}}{1 + x^{2}}) = \cos^{- 1} (\frac{1 - \tan^{2} y}{1 + \tan^{2} y}) = \cos^{- 1} (\cos 2 y) S i n c e 2 y \in [0, \frac{π}{3}] w h i c h i s i n r a n g e o f \cos i n v e r s e f u n c t i o n T h e r e f o r e g (x) = 2 y N o w \lim_{x \to a} \frac{f (x) - f (a)}{g (x) - g (a)} I t s z e r o b y z e r o f o r m a p p l y L h o p s i t a l r u l e \lim_{x \to a} \frac{f (x) - f (a)}{g (x) - g (a)} = \lim_{x \to a} \frac{\frac{d}{d x} \{f (x) - f (a)\}}{\frac{d}{d x} \{g (x) - g (a)\}} = \lim_{x \to a} \frac{\frac{d}{d x} f (x) - 0}{\frac{d}{d x} g (x) - 0} = \lim_{x \to a} \frac{\frac{d}{d x} f (x)}{\frac{d}{d x} g (x)} = \lim_{x \to a} \frac{\frac{d}{d x} \{\frac{π}{2} - 3 y\}}{\frac{d}{d x} \{2 y\}} = \lim_{x \to a} \frac{- 3 \frac{d y}{d x}}{2 \frac{d y}{d x}} = \lim_{x \to a} \frac{- 3 \frac{d y}{d x}}{2 \frac{d y}{d x}} = \lim_{x \to a} \frac{- 3}{2} = - \frac{3}{2}

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