The given equation of the curve is, $16x^2 + 9y^2$ = 145 ... (i) Let $(x_1,y_1)$ lies on the curve ⇒ $16x_1^2 + 9y_1^2$ = 145 ⇒ 16 × $2^2$ + $9y_1^2$ = 145 (Since $x_1$ = 2) ⇒ $9y_1^2$ = 145 - 64 ⇒ $9y_1^2$ = 81 ⇒ $y_1^2$ = 9 ⇒ $y_1$ = 3 ... ($y_1$ > 0) Coordinates of the given point are (2 , 3) $16x^2 + 9y^2$ = 145 Differentiating with respect to x ⇒ 32x + 18y $\frac{dy}{dx}$ = 0 ⇒ $\frac{dy}{dx}$ = $\frac{-16x}{9y}$ ⇒ $\left(\frac{dy}{dx}\right)_{(2,3)}$ = $\frac{-32}{27}$ Equation of tangent at (2 , 3) y - 3 = $\frac{-32}{27}$ (x - 2) ⇒ 32x + 27y - 145 = 0 Equation of normal at (2, 3) y - 3 = $\frac{27}{32}$ (x - 2) ⇒ 27x - 32y + 42 = 0 OR Given that f (x) = $\frac{x^4}{4} - x^3-5x^2$ + 24x + 12 a) Strictly increasing f (x) = $\frac{x^4}{4} - x^3-5x^2$ + 24x + 12 ⇒ f' (x) = $x^3-3x^2$ - 10x + 24 Function is increasing when f ' (x) > 0 ⇒ $x^3 - 3x^2$ - 10x + 24 > 0 ⇒ (x - 4) (x + 3) (x - 2) > 0 ⇒ x = - 3 , 2 , 4 x ∊ (- 3 , 2) ∪ (4 , ∞) b) Strictly decreasing f' (x) = 0 ⇒ $x^3-3x^2$ - 10x + 24 = 0 ⇒ (x - 4) (x + 3) (x - 2) < 0 ⇒ x ∊ (- ∞ , - 3) ∪ (2 , 4)