Applications of Differentiation: past paper questions & worked solutions

$y=2x^2(x-5)=2x^3-10x^2$.

$\dfrac{dy}{dx}=6x^2-20x=2x(3x-10)$.

At a stationary point, $\dfrac{dy}{dx}=0$. Hence $2x(3x-10)=0$, so $x=0$ or $x=\dfrac{10}{3}$.

Since $2x(3x-10)$ is positive outside its two roots and negative between them, $y$ is increasing for $x<0$ and for $x>\dfrac{10}{3}$.

$y=4x^{1/2}+9x^{-1/2}+3$.

$\dfrac{dy}{dx}=2x^{-1/2}-\dfrac92x^{-3/2}=\dfrac{2}{\sqrt{x}}-\dfrac{9}{2x^{3/2}}$.

$\dfrac{dy}{dx}=\dfrac{4x-9}{2x^{3/2}}$. Since $x>0$, the stationary point occurs when $4x-9=0$.

Therefore $x=\dfrac94$.

$\dfrac{d^2y}{dx^2}=-x^{-3/2}+\dfrac{27}{4}x^{-5/2}=-\dfrac1{x^{3/2}}+\dfrac{27}{4x^{5/2}}$.

$\dfrac{d^2y}{dx^2}=\dfrac{27-4x}{4x^{5/2}}$. At $x=\dfrac94$, the numerator is $27-9=18>0$, so the stationary point is a minimum.

For $x>0$, $2x^{3/2}>0$, so $\dfrac{dy}{dx}<0$ when $4x-9<0$. Hence $y$ is decreasing for $0<x<\dfrac94$.

The centre is $(2,5)$ and $P=(8,-3)$. The radius is $\sqrt{(8-2)^2+(-3-5)^2}$.

$r=\sqrt{6^2+(-8)^2}=\sqrt{100}=10$.

With centre $(2,5)$ and radius $10$, the circle is $(x-2)^2+(y-5)^2=100$.

The gradient of the radius from the centre to $P$ is $\dfrac{-3-5}{8-2}=\dfrac{-8}{6}=-\dfrac43$.

The tangent is perpendicular to the radius, so its gradient is $\dfrac34$.

The tangent through $(8,-3)$ is $y+3=\dfrac34(x-8)$.

Multiplying by $4$ gives $4y+12=3x-24$, hence $3x-4y-36=0$.

Dividing by $4x-5$ means the remainder is $f\left(\dfrac54\right)$.

$f\left(\dfrac54\right)=\left(2-\dfrac{5k}{4}\right)^5=\dfrac{243}{32}$.

Since $\dfrac{243}{32}=\left(\dfrac32\right)^5$, we have $2-\dfrac{5k}{4}=\dfrac32$.

$\dfrac{5k}{4}=\dfrac12$, so $k=\dfrac25$.

With $k=\dfrac25$, $f(x)=\left(2-\dfrac25x\right)^5$.

$2^5+5(2^4)\left(-\dfrac25x\right)+10(2^3)\left(-\dfrac25x\right)^2=32-32x+\dfrac{64}{5}x^2$.

For a polynomial, the gradient at $x=0$ is the coefficient of $x$ in the expansion. Therefore $f'(0)=-32$.

The $x$ values increase by $0.5$, so the trapezium-rule width is $h=0.5$.

$\displaystyle \int_0^2 f(x)\,dx\approx \dfrac{0.5}{2}\left[3+1.92+2(2.6833+2.4+2.1466)\right]$.

This gives $4.84495$, so the approximation is $4.845$ to 3 decimal places.

$\displaystyle \int_0^2 \left(2-\dfrac14x^2\right)dx=\left[2x-\dfrac{x^3}{12}\right]_0^2=4-\dfrac{2}{3}=\dfrac{10}{3}$.

The shaded region lies between $C_1$ and $C_2$, so its estimated area is $4.845-\dfrac{10}{3}=1.5116\ldots$.

The rectangle has area $2\times3=6$.

Percentage shaded $=\dfrac{1.5116\ldots}{6}\times100=25.19\ldots\%$, so the estimate is $25.2\%$.

$x^2-6x=(x-3)^2-9$ and $y^2+8y=(y+4)^2-16$.

The circle becomes $(x-3)^2+(y+4)^2=25-k$.

The centre is $(3,-4)$. Since the $x$-axis is tangent, the radius is the distance from the centre to the $x$-axis, which is $4$.

Thus $25-k=4^2=16$, so $k=9$.

With $k=9$, the circle is $(x-3)^2+(y+4)^2=16$. On the $x$-axis, $y=0$, giving $x=3$.

On the $y$-axis, $x=0$, so $9+(y+4)^2=16$. Hence $y=-4\pm\sqrt7$.

The vertical base on the $y$-axis has length $2\sqrt7$, and its perpendicular distance to $(3,0)$ is $3$. The area is $\dfrac12(2\sqrt7)(3)=3\sqrt7$.

For $y=x^3-x^2-16x+2$, $\dfrac{dy}{dx}=3x^2-2x-16$.

Stationary points occur when $3x^2-2x-16=0$.

$3x^2-2x-16=(3x-8)(x+2)$, so $x=\dfrac83$ or $x=-2$.

$\dfrac{d^2y}{dx^2}=6x-2$.

At $x=-2$, $6x-2=-14<0$, so the stationary point is a local maximum.

At $x=\dfrac83$, $6x-2=14>0$, so the stationary point is a local minimum.

Substitute $(2,5)$ into $y=ax^3-3x^2+3x+b$: $5=8a-12+6+b$, so $8a+b=11$.

$\dfrac{dy}{dx}=3ax^2-6x+3$.

At $x=2$, the gradient is $7$, so $12a-12+3=7$. Hence $12a=16$ and $a=\dfrac43$.

Using $8a+b=11$, $b=11-8\left(\dfrac43\right)=\dfrac13$.

With $a=\dfrac43$, $\dfrac{dy}{dx}=4x^2-6x+3$.

For $4x^2-6x+3=0$, the discriminant is $(-6)^2-4(4)(3)=36-48=-12<0$.

The derivative is never zero, so the curve has no turning points.

Let $q=2n-1$ and $p=2n+1$, where $n$ is a positive integer. Then $p>q>0$ and the numbers are consecutive odd numbers.

$p^2-q^2=(2n+1)^2-(2n-1)^2$.

$p^2-q^2=(4n^2+4n+1)-(4n^2-4n+1)=8n$.

Since $n$ is an integer, $8n$ is a multiple of $8$, so $p^2-q^2$ is a multiple of $8$.

For $y=x^3+12x^2+49x+2$, $\dfrac{dy}{dx}=3x^2+24x+49$.

$3x^2+24x+49=3(x+4)^2+1$. This is always positive for real $x$.

Since $\dfrac{dy}{dx}$ is never $0$, the curve has no stationary points.

$y=\dfrac{\sqrt{x}(100-x^2)}{40}=\dfrac52x^{1/2}-\dfrac1{40}x^{5/2}$.

$\dfrac{dy}{dx}=\dfrac54x^{-1/2}-\dfrac1{16}x^{3/2}$.

At $P$, $\dfrac{dy}{dx}=0$. Thus $\dfrac54x^{-1/2}=\dfrac1{16}x^{3/2}$. Multiplying by $16x^{1/2}$ gives $20=x^2$.

Since $x\geq0$, $x=\sqrt{20}=2\sqrt5$.

$\int y\,dx=\int\left(\dfrac52x^{1/2}-\dfrac1{40}x^{5/2}\right)dx=\dfrac53x^{3/2}-\dfrac1{140}x^{7/2}$.

The signed area from $0$ to $k$ is zero because the positive area $R_1$ equals the negative area $R_2$ in magnitude.

$\dfrac53k^{3/2}-\dfrac1{140}k^{7/2}=0$. Since $k>0$, divide by $k^{3/2}$ to get $\dfrac53-\dfrac{k^2}{140}=0$, so $k^2=\dfrac{700}{3}$.

$k=\sqrt{\dfrac{700}{3}}=\dfrac{10\sqrt{21}}{3}$.

$y=2x^{3/2}(4-x)=8x^{3/2}-2x^{5/2}$.

$\dfrac{dy}{dx}=12x^{1/2}-5x^{3/2}=\sqrt{x}(12-5x)$.

For the interior stationary point $P$, $12-5x=0$, so $x=\dfrac{12}{5}$.

$R_1$ is the area from $x=0$ to $x=4$ above the $x$-axis. $R_2$ is the area from $x=4$ to $x=k$ below the $x$-axis.

Equal areas mean $\displaystyle \int_0^4 y\,dx=-\int_4^k y\,dx$, so $\displaystyle \int_0^k y\,dx=0$.

$\displaystyle \int 2x^{3/2}(4-x)dx=\int(8x^{3/2}-2x^{5/2})dx=\dfrac{16}{5}x^{5/2}-\dfrac47x^{7/2}$.

$\dfrac{16}{5}k^{5/2}-\dfrac47k^{7/2}=0$.

$4k^{5/2}\left(\dfrac45-\dfrac{k}{7}\right)=0$.

Since $k>4$, $\dfrac45-\dfrac{k}{7}=0$, so $k=\dfrac{28}{5}$.

The tangent $l$ has gradient $\dfrac52$, so the radius $XN$ is perpendicular to it and has gradient $-\dfrac25$.

The line through $X(4,-3)$ with gradient $-\dfrac25$ is $y+3=-\dfrac25(x-4)$.

Multiplying by $5$ gives $5y+15=-2x+8$, so $2x+5y+7=0$.

Use $y=\dfrac52x-\dfrac{55}{2}$ in $2x+5y+7=0$.

$2x+5\left(\dfrac52x-\dfrac{55}{2}\right)+7=0$. Multiplying by $2$ gives $4x+25x-275+14=0$, so $29x=261$ and $x=9$.

$y=\dfrac52(9)-\dfrac{55}{2}=\dfrac{45-55}{2}=-5$, so $N=(9,-5)$.

$XN^2=(9-4)^2+(-5+3)^2=5^2+(-2)^2=29$.

With centre $(4,-3)$ and radius squared $29$, the circle is $(x-4)^2+(y+3)^2=29$.

$H=\dfrac{1}{20}t^{1/2}(20+6t-t^2)+17$.

$H=\dfrac{1}{20}(20t^{1/2}+6t^{3/2}-t^{5/2})+17$.

$\dfrac{dH}{dt}=\dfrac{1}{20}\left(10t^{-1/2}+9t^{1/2}-\dfrac52t^{3/2}\right)$.

$\dfrac{dH}{dt}=\dfrac{-5t^2+18t+20}{40\sqrt{t}}$.

At $t=\alpha$, $\dfrac{dH}{dt}=0$. Since $\alpha>0$, the denominator is not zero, so $-5\alpha^2+18\alpha+20=0$.

Multiplying by $-1$ gives $5\alpha^2-18\alpha-20=0$.

$\alpha=\dfrac{18\pm\sqrt{18^2+4(5)(20)}}{10}=\dfrac{9\pm\sqrt{181}}{5}$.

Only the positive value lies in $0\le t\le7$, so $\alpha=\dfrac{9+\sqrt{181}}{5}=4.491$ to 3 decimal places.

$\dfrac{d^2H}{dt^2}=\dfrac{-15t^2+18t-20}{80t^{3/2}}$.

At $t=\alpha$, $5\alpha^2=18\alpha+20$. Therefore the numerator of $\dfrac{d^2H}{dt^2}$ is $-15\alpha^2+18\alpha-20=-3(18\alpha+20)+18\alpha-20=-36\alpha-80$.

Since $\alpha>0$, $-36\alpha-80<0$, and the denominator $80\alpha^{3/2}$ is positive. Thus $\dfrac{d^2H}{dt^2}<0$, so $H$ is a maximum.

$y=256x^4-304x-35+27x^{-2}$.

$\dfrac{dy}{dx}=1024x^3-304-54x^{-3}=1024x^3-304-\dfrac{54}{x^3}$.

$1024x^3-304-\dfrac{54}{x^3}=0$.

Multiplying by $x^3$ gives $1024x^6-304x^3-54=0$. Let $u=x^3$.

$1024u^2-304u-54=0$, or $512u^2-152u-27=0=(8u+1)(64u-27)$.

Thus $x^3=-\dfrac18$ or $x^3=\dfrac{27}{64}$, giving $x=-\dfrac12$ or $x=\dfrac34$.

Substitution into $y=256x^4-304x-35+\dfrac{27}{x^2}$ gives $y=241$ at $x=-\dfrac12$ and $y=-134$ at $x=\dfrac34$.

The stationary points are $\left(-\dfrac12,241\right)$ and $\left(\dfrac34,-134\right)$.

For $y=27x^{1/2}-x^{3/2}-20$, $\dfrac{dy}{dx}=\dfrac{27}{2}x^{-1/2}-\dfrac32x^{1/2}$.

$\dfrac{27}{2}x^{-1/2}-\dfrac32x^{1/2}=0$.

Multiplying by $2x^{1/2}$ gives $27-3x=0$, so $x=9$.

At $x=9$, $y=27\sqrt9-9^{3/2}-20=81-27-20=34$.

The stationary point is $(9,34)$.

$\dfrac{d^2y}{dx^2}=-\dfrac{27}{4}x^{-3/2}-\dfrac34x^{-1/2}$.

At $x=9$, both terms in $\dfrac{d^2y}{dx^2}$ are negative, so the stationary point is a local maximum.

The gradient of $QP$ is $\dfrac{14-(-30)}{23-15}=\dfrac{44}{8}=\dfrac{11}{2}$.

The gradient of $QR$ is $\dfrac{-26-(-30)}{-7-15}=\dfrac4{-22}=-\dfrac2{11}$.

The product of the two gradients is $-1$, so $QP$ is perpendicular to $QR$ and $\angle PQR=90^\circ$.

Since the angle at $Q$ is a right angle, $PR$ is a diameter of the circle.

The centre is the midpoint of $P(23,14)$ and $R(-7,-26)$, so it is $(8,-6)$.

$PR=\sqrt{30^2+40^2}=50$, so the radius is $25$.

The greatest distance from $Q$ occurs at the point opposite $Q$ on the diameter through the centre. Thus $S=2(8,-6)-(15,-30)=(1,18)$.

The radius from the centre to $S$ has gradient $\dfrac{18-(-6)}{1-8}=-\dfrac{24}{7}$, so the tangent gradient is $\dfrac7{24}$.

$y-18=\dfrac7{24}(x-1)$, which rearranges to $7x-24y+425=0$.

$x^2+6x=(x+3)^2-9$ and $y^2-4y=(y-2)^2-4$.

The circle becomes $(x+3)^2+(y-2)^2=27$. Hence the centre is $(-3,2)$ and the radius is $\sqrt{27}=3\sqrt3$.

A horizontal tangent has equation $y=k$. The possible tangent levels are one radius above and below the centre, so $k=2\pm3\sqrt3$.

The chord has length $4$, so half the chord is $2$. If the chord is $y=p$, its perpendicular distance from the centre is $|p-2|$.

$(p-2)^2+2^2=(3\sqrt3)^2=27$, so $(p-2)^2=23$.

Since $p$ is negative, $p=2-\sqrt{23}$.

$y=\dfrac12x^2+1458x^{-3/2}-74$.

$\dfrac{dy}{dx}=x+1458\left(-\dfrac32\right)x^{-5/2}=x-2187x^{-5/2}$.

At a stationary point, $x-2187x^{-5/2}=0$.

$x=2187x^{-5/2}$, so $x^{7/2}=2187=3^7$.

Raising both sides to the power $\dfrac27$ gives $x=3^2=9$. Since $x>0$, this is the only stationary point.

At $x=9$, $y=\dfrac12(9)^2+\dfrac{1458}{\sqrt{9^3}}-74=\dfrac{81}{2}+54-74=\dfrac{41}{2}$.

The region lies from $x=4$ to $x=9$ between the curve and the horizontal line $y=\dfrac{41}{2}$.

$\displaystyle R=\int_4^9\left(\dfrac12x^2+1458x^{-3/2}-74-\dfrac{41}{2}\right)dx$.

$\displaystyle \int\left(\dfrac12x^2+1458x^{-3/2}-\dfrac{189}{2}\right)dx=\dfrac{x^3}{6}-2916x^{-1/2}-\dfrac{189}{2}x$.

$\displaystyle R=\left[\dfrac{x^3}{6}-2916x^{-1/2}-\dfrac{189}{2}x\right]_4^9=\dfrac{373}{3}$.

Since $(2x+3)$ is a factor, $x=-\dfrac32$ is a root.

$4\left(-\dfrac32\right)^3+p\left(-\dfrac32\right)^2+8\left(-\dfrac32\right)+q=0$.

$-\dfrac{27}{2}+\dfrac94p-12+q=0$, so $\dfrac94p+q=\dfrac{51}{2}$.

The remainder on division by $(x+2)$ is $f(-2)$.

$f(-2)=4(-8)+4p-16+q=4p+q-48$.

$4p+q-48=-5$, so $4p+q=43$.

Subtract $\dfrac94p+q=\dfrac{51}{2}$ from $4p+q=43$: $\dfrac74p=\dfrac{35}{2}$, so $p=10$. Then $4(10)+q=43$, so $q=3$.

With $p=10$, $f(x)=4x^3+10x^2+8x+3$, so $f'(x)=12x^2+20x+8$.

$f'(x)=4(3x+2)(x+1)$.

The quadratic derivative is negative between its roots, so $f(x)$ is decreasing for $-1<x<-\dfrac23$.

$y=\dfrac{9x-x^2}{2\sqrt{x}}=\dfrac12(9x^{1/2}-x^{3/2})$.

$\dfrac{dy}{dx}=\dfrac12\left(\dfrac92x^{-1/2}-\dfrac32x^{1/2}\right)=\dfrac{3(3-x)}{4\sqrt{x}}$.

For $x>0$, the denominator is not zero. So $\dfrac{dy}{dx}=0$ when $3-x=0$.

Therefore the $x$ coordinate of $P$ is $3$.

$y=0$ when $9x-x^2=0$, so $x=9$ for the positive intercept.

The region is under the curve from $x=1$ to $x=9$, so $R=\displaystyle \int_1^9 \dfrac{9x-x^2}{2\sqrt{x}}dx$.

$R=\displaystyle \int_1^9 \dfrac12(9x^{1/2}-x^{3/2})dx=\left[3x^{3/2}-\dfrac15x^{5/2}\right]_1^9$.

$R=\left(3(27)-\dfrac15(243)\right)-\left(3-\dfrac15\right)=\dfrac{148}{5}$.

The cuboid has dimensions $x$, $3x$ and $h$, so its volume is $3x^2h=972$.

Hence $h=\dfrac{972}{3x^2}=\dfrac{324}{x^2}$.

The surface area is $S=2(3x^2+3xh+xh)=6x^2+8xh$.

$S=6x^2+8x\left(\dfrac{324}{x^2}\right)=6x^2+\dfrac{2592}{x}$.

$S=6x^2+2592x^{-1}$, so $\dfrac{dS}{dx}=12x-2592x^{-2}=12x-\dfrac{2592}{x^2}$.

For a stationary value, $12x-\dfrac{2592}{x^2}=0$.

Multiplying by $x^2$ gives $12x^3=2592$, so $x^3=216$ and $x=6$.

$\dfrac{d^2S}{dx^2}=12+\dfrac{5184}{x^3}$.

At $x=6$, $\dfrac{d^2S}{dx^2}=12+\dfrac{5184}{216}=36>0$, so the stationary value is a minimum.

$S=6(6)^2+\dfrac{2592}{6}=216+432=648$, so the minimum surface area is $648\text{ cm}^2$.

$12x^3(x-7)+14x(13x-15)=12x^4-84x^3+182x^2-210x$.

Since $\sqrt{x}=x^{1/2}$, divide each term by $21x^{1/2}$.

$y=\dfrac{12}{21}x^{7/2}-\dfrac{84}{21}x^{5/2}+\dfrac{182}{21}x^{3/2}-\dfrac{210}{21}x^{1/2}=\dfrac47x^{7/2}-4x^{5/2}+\dfrac{26}{3}x^{3/2}-10x^{1/2}$.

$\dfrac{dy}{dx}=2x^{5/2}-10x^{3/2}+13x^{1/2}-5x^{-1/2}$.

$\dfrac{dy}{dx}=x^{-1/2}(2x^3-10x^2+13x-5)$.

At a turning point, $\dfrac{dy}{dx}=0$. Since $x>0$, $x^{-1/2}\ne0$, so the $x$ coordinates satisfy $2x^3-10x^2+13x-5=0$.

One root is $x=1$, so divide $2x^3-10x^2+13x-5$ by $(x-1)$.

$2x^3-10x^2+13x-5=(x-1)(2x^2-8x+5)$.

$2x^2-8x+5=0$ gives $x=\dfrac{8\pm\sqrt{64-40}}{4}=\dfrac{8\pm2\sqrt6}{4}=2\pm\dfrac{\sqrt6}{2}$.

For $y=x^3-4x^{5/2}-kx^{1/2}+28x-44$, $\dfrac{dy}{dx}=3x^2-10x^{3/2}-\dfrac{k}{2}x^{-1/2}+28$.

$\dfrac{dy}{dx}=3x^2-10x^{3/2}-\dfrac{k}{2\sqrt{x}}+28$.

At $T$, $x=9$ and $\dfrac{dy}{dx}=0$.

$0=3(9)^2-10(9)^{3/2}-\dfrac{k}{2\sqrt9}+28=243-270-\dfrac{k}{6}+28$.

This simplifies to $0=1-\dfrac{k}{6}$, so $k=6$.

With $k=6$, the $y$ coordinate of $T$ is $9^3-4(9)^{5/2}-6\sqrt9+28(9)-44=-53$.

The line $NT$ is $y=-53$. The region is between the curve and this line from $x=0$ to $x=9$.

$R=\displaystyle\int_0^9\left(x^3-4x^{5/2}-6x^{1/2}+28x-44+53\right)dx$.

$R=\left[\dfrac{x^4}{4}-\dfrac87x^{7/2}-4x^{3/2}+14x^2+9x\right]_0^9$.

$R=\dfrac{6939}{28}=247.821\ldots$, so the area is $248$ to $3$ significant figures.

The volume is $(3x)(x)(y)=3x^2y$. Since the capacity is $120$, $3x^2y=120$, so $y=\dfrac{40}{x^2}$.

The open top means the metal area is the base plus four sides: $A=3x^2+2(3xy)+2(xy)$.

$A=3x^2+8xy=3x^2+8x\left(\dfrac{40}{x^2}\right)=3x^2+\dfrac{320}{x}$. Hence $P=3$ and $Q=320$.

$\dfrac{dA}{dx}=6x-320x^{-2}=6x-\dfrac{320}{x^2}$.

Set $\dfrac{dA}{dx}=0$: $6x=\dfrac{320}{x^2}$, so $6x^3=320$.

$x^3=\dfrac{160}{3}$, so $x=\sqrt[3]{\dfrac{160}{3}}=3.764\ldots$. To $3$ significant figures, $x=3.76$.

$\dfrac{d^2A}{dx^2}=6+640x^{-3}=6+\dfrac{640}{x^3}$.

For $x>0$, $6+\dfrac{640}{x^3}>0$, so the stationary value found in part (b) is a minimum.

$x^2+y^2+4x-30y+209=0$ becomes $(x+2)^2+(y-15)^2=20$.

The centre is $(-2,15)$ and the radius is $\sqrt{20}=2\sqrt5$.

The line $y=mx+1$ can be written as $mx-y+1=0$. Its distance from $(-2,15)$ must be $2\sqrt5$.

$\dfrac{|m(-2)-15+1|}{\sqrt{m^2+1}}=2\sqrt5$, so $\dfrac{|-2m-14|}{\sqrt{m^2+1}}=2\sqrt5$.

Squaring gives $4(m+7)^2=20(m^2+1)$. Hence $(m+7)^2=5(m^2+1)$, so $2m^2-7m-22=0$.

$2m^2-7m-22=0$ gives $m=-2$ or $m=\dfrac{11}{2}$.

A tangent is perpendicular to the radius at the point of contact.

For $m=-2$, the radius gradient is $\dfrac12$. The radius through the centre is $y-15=\dfrac12(x+2)$. Solving with $y=-2x+1$ gives $x=-6$ and $y=13$.

For $m=\dfrac{11}{2}$, the radius gradient is $-\dfrac2{11}$. The radius through the centre is $y-15=-\dfrac2{11}(x+2)$. Solving with $y=\dfrac{11}{2}x+1$ gives $x=\dfrac{12}{5}$ and $y=\dfrac{71}{5}$.

$\dfrac{(x-k)^2}{\sqrt{x}}=\dfrac{x^2-2kx+k^2}{x^{1/2}}=x^{3/2}-2kx^{1/2}+k^2x^{-1/2}$.

$\displaystyle \int \left(x^{3/2}-2kx^{1/2}+k^2x^{-1/2}\right)dx=\dfrac25x^{5/2}-\dfrac{4k}{3}x^{3/2}+2k^2x^{1/2}$.

From $1$ to $16$, this gives $\left(\dfrac{2048}{5}-\dfrac{256k}{3}+8k^2\right)-\left(\dfrac25-\dfrac{4k}{3}+2k^2\right)$.

The result is $6k^2-84k+\dfrac{2046}{5}$, so $a=6$ and $b=-84$.

Since $A(1,9)$ lies on $C$, $9=\dfrac{(1-k)^2}{\sqrt1}=(1-k)^2$.

Thus $1-k=\pm3$, so $k=4$ or $k=-2$. Since $k$ is positive, $k=4$.

When $k=4$ and $x=16$, $q=\dfrac{(16-4)^2}{\sqrt{16}}=\dfrac{144}{4}=36$, so $B=(16,36)$.

The line through $A(1,9)$ and $B(16,36)$ has gradient $\dfrac{36-9}{16-1}=\dfrac95$, so $l$ has equation $y=\dfrac95x+\dfrac{36}{5}$.

$\displaystyle \int_1^{16}\left(\dfrac95x+\dfrac{36}{5}\right)dx=\left[\dfrac{9}{10}x^2+\dfrac{36}{5}x\right]_1^{16}=\dfrac{675}{2}$.

From part (a), with $k=4$, $\displaystyle \int_1^{16}\dfrac{(x-4)^2}{\sqrt{x}}dx=6(4)^2-84(4)+\dfrac{2046}{5}=\dfrac{846}{5}$.

The shaded area is $\dfrac{675}{2}-\dfrac{846}{5}=\dfrac{1683}{10}$.

$x^2+y^2+8x-4y=0$ becomes $(x+4)^2-16+(y-2)^2-4=0$.

So $(x+4)^2+(y-2)^2=20$.

The centre is $(-4,2)$ and the radius is $\sqrt{20}=2\sqrt5$.

The tangent $x+2y+10=0$ has gradient $-\dfrac12$.

The radius to the point of contact is perpendicular to the tangent, so its gradient is $2$.

The radius line through the centre $(-4,2)$ is $y-2=2(x+4)$, so $y=2x+10$.

Substitute $y=2x+10$ into $x+2y+10=0$: $x+2(2x+10)+10=0$.

$5x+30=0$, so $x=-6$. Then $y=2(-6)+10=-2$, hence $P=(-6,-2)$.

The normal at $P$ is the radius line already found, so its equation is $y=2x+10$.

At an intersection, $x-x^2=mx$, so $x(1-m-x)=0$.

The non-zero intersection has $x=1-m$, and then $y=m(1-m)$, so $P=(1-m,m(1-m))$.

Region $R_1$ is between the curve and the line from $x=0$ to $x=1-m$.

$A_{R_1}=\int_0^{1-m}\{(x-x^2)-mx\}\,dx=\int_0^{1-m}\{(1-m)x-x^2\}\,dx$.

$A_{R_1}=\left[\dfrac{1-m}{2}x^2-\dfrac13x^3\right]_0^{1-m}=\dfrac{(1-m)^3}{2}-\dfrac{(1-m)^3}{3}=\dfrac{(1-m)^3}{6}$.

The total area under $y=x-x^2$ from $x=0$ to $x=1$ is $\int_0^1(x-x^2)\,dx=\dfrac16$.

Since $R_1$ and $R_2$ are equal and together make this total area, each has area $\dfrac{1}{12}$.

$\dfrac{(1-m)^3}{6}=\dfrac1{12}$, so $(1-m)^3=\dfrac12$ and $m=1-\dfrac1{\sqrt[3]{2}}$.

For $y=\dfrac43x^3-11x^2+kx$, $\dfrac{dy}{dx}=4x^2-22x+k$.

At the turning point $M$, $x=2$ and $\dfrac{dy}{dx}=0$.

$4(2)^2-22(2)+k=0$, so $16-44+k=0$ and $k=28$.

With $k=28$, $\dfrac{dy}{dx}=4x^2-22x+28=2(2x-7)(x-2)$.

The derivative is positive outside its roots, so $y$ is increasing for $x<2$ or $x>\dfrac72$.

At $x=2$, $y=\dfrac43(8)-11(4)+28(2)=\dfrac{68}{3}$. Hence line $l$ has equation $y=\dfrac{68}{3}$.

The region is between the line and the curve from $x=0$ to $x=2$, so $A=\int_0^2\left(\dfrac{68}{3}-\left(\dfrac43x^3-11x^2+28x\right)\right)dx$.

$A=\left[\dfrac{68}{3}x-\dfrac13x^4+\dfrac{11}{3}x^3-14x^2\right]_0^2$.

Substituting $x=2$ gives $\dfrac{136}{3}-\dfrac{16}{3}+\dfrac{88}{3}-56=\dfrac{40}{3}$.

The volume is $l^2h=250000$, so $l^2=\dfrac{250000}{h}$.

The open box has one square base and four side faces, so $S=l^2+4lh$.

Since $l=\sqrt{\dfrac{250000}{h}}=\dfrac{500}{\sqrt h}$, $4lh=4\cdot\dfrac{500}{\sqrt h}\cdot h=2000\sqrt h$.

Therefore $S=\dfrac{250000}{h}+2000\sqrt h$.

$S=250000h^{-1}+2000h^{1/2}$, so $\dfrac{dS}{dh}=-250000h^{-2}+1000h^{-1/2}$.

$-250000h^{-2}+1000h^{-1/2}=0$, so $1000h^{-1/2}=250000h^{-2}$.

Multiplying by $h^2$ gives $1000h^{3/2}=250000$, so $h^{3/2}=250$ and $h=250^{2/3}$.

Since $h=250^k$, we have $k=\dfrac23$.

$\dfrac{d^2S}{dh^2}=500000h^{-3}-500h^{-3/2}$.

At $h=250^{2/3}$, $h^{3/2}=250$, so the second derivative is $\dfrac{500000}{250^2}-\dfrac{500}{250}=8-2=6>0$.

The stationary point is therefore a minimum.

From $(x-k)^2+(y-2k)^2=k+7$, the centre is $(k,2k)$ and the radius is $\sqrt{k+7}$.

Since $P(2,3)$ lies on the circle, $(2-k)^2+(3-2k)^2=k+7$.

Expanding gives $k^2-4k+4+4k^2-12k+9=k+7$, hence $5k^2-17k+6=0$.

$5k^2-17k+6=(5k-2)(k-3)$, so $k=\dfrac25$ or $k=3$.

For the tangent calculation, $k<2$, so the centre is $\left(\dfrac25,\dfrac45\right)$.

The radius from the centre to $P$ has gradient $\dfrac{3-4/5}{2-2/5}=\dfrac{11}{8}$, so the tangent gradient is $-\dfrac8{11}$.

The tangent is $y-3=-\dfrac8{11}(x-2)$. Setting $y=0$ gives $x=\dfrac{49}{8}$, so $T=\left(\dfrac{49}{8},0\right)$.

Triangle $OPT$ has base $\dfrac{49}{8}$ on the $x$-axis and height $3$, so its area is $\dfrac12\cdot\dfrac{49}{8}\cdot3=\dfrac{147}{16}$.

Write $y=4x^3-9x+kx^{-1}$. Then $\dfrac{dy}{dx}=12x^2-9-kx^{-2}$.

At $x=\dfrac12$, $\dfrac{dy}{dx}=0$, so $0=12\left(\dfrac12\right)^2-9-k\left(\dfrac12\right)^{-2}=3-9-4k$.

$-6-4k=0$, hence $k=-\dfrac32$.

With $k=-\dfrac32$, $\dfrac{dy}{dx}=12x^2-9+\dfrac{3}{2x^2}$ and $\dfrac{d^2y}{dx^2}=24x-3x^{-3}$.

At $x=\dfrac12$, $\dfrac{d^2y}{dx^2}=12-24=-12<0$, so $P$ is a local maximum.

Set $12x^2-9+\dfrac{3}{2x^2}=0$. Multiplying by $2x^2$ gives $24x^4-18x^2+3=0$, or $8x^4-6x^2+1=0$.

$(4x^2-1)(2x^2-1)=0$. The given point has $x=\dfrac12$, so the second stationary point has $x^2=\dfrac12$.

Since $x>0$, the second stationary point has $x=\dfrac1{\sqrt2}$.

Each end is a right-angled triangle with perpendicular sides $3x$ and $2x$, so its area is $\dfrac12(3x)(2x)=3x^2$.

The volume is cross-section area times length, so $3x^2L=12$ and $L=\dfrac4{x^2}$.

There are two triangular ends with total area $6x^2$, and two rectangular sides with total area $3xL+2xL=5xL$.

$S=6x^2+5x\left(\dfrac4{x^2}\right)=6x^2+\dfrac{20}{x}$. Hence $P=6$ and $Q=20$.

$\dfrac{dS}{dx}=12x-20x^{-2}=12x-\dfrac{20}{x^2}$.

Set $\dfrac{dS}{dx}=0$: $12x=\dfrac{20}{x^2}$, so $x^3=\dfrac53$.

At $x=\sqrt[3]{\dfrac53}$, $S=6x^2+\dfrac{20}{x}=25.3029\ldots$, so the minimum value is $25.3$ to one decimal place.

$\dfrac{d^2S}{dx^2}=12+40x^{-3}=12+\dfrac{40}{x^3}$.

Since $x>0$, $12+\dfrac{40}{x^3}>0$, so the stationary value found is a minimum.

$4x^3+13x^2-10x+8=(x-2)(4x^2+21x+32)+72$.

Hence $Q(x)=4x^2+21x+32$ and $R=72$.

$f(-4)=4(-4)^3+13(-4)^2-10(-4)+8=-256+208+40+8=0$.

Since $f(-4)=0$, $(x+4)$ is a factor of $f(x)$.

Dividing by $(x+4)$ gives $f(x)=(x+4)(4x^2-3x+2)$.

The discriminant of $4x^2-3x+2$ is $(-3)^2-4(4)(2)=9-32=-23<0$, so this quadratic has no real roots.

Therefore $f(x)=0$ has only one real solution, $x=-4$.

$f'(x)=12x^2+26x-10=2(3x-1)(2x+5)$.

$f(x)$ is decreasing when $f'(x)<0$, which is between the roots. Hence $-\dfrac52<x<\dfrac13$.

$y=\dfrac{9x^2(5-\sqrt{x})}{5}=9x^2-\dfrac95x^{5/2}$.

$\dfrac{dy}{dx}=18x-\dfrac92x^{3/2}=\dfrac92x(4-\sqrt{x})$.

For the turning point shown away from the origin, set $4-\sqrt{x}=0$. Hence $x=16$.

At $x=16$, $y=\dfrac{9(16)^2(5-4)}{5}=\dfrac{2304}{5}$.

Therefore $M\left(16,\dfrac{2304}{5}\right)$.

For the positive intercept, $\dfrac{9x^2(5-\sqrt{x})}{5}=0$ gives $5-\sqrt{x}=0$, so $x=25$.

The top boundary is $y=\dfrac{2304}{5}$ and the curve is below it from $x=16$ to $x=25$. So $R=\displaystyle\int_{16}^{25}\left(\dfrac{2304}{5}-9x^2+\dfrac95x^{5/2}\right)dx$.

$R=\left[\dfrac{2304}{5}x-3x^3+\dfrac{18}{35}x^{7/2}\right]_{16}^{25}$.

$R=\dfrac{9189}{7}=1312.714\ldots$, so the area is $1312.7$ to one decimal place.

For $y=\dfrac23x^2-9\sqrt{x}+13$, $\dfrac{dy}{dx}=\dfrac43x-\dfrac{9}{2\sqrt{x}}$.

$y$ is increasing when $\dfrac{dy}{dx}>0$.

For $x>0$, multiply by $6\sqrt{x}$: $8x\sqrt{x}-27>0$.

$x^{3/2}>\dfrac{27}{8}=\left(\dfrac32\right)^3$, so $\sqrt{x}>\dfrac32$ and $x>\dfrac94$.

At $P(9,40)$, $\dfrac{dy}{dx}=\dfrac43(9)-\dfrac{9}{2\cdot3}=12-\dfrac32=\dfrac{21}{2}$.

$y-40=\dfrac{21}{2}(x-9)$, so $y=\dfrac{21}{2}x-\dfrac{109}{2}$.

Set $y=0$: $\dfrac{21}{2}x-\dfrac{109}{2}=0$, so $x=\dfrac{109}{21}$.

From $x=0$ to $x=\dfrac{109}{21}$, the lower boundary is the $x$-axis. From $x=\dfrac{109}{21}$ to $x=9$, the lower boundary is the tangent.

$\displaystyle R=\int_0^{109/21}\left(\dfrac23x^2-9\sqrt{x}+13\right)dx+\int_{109/21}^{9}\left(\dfrac23x^2-9\sqrt{x}+13-\left(\dfrac{21}{2}x-\dfrac{109}{2}\right)\right)dx$.

$\displaystyle \int\left(\dfrac23x^2-9\sqrt{x}+13\right)dx=\dfrac{2}{9}x^3-6x^{3/2}+13x$, and $\displaystyle \int\left(\dfrac{21}{2}x-\dfrac{109}{2}\right)dx=\dfrac{21}{4}x^2-\dfrac{109}{2}x$.

Substituting the limits and simplifying gives $R=\dfrac{857}{21}$.

For $y=x^3-5x^2+3x+14$, $\dfrac{dy}{dx}=3x^2-10x+3$.

$3x^2-10x+3=0$ factors as $(3x-1)(x-3)=0$, so $x=\dfrac13$ or $x=3$.

$\dfrac{d^2y}{dx^2}=6x-10$. At $x=3$, this is positive, so the minimum turning point has $x=3$.

Substitute $x=3$: $y=27-45+9+14=5$. Hence $T=(3,5)$.

At $x=2$ on $C_1$, $y=8-20+6+14=8$, so $A=(2,8)$.

The centre of $C_2$ is $T=(3,5)$ and it passes through $A=(2,8)$. Thus $r^2=(2-3)^2+(8-5)^2=10$.

Therefore $C_2$ has equation $(x-3)^2+(y-5)^2=10$.

The radius $TA$ has gradient $\dfrac{8-5}{2-3}=-3$, so the tangent at $A$ has gradient $\dfrac13$.

Using $A=(2,8)$, $y-8=\dfrac13(x-2)$, so $y=\dfrac13x+\dfrac{22}{3}$.

From $x=0$ to $x=2$, the cubic is above the line. The required area is $\displaystyle \int_0^2 \left[x^3-5x^2+3x+14-\left(\dfrac13x+\dfrac{22}{3}\right)\right]dx$.

The integrand is $x^3-5x^2+\dfrac83x+\dfrac{20}{3}$.

$\displaystyle \left[\dfrac{x^4}{4}-\dfrac{5x^3}{3}+\dfrac{4x^2}{3}+\dfrac{20x}{3}\right]_0^2=4-\dfrac{40}{3}+\dfrac{16}{3}+\dfrac{40}{3}=\dfrac{28}{3}$.