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

Let y=f(x) be defined parametrically as y=t2+t|t|, x=2t|t|,tR and f(x)=k has at least one real solution, then

A
kR
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
kR{0}
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
kR+{0}
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
kR
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C kR+{0}
y=t2+t|t|,x=2t|t|
Case 1: when t<0
x=3t and y=0, x<0

Case 2: when t0
x=t and y=2t2, x0
f(x)={2x2; x00; x<0

Since, f(x) is polynomial and continuous at x=0, so continuous in R.
Range of f(x) is [0,)
From I.V.T. : we can say f(x)=k has at least one solution if k[0,)

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
Join BYJU'S Learning Program
CrossIcon