Questions
a
) one real root (2) (3) (4)
b
two real roots
c
three real roots
d
four real roots
detailed solution
Correct option is A
Clearly, x = 1 satisfies the given equation. Assume that f(x)= ex−1 + x−2 = 0 has a real root α other than x = 1. We may suppose that α>1 (the case α<1 is exactly similar). Applying Rolle's theorem on [1,α] (if α<1 apply the theorem on [α,1], we get β∈(1,α) such that f′(β)=0. But f′(β)=eβ−1+1, so that eβ−1=−1, which is not possible. Hence there is noreal root other than 1.Talk to our academic expert!
Similar Questions
If are real numbers such that , then the quadratic equation has
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