The real number k for which the equation, 2x^3+3x + k = 0 has two distinct real roots in [0,1]
(1) lies between 2 and 3
(2) lies between −1and 0
(3) does not exist
(4) lies between 1 and 2
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how can a cubic have two real roots? i thought it is either all real or 1real + 1 conjugate complex?
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\[\Huge 6x^2+3 \]
6x^2=>always positive since its a square
3 can't be negative either
so I can say the slope of the graph is Continuoulsy increasing
therefore it will cut the x-axis only ONCE..
It can't cut twice,so it can never have 2 values..so option C :)
@yrelhan4 ?check it please
do u know the answer?
@Math2400 Madam can you suggest some different way ?
isn't it a question from JEE main?
nope it is correct,and my solution is perfect
how can a cubic expression have 2 real roots? its either 3 real, or 1 real and 2 complex.
yo so C
You all confusing me, request yo'll to give a definite solution
Here's a simple argument based on Descartes' Rule of Signs.
Let us suppose there is a number \(k\) such that \(P(x)=2x^3+3x + k = 0\) has two distinct real roots in \([0,1]\). For this to be true in a real polynomial of degree 3, we must have at least two positive real roots. For that to be the case, the Rule of Signs requires that we have 2 or more sign changes in P(x). All of the terms with the possible exception of \(k\) are positive, so we can have at most 1 sign change. Therefore, it is not possible to have more than 1 positive real root, and we thus cannot have 2 positive real roots in [0,1].