    Question

# The curve passes through (1,1) satisfies the differential equation dydx+√(1−1x2)(1−1y2)=0 If it passes through (√2,k) then the possible value(s) of [k] is/are (where [.] denote the greatest integer function)

A
1
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B
1
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C
2
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D
2
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Solution

## The correct option is D −2dydx+√(1−1x2)(1−1y2)=0 ⇒dy√(1−1y2)=−√(1−1x2)dx Let y=secθ and x=secϕ dy=secθtanθ dθ, dx=secϕtanϕ dϕ ⇒sec2θ dθ=−(sec2ϕ−1)dϕ Integrating both the sides tanθ=−tanϕ+ϕ+C⇒√y2−1=−√x2−1+sec−1x+C It passes through (1,1) so, ⇒C=0 Putting (√2,k) ⇒√k2−1=−1+π4⇒k2=(π4−1)2+1⇒1<k2<2 [∵0<(π4−1)2<1]⇒[k]=1,−2  Suggest Corrections  0      Similar questions  Related Videos   Geometrical Interpretation of a Derivative
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