Saved Bookmarks
| 1. |
Let `x_(i) epsilonR,i=1,2,3……….n` are numbers such that `sum_(i=1)^(n)isqrt(x_(i)-i^(2))=(sum_(i=1)^(n)x_(i))/2` and `x_(1)+x_(2)+……….+x_(n)=280` Probability that a randomly selected triangle formed by vertices of a `2n+1` sided regular polygon is isosceles isA. `3/13`B. `5/13`C. `7/13`D. `9/13` |
|
Answer» Correct Answer - A `sum_(i=1)^(n-1)(x_(i)-2isqrt(x_(i)-i^(2)))=0` `sum_(i=1)^(n-1)(sqrt(x_(i)-i^(2)))^(2)-2isqrt(x_(i)-i^(2))+i^(2)=0` `sum_(i=1)^(n-1)(sqrt(x_(i)-i^(2))-i)^(2)=0` so, `x_(i)=2i^(2)` Now, `x_(1)^(2)+….+x_(n)^(2)=280` `2[1^(2)+2^(2)+........n^(2)]=280` `n=7` `y_(1)+y_(2)+y_(3)=7` `y_(1)^(1)+y_(2)^(1)+y_(3)^(1)=4` `.^(4+3-1)C_(3)=.^(6)C_(3)=20` Total triangles formed `=.^(15)C_(3)=(15xx14xx13)/6` `N` of isosceles triangles formed `=15xx7` probability `=(15xx7)/(15xx14xx13)xx6` `3/13` |
|