Curve y = $\sqrt{3x-2}$ $\frac{dy}{dx}$ = $\frac{1}{2}(3x-2)^{-1/2} \times 3$ ⇒ $\frac{dy}{dx}$ = $\frac{3}{2\sqrt{3x-2}}$...(1) Since, the tangent is parallel to the line $4x - 2y$ = - 5 Therefore, slope of tangent can be obtained from equation y = $\frac{4x}{2} + \frac{5}{2}$ Slope = 2 ⇒ $\frac{dy}{dx}$ = 2 Comparing equations (1) and (2), we have, $\frac{3}{2} \times \frac{1}{\sqrt{3x-2}}$ = 2 $\frac{1}{\sqrt{3x-2}}$ = $\frac{4}{3}$ ⇒ $\frac{1}{3x-2}$ = $\frac{16}{9}$ ⇒ 9 = 48x - 32 ⇒ x = $\frac{41}{48}$ We have y = $\sqrt{3x-2}$ Thus, substituting the value of x in the above equation, y = $\sqrt{3 \times \frac{41}{48} - 2}$ ⇒ y = $\sqrt{\frac{41}{16} - 2}$ ⇒ y = $\sqrt{\frac{41-32}{16}}$ ⇒ y = $\sqrt{\frac{9}{16}}$ ⇒ y = $\frac{3}{4}$ Equation of tangent is $\left(y - \frac{3}{4}\right)$ = 2$\left(x - \frac{41}{48}\right)$ ⇒ $\left(y - \frac{3}{4}\right)$ = $2x - \frac{41}{24}$ ⇒ y = $2x - \frac{41}{24} + \frac{3}{4}$ ⇒ y = $2x - \frac{41}{24} + \frac{18}{24}$ ⇒ y = $2x - \frac{23}{24}$ ⇒ $24y = 48x - 23$ ⇒ $48x - 24y - 23 + 0$ OR $f(x)$ = $x^3 + \frac{1}{x^3}, x \neq 0$ ⇒ $f'(x) = 3x^2 - 3x^{-4} = 3\left(x^2 - \frac{1}{x^4}\right)$ ⇒ $f'(x) = 3x^2 - 3x^4 = \frac{3}{x^4}(x^6 - 1)$ ⇒ $f'(x) = \frac{3}{x^4}(x^2-1)(x^4+x^2+1)$ ⇒ f'(x) = 3 $\left(\frac{x^4+x^2+1}{x^4}\right)(x^2-1)$ (i) For an increasing function, we should have, f ' (x) > 0 ⇒ 3 $\left(\frac{x^4+x^2+1}{x^4}\right)(x^2-1)$ > 0 ⇒ $(x^2-1)$ > 0 [Since 3 $\left(\frac{x^4+x^2+1}{x^4}\right)$ > 0] ⇒ (x - 1) (x + 1) > 0 ⇒ x ∊ (- ∞ , - 1) ∪ x ∊ (1 , ∞) So, f(x) is increasing on (- ∞ , - 1) ∪ (1 , ∞) (ii) For a decreasing function, we should have f’(x) < 0 f ' (x) < 0 ⇒ 3 $\left(\frac{x^4+x^2+1}{x^4}\right)(x^2-1)$ < 0 ⇒ $(x^2-1)$ < 0 [Since 3 $\left(\frac{x^4+x^2+1}{x^4}\right)(x^2-1)$ > 0] ⇒ (x - 1) (x + 1) < 0 ⇒ ∊ (- 1 , 0) ∪ x ∊ (0 , 1) So f(x) is decreasing on (- 1 , 0) ∪ (0 , 1)