Question

prove that [1÷(3x+1)]+[1÷(x+1)]-[1÷(3x+1)(x+1)] does not lie between 1and4 if x is real?

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Answer 1

Answer:

here is your answer

Step-by-step explanation:

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