Page 6 - Real Number System_Neat

∴ The required solution is ∈ (−∞, 0) ∪ (2, ∞)

2
f) 6 + 5 − ≥ 0

Solution:

Here,

2
6 + 5 − ≥ 0
The corresponding equation of given inequality is

2
6 + 5 − = 0
2
or, 6 + 6 − − = 0

or, 6(1 + ) − (1 + ) = 0

or, (1 + )(6 − ) = 0

∴ = −1, 6 − ∞ -1 6 ∞

These two points divide the real line into 3 sub-intervals as shown

in figure.

Now,

Sign of
Interval
1 + 6 − (1 + )(6 − )
(-∞, -1) − + −
(-1, 6) + + +

(6, ∞) + − −

2
Also, for = −1 6, 6 + 5 − = 0.

∴ The required solution is ∈ [−1, 6]