goformit100
  • goformit100
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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goformit100
  • goformit100
@zzr0ck3r
DLS
  • DLS
Differentiate it
anonymous
  • anonymous
how can a cubic have two real roots? i thought it is either all real or 1real + 1 conjugate complex?

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More answers

DLS
  • DLS
\[\Huge 6x^2+3 \] 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.. |dw:1371278556282:dw| It can't cut twice,so it can never have 2 values..so option C :)
DLS
  • DLS
@yrelhan4 ?check it please
goformit100
  • goformit100
ok
DLS
  • DLS
do u know the answer?
goformit100
  • goformit100
@Math2400 Madam can you suggest some different way ?
goformit100
  • goformit100
@dan815
DLS
  • DLS
isn't it a question from JEE main?
goformit100
  • goformit100
How ?
DLS
  • DLS
nope it is correct,and my solution is perfect
anonymous
  • anonymous
how can a cubic expression have 2 real roots? its either 3 real, or 1 real and 2 complex.
DLS
  • DLS
yo so C
goformit100
  • goformit100
You all confusing me, request yo'll to give a definite solution
DLS
  • DLS
@goformit100 http://www.youtube.com/watch?v=P-RlgYTeI-8
yrelhan4
  • yrelhan4
sahi hai.
goformit100
  • goformit100
What is @yrelhan4 Saying ?
DLS
  • DLS
@goformit100 he is saying my solution is perfect
goformit100
  • goformit100
ok
goformit100
  • goformit100
What ?
goformit100
  • goformit100
|dw:1371292050582:dw|
DLS
  • DLS
do u understand hindi?
goformit100
  • goformit100
Thankyou
Nurali
  • Nurali
i think does not exist
whpalmer4
  • whpalmer4
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].
goformit100
  • goformit100
Thankyou

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