(x-1/3 x+4)

Question

(x-1/3 x+4)<(x-3/3 x-2) holds, for all x in the interval (A) ((-4/3), (2/3)) (B) (-∞, (5/4)) (C) ((3/3), ∞) (D) (-∞, (-5/4)) ∪((3/4),-∞)

Tardigrade Admin · Accepted Answer

We have, (x-1/3 x+4)-(x-3/3 x-2)<0 ⇒ ((x-1)(3 x-2)-(x-3)(3 x+4)/(3 x+4)(3 x-2))<0 ⇒ ((3 x2-2 x-3 x+2)-(3 x2+4 x-9 x-12)/(3 x+4)(3 x-2))<0 ⇒ (-5 x+2+5 x+12/(3 x+4)(3 x-2))<0 ⇒ (14/(3 x+4)(3 x-2))<0 ⇒ (3 x + 4) (3 x -2)<0 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/mathematics/12bf861801bb0d339c0a21d9521c30e9-.png" /> ⇒ x ∈(-(4/3), (2/3))

Q. $\frac{x - 1}{3 x + 4} < \frac{x - 3}{3 x - 2}$ holds, for all $x$ in the interval

$(\frac{- 4}{3}, \frac{2}{3})$

$(- \infty, \frac{5}{4})$

$(\frac{3}{3}, \infty)$

$(- \infty, \frac{- 5}{4}) \cup (\frac{3}{4}, - \infty)$

Q. 3x+4x−1​<3x−2x−3​ holds, for all x in the interval

(3−4​,32​)

(−∞,45​)

(33​,∞)

(−∞,4−5​)∪(43​,−∞)

We have, 3x+4x−1​−3x−2x−3​<0 ⇒(3x+4)(3x−2)(x−1)(3x−2)−(x−3)(3x+4)​<0 ⇒(3x+4)(3x−2)(3x2−2x−3x+2)−(3x2+4x−9x−12)​<0 ⇒(3x+4)(3x−2)−5x+2+5x+12​<0 ⇒(3x+4)(3x−2)14​<0 ⇒(3x+4)(3x−2)<0 ⇒x∈(−34​,32​)

Q. $\frac{x - 1}{3 x + 4} < \frac{x - 3}{3 x - 2}$ holds, for all $x$ in the interval

$(\frac{- 4}{3}, \frac{2}{3})$

$(- \infty, \frac{5}{4})$

$(\frac{3}{3}, \infty)$

$(- \infty, \frac{- 5}{4}) \cup (\frac{3}{4}, - \infty)$