If x is real, then show that
1/(3x+1)+1/(X+1)-1/(3x+1)(X+1)
does not lie between 1 and 4
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Answer:
Let y=
3x+1
1
+
x+1
1
−
(3x+1)(x+1)
1
⇒y=
(3x+1)(x+1)
(x+1)+(3x+1)
−
(3x+1)(x+1)
1
⇒y=
(3x+1)(x+1)
(4x+2)−1
=
(3x+1)(x+1)
4x+1
=
3x
2
+4x+1
4x+1
⇒y(3x
2
+4x+1)=4x+1
⇒(3y)x
2
+4(y−1)x+(y−1)=0
No since x is real,
So the discriminant of above quadratic equation should be no-negative
⇒16(y−1)
2
−12y(y−1)≥0
⇒4(y−1)
2
−3y(y−1)≥0
⇒(y−1)[4(y−1)−3y]≥0
⇒(y−1)(y−4)≥0
⇒y∈(−∞,1]∪[4,∞)
That means range of given expression does not lie between 1 and 4
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