dhirendra
Last Activity: 7 Years ago
x=1 , x=3 and x=–1 are the points where modulus changes its sign.
therefore,
i] when x
we get , 1 – x + 2x – 6 – 3 – 3x = 10
x = – 9 (on solving)
x = – 9 lies in the above chosen interval thus can be accepted.
ii] when –1
we get , –x – 1 + 2x – 6 + 3x + 3 = 10
x = 7/2 (on solving)
x = 7/2 does not lie in the chosen interval thus is rejected.
iii] when 1
we get , x – 1 + 2x – 6 + 3x + 3 = 10
x = 7/3 (on solving)
x = 7/3 lies in the chosen interval thus can be accepted.
iv] when x>3
we get , x – 1 – 2x + 6 + 3x + 3 = 10
x = 1 (on solving)
x = 1 does not lie in the interval thus rejected.
the acceptable values of x are the answer, ie; x = –9 and x = 7/3
