1. Problem Q.3(a): Find $\frac{\partial y}{\partial x}$ if $y = x^3 \ln \sqrt{x}$. 2. Use the property $\ln \sqrt{x} = \frac{1}{2} \ln x$ to rewrite $y = x^3 \cdot \frac{1}{2} \ln x = \frac{1}{2} x^3 \ln x$. 3. Differentiate using the product rule: $\frac{d}{dx} (x^3 \ln x) = 3x^2 \ln x + x^3 \cdot \frac{1}{x} = 3x^2 \ln x + x^2$. 4. Therefore, $\frac{\partial y}{\partial x} = \frac{1}{2} (3x^2 \ln x + x^2) = \frac{3}{2} x^2 \ln x + \frac{1}{2} x^2$. --- 5. Problem Q.3(b): Find the angle between curves $y = x^3 - 2x + 1$ and $y = x^2 + 1$ at point $(2,5)$. 6. Find derivatives: $y_1' = 3x^2 - 2$, $y_2' = 2x$. 7. At $x=2$, $y_1' = 3(4) - 2 = 12 - 2 = 10$, $y_2' = 4$. 8. The angle $\theta$ between curves is given by $\tan \theta = \left| \frac{m_1 - m_2}{1 + m_1 m_2} \right|$ where $m_1, m_2$ are slopes. 9. Substitute: $\tan \theta = \left| \frac{10 - 4}{1 + 10 \times 4} \right| = \frac{6}{41}$. 10. So, $\theta = \arctan \frac{6}{41}$ radians. --- 11. Problem Q.4(a): Evaluate $\int \frac{x^2 + 16}{x} dx$. 12. Simplify integrand: $\frac{x^2}{x} + \frac{16}{x} = x + \frac{16}{x}$. 13. Integrate term-wise: $\int x dx + \int \frac{16}{x} dx = \frac{x^2}{2} + 16 \ln |x| + C$. --- 14. Problem Q.4(b): Given points $A(-1,3)$, $B(2,1)$, $C(5,-1)$, show $|AB| + |BC| = |AC|$. 15. Calculate distances: $|AB| = \sqrt{(2+1)^2 + (1-3)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{9 + 4} = \sqrt{13}$. $|BC| = \sqrt{(5-2)^2 + (-1-1)^2} = \sqrt{3^2 + (-2)^2} = \sqrt{13}$. $|AC| = \sqrt{(5+1)^2 + (-1-3)^2} = \sqrt{6^2 + (-4)^2} = \sqrt{36 + 16} = \sqrt{52} = 2\sqrt{13}$. 16. Since $|AB| + |BC| = \sqrt{13} + \sqrt{13} = 2\sqrt{13} = |AC|$, the relation holds. --- 17. Problem Q.5(a): Find $c$ so line $y = -x + c$ touches circle $x^2 + y^2 = 16$. 18. Substitute $y = -x + c$ into circle: $x^2 + (-x + c)^2 = 16$. 19. Expand: $x^2 + x^2 - 2cx + c^2 = 16 \Rightarrow 2x^2 - 2cx + c^2 - 16 = 0$. 20. For tangency, discriminant $D=0$: $D = (-2c)^2 - 4 \times 2 \times (c^2 -16) = 4c^2 - 8(c^2 -16) = 4c^2 - 8c^2 + 128 = -4c^2 + 128$. 21. Set $D=0$: $-4c^2 + 128 = 0 \Rightarrow 4c^2 = 128 \Rightarrow c^2 = 32 \Rightarrow c = \pm 4\sqrt{2}$. --- 22. Problem Q.5(b): Find parabola equation with focus $F(4,0)$ and directrix $x = -4$. 23. Parabola definition: points equidistant from focus and directrix. 24. Let $(x,y)$ be a point on parabola. 25. Distance to focus: $\sqrt{(x-4)^2 + y^2}$. 26. Distance to directrix: $|x + 4|$. 27. Set equal: $\sqrt{(x-4)^2 + y^2} = |x + 4|$. 28. Square both sides: $(x-4)^2 + y^2 = (x+4)^2$. 29. Expand: $x^2 - 8x + 16 + y^2 = x^2 + 8x + 16$. 30. Simplify: $y^2 = 16x$. --- 31. Problem Q.6(a): Find tangent and normal at $(3, \frac{12}{5})$ to ellipse $\frac{x^2}{25} + \frac{y^2}{9} = 1$. 32. Differentiate implicitly: $\frac{2x}{25} + \frac{2y}{9} \frac{dy}{dx} = 0 \Rightarrow \frac{dy}{dx} = - \frac{9x}{25y}$. 33. At $(3, \frac{12}{5})$, slope $m = - \frac{9 \times 3}{25 \times \frac{12}{5}} = - \frac{27}{25 \times \frac{12}{5}} = - \frac{27}{60} = - \frac{9}{20}$. 34. Equation of tangent: $y - \frac{12}{5} = - \frac{9}{20} (x - 3)$. 35. Equation of normal slope $m_n = \frac{20}{9}$: $y - \frac{12}{5} = \frac{20}{9} (x - 3)$. --- 36. Problem Q.6(b): Approximate $\int_0^1 e^{2x} dx$ using Simpson's rule with $n=4$. 37. Step size $h = \frac{1-0}{4} = 0.25$. 38. Points: $x_0=0$, $x_1=0.25$, $x_2=0.5$, $x_3=0.75$, $x_4=1$. 39. Evaluate $f(x) = e^{2x}$: $f_0 = e^0 = 1$, $f_1 = e^{0.5} \approx 1.6487$, $f_2 = e^{1} \approx 2.7183$, $f_3 = e^{1.5} \approx 4.4817$, $f_4 = e^{2} \approx 7.3891$. 40. Simpson's rule: $\int_0^1 f(x) dx \approx \frac{h}{3} [f_0 + 4f_1 + 2f_2 + 4f_3 + f_4]$ $= \frac{0.25}{3} [1 + 4(1.6487) + 2(2.7183) + 4(4.4817) + 7.3891]$ $= \frac{0.25}{3} [1 + 6.5948 + 5.4366 + 17.9268 + 7.3891]$ $= \frac{0.25}{3} [38.3473] = 0.0833 \times 38.3473 \approx 3.1956$. --- Final answers: Q.3(a): $\frac{\partial y}{\partial x} = \frac{3}{2} x^2 \ln x + \frac{1}{2} x^2$. Q.3(b): Angle $\theta = \arctan \frac{6}{41}$ radians. Q.4(a): $\int \frac{x^2 + 16}{x} dx = \frac{x^2}{2} + 16 \ln |x| + C$. Q.4(b): $|AB| + |BC| = |AC|$ holds true. Q.5(a): $c = \pm 4 \sqrt{2}$. Q.5(b): Parabola equation: $y^2 = 16x$. Q.6(a): Tangent: $y - \frac{12}{5} = - \frac{9}{20} (x - 3)$, Normal: $y - \frac{12}{5} = \frac{20}{9} (x - 3)$. Q.6(b): Approximate integral $\approx 3.1956$.