wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

The real values of k for which the equation 4x3+3x26x+k=0 has two distinct real roots in [0,1] lie in the inteval ?

Open in App
Solution

f(x)=4x3+3x26x+k
f(x)=12x2+6x6=0
6(2x2+x1)=0
2x2+x1=0
2x2+2xx1=0
2x(x+1)1(x+1)=0
(x+1)(2x1)=0
x=1,12
Thus,x=12 lies in the interval [0,1]
Put x=12 in 4x3+3x26x+k=0
4×(12)3+3×(12)26×12+k=0
4×18+3×146×12+k=0
12+343+k=0
24+34124+k=0
2+3124+k=0
74+k=0
k=74

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Composite Functions
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon