Let a,b,c be the sides of a triangle. Now two of them are equal to `lamda epsilon R`. If the roots of the equation `x^(2)+2(a+b+c)x+3lamda(ab+bc+ca)=0` are real then

Options

A. `lambda lt (4)/(3)`
B. `lambda gt (5)/(3)`
C. `lambda in ((1)/(3), (5)/(3))`
D. `lambda in ((4)/(3), (5)/(3))`

Answer

A

Explanation

The equation `x^(2) - 2(a+b+c) x + 3 lambda (ab+bc+ca)=0` has real roots.
`therefore" "4(a+b+c)^(2)-12 lambda (ab + bc + ca) ge 0`
`rArr" "lambda le ((a+b+c)^(2))/(3(ab+bc+ca))rArr lambda le (a^(2)+b^(2)+c^(2))/(3(ab+bc+ca))+(2)/(3)`
Since a, b, c denote the sides of a triangle. Therefore,
`|a-b|lt c rArr a^(2) + b^(2) - 2ab lt c^(2)`
`|b-c|lt a rArr b^(2)+c^(2) - 2bc lt a^(2)`
and, `|c-a| lt b rArr c^(2) + a^(2) - 2ca lt b^(2)`
Adding these, we get `a^(2) + b^(2) + c^(2) lt 2ab + 2bc + 2ca`
`rArr" "(a^(2)+b^(2)+c^(2))/(ab+bc+ca)lt 2 rArr (a^(2)+b^(2)+c^(2))/((ab+bc+ca))lt (2)/(3)`
`therefore" "lambda lt (2)/(3)+(2)/(3)=(4)/(3)`