Questions of the Week - StudyWell

Question of the Week – Week 43

Find, in terms of m, the coordinates of the point (1,0) after being reflected in the line $y=mx$ .

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Question of the Week – Week 42

Find rational numbers x and y such that $4^x\times 3^y=24$ .

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Question of the Week – Week 41

Solve $\frac{1}{3^{x}-1}+3^{1-x}=1.5$

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Question of the Week – Week 40

The points (0,6), (9,0), (0,-6) and (-k,0) lie on the circumference of a circle. Find the value of k.

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Question of the Week – Week 39

The diagram shows the curve with equation $y=-x(x-2)(x+1)$ . Find the area shaded shown in green.

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Question of the Week – Week 38

Given that $f(x)=x^{\frac{1}{2}}+9x^{-\frac{1}{2}}$ (and $x\ne 0$ ), find the values of x for which f(x) is a decreasing function.

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Question of the Week – Week 37

Prove that $\frac{\cos^4(\theta)-\sin^4(\theta)}{\cos^2(\theta)}\equiv 1-\tan^2{\theta}$ .

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Question of the Week – Week 36

The points A and B have position vectors $3{\bf i}-4{\bf j}$ and $2{\bf j}-7{\bf i}$ respectively. The point P splits the line vector $\overrightarrow{AB}$ in the ratio 2:3. Find the position vector $\overrightarrow{OP}$ .

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Question of the Week – Week 35

A ship leaves a port and heads to its destination for 65 km on a bearing of $071^\circ$ before receiving a distress signal. The distress signal is sent from a vessel with a position that is 54km away from and at a bearing of $122^\circ$ relative to the port. Calculate the distance between the vessel in distress and the ship.

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Question of the Week – Week 34

From C2 Edexcel Paper June 2013.

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Question of the Week – Week 33

From C2 Edexcel Paper May 2014.

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Question of the Week – Week 32

Point P with x-coordinate 0.5 lies on the curve with equation $y=2x^2$ . The normal to the curve at point P intersects the curve at points P and Q. Find the coordinates of Q.

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Question of the Week – Week 31

From Edexcel C2 June 2017 exam.

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Question of the Week – Week 30

From Edexcel C2 January 2010 exam.

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Question of the Week – Week 29

The gradient of a curve is given by $\frac{dy}{dx}=\frac{\left(3-\sqrt{x}\right)^2}{\sqrt{x}}$ for $x\textgreater 0$ . Given that $y=\frac{2}{3}$ when $x=1$ , find y in terms of x.

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Question of the Week – Week 28

Given that the equation $20x^2=4kx-14kx^2+2$ , where k is a constant, has no real roots, find the set of possible values of k.

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Question of the Week – Week 27

Simplify $x\left(2x^{-\frac{1}{4}}\right)^4$ .

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Question of the Week – Week 26

Given that $f$ for $x\textgreater 0$ , and the point $(4,-8)$ lies on the graph of $y=f(x)$ , find $f(x)$ simplifying each term.

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Question of the Week – Week 25

Solve $3^xe^{4x-1}=5$ , giving your answer in the form $\frac{a+\ln(b)}{c+\ln(d)}$ .

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Question of the Week – Week 24

Differentiate $y=2x^3$ from first principles.

Answer: In order to differentiate from first principles, one must identify the function f(x) then calculate as far as possible:

$f$

See more on differentiation from first principles. In this case, $y=f(x)=2x^3$ noting that $f(x+h)$ is found by replacing x, in the expression for f(x), with x+h. It follows that

$\frac{dy}{dx}=\lim_{h\rightarrow 0}\frac{2(x+h)^3-2x^3}{h}=\lim_{h\rightarrow 0}\frac{2x^3+6x^2h+6xh^2+2h^3-2x^3}{h}$
$=\lim_{h\rightarrow 0}(6x^2+6xh+2h^2)=6x^2$

Question of the Week – Week 23

Solve $2\tan(2x)-3\sin(2x)=0$ on the interval $-180^\circ\leq x\leq 180^\circ$ .

Write the equation as $2\frac{\sin(2x)}{\cos(2x)}-3\sin(2x)=0$ and factorise to $\sin(2x)\left(\frac{2}{\cos(2x)}-3\right)=0$ . It is important to factor out sin(2x) here rather than divide by sin(2x) because sin(2x) may be zero and division by zero is not allowed. $\sin(2x)=0$ actually provides some solutions to this equation. Inverting sin gives $2x=-360^\circ, -180^\circ, 0^\circ, 180^\circ, 360^\circ$ . See Trigonometric Graphs and Trigonometric Equations. Note that if $-180^\circ\leq x\leq 180^\circ$ then $-360^\circ\leq 2 x\leq 360^\circ$ .

The remaining solutions are given by solving $\frac{2}{\cos(2x)}=3$ or $\cos(2x)=\frac{2}{3}$ . Hence, $2x=,-311.81^\circ, -48.19^\circ, 48.19^\circ, 311.81^\circ$ .

The full set of solutions to the original equation are $x=-180^\circ, -155.91^\circ, -90^\circ, -24.09^\circ, 0^\circ, 24.09^\circ, 90^\circ, 155.91^\circ, 180^\circ$ .

Question of the Week – Week 22

See Solution.

Question of the Week – Week 21

The line $l_1$ has equation $4y+3=2x$ . Points $A\left(\frac{19}{2},4\right)$ and B (to the left of A) lie on $l_1$ such that $AB=\sqrt{80}$ . The line $l_2$ passes through $C(2,4)$ , is perpendicular to $l_1$ and intersects $l_1$ at the point $D$ . The point $E$ lies on $l_2$ such that the length of $CDE=3$ times the length of $CD$ . Find the area of the quadrilateral $ACBE$ .

The best thing to do first is sketch a diagram containing all of the important information given in the question. We can’t begin to find the area of the given quadrilateral without this.

The quadrilateral can be seen here (dashed) and the area is given by half of the area of the rectangle containing it (dotted). Since we know that $AB=\sqrt{80}$ , the only thing to find is the length $CE$ which we are told is $3\times CD$ . We already know the coordinates of C and so we need the coordinates of D. This is found by solving the simultaneous equations $4y+3=2x$ and $2x+y-8=0$ . The latter equation is that of $l_2$ found using standard techniques. See Straight Lines Example 1 part 2. It follows that the coordinates of D are (3.5,1). Using Pythagoras, the length of CD is $\frac{3}{2}\sqrt{5}$ and so CE is $\frac{9}{2}\sqrt{5}$ . The area of the rectangle is thus $\frac{9}{2}\sqrt{5}\times\sqrt{80}=90$ – see Surds. It follows that the area of the given quadrilateral is 45 square units.

Question of the Week – Week 20

Use the first 3 terms of the binomial expansion of $\left(1+\frac{x}{4}\right)^8$ (in ascending powers of x) to find an approximation for $1.025^8$ .

The first 3 terms of the binomial expansion are given by
$\binom{8}{0}\times 1^8\times \left(\frac{x}{4}\right)^0+\binom{8}{1}\times 1^7 \times\left(\frac{x}{4}\right)^1+\binom{8}{2}\times 1^6 \times\left(\frac{x}{4}\right)^2=1+2x+\frac{7}{4}x^2.$
See more on binomial expansion. To find an approximation for $1.025^8$ we select $\frac{x}{4}=0.025$ , i.e. $x=0.1$ . Hence,

$1.025^8\approx 1+2\times 0.1 +\frac{7}{4}\times 0.1^2=1.2175$ . This is not a bad approximation comparing it with the answer a calculator gives of 1.218402898.

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Question of the Week – Week 19

Sketch the graph of $y=(x+1)(x-k)^2$ where $k\textgreater 0$ , labelling any points of intersection with the axes.

The graph is of a positive cubic. There is a root at x=-1 and a repeated root at x=k where k is positive so on the left of the y-axis. The y-intercept can be found by substituting x=0 into the equation for y.

See more on Cubics.

Question of the Week – Week 18

Prove by contradiction that $\sqrt{2}$ is irrational.

Proof by contradiction usually starts with an assumption that opposes the statement in the question. Going from there, one hopes to meet a contradiction so that the original assumption is ruled out. In this question, we assume that the $\sqrt{2}$ is rational. A rational number can be expressed $\frac{a}{b}$ where a and b are integers with no common factors. It follows that

$\sqrt{2}=\frac{a}{b}\hspace{10pt}\Rightarrow\hspace{10pt} 2=\frac{a^2}{b^2}\hspace{10pt}\Rightarrow\hspace{10pt}a^2=2b^2$

From this last expression we can see that 2 is a factor of $a^2$ and so it must be a factor of a (the factors of a will appear in factors of $a^2$ in pairs). We can write $a=2m$ and so $a^2=4m^2=2b^2$ or $b^2=2m^2$ and so 2 is also a factor of $b^2$ and also b. The original assumption that $\sqrt{2}$ can be written as a rational number a/b where a and b are integers with NO COMMON FACTORS has been violated. Our original assumption must be wrong. We may conclude that $\sqrt{2}$ is irrational.

This proof is quite complicated so you would be expected to memorise it having seen it before.

Question of the Week – Week 17

Given that $8^x\times5^{2x}=2^{2x+2}\times5^{x-1}$ , find the value of $10^x$ .

In order to solve this, one must attempt to express the given powers as a power of 10. $8^x$ can be written as $\left(2^3\right)^x=2^{3x}=2^x\times 2^x\times 2^x$ and we can write $5^{2x}$ as $5^x\times 5^x$ . This gives the left hand side as $2^x\times2^x\times 2^x\times 5^x\times 5^x$ or $2^x\times (2\times 5)^x\times (2\times 5)^x=2^x\times10^x\times 10^x$ . Similarly the right hand side can be written as $2^2\times 2^x\times 2^x\times 5^x\times 5^{-1}=\frac{4}{5}\times 2^x\times 10^x$ . Equating left and right hand sides and moving over to one side gives

$2^x\times 10^x\times 10^x-\frac{4}{5}\times 2^x\times 10^x=0$

$2^x10^x\left(1-\frac{4}{5}\times 10^x\right)=0$

once factorised. $2^x 10^x\ne 0$ and so the term in brackets is zero. It follows that the value of $10^x=\frac{5}{4}$ .

Question of the Week – Week 16

Given that (x-3) is a factor of $f(x)=2x^3-5x^2-9x+18$ , find all the exact solutions of $g(y)=0$ where $g(y)=2\left(3^{3y}\right)-5\left(3^{2y}\right)-9\left(3^y\right)+18$ .

If x-3 is a factor then $f(x)$ can be written as $f(x)=(x-3)(2x^2+x-6)$ . Click here to find out more on how to do this. Note that the question could also say $f(3)=0$ instead of x-3 is factor. These are equivalent statements according to Factor Theorem. The quadratic can also be factorised so that $f(x)=(x-3)(2x-3)(x+2)$ . Noting that $g(y)$ is the same as $f(x)$ if we let $x=3^y$ . Hence, $g(y)=(3^y-3)(2\left(3^y\right)-3)(3^y+2)=0$ gives the roots $3^y=3$ , $3^y=\frac{3}{2}$ and $3^y=-2$ . It follows that the solutions are $y=$ and $\ln_3\left(\frac{3}{2}\right)$ – see Logs. The third equation has no solution for y.

Question of the Week – Week 15

Show that $\int_{k}^{2k}\frac{2}{(2x-k)^2}\,dx$ is inversely proportional to k.

This one is for the students studying calculus and for those who would like a taste of it. Integration by substitution can be used or the reverse chain rule. You could also write $\frac{2}{(2x-k)^2}$ as $2(2x-k)^{-2}$ and guess its integral as follows. We know that $(2x-k)^{-1}$ differentiates to $-2(2x-k)^{-2}$ using the chain rule. Hence, $2(2x-k)^{-2}$ integrates to $-(2x-k)^{-1}+c=-\frac{1}{2x-k}+c$ . For definite integration,

$\int_{k}^{2k}\frac{2}{(2x-k)^2}\,dx=\left[-\frac{1}{2x-k}\right]_k^{2k}=-\frac{1}{4k-k}+\frac{1}{2k-k}=-\frac{1}{3k}+\frac{1}{k}=-\frac{1}{3k}+\frac{3}{3k}=\frac{2}{k}\propto\frac{1}{k}$

Hence, the integral is inversely proportional to k.

Question of the Week – Week 14

The points P(-3,2), Q(9,10) and R(a,4) lie on a circle. Given that PR is a diameter of the circle, show that a=13.

Begin by drawing a rough sketch of the circle. Since PR is a diameter of the circle, angle PQR is a right-angle and PQ is perpendicular to QR. This means that if the gradient of PQ is m then the gradient of QR =-1/m. The gradient of PQ is $\frac{rise}{run}=\frac{10-2}{9+3}=\frac{8}{12}=\frac{2}{3}$ . Hence the gradient of QR=-3/2. It follows that $\frac{10-4}{9-a}=-\frac{3}{2}$ . Solving for a gives a=13.

See more on Circles.

Question of the Week – Week 13

Solve the simultaneous equations:

$y=x-4,\hspace{20pt}2x^2-xy=8$

giving your answers exactly.

The expression for y in the first equation can be substituted into the second then simplified to give:

$2x^2-x(x-4)=8\Rightarrow x^2+4x=8$

Rearranging and completing the square to solve the resulting quadratic as follows:

$x^2+4x-8=0\Rightarrow (x+2)^2-12=0\Rightarrow x=-2\pm\sqrt{12}$

Simplifying the surd gives $x=-2\pm2\sqrt{3}$ with corresponding y solutions given by $y=-6\pm2\sqrt{3}$ . Note that these y solutions are best found by substituting x into the most basic of the original equations.

Question of the Week – Week 12

Given that

$(5-c)^2\textgreater 12,$

find the set of all possible values of c.

Consider $x^2\textgreater 9$ , for example, x has to be greater than 3 or less than -3. Similarly, if $(5-c)^2\textgreater 12$ , 5-c has to be greater than root 12 or less than minus root 12. Subtracting 5 suggests that $-c\textgreater -5+\sqrt{12}$ or $-c\textless-5-\sqrt{12}$ . Multiplying both sides of the inequalities by -1 reverses the sign of the inequality (click here for more on inequalities). Finally, we have $c\textless 5-\sqrt{12}$ or $c\textgreater 5+\sqrt{12}$ .

Question of the Week – Week 11

Find all the values for x such that

$\frac{\log_2(32)+\log_2(16)}{\log_2(x)}=\log_2{(x)}$

Solution:

First think about reading the logs – see the logarithms for more details. $\log_2(32)$ can actually be read as ‘the power of 2 that gives 32’ – we know this to be 5 since $2^5=32$ . Similarly, $\log_2(16)=4$ and so

$\frac{5+4}{\log_2(x)}=\log_2{(x)}.$

Multiply both sides of the equation by $\log_2(x)$ to get

$9=\left(\log_2{x}\right)^2$ .

Note that there is no rule for simplifying a log that is squared – see log rules. Square rooting gives $\log_2{(x)}=\pm 3$ giving the solutions $x=2^{3}=8$ and $x=2^{-3}=\frac{1}{8}$ .

Question of the Week – Week 10

Solve the following equation:

$y-10y^{0.5}+23=0$

If we let $y=x^2$ , we can write the equation as $x^2-10x+23=0$ . We are not able to factorise this quadratic and so we complete the square (you could also use the quadratic formula):

$x^2-10x+23=0\Rightarrow(x-5)^2-2=0\Rightarrow x=5\pm\sqrt{2}$

It follows that

$y=\left(5\pm\sqrt{2}\right)^2=27\pm10\sqrt{2}$

See surds for the final stage of working – make sure to do the full calculations for yourself.

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Question of the Week – Week 9

Simplify

$(2\sqrt{5})^2$
$\frac{\sqrt{2}}{2\sqrt{5}-3\sqrt{2}}$

Solutions:

$(2\sqrt{5})^2=2\sqrt{5}\times2\sqrt{5}=2\times\sqrt{5}\times 2\times\sqrt{5}=2\times 2\times \sqrt{5}\times\sqrt{5}=4\times 5=20$
$\frac{\sqrt{2}}{2\sqrt{5}-3\sqrt{2}}\times \frac{2\sqrt{5}+3\sqrt{2}}{2\sqrt{5}+3\sqrt{2}}=\frac{\sqrt{2}\times2\sqrt{5}+\sqrt{2}\times3\sqrt{2}}{\left(2\sqrt{5}\right)^2-\left(3\sqrt{2}\right)^2}=\frac{2\sqrt{10}+6}{20-18}=\sqrt{10}+3$

This question was used in a recent exam and students did not like question 2 – they were not used to seeing two surds in the denominator. The same principles of rationalising the denominator apply – multiplying the top and bottom of the fraction with the same expression where the minus sign is replaced with a plus eliminates the unwanted surds.

Question of the Week – Week 8

$4x-5-x^2=q-(x+p)^2$

where p and q are integers. Find the values of p and q.

When you are confronted with a question that looks unusual, you should always try to make it look as familiar as possible so you can use the usual methods to tackle it. Start by writing:

$4x-5-x^2=-x^2+4x-5=-\left(x^2-4x+5\right).$

You can then isolate the inside of the brackets and complete the square, i.e.

$x^2-4x+5=(x-2)^2+1$

and it follows that

$-\left(x^2-4x+5\right)=-\left((x-2)^2+1\right)=-1-(x-2)^2$

and so q=-1 and p=-2.

Question of the Week – Week 7

Solve

$\sqrt{2}\left(\cos(x)+\sin(x)\right)=\frac{1}{\cos(x)-\sin(x)}$

for $0\leq x\leq 2\pi$ .

Rearranging to get all trigonometric terms on one side:

$\left(\cos(x)+\sin(x)\right)\left(\cos(x)-\sin(x)\right)=\frac{1}{\sqrt{2}}$

$\cos^2(x)-\sin^2(x)=\frac{1}{\sqrt{2}}.$

Students often get stuck at this point – if they don’t know their double angle formulae very well. Using $\cos^2(x)-\sin^2(x)=\cos(2x)$ we can convert the equation into something solvable:

$2x=\cos^{-1}\left(\frac{1}{\sqrt{2}}\right)=\frac{\pi}{4},\frac{7\pi}{4}, \frac{9\pi}{4}, \frac{15\pi}{4}$

see the graph if cos noting that if $0\leq x\leq 2\pi$ then we need all the solutions for $0\leq 2x\leq 4\pi$ . It follows that the solutions are

$x=\frac{\pi}{8},\frac{7\pi}{8}, \frac{9\pi}{8}, \frac{15\pi}{8}$ .

Question of the Week – Week 6

The equation $kx^2+4kx+3=0$ , where k is a constant, has no real roots.

Prove that $\hspace{30pt}0\textless k \hspace{3pt} \textless \hspace{3pt}\frac{3}{4}.$

The question indicates that the discriminant is negative since we are told there are no roots – see discriminants. It follows that

$(4k)^2-4\times k\times 3\textless 0$

$16k^2-12k\textless 0$

At this point we cannot divide both sides by k as we do not know what the sign of k is and so don’t know whether to reverse the inequality or not. You could consider k positive and negative separately but it is easier to factorise here:

$4k(4k-3)\textless 0$

This quadratic in k can easily be sketched (by considering the roots) to see that the quadratic is only strictly negative between 0 and 3/4 (not inclusive).

Question of the Week – Week 5

Simplify $\log_x(81)\times\log_3(x)$ .

Questions that involve a change of base are quite rare but it is still possible that they will come up in your exam. It should be obvious that a change of base is required because the base is 3 on one of the log expressions whereas it is x on the other. The change of base formula is given (in the Edexel formula booklet) as:

$\log_a(x)=\frac{\log_b(x)}{\log_b(a)}.$

Choosing a=x, x=81 and b=3 in this formula gives:

$\log_x(81)=\frac{\log_3(81)}{\log_3(x)}.$

Substituting this into the original expression gives:

$\log_x(81)\times\log_3(x)=\frac{\log_3(81)}{\log_3(x)}\times\log_3(x)=\log_3(81).$

The final log expression can be evaluated as the power of 3 that gives 81 and so the final answer is 4.

Question of the Week – Week 4

The first three terms of a geometric sequence are

$7k-5, 5k-7, 2k+10,$

where k is a constant. Show that

$11k^2-130k+99=0.$

Given that k is not an integer, show that $\hspace{10pt}k=\frac{9}{11}$ .

Since the sequence is a geometric one the ratio between each of the terms must be the same i.e. $\frac{5k-7}{7k-5}=\frac{2k+10}{5k-7}$ . It follows that $(5k-7)(5k-7)=(2k+10)(7k-5)$ . Expanding gives $25k^2-70k+49=14k^2+60k-50$ and so $11k^2-130k+99=0$ as required.

Question of the Week – Week 3

Prove that for all positive x and y:

$\sqrt{xy}\leq \frac{x+y}{\sqrt{2}}.$

The trick is to spot the square root on the left and hence consider the square of the right. First note that $\left(\frac{x+y}{\sqrt{2}}\right)^2=\frac{(x+y)^2}{2}=\frac{x^2+2xy+y^2}{2}$ . Secondly, $x^2$ and $y^2$ are both positive and so $\frac{2xy}{2}=xy\leq \frac{(x+y)^2}{2}$ . By square rooting both sides, it follows that $\sqrt{xy}\leq \frac{x+y}{\sqrt{2}}$ .

Question of the Week – Week 2

Solve $2^{2x+1}-17\left(2^x\right)=-8.$

Although it may not be obvious at first, the above equation is actually a quadratic. It can be written as

$2\times 2^{2x} -17\left(2^x\right)=-8$

and then as

$2\left( 2^{x}\right)^2 -17\left(2^x\right)+8=0.$

Letting $y=2^x$ , this can be factorised to

$\left( 2y-1\right)\left( y-8\right)=0.$

It follows that $y=2^x=\frac{1}{2}$ or $y=2^x=8$ giving the solutions $x=-1$ or $x=3$ .

Question of the Week – Week 1

Do you know the correct way to solve the following inequality?

$\frac{2}{x-5}\geq 6,\hspace{15pt}x\ne 5$

Most students will instantly multiply both sides by $x-5$ but it is important to consider if $x-5$ is positive or not. This is because if $x-5$ is negative, the inequality sign will be reversed.