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Mathematics Revision Solutions

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TSR Wiki > Study Help > Subjects and Revision > Subject Guides > Mathematics > Example Questions and Exam Papers > Revision Solutions

This page contains worked solutions to all Mathematics Revision Questions.

Solutions

M0001

\frac{x}{(x+1)(x+3)} + \frac{x+12}{x^{2}-9}

Recall that x^{2}-9 factorises to (x+3)(x-3).

=\frac{x}{(x+1)(x+3)} + \frac{x+12}{(x+3)(x-3)}

Put both fractions over a common denominator.

=\frac{x(x-3)+(x+12)(x+1)}{(x+1)(x+3)(x-3)}

Expand out all terms on the top.

=\frac{2(x^{2}+5x+6)}{(x+1)(x+3)(x-3)}

Factorise the numerator.

=\frac{2(x+2)(x+3)}{(x+1)(x+3)(x-3)}

Cancel the factor of (x+3).

=\frac{2(x+2)}{(x+1)(x-3)}

Discussion on TSR Mathematics Forum

M0002

x^2+2x \ge 3

Subtract 3 from both sides and factorise.

(x+3)(x-1) \ge 0

Now sketch the graph of y=(x+3)(x-1). This is a parabola through (-3,0) and (1,0), with a minimum at x=-1.

You are trying to find all the regions in which y \ge 0. These should be x \le -3 and x \ge 1.

Discussion on TSR Mathematics Forum

M0003

x\frac{d^2y}{dx^2}-2\frac{dy}{dx} + x = 0

We let \frac{dy}{dx}=v as the question suggests. Then \frac{d^2y}{dx^2}=\frac{dv}{dx}.

Now substitute these into the differential equation,

x\frac{dv}{dx}-2v=-x and divide through by x to get \frac{dv}{dx}-\frac{2}{x}v=-1.

Now calculate the Integrating Factor: e^{\small\int \frac{-2}{x}\,dx} = e^{-2\ln(x)} = e^{\ln(x^{-2})} = x^{-2}

Multiplying the whole differential equation by the IF, we get (x^{-2})\frac{dv}{dx} - 2(x^{-3})v = -x^{-2}

Now we spot that the LHS of the equation is equal to \frac{d}{dx}(vx^{-2}).

So vx^{-2}=\normalsize\int -x^{-2} \,dx and hence vx^{-2}=\frac{1}{x}+c.

Now we can multiply through by x^2 to see that v=x+cx^2.

Recalling that v=\frac{dy}{dx}, we now write \frac{dy}{dx}=x+cx^2

Integrating, we arrive at the solution.

y=\frac{x^2}{2}+\frac{cx^3}{3}+d

Discussion on TSR Mathematics Forum

M0004

y=\frac{e^{x}}{x-1}

Differentiate once by using the quotient rule.

\frac{dy}{dx}=\frac{e^{x}(x-1)-e^x}{(x-1)^2}

Now simplify:

\frac{dy}{dx}=\frac{e^{x}(x-2)}{(x-1)^2}

And differentiate again, using the quotient rule.

\frac{d^{2}y}{dx^2}=\frac{\frac{d}{dx}\left(e^{x}(x-2)\right)(x-1)^2-\frac{d}{dx}\left((x-1)^2\right)e^{x}(x-2)}{(x-1)^4}

Note that this differentiation also requires the use of the product rule:

\frac{d^{2}y}{dx^2}=\frac{\left(e^x(x-2)+e^x\right)(x-1)^2-(2x-2)\left(e^{x}(x-2)\right)}{(x-1)^4}

Finally, simplify and cancel a factor of (x-1).

\frac{d^{2}y}{dx^2}=\frac{e^x(x^2+4x+5)}{(x-1)^3}

M0005

The trick here is to use multiple applications of the chain rule, very carefully.

\frac{d}{dx}\left(\ln(x + \sqrt{x^{2} +1})\right) =\frac{\frac{d}{dx}\left(x+\sqrt{x^{2}+1}\right)}{x+\sqrt{x^{2} +1} =\frac{1+\frac{d}{dx}(x^2+1)^{\frac{1}{2}}}{x+\sqrt{x^{2} +1} =\frac{1+\frac{1}{2}\frac{1}{\sqrt{x^2+1}}(2x)}{x+\sqrt{x^2+1}}

Now we simplify.

=\frac{1+\frac{x}{\sqrt{x^2+1}}}{x+\sqrt{x^2+1}}

Both the numerator and denominator should be multiplied by \sqrt{x^2+1}.

=\frac{\sqrt{x^2+1} + x}{x\sqrt{x^2+1} + x^2+1}

On the denominator there is a factor of \sqrt{x^2+1} that can be taken out, leaving a common factor to be cancelled through.

=\frac{(\sqrt{x^2+1} + x)}{\sqrt{x^2+1}(x + \sqrt{x^2+1})} =\frac{1}{\sqrt{x^2+1}}

Discussion on TSR Mathematics Forum

M0006

We are told that H(H(y))=y.

Applying H to both sides, we see that H(H(H(y)))=H(y). Similarly, H(H(H(H(y))))=H(H(y))=y.

You are expected to spot that if H is applied an odd number of times, it is equivalent to H(y), and if it is applied an even number of times, it is equivalent to y.

Thus H(H(H(...(H(y))...) 80 times is y, and 81 times is H(y).

Discussion on TSR Mathematics Forum

M0007

Initially the particle has acceleration a = (6t - 10).

We can integrate to find velocity: v = 3t2 - 10t + c.

Initially the velocity is 3 ms-1, so c = 3.

Now integrating again, we calculate distance: x = t3 - 5t2 + 3t + d

The particle begins at O, so d = 0.

We need to find the distance travelled in the first 2 seconds. However, this is not as simple as substituting t = 2 into the equation for x.

If we factorise the velocity, v = (3t-1)(t-3), we find that v=0 when t= 1/3 or 3.

So having travelled a certain distance, at t= 1/3 the particle turns around and comes back.

The total distance it travels is thus the distance travelled when t = 1/3, added to the distance travelled between then and when t = 2.

Dist = [x]_{t=0}^{t=1/3} - [x]_{t=1/3}^{t=2} = 2[x]_{t=1/3} - [x]_{t=2} = 2[t^3-5t^2+3t]_{t=1/3} - [t^3-5t^2+3t]_{t=2} = \frac{188}{27}

Discussion on TSR Mathematics Forum

M0008

\sin(x)+\cos(x)=1

Recall that \sin(x+\alpha)\equiv\sin(x)\sin(\alpha)+\cos(x)\cos(\alpha)

So we need to find some R,\,\alpha such that \sin(x)+\cos(x)\equiv R\sin(x+\alpha)\equiv R\sin(x)\sin(\alpha)+R\cos(x)\cos(\alpha).

Equating coefficients:

sin(x): 1=R\sin(\alpha) [1]

cos(x): 1=R\cos(\alpha) [2]

[1]2+[2]2: R(\sin^2(\alpha)+\cos^2(\alpha))=2 so R=\sqrt{2}.

[1] / [2]: 1=\tan(\alpha) so \alpha = \frac{\pi}{4}

So we need to solve \sqrt{2}\sin(x+\frac{\pi}{4})=1

This is easy, provided we remember not to lose any solutions.

\sin(x+\frac{\pi}{4})=\frac{1}{\sqrt{2}} so x+\frac{\pi}{4}=\left(\frac{\pi}{2} \pm \frac{\pi}{4}\right) + 2\pi n for any n\in \mathbb{Z}.

Thus x=\frac{\pi}{4}(1 \pm 1) + 2\pi n

There are many other ways of approaching this question, and these are listed on the thread linked below.

Discussion on TSR Mathematics Forum

M0009

\frac{dy}{dx} - 2y\,cosec(x) = \tan(\frac{x}{2})

This is an exact first order differential equation of the form \frac{dy}{dx} + P(x)y = f(x). It should be solved using the Integrating Factor method.

IF: e^{\normalsize\int -2cosec(x)\,dx} = e^{-2ln|\tan\frac{x}{2}|} = \frac{1}{(\tan^2 \frac{x}{2})}

Multiply both sides of the differential equation by the integrating factor.

\frac{1}{(\tan^2 \frac{x}{2})}\left(\frac{dy}{dx}\right) - \frac{2ycosecx}{(\tan^2 \frac{x}{2})} = \frac{\tan\frac{x}{2}}{(\tan^2\frac{x}{2})}

\frac{d}{dx}\left(\frac{y}{(\tan^2\frac{x}{2})}\right) = \frac{1}{{\tan\frac{x}{2}}

Now integrate both sides, and simplify:

\frac{y}{(\tan^2\frac{x}{2})} = \int \cot\frac{x}{2}\, dx

\frac{y}{(\tan^2\frac{x}{2})} = 2\ln\left|\sin\frac{x}{2}\right| + c

y\cot^2\left(\frac{x}{2}\right) = 2\ln\left|\sin\left(\frac{x}{2}\right)\right| + c

Discussion on TSR Mathematics Forum

M0010

\large 3x^2 + 4y^2 = 36

\large \frac{x^2}{12} + \frac{y^2}{9} = 1

\large a^2 = 12

\large b^2 = 9

\large b^2 = a^2(1-e^2)

\large 9 = 12(1-e^2)

\large \frac{3}{4} = 1 - e^2

\large e^2 = \frac{1}{4}

\large e = \frac{1}{2}

Foci coordinates is (ae,0)

\large a = 2\sqrt3

\large ae = \frac{2\sqrt3}{2} = \sqrt3

Therefore (rt3,0) is the foci.

Okay so Q is the intersection of the line SP and the line perpendicular to the tangent at P.

Equation of SP

\large \frac{dy}{dx} = \frac{bsin\theta}{acos\theta - \sqrt3} = \frac{3sin\theta}{2\sqrt3cos\theta - \sqrt3}

Passes through (rt3,0)

\large y = \frac{3sin\theta}{2\sqrt3cos\theta - \sqrt3}(x - \sqrt3)

Equation of line perpendicular to the tangent at P.

The gradient of any tangent to an ellipse is \large -\frac{bcos\theta}{asin\theta}

Therefore

\large -\frac{3cos\theta}{2\sqrt3sin\theta}

So the perpendicular gradient is [tex]\large \frac{2\sqrt3sin\theta}{3cos\theta} </tex>

Passes through (0,0)

Therefore

\large y = \frac{2\sqrt3sin\theta x}{3cos\theta}

Okay so Q is the intersection of these two equations

\large \frac{2\sqrt3sin\theta x}{3cos\theta} = \frac{3sin\theta(x - \sqrt3)}{\sqrt3(2cos\theta - 1)}

6sin\theta(2cos\theta-1)x = 9sin\theta cos\theta(x-\sqrt3)

12xsin\theta cos\theta -6xsin \theta = 9sin\theta cos\theta x -9\sqrt3sin\theta cos \theta

3xsin\theta cos\theta -6xsin\theta = -9\sqrt3 sin\theta cos \theta

x(3sin\theta cos\theta - 6sin\theta = -9\sqrt3 sin\theta cos \theta

\large x = \frac{-9\sqrt3 sin \theta cos \theta}{3sin\theta cos\theta - 6sin \theta}

x = \frac{-9\sqrt3 sin \theta cos \theta}{3sin\theta(cos\theta -2)}

x = \frac{-3\sqrt3cos\theta}{cos\theta-2}

x = \frac{3\sqrt3cos\theta}{2-cos\theta} First part of the parametric equation.

Rearrange \large y = \frac{2\sqrt3sin\theta x}{3cos\theta} to get x on its own.

\large x = \frac{3cos\theta y}{2\sqrt3sin\theta}

Sub this into \large y = \frac{3sin\theta}{\sqrt3(2cos\theta-1)}(\frac{3cos\theta y}{2\sqrt3sin\theta} - \sqrt3)

\large y = \frac{9\sin\theta cos\theta y}{6\sin\theta(2\cos\theta-1)} - \frac{3\sin\theta}{2\cos\theta-1}

y = \frac{9cosy}{6(2cos\theta-1)} - \frac{3sin\theta}{2cos\theta-1}

y(1-\frac{9cos\theta}{6(2cos\theta-1)} = - \frac{3sin\theta}{2cos\theta-1}

y(1-\frac{3cos\theta}{2(2cos\theta-1)} = -\frac{3sin\theta}{2cos\theta-1}

y = -\frac{\frac{3sin\theta}{2cos\theta-1}}{1-\frac{3cos\theta}{2(2cos\theta-1)}

y = -\frac{3sin\theta}{2cos\theta-1 -\frac{3cos\theta}{2}}

So!

y= -\frac{3sin\theta}{\frac{cos\theta}{2}-1}

y = -\frac{6sin\theta}{cos\theta-2}

Finally Locus of Q

\large x = \frac{-3\sqrt3cos \theta}{(cos\theta - 2)}

y = \frac{6sin\theta}{2-cos\theta}

Okay if the parametric equation gives us a circle,

The diameter is going to cross the x axis twice and therefore when y = 0

so

\frac{6sin\theta}{2-cos\theta} = 0

6sin\theta = 0

theta = 0,\pi

At these point

x = \frac{3\sqrt3cos0}{cos0-2} = 3\sqrt3

x = \frac{3\sqrt3cos\pi}{cos\pi-2} = -\sqrt3

Therefore diameter is 4\sqrt3 and therefore radius is 2\sqrt3

Therefore r^2 = 12

Its a circle shifted root 3 to the left.

(x-\sqrt3)^2 + y^2 = 12

Now to test that the parametric equation does satisfy this cartesian equation. We sub the values in.

x-\sqrt3 = \frac{3\sqrt3cos \theta}{(2-cos\theta)} - \sqrt3

= \frac{3\sqrt3cos\theta-\sqrt3(2-cos\theta}{2-cos\theta}

= \frac{3\sqrt3cos\theta +\sqrt3cos\theta -2\sqrt3}{2-cos\theta}

= \frac{4\sqrt3cos\theta -2\sqrt3}{2-cos\theta}

= \frac{2\sqrt3(2cos\theta -1)}{2-cos\theta}

therefore

Subbing x and y into the cartesian equation.

(\frac{2\sqrt3(2cos\theta -1)}{2-cos\theta})^2 + (\frac{6sin\theta}{2-cos\theta})^2 = 12

\frac{12(2cos\theta-1)^2}{(2-cos\theta)^2} + \frac{36sin^2\theta}{(2-cos\theta)^2} = 12

12(2cos\theta-1)^2 + 36sin^2\theta = 12(2-cos\theta)^2

12(4cos^2\theta - 4cos\theta +1) + 36(1-cos^2\theta) = 12(cos^2\theta - 4cos\theta + 4)

48cos^2\theta - 48cos\theta + 12 + 36 -36cos^2\theta = 12cos^2\theta - 48cos\theta + 48

12cos^2\theta - 48cos\theta + 48 = 12cos^2\theta - 48cos\theta + 48

This means that it can be clearly seen that

0 = 0

Thus, the parametric equation

\large x = \frac{-3\sqrt3cos \theta}{(cos\theta - 2)}

\large y = -\frac{6sin\theta}{cos\theta-2}

Does satisfy the equation (x-\sqrt3)^2 + y^2 = 12 and therefore must represent a circle. This means that the locus of Q must be a circle.

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