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Question

# if y=(sinâˆ’1x)2+k sinâˆ’1x, then (1âˆ’x2)d2ydx2âˆ’xdydx =2

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Solution

## The correct option is C 2We have, y=(sin−1x)2+ksin−1x On differentiation this equation with respect to x and we get, dydx=ddx[(sin−1x)2+ksin−1x] dydx=2sin−1xddx(sin−1x)+kddx(sin−1x) dydx=2sin−1x×1√1−x2+k×1√1−x2 dydx=2sin−1x+k√1−x2 Again differentiation and we get, d2ydx2=√1−x2ddx(2sin−1x+k)−(2sin−1x+k)ddx(√1−x2)(√1−x2)2 d2ydx2=√1−x2(2×1√1−x2+0)−(2sin−1x+k)12√1−x2×(0−2x)1−x2 d2ydx2=2+x(2sin−1x+k)√1−x21−x2 d2ydx2=2√1−x2+x(2sin−1x+k)(1−x2)√1−x2 Then, (1−x2)d2ydx2−xdydx =(1−x2)2√1−x2+x(2sin−1x+k)(1−x2)√1−x2−x(2sin−1x+k√1−x2) =2√1−x2+x(2sin−1x+k)√1−x2−x(2sin−1x+k)√1−x2 =2√1−x2+x(2sin−1x+k)−x(2sin−1x+k)√1−x2 =2√1−x2√1−x2 =2 Hence, this is the answer.

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