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

If x,y,z are non-zero real number and t=|x+y||x|+|y|+|x+z||x|+|z|+|y+z||y|+|z|, then range of t must be

A
0t1
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
0t3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
1t2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
1t3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D 1t3
|x|+|y|=|x+y| if x and y are of same sign, i.e. both 0 or both 0
Therefore,
|x+y|(|x|+|y|)=1 if x and y are of same sign.
From above, it follows that if x,y and z are all of same sign, then the equation is equals to 3.
The other possibility is that two are of same sign and the third with the opposite sign.
Let x and y are of same sign with losing generality.
In that case,
|x+y|(|x|+|y|)=1,
while 0|x+z|(|x|+|z|)<1 and
0|y+z|(|y|+|z|)<1.
Thus established that:
(a) The upper limit is 3 i.e. the value when x,y and z are all of same sign.
(b) The lower limit is at least 1, since the same signed pair yield 1, while any opposite signs pair yield at least 0.
To establish 1 as the lower limit, let x=y=z.
Then, the equation resolves to a value of 1.
Hence, the answer is 1t3.

flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Framing a Linear Equation
MATHEMATICS
Watch in App
Join BYJU'S Learning Program
CrossIcon