If the tangent to the curve y=1-x^2 at x= (0 1) meets the axes at P and Q . Also varies, the minimum value of the area of the triangle OPQ is k times the area bounded by the axes and the part of the curve for which 0 x 1 , then k is

Step-by-step solution

Area under curve and axis = ₀^1 (1- x ^2 ) dx = 2 3 at point ( , 1- ^2 ) tangent is [y- (1- ^2 ) ]=-2 (x- ) 2 x+y= ^2+1 P ( ^2+1 2 , 0 ), Q (0, ^2+1 ) Hence, Q (2,1+4 x₁-x₁^2 ) A= [ (1+2 x₁-x₁^2 )+ (1+4 x₁-x₁^2 ) ]=1-3 x₁-x₁^2 dA dx =-3-2 x ₁ x ₁=- 3 2