Question

# Find the co-ordinates of the point (s) on the curve y=x2−1x2+1,x>0 such that tangent at these point (s)have the greatest slope.

A
(13,12).
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B
(13,12).
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C
(13,45).
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D
(3,12).
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Solution

## The correct option is B (1√3,−12).y=x2−1x2+1=x2+1−2x2+1=1−2x2+1⇒y=1−2x2+1S=slope =dydx=4x(x2+1)2For maximum and minimum of S, dSdx=0dSdx=4(x2+1)2.1−x.2(x2+1).2x(x2+1)4dSdx=4x2+1−4x2(x2+1)3or dSdx=41−3x2(1+x2)3=0∴x=1√3,−1√3Now consider change of sign at x=1√3,dSdx changes from +ve to -ve.Hence S has maximum at x=1√3. When x=1√3 the value of y=−12Hence the point is (1√3,−1√2).

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