Question

Prove that
1÷3x+1+1÷x+1-1÷(3x+1)(x+1)
does not lie between 1 and 4 if x is real.​

Answer 1

we have to prove that 1/(3x + 1) + 1/(x + 1) - 1/(3x + 1)(x + 1) doesn't lie between 1 and 4 if x is real.

solution : 1/(3x + 1) + 1/(x + 1) - 1/(3x + 1)(x + 1)

= {(x + 1) + (3x + 1)}/(3x + 1)(x + 1) - 1/(3x + 1)(x + 1)

= (4x + 2)/(3x + 1)(x + 1) - 1/(3x + 1)(x + 1)

= {4x + 2 - 1}/(3x + 1)(x + 1)

= (4x + 1)/(3x + 1)(x + 1)

now if we assume, given expression lies between 1 and 4.

i.e., 1 ≤ (4x + 1)/(3x + 1)(x + 1) ≤ 4

case 1 : 1 ≤ (4x + 1)/(3x + 1)(x + 1)

⇒(4x + 1)/(3x + 1)(x + 1) - 1 ≥ 0

⇒{(4x + 1) - (3x² + 4x + 1)}/(3x + 1)(x + 1) ≥ 0

⇒-3x²/(3x + 1)(x + 1) ≥ 0

⇒3x²/(3x + 1)(x + 1) ≤ 0

but x² always positive in any real no

so, (3x + 1)(x + 1) < 0.........(1)

case 2 : (4x + 1)/(3x + 1)(x + 1) ≤ 4

⇒(4x + 1)/(3x + 1)(x + 1) - 4 ≤ 0

⇒(4x + 1 - 12x² - 16x - 4)/(3x + 1)(x + 1) ≤ 0

⇒-(12x² + 12x + 3)/(3x + 1)(x + 1) ≤ 0

⇒(4x² + 4x + 1)/(3x + 1)(x + 1) ≥ 0

⇒(2x + 1)²/(3x + 1)(x + 1) ≥ 0

but (2x + 1)² ≥ 0 for all real no

so, (3x + 1)(x + 1) > 0.......(2)

we see equations (1) and (2) both are contradictions to each other.

it means, given expression doesn't lie between 1 to 4, if x is real no.

hence prove.