Questions of the Week - StudyWell

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.

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Question of the Week – Week 23

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

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Question of the Week – Week 22

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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 $ACBE$ .

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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 $1.025^8$ .

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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.

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Question of the Week – Week 18

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

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Question of the Week – Week 17

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

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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 $g(y)=2\left(3^{3y}\right)-5\left(3^{2y}\right)-9\left(3^y\right)+18$ .

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Question of the Week – Week 15

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

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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.

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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.