If tangents are drawn to the curve y=x1+x2 such that the tangents are concurrent at (1,12), then
A
Number of such tangents is 3.
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B
Each of the tangents has exactly two points in common with the curve.
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C
Abscissa of point of contact for one of the tangents is 1+√2.
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D
Abscissa of point of contact for one of the tangents is 1−√2.
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Solution
The correct options are A Number of such tangents is 3. C Abscissa of point of contact for one of the tangents is 1+√2. D Abscissa of point of contact for one of the tangents is 1−√2. y=x1+x2 dydx=1−x2(1+x2)2 Let point of contact be P(a,b). A(1,12) and b=a1+a2 Then slope of PA= slope of tangent at P ⇒b−12a−1=1−a2(1+a2)2 ⇒(a1+a2−12)(1+a2)2=(a−1)(1−a2) ⇒(a−1)2(a2−2a−1)=0 ⇒a2−2a−1=0 or (a−1)2=0 ⇒a=1±√2 or a=1 Thus, there are three possible tangents.