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`min[(x_1-x^2)^2+(3+sqrt(1-x1 2)-sqrt(4x_2))],AAx_1,x_2 in R ,`is`4sqrt(5)+1`(b) `3-2sqrt(2)``sqrt(5)+1`(d) `sqrt(5)-1`A. `4sqrt(1)`B. `3-2sqrt(2)`C. `sqrt(5)+1`D. `sqrt(5)-1` |
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Answer» Correct Answer - B (2) Let `y_(1)=3+sqrt(1-x_(1)^(2))andy_(2)=sqrt(4x_(2))` `orx_(1)^(2)+(y_(1)-3)^(2)=1andy_(2)^(2)=4x_(2)` Thus, `(x_(1),y_(1))` lies on the circle `x^(2)+(y-3)^(2)=1`. Also, `(x_(2),y_(2))` lies on the upper half of the parabola `y^(2)=4x`. Thus, the given expression is square of the shortest distance between the curves `x^(2)+(y-3)^(2)=1andy^(2)=4x`. Now, the shortest distance always occurs along the common normal to the curves and the normal to the circle passes through the center of the circle. Normal to the parabola `y^(2)=4x" is "y-mx-2m-m^(3)`. It passes through (0,3). Therefore, `m^(3)+2m+3=0`, which has only one root, m=-1. Hence, the required minimum distance is `sqrt(1+1)-1=sqrt(2)-1`. |
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