Consider the equation sin−1x2−6x+172+cos−1k=π2 then

sin−1x2−6x+172=sin−1kThe above equation has solutions only when -1≤k≤1The graph of the curve y=x2-6x+172 is as below

Consider the equation sin−1x2−6x+172+cos−1k=π2 then

sin−1x2−6x+172=sin−1kThe above equation has solutions only when -1≤k≤1The graph of the curve y=x2-6x+172 is as below

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$C o n s i d e r t h e e q u a t i o n$ $\sin^{- 1} (x^{2} - 6 x + \frac{17}{2}) + \cos^{- 1} k = \frac{π}{2}$ $t h e n$

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$T h e e q u a t i o n h a s u n i q u e s o l u t i o n i f k = - \frac{1}{2}$

$T h e e q u t i o n m u s t h a v e r e a l r o o t s i f k \in (- \frac{1}{2}, 1)$

$t h e e q u a t i o n m u s t h a v e r e a l r o o t s i f k \in (- 1, \frac{1}{2})$

$T h e l a r g e s t v a l u e o f k f o r w h i c h t h e e q u a t i o n h a s 2 d i s t i n c t s o l u t i o n s i s 1$

answer is A, B, D.

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$\sin^{- 1} (x^{2} - 6 x + \frac{17}{2}) = \sin^{- 1} k$

$T h e a b o v e e q u a t i o n h a s s o l u t i o n s o n l y w h e n - 1 \leq k \leq 1$

$T h e g r a p h o f t h e c u r v e y = x^{2} - 6 x + \frac{17}{2}$ $i s a s b e l o w$

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Consider the equation sin−1x2−6x+172+cos−1k=π2 then

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$C o n s i d e r t h e e q u a t i o n$ $\sin^{- 1} (x^{2} - 6 x + \frac{17}{2}) + \cos^{- 1} k = \frac{π}{2}$ $t h e n$