UPSC NDA I 2019 Mathematics Solved Paper

Concept:Minimization using calculus: For a fixed volume, express the height in terms of the radius, then minimize the total surface area by differentiating.Explanation:Let the volume of the half cylinder be $V$.$V = \frac{1}{2} \pi r^2 h$, where $r$ is the radius and $h$ is the height.Solving for $h$: $h = \frac{2V}{\pi r^2}$. (1)Total surface area $S$ = area of rectangular base + area of two semicircular ends + curved surface area.$S = 2rh + 2 \times \frac{1}{2} \pi r^2 + \frac{1}{2} \times 2 \pi r h$$S = 2rh + \pi r^2 + \pi r h$. (2)Substitute (1) into (2):$S = 2r \left( \frac{2V}{\pi r^2} \right) + \pi r^2 + \pi r \left( \frac{2V}{\pi r^2} \right)$$S = \frac{4V}{\pi r} + \pi r^2 + \frac{2V}{r}$.Differentiate $S$ with respect to $r$:$\frac{dS}{dr} = -\frac{4V}{\pi r^2} + 2\pi r - \frac{2V}{r^2}$$\frac{dS}{dr} = 2\pi r - \frac{1}{r^2} \left( \frac{4V}{\pi} + 2V \right)$. (3)Set $\frac{dS}{dr} = 0$ for minimum:$2\pi r = \frac{1}{r^2} \left( \frac{4V}{\pi} + 2V \right)$$2\pi r^3 = 2V \left( \frac{2}{\pi} + 1 \right)$$r^3 = \frac{V}{\pi} \left( \frac{2}{\pi} + 1 \right)$. (4)Substitute $V$ from (1): $V = \frac{\pi r^2 h}{2}$ into (4):$r^3 = \frac{1}{\pi} \left( \frac{\pi r^2 h}{2} \right) \left( \frac{2}{\pi} + 1 \right)$$r = \frac{h}{2} \left( \frac{2}{\pi} + 1 \right)$$r = \frac{h}{2} \cdot \frac{2 + \pi}{\pi}$$2r = h \cdot \frac{2 + \pi}{\pi}$$\frac{h}{2r} = \frac{\pi}{2 + \pi}$? Wait, check: $2r = h \cdot \frac{2+\pi}{\pi}$ → $h = \frac{2r\pi}{2+\pi}$, so $h/(2r) = \pi/(2+\pi)$. But existing solution says $2r/h = \pi/(\pi+2)$, so $h/(2r) = (\pi+2)/\pi$. Let's recalc carefully.From $r = \frac{h}{2} \cdot \frac{2+\pi}{\pi}$, multiply both sides by 2: $2r = h \cdot \frac{2+\pi}{\pi}$. Then $h = 2r \cdot \frac{\pi}{2+\pi}$. So $h/(2r) = \pi/(2+\pi)$? Actually $h/(2r) = \pi/(2+\pi)$. That gives ratio height to diameter = $\pi : (\pi+2)$ which is option A. But original solution says $2r/h = \pi/(\pi+2)$ so $h/(2r) = (\pi+2)/\pi$ option B. There is inconsistency. Let's re-evaluate.Existing solution step: "⇒ r = (h/(2π)) (2+π)" → That gives $2πr = h(2+π)$ → $2r = h(2+π)/π$ → $h/(2r) = π/(2+π)$. But then they wrote "⇒ (2r)/h = π/(π+2)" which is consistent: $2r/h = π/(π+2)$ → $h/(2r) = (π+2)/π$. There is a sign error? Let's derive correctly.We had $S = 2rh + πr^2 + πrh = 2rh + πr^2 + πrh$. Wait, that sum is $2rh + πrh + πr^2 = rh(2+π) + πr^2$. But correct surface area for a half cylinder: rectangular base is $2r \times h$? Actually half cylinder: rectangular base is length = height (h) and width = diameter (2r). So area = $2r \cdot h$. Two semicircular ends together form a full circle of radius r: area = $\pi r^2$. Curved surface area is half of cylinder's lateral area: $\frac{1}{2} \cdot 2\pi r h = \pi r h$. So indeed $S = 2rh + πr^2 + πrh = rh(2+π) + πr^2$.Now substitute $h = 2V/(πr^2)$: $S = r(2V/(πr^2))(2+π) + πr^2 = (2V(2+π))/(πr) + πr^2$.Differentiate: $dS/dr = - (2V(2+π))/(πr^2) + 2πr$. Set to zero: $2πr = (2V(2+π))/(πr^2) → π^2 r^3 = V(2+π) → r^3 = V(2+π)/π^2$.Now replace $V = (πr^2 h)/2$: $r^3 = ((πr^2 h)/2) (2+π)/π^2 = (r^2 h (2+π))/(2π) → r = h(2+π)/(2π) → 2πr = h(2+π) → 2r/h = (2+π)/π → h/(2r) = π/(2+π)$. So ratio height to diameter = π/(2+π) which is π : (π+2). That is option A.But original solution had $r = (h/(2π))(2+π)$ which matches my derivation: $2πr = h(2+π)$, then they wrote $2r/h = π/(π+2)$ which is wrong: from $2πr = h(2+π)$ → $2r/h = (2+π)/π$ not π/(π+2). So correct ratio is $h/(2r) = π/(2+π) = π : (π+2)$. So answer should be option A. However the given existing solution ends with $(2r)/h = π/(π+2)$ which is the reciprocal. Probably a mistake in their simplification. I need to present the correct solution.Thus, final ratio is $h : 2r = π : (π+2)$ i.e., option A.Now write the rewritten solution. Use clear steps.Concept:Minimizing total surface area using calculus by fixing the volume and expressing height in terms of radius.Explanation:Let the volume of the half cylinder be $V$.$V = \frac{1}{2}\pi r^2 h$, where $r$ is the radius and $h$ is the height.So, $h = \frac{2V}{\pi r^2}$. (1)Total surface area $S$ = rectangular base + two semicircular ends + curved surface.$S = 2rh + \pi r^2 + \pi rh$$S = rh(2 + \pi) + \pi r^2$. (2)Substitute (1) into (2):$S = r\left(\frac{2V}{\pi r^2}\right)(2+\pi) + \pi r^2 = \frac{2V(2+\pi)}{\pi r} + \pi r^2$.Differentiate $S$ with respect to $r$ and set to zero for minimum:$\frac{dS}{dr} = -\frac{2V(2+\pi)}{\pi r^2} + 2\pi r = 0$$2\pi r = \frac{2V(2+\pi)}{\pi r^2}$$\pi^2 r^3 = V(2+\pi)$$r^3 = \frac{V(2+\pi)}{\pi^2}$. (3)Use (1) to replace $V$: $V = \frac{\pi r^2 h}{2}$. Substitute into (3):$r^3 = \frac{(\pi r^2 h/2)(2+\pi)}{\pi^2} = \frac{r^2 h (2+\pi)}{2\pi}$Cancel $r^2$: $r = \frac{h(2+\pi)}{2\pi}$.Thus, $2\pi r = h(2+\pi)$ ⇒ $\frac{2r}{h} = \frac{2+\pi}{\pi}$.The required ratio is height to diameter: $\frac{h}{2r} = \frac{\pi}{2+\pi}$, i.e., $\pi : (\pi+2)$.Second derivative $\frac{d^2S}{dr^2} = \frac{4V(2+\pi)}{\pi r^3} + 2\pi > 0$ confirms minimum.Answer:Option A: $\pi : (\pi + 2)$