Rational Algebra: past paper questions & worked solutions

First split the fraction into powers:

Integrate term by term. Since $x\gt 0$, use $\ln x$ without absolute-value notation:

Factor $x^2+x-12=(x+4)(x-3)$ and combine the fractions:

The cancellation is valid because $x\gt 3$. In particular, the numerator simplifies as $49x+7x^2-21x=7x^2+28x=7x(x+4)$.

Rewrite $f(x)=7+21/(x-3)$. Differentiating gives

Use the product rule and $\frac{d}{dx}\ln(2x)=1/x$:

Write $f(x)=e^{2x}(4x+3)^{-2}$. Product and chain rules give

Factoring out $2e^{2x}(4x+3)^{-3}$,

Thus $a=4$ and $b=-1$.

Apply the quotient rule, retaining the restriction $x\ne-3$:

The numerator vanishes at $10-4x=0$, so $x=5/2$; the denominator changes sign at the excluded value $x=-3$. For $x\lt -3$, the numerator is positive and denominator negative, so $y'\lt 0$. For $-3\lt x\lt 5/2$, both are positive, so $y'\gt 0$. For $x\gt 5/2$, the numerator is negative and denominator positive, so $y'\lt 0$. Thus the positive-gradient interval is as follows.

For $f(x)=x^4-5x^2+4x-7$,

The polynomial is continuous on $[2,3]$ and the endpoint values have opposite signs. Therefore a root lies strictly between 2 and 3.

Rearrange $x^4-5x^2+4x-7=0$ to $x^4=5x^2-4x+7$. Since $f(0)=-7$, zero is not a root, so division by $x$ is valid at a root:

Taking the real cube root gives the requested form.

With $x_1=2$,

Continue using unrounded intermediate values: $x_3=2.146846021\ldots$, $x_4=2.154063147\ldots$, $x_5=2.155869187\ldots$, $x_6=2.156321484\ldots$. Further iteration gives $x_{10}=2.156472046\ldots$ and $x_{13}=2.156472632\ldots$.

For $x\ge0$, $f(x)=9-x^2$ has maximum $f(0)=9$ and decreases without a lower bound.

Apply $g$ before $f$:

The intermediate value $3/4$ is in the nonnegative domain of $f$.

Let $y=3/(2x+1)$. Then $y(2x+1)=3$, so $2xy=3-y$ and $x=(3-y)/(2y)$. Exchange the variables. Since $g$ decreases from 3 towards 0 on $[0,\infty)$, its range is $(0,3]$, which is the inverse domain.

Since $d(3x^2+k)/dx=6x$, the numerator is $3/2$ times this derivative. The denominator is positive because $k\gt 0$, so

Apply the limits to (a) and combine the logarithms:

Hence $\ln((75+k)/(12+k))=(2/3)\ln8=\ln4$, so $75+k=4(12+k)$. This gives $27=3k$ and $k=9$, consistent with $k\gt 0$.

At the start of the study, $t=0$, so

There are initially $300$ toads.

Set $N=420$ and clear the denominator:

Taking natural logarithms gives

Thus $\boxed{t=16.22\text{ years}}$ to 2 decimal places.

Divide the numerator and denominator by $e^{0.12t}$:

Since $e^{-0.12t}\gt 0$ for every finite $t\geq0$, the denominator is greater than $2$. Hence

The population approaches $450$ as $t$ increases without bound, so it can never reach $500$ according to this model.

Expanding the compound angle,

so $\cos\theta+\sqrt3\sin\theta=3\sin\theta$ and $\cos\theta=(3-\sqrt3)\sin\theta$. If $\cos\theta=0$, this equation would require $\sin\theta=0$, which is impossible. We may therefore divide by $\cos\theta$:

Set $\theta=2x-10^\circ$, so $2x-70^\circ=\theta-60^\circ$ and $-10^\circ\le\theta\lt 350^\circ$.

The admissible values are $38.2619662\ldots^\circ$ and $218.2619662\ldots^\circ$. Hence $x=(\theta+10^\circ)/2$, giving $x=24.1^\circ$ or $114.1^\circ$ (1 decimal place).

The function is continuous on $[1,2]$ and the endpoint values have opposite signs, so a root $\alpha$ lies in $(1,2)$.

Multiplying $f(x)=0$ by $x^2e^x$, which is nonzero on the stated interval,

As $2x+3\gt 0$ here, divide and take the real cube root:

Starting with $x_1=1$ and retaining unrounded values,

Repeated iteration converges to $1.181331\ldots$, giving $\alpha=1.1813$ to four decimal places.

Using the chain rule,

Treat $k$ as a constant when integrating:

The stated inequality therefore implies

Since the exponential is increasing and $k+3\gt 0$,

The roots are $(21\pm\sqrt{697})/2$, so the positive upper bound is $k\lt (21+\sqrt{697})/2=23.700378782\ldots$. With $k\gt 1$ an integer, the greatest possible value is $k=23$.

\paragraph{Solution} Put $u=x^2+(3k-2)x$, so $u'=2x+3k-2$. By the product and chain rules,

As $2e^u\gt 0$, stationary points satisfy

For two distinct real roots, the discriminant must be positive:

Consequently

For these values the quadratic changes sign at both simple roots, so both stationary points are genuine turning points. Equality gives a repeated root with no derivative sign change and is excluded.

Multiplying the proposed identity by $x-2$ gives

Thus $A=2$, $B-4=-5$ and $C-2(-1)=8$, giving $B=-1$, $C=6$. Hence $g(x)=2x-1+6/(x-2)$, for $x\ne2$.

Since $x-2\gt 0$ throughout $[4,8]$,

Multiply the proposed identity by $(x+3)^2=x^2+6x+9$ and compare coefficients:

The coefficient of $x$ checks: $9B+6C=27-6=21$. The constant equation is $9C+D=-4$, so $D=5$. Hence $f(x)=2x^2+3x-1+5(x+3)^{-2}$.

Integrate each term in the decomposition:

Here $K$ is an arbitrary constant; the domain $x\gt -3$ avoids the denominator zero.

For $x\ge0$, $0\lt 4/(3x+2)\le2$. Thus $f(0)=3$ is attained, while $f(x)$ increases towards 5 without reaching it.

(i) Put $y=5-4/(3x+2)$. Rearranging gives $(5-y)(3x+2)=4$, hence

Exchange variables. (ii) The inverse domain is the range of $f$ found above.

The original function has an absolute value and angle $x/3+\pi/6$:

Then $fg(-\pi)=f(2)=5-4/(3\cdot2+2)=5-1/2=9/2$.

At $t=0$, the model gives $30=80p/(p+4)$. Thus $30p+120=80p$, so $50p=120$ and $p=2.4\gt 0$.

Put $E=e^{0.15T}$ and use $p=2.4$. At area $50\,\mathrm{m^2}$,

Taking natural logarithms gives $0.15T=\ln(25/9)$, so $T=\ln(25/9)/0.15=6.8110083\ldots$ days.

Divide numerator and denominator by $pe^{0.15t}$: $A=80/(1+(4/p)e^{-0.15t})\longrightarrow80$ as $t\longrightarrow\infty$. Thus the model's limiting pond surface area is $80\,\mathrm{m^2}$; the finite-time covered area approaches this value from below.

Write $f(x)=(2x+3)(4x-1)^{-1/2}$. The product and chain rules give

The denominator of $f'(x)$ is positive throughout the domain, so the derivative is negative for $1/4\lt x\lt 2$ and positive for $x\gt 2$. Consequently $x=2$ is the global minimum, with $f(2)=7/\sqrt7=\sqrt7$. Also $f(x)\to\infty$ as $x\to1/4^+$ and as $x\to\infty$. By continuity all values at least $\sqrt7$ are attained, including the minimum.

Using the quotient rule,

Thus $a=4$, $b=-24$ and $c=27$.

A stationary point requires

The roots are $3/2$ and $9/2$. Reject $9/2$ because the stated domain is $x\lt 3$. At $x=3/2$, $f(x)=-2$, so the turning point is $\boxed{(3/2,-2)}$. The derivative changes from positive to negative there, confirming a maximum.

The inside factor 2 halves the horizontal coordinate; adding 3 raises the vertical coordinate by 3. Therefore

Substitute $f(x)$ into $f$ and multiply numerator and denominator by $x-4$:

The composition requires $x\ne4$ and $f(x)\ne4$, hence also $x\ne16$.

To find $f^{-1}(12)$, solve $f(x)=12$:

Thus $f^{-1}(12)=\frac{32}{5}$.

Put $u=\ln a$. Then $fg(a)=g(a)$ means $f(u)=u$, where $u\ne4$:

Therefore $a=e^8$ or $a=e^{-2}$. Both are positive and neither is $e^4$, so both satisfy the domain restrictions.

At $t=0$, $N=4000/(19+1)=200$ squirrels.

The quotient rule gives

At $t=T$, the derivative is zero. The exponential and denominator are positive, so $19-e^{0.2T}=0$. Therefore $e^{0.2T}=19$ and $A=19$. The derivative is positive before $T$ and negative after $T$, confirming a maximum.

Since $e^{0.2T}=19$, we have $e^{0.1T}=\sqrt{19}$. Substitution into the model gives

The maximum number is $459$ squirrels to the nearest whole number.

First apply $f$, then $g$:

For $x\geq0$,

Thus $f(0)=0$ and $f(x)\lt 2/3$. Also $f'(x)=2/(3x+1)^2\gt 0$, so the range is

Therefore

On the common domain $0\leq x\lt 2/3$, both denominators below are nonzero:

Both candidates lie in the common domain, giving

Let $y=(4x+3)/(x-2)$. Then $yx-2y=4x+3$, so

Swap the input and output names. Since $f(x)=4+11/(x-2)$ can never equal 4, the inverse domain excludes 4.

Substitute $f(x)$ into both occurrences of the argument:

Thus $a=19$, $b=6$, $c=2$, $d=7$. The composite retains $x\ne2$ from the inner function and also requires $f(x)\ne2$, that is $x\ne-7/2$.

For $y=2f(3x)+8$, the point mapping is $(x,y)\mapsto(x/3,2y+8)$. Hence $(3,15)\mapsto(3/3,2(15)+8)=(1,38)$.

First apply $f$: $f(2)=6-21/(2\cdot2+3)=3$. Then apply $g$: $gf(2)=g(3)=3^2+5$.

Let $y=6-21/(2x+3)$. Then

Interchange $x,y$. On the original domain $x\ge0$, $f$ increases from $f(0)=-1$ towards 6, never attaining 6. Thus the inverse domain is $-1\le x\lt 6$.

Since $x^2+5\gt 0$, take $x^2+5=11$, not $-11$. Therefore $x^2=6$.

Expand both compound angles:

Thus $(2-\sqrt3)\sin x=-\cos x$. Here $\cos x\ne0$: otherwise this equation would require $\sin x=0$ as well, which is impossible. Dividing by $(2-\sqrt3)\cos x$ gives

Set $x=2\theta-45^\circ$. Then $x+45^\circ=2\theta$ and $x-60^\circ=2\theta-105^\circ$, so part (a) gives $\tan(2\theta-45^\circ)=-2-\sqrt3$. Since $\tan75^\circ=2+\sqrt3$,

In $0\le\theta\lt 180^\circ$, only $k=1,2$ are allowed. Substitution in the original equation confirms both values.

The original equation is defined only when $\cos\theta+\sin\theta\ne0$ and $\cos2\theta\ne0$. Multiplying by $\cos\theta+\sin\theta$ gives

Using $2\sin\theta\cos\theta=\sin2\theta$, multiply through by 2: $3\sin2\theta=4\cos2\theta+2$.

Write $3\sin2x-4\cos2x=5\sin(2x-\phi)$, where $\cos\phi=3/5$, $\sin\phi=4/5$ and $\phi=\arctan(4/3)$. Then $\sin(2x-\phi)=2/5$. Because $\pi\lt x\lt 3\pi/2$, $2\pi-\phi\lt 2x-\phi\lt 3\pi-\phi$, approximately $(5.356,8.497)$. With $\gamma=\arcsin(2/5)$, the only angle in this interval is $2\pi+\gamma$; $3\pi-\gamma\approx9.013$ is outside it. Thus

At this value $\cos x+\sin x\ne0$ and $\cos2x\ne0$; direct substitution satisfies the original equation.

Apply $f$ twice, in order:

Let $y=(x+3)/(x-4)$. Then

Interchanging $x$ and $y$ gives the inverse. The value 1 is never an output of $f$, since $(x+3)/(x-4)=1$ would require $3=-4$.

Since $g$ is defined for positive inputs, $gf(a)=7$ requires $f(a)\gt 0$.

The negative square root is not in the domain of $g$. Hence

This value is not 4 and gives $f(a)=\sqrt2\gt 0$, as required.

The quotient rule gives, for $y\ne1$,

A tangent parallel to the $y$-axis is vertical, so $dx/dy=0$. The numerator gives $(y-3)(y+1)=0$, hence $y=3$ or $y=-1$; neither makes the denominator zero. Then

Thus $Q=(4,3)$ and $P=(-4/3,-1)$, with the vertical tangent equations below.

Differentiate to obtain $y'=6x/(x^2-5)-8x$. At a stationary point in $x\gt \sqrt5$, x is nonzero, so $6/(x^2-5)=8$, giving $6=8x^2-40$ and $x^2=23/4$. The original positive domain selects $x=\sqrt{23}/2$, which exceeds $\sqrt5$. Hence $p=23$.

(a) Differentiate the squared sine with the chain rule: $y'=4-24\sin x\cos x=4-12\sin2x$. Thus $A=4$, $B=-12$. (b) Since $-1\le\sin2x\le1$, the largest derivative is attained when $\sin2x=-1$: $4-12(-1)=16$.

Write the integrand as $12(2x-1)^{-2}$. Reverse the chain rule, dividing by the inner derivative 2:

(a) Since $4x+3=4(x+2)-5$,

(b) A primitive is $4x-5\ln|x+2|$. The absolute value is essential here: $x+2$ is negative throughout $[-8,-5]$.

Factor $2x^2-3x-5=(2x-5)(x+1)$. Then

so

On $x\gt 4$ the cancelled factor is nonzero. One integer choice is $a=4,b=1,c=2,d=-5$.

Let $y=(4x+1)/(2x-5)$. Rearranging gives $2xy-5y=4x+1$, hence $x(2y-4)=5y+1$ and $x=(5y+1)/(2y-4)$. Swap variable names to obtain the inverse rule; its restricted domain is determined in (c).

Rewrite $f(x)=2+11/(2x-5)$. For $x\gt 4$, the positive denominator is greater than 3 and grows without bound. Thus $0\lt 11/(2x-5)\lt 11/3$, so $2\lt f(x)\lt 17/3$. Neither bound is attained: $x=4$ is excluded and $f(x)=2$ is impossible. The range of f is the domain of its inverse.

First evaluate the inner function:

Then apply $f$ to this result:

Put $y=f(x)$ and rearrange to make $x$ the subject:

For $x\gt 0$, the denominator $x+1\gt 1$, so $0\lt f(x)\lt 12$. Conversely, every $y$ with $0\lt y\lt 12$ gives $x=12/y-1\gt 0$. Thus the range of $f$, and hence the domain of its inverse, is $(0,12)$. Interchanging $x$ and $y$ gives

Both functions must be defined, so $0\lt x\lt 12$. Substitute the formula for the inverse and multiply by the nonzero quantity $x(x+1)$:

The algebraic candidates are $x=-4$ and $x=3$. Reject $x=-4$ because it lies outside $0\lt x\lt 12$. For $x=3$, $f^{-1}(3)=12/3-1=3=f(3)$, so the only real solution is

Let $t=\tan x$. For the original equation to be defined, $\sin x\ne0$, $\cos x\ne0$ and $\cos 2x\ne0$; hence $t\ne0$ and $t^2\ne1$. Using the double-angle and reciprocal identities,

Substituting into the equation gives

Multiply by the nonzero quantity $t(1-t^2)$ and expand:

Therefore

Factor the quadratic in $t^2$:

Since $t$ is real, $t^2\ge0$, so discard $t^2=-1/5$. Thus

In degrees, let

Tangent is positive in quadrants I and III and negative in quadrants II and IV. The four solutions in $0\le x\lt 360^\circ$ are therefore

These have finite, nonzero tangent with $t^2=5\ne1$, so every transformation in part (a) is reversible and all four satisfy the original equation. To one decimal place,

For angles where the original expression is defined,

Divide the given equation by $4\operatorname{cosec}4\theta$ and use (a):

Use $\cos^2\theta=(1+\cos2\theta)/2$ and multiply by $2\cos2\theta$:

Therefore

The divisions are valid on the domain of the original equation.

Set $u=\cos2x$:

Reject $u=3$ because cosine lies between $-1$ and $1$. Thus $\cos2x=-2/5$. Since $0\lt 2x\lt 360^\circ$,

Hence $\boxed{x=56.8^\circ\text{ or }123.2^\circ}$ to one decimal place. Neither is an excluded multiple of $45^\circ$; these are all the solutions in the interval.

Since $x^2+3\ge3$, with equality at $x=0$, and the logarithm is increasing, the range of $f$ is $[\ln3,\infty)$.

Let $y=(3+5x)/(x+2)$. Then

Moreover, $g(x)=5-7/(x+2)$ with $x\gt -2$ has range $(-\infty,5)$. Thus

Therefore

Taking natural logarithms gives $a=\frac12\ln5$. The input $e^{2a}=5$ satisfies the domain of $g$.

Multiply the identity by $x^2+4$:

Comparing the cubic, quadratic and linear coefficients gives $A=4$, $B=2$ and $4A+C=3$, hence $C=3-16=-13$.

The $x^2$ coefficients give $B=2$, while the constant terms give $4B+D=8$. Substitute $B=2$: $4(2)+D=8$, so $D=0$.

Using the decomposition, the integrand is $4x+2-13x/(x^2+4)$. Since the derivative of $x^2+4$ is $2x$, an antiderivative is

Therefore

Combine the logarithms: $36-\frac{13}{2}\ln4=36-13\ln2$.

At $x=1/4$, $\sin^2 4y=1/4$. Since $0\le4y\le\pi/2$,

Only this branch lies in the given range of $y$.

Differentiate $x=(\sin4y)^2$ with respect to $y$:

On the interior of the given interval, $\sin4y=\sqrt{x}$ and $\cos4y=\sqrt{1-x}$ are positive. Taking the reciprocal,

Complete the square:

Thus $dy/dx=(q+r(x+s)^2)^{-1/2}$ has the constants below, with $0\lt x\lt 1$ for a finite derivative; the endpoints have vertical tangents.

The derivative is positive. Its minimum occurs when the square root in the denominator is largest. Since $(x-1/2)^2\ge0$, the largest radicand is 16, attained at $x=1/2$. Hence the smallest gradient is $1/\sqrt{16}$.

The product rule gives

Comparison with the stated form gives $A=2$.

Since $e^{-4x}\gt 0$, $f'(x)=0$ means $2x+1=0$ or $1-4x=0$. Thus $x=-1/2$ or $x=1/4$. Substitution gives $f(-1/2)=0$ and $f(1/4)=(3/2)^3e^{-1}=27/(8e)$.

The derivative changes from positive to negative at $x=1/4$, so this is the maximum; at $x=-1/2$ it does not change sign. The transformation $g(x)=8f(x-2)$ sends $(x,y)$ to $(x+2,8y)$. Hence the maximum stationary point is $(1/4+2,8\cdot27/(8e))=(9/4,27/e)$.

For $x\ne n\pi/2$, both sine and cosine are nonzero. Then

By part (a), the equation is $\cot^2\theta-\tan^2\theta=2\tan^2\theta$, so $\cot^2\theta=3\tan^2\theta$. The original expression excludes $\theta=0$ in the stated interval, so $\tan\theta\ne0$ and multiplication by $\tan^2\theta$ gives

Tangent is one-to-one on $(-\pi/2,\pi/2)$; the two angles are $\pm\arctan(3^{-1/4})=\pm0.6497662865\ldots$ radians. Both are valid and nonzero. To two decimal places:

At the start $t=0$, so $e^{kt}=1$ and the model gives $F(0)=350/(9+1)=35$ fruit flies.

Using $F(15)=200$,

Take natural logs to obtain $15k=\ln12$, as required.

By the quotient rule,

Set $F'(T)=10$ and let $u=e^{kT}\gt 0$. Then

With the exact $k=\ln12/15$, the quadratic formula gives

Both values are positive and at least 1, hence both correspond to $T\ge0$. Finally $T=\ln u/k$ gives $5.6779528361\ldots$ or $20.8488936858\ldots$ days. To one decimal place the two times are as follows.

Clear the denominator in the proposed identity:

Matching coefficients gives $A=2$, $3A+B=0$, hence $B=-6$. Then $4A+3B+2C=-4$ gives $C=3$; the constant checks as $4B+3C=-24+9=-15$. Therefore $f(x)=2x-6+3(2x+3)/(x^2+3x+4)$.

The numerator $2x+3$ is the derivative of $x^2+3x+4$. This quadratic is $(x+3/2)^2+7/4\gt 0$, so

Using both limits,

Thus $p=4$ and $q=8$ are integers, as required.

Write $s=\sin\theta$. The compound-angle formula gives

Hence

The given nonzero denominator ensures that $1-s\ne0$ and the other factor is nonzero. Cancel the common factor to obtain the required identity.

Use (a) and $\csc\theta=1/\sin\theta$. With $s=\sin\theta\ne0$,

Thus $s=(-5\pm\sqrt{97})/18$, approximately $0.269380989$ and $-0.824936545$. Both lie in $[-1,1]$, are nonzero, and satisfy the original nonzero-denominator condition. The positive value gives $15.6274\ldots^\circ$ and $164.3726\ldots^\circ$; the negative value gives $235.5821\ldots^\circ$ and $304.4179\ldots^\circ$. All four angles lie in $0\lt \theta\le360^\circ$.

$g(5)=2(5)-7=3$, then $fg(5)=f(3)=(5-3)/(9+2)=2/11$.

Let $y=(5-x)/(3x+2)$. Then $3xy+2y=5-x$, so $x(3y+1)=5-2y$ and $x=(5-2y)/(3y+1)$. Exchange x and y. The inverse cannot have input $-1/3$, since this is not an output of f: $(5-x)/(3x+2)=-1/3$ would imply $15=-2$.

Here $a\ne0$ and $a\ne-3/2$ are required by the input and denominator. Substitute into both functions:

Thus $5a-1=(2a-1)(2a+3)=4a^2+4a-3$, giving $4a^2-a-2=0$. By the quadratic formula, $a=(1\pm\sqrt{33})/8$. Neither value is excluded.

Let $F(x)=6\ln(2x+3)-x^2/2+4$. Then $F(-1.25)=-0.94013308\ldots$ and $F(-1.2)=0.21504626\ldots$. The log argument is positive on this interval, so F is continuous there. The change of sign proves a zero between these endpoints, giving the stated bracket for P.

Apply the given recurrence with $x_1=6$:

Continuing without rounding the internal values,

Further iteration settles at $6.41417418\ldots$. The positive square-root branch corresponds to Q.

Differentiate: $F'(x)=12/(2x+3)-x$. For a stationary point, $12=x(2x+3)$, or $2x^2+3x-12=0$. The quadratic formula gives $x=(-3\pm\sqrt{105})/4$. The negative choice is less than $-3/2$ and outside the original domain. Also $F”(x)=-24/(2x+3)^2-1\lt 0$, confirming the admissible point is a maximum.

Factor $x^2+7x+12=(x+3)(x+4)$. A common denominator gives

The cancelled factor is nonzero throughout the original domain $x\gt 0$.

Let $y=5x/(x+3)$. Then $yx+3y=5x$, so $x=3y/(5-y)$. Exchange variable names. Since $f(x)=5-15/(x+3)$ for $x\gt 0$, its range is $0\lt f(x)\lt 5$. This becomes the inverse domain.

(i) By the quotient rule, $f'(x)=[5(x+3)-5x]/(x+3)^2=15/(x+3)^2$. (ii) The numerator and denominator are positive on $x\gt 0$, so $f'(x)\gt 0$ throughout its domain. Therefore f is increasing.

Write $f(x)=(x^3-4x)e^{-x/2}$. By the product rule,

At O, $f'(0)=-4$, so the normal gradient is $1/4$ and its equation is $y=x/4$. At the other intersection P, x is nonzero; equating curve and normal and dividing by x gives $(x^2-4)e^{-x/2}=1/4$. Thus $x^2=4+e^{x/2}/4=(16+e^{x/2})/4$. Since P lies in the third quadrant, select the negative square root.

Starting from $x_1=-2$, $x_2=-\tfrac12\sqrt{16+e^{-1}}=-2.0228617996\ldots$. Further unrounded iterates are $x_3=-2.0226034084\ldots$, $x_4=-2.0226063126\ldots$ and $x_5=-2.0226062799\ldots$. The continuing sequence settles at $-2.0226062803\ldots$.

Expand $R\cos(x-\alpha)=R\cos\alpha\cos x+R\sin\alpha\sin x$. Matching coefficients gives $R\cos\alpha=1$, $R\sin\alpha=4$. Thus $R=\sqrt{1^2+4^2}=\sqrt{17}$ and $\tan\alpha=4$. Since $0\lt \alpha\lt \pi/2$, $\alpha=\arctan4=1.32581766\ldots$ radians.

The model becomes $H=24/[3+\sqrt{17}\cos(t/2-\alpha)]$. For $0\le t\le6.5$, the phase lies between $-\alpha$ and $3.25-\alpha$; throughout this interval the denominator remains positive. Its maximum $3+\sqrt{17}$ is attained at $t=2\alpha=2.651635\ldots$, inside the time interval. The height is then smallest:

To the nearest centimetre this is $3.37\ \mathrm m$ (337 cm).

For $H=10$, $3+\sqrt{17}\cos(t/2-\alpha)=24/10$, so

Let $\gamma=\arccos[-3/(5\sqrt{17})]$. In the permitted phase interval $[-\alpha,3.25-\alpha]$, only $+\gamma$ occurs: $-\gamma\lt -\alpha$, and shifts by $2\pi$ are outside it. Using the unrounded $\alpha=\arctan4$,

At the vertex, the expression inside the modulus is zero:

Hence \(\boxed{P=\left(\frac52,3\right)}\).

Write the function in its two branches:

On the left branch,

This lies in \(0\leq x\leq\frac52\), so it is valid. On the right branch,

This satisfies \(x\geq\frac52\), so it is also valid. Thus

Each straight branch can have at most one intersection with \(y=kx+2\). Two distinct roots therefore require one intersection strictly to each side of the vertex.

For the left branch,

The condition \(0\leq x_L\lt \frac52\) first requires \(k+4\gt 0\), then gives

For the right branch,

The condition \(x_R\gt \frac52\) requires \(k\lt 4\), and then

At \(k=\frac25\), both intersections coincide at the vertex, giving only one root. At \(k=4\), the line is parallel to the right branch and has no intersection with that branch, again leaving only one root. Hence

Increase the power to $8$ and divide by both $8$ and the inner derivative $3$:

The denominator is $x^2+x-2$. The first quotient term is $4x$. Subtracting $4x(x^2+x-2)$ leaves $-3x^2+x+14$. The next quotient term is $-3$. Subtracting $-3(x^2+x-2)$ leaves $4x+8$. Thus polynomial division gives

Since $4x+8=4(x+2)$,

Hence $a=4$, $b=-3$, $c=4$.

For $x\gt 1$, an antiderivative is $2x^2-3x+4\ln(x-1)$, so

Thus the exact integral is $126+8\ln2$.

For $x\gt -7/4$, apply the product and chain rules:

The interior stationary values satisfy $x^2(2x+3)=0$, so $x=0$ or $x=-3/2$. Near $x=-3/2$, the derivative changes from negative to positive; near $x=0$ it stays positive on both sides. Thus the minimum turning point is at $x=-3/2$.

The endpoint has $f(-7/4)=0$, the global minimum is $-27/8$ and $f(x)$ tends to infinity as $x$ increases. Hence the range of $f$ is $[-27/8,\infty)$. Multiplication by $-4$ reverses the inequality and makes the upper endpoint $27/2$.

The transformation is a shift right by $3/2$, vertical stretch by 40, then shift down by 8:

Therefore $Q=(1/2,3/8)$ maps to $(1/2+3/2,\;40\cdot3/8-8)$.

To find $f^{-1}(22)$, solve $f(u)=22$: $2+5\ln u=22$, so $\ln u=4$ and $u=e^4\gt 0$.

Using the quotient rule,

For every $x\gt 1/3$, the denominator is positive, so $g'(x)\gt 0$. Thus g is strictly increasing on its entire domain.

Set $y=(6x-2)/(2x+1)$. Then

As $x\downarrow1/3$, $g(x)\to0$; as $x\to\infty$, $g(x)\to3$. Since g is strictly increasing, its range is $(0,3)$; neither limiting value is attained. Interchange the input and output symbols.

The inner function g takes every value in $(0,3)$. Since $f(u)=2+5\ln u$ is increasing for $u\gt 0$, its values on this interval run from $-\infty$ to, but not including, $2+5\ln3$.

The exponential is positive and the squared factor is nonnegative, so $f(x)\ge0$. The value zero is attained at $x=\pm\sqrt{3/2}$. The graph has its highest point at $x=0$, where $f(0)=9$; the outer maxima are lower. Thus both endpoints of the range are included.

For $f(x)=e^{-x^2}(2x^2-3)^2$, product and chain rules give

Factoring yields

The stationary abscissae are $0$, $\pm\sqrt{3/2}$ and $\pm\sqrt{7/2}$. At the outer local maxima,

The central maximum is 9 and the two minima are zero. Also $f(x)\to0$ as $x\to\pm\infty$. A positive horizontal line above both outer maxima but below 9 meets only the two sides of the central peak. At $k=16e^{-7/2}$ it also touches both outer peaks, giving four distinct intersections; at $k=9$ there is only one. Both bounds must be excluded.

Use product and chain rules:

With a common denominator,

This derivative formula applies for $x\gt -9/4$; the endpoint derivative is not finite.

The denominator is positive in the derivative domain. Thus $5x^2+9x-2=(5x-1)(x+2)=0$.

The derivative changes from positive to negative at $x=-2$, so this is the local maximum P. Substitute into f: $f(-2)=5(4-2)\sqrt{-8+9}=10$.

On $[-9/4,0]$, f rises from $f(-9/4)=0$ to $f(-2)=10$, then decreases to $f(0)=-30$. The other stationary point $1/5$ is outside this restricted interval. Thus the minimum and maximum of f on it are $-30$ and 10. Applying the positive vertical transformation $g=2f+4$ gives endpoints $2(-30)+4=-56$ and $2(10)+4=24$, both attained.

Substitute $g(x)=5-2x^2$ into f:

The composite requires $x\le0$ from g and $g(x)\ne5$ from f; therefore $x\lt 0$. Now $4-13/x^2=3$ gives $x^2=13$. Only the negative root belongs to the composite domain.

Let $y=(4x+6)/(x-5)$. Then $yx-5y=4x+6$, so $x(y-4)=5y+6$ and $x=(5y+6)/(y-4)$. Exchange the variable names. The original function is $4+26/(x-5)$, whose range excludes 4.

The graph of g is only the left half of the downward parabola: $y=5-2x^2$, $x\le0$. Its intercepts are $(0,5)$ and $(-\sqrt{5/2},0)$. For the inverse, $x=5-2y^2$ and $y\le0$, hence $g^{-1}(x)=-\sqrt{(5-x)/2}$, $x\le5$. Reflect the original branch in $y=x$. The inverse intercepts are $(5,0)$ and $(0,-\sqrt{5/2})$.

Since $3x-1\gt 0$ throughout $3\le x\le42$, an antiderivative is

The factor $1/3$ compensates for the derivative of $3x-1$. Thus

Using $\ln(a^3)=3\ln a$ for $a\gt 0$, the exact value in simplest form is

Multiply the given decomposition by $(x-1)^2$:

Comparing coefficients gives

The coefficient of $x$ also agrees, since $A-2B=2-2(-3)=8$. Hence

Integrating each term,

Therefore, for $x\gt 1$,

where $K$ is an arbitrary constant of integration, distinct from the coefficient $C=4$.

Differentiate $y=(3+4e^{x^2})^{1/2}$ using the chain rule:

Since the tangent at $P=(\alpha,\sqrt{3+4e^{\alpha^2}})$ passes through the origin, its gradient is $y_P/\alpha$, with $\alpha\gt 0$. Equate this to the derivative:

Consequently $x=\alpha$ solves the displayed equation.

Let $h(x)=4x^2e^{x^2}-4e^{x^2}-3$, a continuous function. Then $h(1)=-3\lt 0$ and $h(2)=12e^4-3=652.1778\ldots\gt 0$, so a root lies between 1 and 2. Moreover $h'(x)=8x^3e^{x^2}\gt 0$ for $x\gt 0$, so this is the unique positive root corresponding to the tangent abscissa $\alpha$.

From $4x^2e^{x^2}=4e^{x^2}+3$, divide by the positive $e^{x^2}$ to get $4x^2=4+3e^{-x^2}$. The source domain has $x\ge0$, so take the positive square root.

With $x_1=1$, the unrounded iterates start

Continue: $x_6=1.1051212304\ldots$, $x_7=1.1050504628\ldots$, $x_8=1.1050661137\ldots$, $x_9=1.1050626523\ldots$; further iteration settles at $1.1050632792\ldots$. Therefore (i) $x_3=1.0997$ and (ii) $\alpha=1.1051$, both to four decimal places.

By the quotient rule,

For every $x\gt 4$, the denominator is positive and the numerator is negative. Therefore $f'(x)\lt 0$ throughout its domain, so f is decreasing.

Set $y=(5x-3)/(x-4)$. Then $yx-4y=5x-3$, giving $x(y-5)=4y-3$ and $x=(4y-3)/(y-5)$. Exchange the variable names. Since $f(x)=5+17/(x-4)$ on $x\gt 4$, its range is $(5,\infty)$, which becomes the inverse's domain.

(i) For $x\gt 4$, $f(x)\gt 5\gt 4$, so the second application is defined.

Thus $a=22$, $b=-3$, $c=13$. (ii) Rewrite $ff(x)=22-289/(x+13)$ with the original domain $x\gt 4$. It is strictly increasing, tends to 5 as $x\to4^+$ and to 22 as $x\to\infty$. Neither limiting value is attained.

Using the quotient rule and the chain rule,

Thus $P(x)=(2x+5)(2x-17)$ and $Q(x)=(x-3)^2$ are fully factorised quadratics.

For $x\ne3$, the denominator $(x-3)^2$ is positive, so the sign of $f'(x)$ is the sign of $(2x+5)(2x-17)$. For an increasing function we require

Both factors are negative when $x\lt -\frac52$, and both are positive when $x\gt \frac{17}{2}$. Therefore the required values are

For $0\lt x\lt \frac{\pi}{4}$, we have $\sin4x\gt 0$. Applying the product rule and the chain rule gives

Since $g(0)=0$ while $g(x)\gt 0$ in the interior, the stated maximum $M$ does not occur at $x=0$. Its coordinate therefore lies in $0\lt x\lt \frac{\pi}{4}$, where $g$ is differentiable, so $g'(x)=0$ at $M$. Consequently,

At such a point $\cos4x\ne0$: the only interior zero of $\cos4x$ is $x=\frac{\pi}{8}$, where the left-hand side above is $1$, not $0$. Dividing by $\cos4x$ therefore gives

Prices in the model are in thousands of pounds, so

Taking logarithms gives $T=10\ln2=6.9314718\ldots$, hence $\boxed{T=6.93\text{ years}}$.

Thus the answer is $\boxed{-0.97\text{ thousand pounds per year}}$ to 2 decimal places.

At $t=0$, $36=4+m$, so $\boxed{m=32}$.

The initial car price is $7+16=23$ thousand pounds. The difference is $36-23=13$ thousand pounds, or $\boxed{\pounds13\,000}$.

Equating the models with $m=32$ gives

Let $u=e^{-0.1\alpha}\gt 0$. Then $32u^2-16u-3=0$, so

Reject the negative root because an exponential is positive. Hence

This value is inside the model interval $0\leq t\leq15$.

Write $y=(16/9)(3x-k)^{-1}$. The chain rule gives

At $x=1$, the given gradient yields

Hence $3-k=\pm2/3$. Both resulting k-values are positive and different from 3.

Since $k\lt 3$, use $k=7/3$. At $x=1$, $y=16/[9(3-7/3)]=8/3$. The tangent gradient is $-12$, so the normal gradient is $1/12$:

With $k=7/3$, the denominator does not vanish on $[1,3]$. Using the reciprocal-linear antiderivative,

Point $P$ lies on the descending branch of $C_2$, where $5x+b\lt 0$. At $P=(-2,25)$,

hence $a=15+b$.

Point $R=(2,9)$ lies on the ascending branch, so

Combining with $a-b=15$ gives $2a=14$, hence $a=7$ and $b=-8$.

The vertex occurs at $5x-8=0$, giving $(x,y)=(8/5,7)$.

The diagram places $Q$ on the descending branch $y=7-(5x-8)=15-5x$, with $x\lt 8/5$. Equate this with the cubic:

Since $x=-2$ is the known intersection $P$, divide by $x+2$:

The candidates are $-2,3/2,3$. Reject $x=3$ because it is outside the descending branch, and $x=-2$ is $P$, not $Q$. Thus $x_Q=3/2$ and $y_Q=15-5(3/2)=15/2$, giving $Q=(3/2,15/2)$.

Factor the first denominator:

On the stated domain $x\gt 2$, both denominators and $x+4$ are nonzero. Hence

The quotient rule gives

For every $x\gt 2$, the squared denominator is positive; hence $f'(x)\lt 0$. Therefore f is strictly decreasing on its whole domain.

Set $y=3+2\ln x$. Then $\ln x=(y-3)/2$, so $x=e^{(y-3)/2}$. For $x\ge1$, the original range is $y\ge3$. Interchanging the input and output symbols gives the inverse with its domain.

The equation $g(f(a))=5$ gives $3+2\ln(f(a))=5$, so $\ln(f(a))=1$ and $f(a)=e$. Using the simplified f,

The resulting value is approximately $2.2083\gt 2$, and the inner output $e\ge1$ is admissible for g.

At the start, $t=0$ and $e^{0.3t}=1$. Thus $N(0)=600/(2+1)=200$.

Divide numerator and denominator by $e^{0.3t}$ to write $N=600/(1+2e^{-0.3t})$. As $t\to\infty$, the denominator tends to 1, so the upper limit is 600 fish.

Set $N=500$ and multiply through: $1000+500e^{0.3t}=600e^{0.3t}$, so $e^{0.3t}=10$. Hence $t=\ln10/0.3=7.675283643\ldots$ years. The fractional year is $12(0.675283643\ldots)=8.10340372\ldots$ months. Round only the months to the nearest whole month.

The quotient rule gives

The two $e^{0.6t}$ terms cancel exactly.

Let $z=e^{0.3T}$. Since $T\ge0$, require $z\ge1$. Then $360z/(2+z)^2=8$ gives $45z=(2+z)^2$, hence

The smaller root lies between 0 and 1 and would give $T\lt 0$, so reject it. For the larger root,

With $u=6x^2+4x-2$ and $v=2x+1$, the quotient rule gives

At $x=2$, $f'(2)=80/25=16/5$, so the normal gradient is $-5/16$. Through $P=(2,6)$,

Multiplying the proposed identity by $2x+1$ gives

Compare coefficients: $2A=6$, $A+2B=4$, $B+D=-2$. Thus $f(x)=3x+\frac12-\frac{5}{2(2x+1)}$.

The curve meets the $x$-axis when $6x^2+4x-2=2(3x-1)(x+1)=0$. The root $x=-1$ is outside $x\gt -1/2$, leaving $x=1/3$. The normal meets the axis at $x=106/5$. Split the shaded region at $P$: a curve area from $1/3$ to 2, then a triangle.

The triangle has area $\frac12(106/5-2)\cdot6=288/5$. Adding gives the exact area of $R$.

Factor the denominator as $(x-4)(x+3)$ and multiply the proposed identity through by it:

The right side expands to $x^4-x^3+(P-12)x^2+(Q-P)x+3Q-12P$. Comparing coefficients gives $P-12=-10$, so $P=2$, and $Q-P=3$, so $Q=5$. The constant check is $3(5)-12(2)=-9$, as required. The identity holds where the original fraction is defined: the stated $x\gt -3$ still excludes $x=4$.

From (a), $g(x)=x^2+2+5/(x-4)$ on the curve domain. Thus $g'(x)=2x-5/(x-4)^2$. At $x=2$, $g(2)=4+2-5/2=7/2$ and $g'(2)=4-5/4=11/4$. The tangent through $(2,7/2)$ is

The source shading is above the x-axis from $x=0$ to $x=2$, so integrate g over those bounds.

The absolute value is essential because $x-4\lt 0$ throughout this interval. Evaluating gives

Thus $a=20/3$ and $b=-5$ in the requested form.