The equation of the given curve is y = $x^3$ – 11x + 5 The equation of the tangent to the given curve is given as y = x − 11 (which is of the form y = mx + c). ∴ Slope of the tangent = 1 Now, the slope of the tangent to the given curve at the point (x, y) is given by, $\frac{dy}{dx}$ = $3x^2$ - 11 Then, we have : $3x^2$ - 11 ⇒ $3x^2$ = 12 ⇒ $x^2$ = 4 ⇒ x = ± 2 When x = 2, y = $(2)^3$ − 11 (2) + 5 = 8 − 22 + 5 = −9. When x = −2, y = $(-2)^3$ − 11 (−2) + 5 = −8 + 22 + 5 = 19. Hence, the required points are (2, −9) and (−2, 19). OR Consider y = $\sqrt{x}$ , Let x = 49 andΔx = 0.5 Then, Δy = $\sqrt{x+\Delta x} - \sqrt{x}$ = $\sqrt{49.5} - \sqrt{49}$ = $\sqrt{49.5} - 7$ ⇒ $\sqrt{49.5}$ = 7 + Δy Now, dy is approximately equal to Δy and is given by, dy = $\left(\frac{dy}{dx}\right)$ Δx = $\frac{1}{22\sqrt{x}}$ (05) [Since y = $\sqrt{x}$] = $\frac{1}{2\sqrt{49}}$ (0.5) = $\frac{1}{14}$ (0.5) = 0.035 Hence the approximate value of $\sqrt{49.5}$ is 7 + 0.035 = 7.035