The values of 'a' for which the roots of the equation x2+x+a=0 are real and exceed ‘a’ are
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
0
b
a<1/4
c
a<−2
d
−2
answer is C.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Let f(x)=x2+x+a Both the roots of f(x)=0 will exceed a, if(i) Discriminant > 0(ii) a lies outside the roots i.e. f(a)>0(iii) a0 and a<−1/2⇒ a<−1/2 and a2+2a>0⇒ a<−1/2 and a(a+2)>0⇒ a<−1/2 and a+2<0 [∵a<0]⇒a<−1/2 and a<−2⇒a<−2