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

Number of values of x which satisfies the relation |x−1| + |x−2| =1


A

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

0

No worries! We‘ve got your back. Try BYJU‘S free classes today!
D

Infinite

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is D

Infinite


We will first remove the modulus and split the function.

Let f(x) = |x1| +|x2|, it attains different values in different intervals.

x 2

if x 2,|x1| = x-1 and |x2|= x - 2

f(x) = x -1 + x - 2

= 2x - 3
2x-3 = 1 gives x=2.............................(1)

1 x 2


f(x) = x-1+2-x =1

So for all the values of x between 1 and 2 the equation is satisfied.

There are infinite values of x between 1 and 2..............................(2)

For x less than or equal to 1

f(x) = 1-x +2-x
= 3-2x
3-2x = 1 gives x = 1....................(3)

From 1 ,2 and 3 we can conclude that there are infinite values of x which satisfy the equation


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