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4a) u = 2x+7 implies du = 2dx and x= \frac{u-7}{2} Then \int x\sqrt{2x+7}dx = \int (\frac{u-7}{2})\sqrt{u} (\frac{du}{2})=\frac{1}{4} \int (u-7)\sqrt{u} du = \frac{1}{4} \int (u^{\frac{3}{2}}-7u^{\frac{1}{2}}) = \frac{1}{10}u^{\frac{5}{2}} - 7\frac{1}{6}u^{\frac{3}{2}} To get the...

I thought to share the answer, or at least give a rough idea of how to go about the question. We use induction on n. In the n+1 case, if the number of classes, k , is n+1 then then the statement follows trivially. We need to consider the case where k < n+1 in that case at least there are...

Another idea is to use de moivre's theorem. \textrm{if} z = cos\theta + i sin\theta , z^{-1} = cos\theta - i sin\theta

The deducing part: \textrm{from a)ii } \ \ b^{2} - a^{2} \leq 2\sqrt{ab}(b-a), c^{2} - b^{2} \leq 2\sqrt{bc}(c-b) \textrm{adding them together} \ c^{2} - a^{2} \leq 2(\sqrt{ab}(b-a) + \sqrt{bc}(c-b) ) = 2\sqrt{b}(\sqrt{a}(b-a) + \sqrt{c}(c-b))

\textrm{well, the roots of -1 are} [cis(\frac{\pi}{8}) , cis(\frac{-\pi}{8}) ] ,[ cis(\frac{3\pi}{8}),cis(\frac{-3\pi}{8})] , [cis(\frac{5\pi}{8}), cis(\frac{-5\pi}{8})] , [cis(\frac{7\pi}{8}), cis(\frac{-7\pi}{8})] \textrm{by paring roots in the square brackets you get the first...

ii) First method, start from LHS: \frac{1}{z-1} + \frac{1}{\omega z -1} + \frac{1}{\omega^{2} z-1} \frac{1}{z-1} + \frac{1}{\omega (z -\omega^{2})} + \frac{1}{\omega^{2} ( z-\omega)} \frac{1}{z-1} + \frac{\omega^{2}}{ (z -\omega^{2})} + \frac{\omega}{ ( z-\omega)} \ \ \...

By triangle inequality: |z_{1} + z_{2}| \leq |z_{1}| + |z_{2}| \textrm{letting} \ z_{2} = - z_{2} |z_{1} - z_{2}| - |z_{2}| \leq |z_{1}| \ \ \ \ \ (1) \textrm{letting} \ \ \omega_{1} = z_{1} - z_{2} , \ \ \omega_{2} = z_{2} \textrm{then} \ \omega_{1} + \omega_{2} = z_{1}...

Q24: Total distance travelled is: x+80-\sqrt{x^{2}-400} \textrm{time elapsed} \ \ \ \ \frac{x}{7000} + 80 - \frac{\sqrt{x^{2}-400}}{11000} \textrm{note the speeds are in km/h and the distances are in meters} \textrm{differentiate it and equate it to zero} \ \ \frac{1}{7000} -...

Re: How do I get the exact gradient, with rational denominator, of the normal of para This is slightly more general method: let's deferential with respect to x: \frac{d}{dx}(y^{2} ) = \frac{d}{dx}(12x) 2y \frac{d}{dx}y = 12 \ \ \textrm{(using chain rule) that implies } \textrm{at}...

I haven't posted for a long time, I decided to post another unique induction question: Suppose there are n students in a school with k classes.If for every two classes there is at least one student from each class who are friend with each other, i.e for every class A and B there is one student...

Re: Year 11 Mathematics 3 Unit Cambridge Question & Answer Thread I believe that is the question you are referring to: a) ln(x^{2} + 5x) = 2ln(x+1) = ln((x+1)^{2}) x^{2} + 5x = (x+1)^{2} = x^{2} + 2x+1 3x = 1 implies x = \frac{1}{3} b) log(7x-12) = 2 log(x) = log(x^{2}) x^{2} =...

This is a more trigonometry based solution: letting z_{1} = A_{1} e^{i\theta_{1}} , z_{2}= e^{i\theta_{2}} and using our assumption we get z_{1} + z_{2} = (A_{1} + A_{2})e^{i\phi} z_{1} + z_{2} = A_{1} e^{i\theta_{1}} + A_{2} e^{i\theta_{2}} = (A_{1} + A_{2} ) e^{i\phi} dividing by...

This questions involves some presumed knowledge(You can easily derive them) First: if PQ is a focal length then pq =-1 (Find the question passing through P and Q and use the fact that it passes through focal point (0,a) ) and the slope of tangent at P and Q is p and q respectively...

You don't need to use fancy trig identity. The only thing is to use definition of tanx in terms of sines and cosines. \frac{tan^{2}x}{sin^{2}x}+ \frac{sin \ x}{tan^{2}x} = \frac{\frac{sin^{2}x}{cos^{2}x}}{sin^{2}x} + \frac{sin \ x}{\frac{sin^{2}x}{cos^{2}x}}...

Hi frog1944, well, there is no set general rule that we can assign to every situation. For the most part, it comes from experience. Solving many difficult questions, even beyond extension2, will give you enough experience to tackle most questions.In the nut shell, choose methods that order your...

The answer is nk. $$lim_{x\to \1 }\frac{f(x^{n}-1)(x^{n-1}+...+ 1)}{(x-1)(x^{n-1}+...+1)} = lim_{x\to \ 1}\frac{f(x^{n}-1)(x^{n-1}+..1)}{x^{n}-1}$$ by letting $y = x^{n}-1$ and using assumption we get nk.

Geometric progression.

Yes you're right I checked the matrix and the determinant is indeed zero. I do agree the function is decreasing in each axis but I think the first partial in each direction is strictly positive.

did you evaluate the Hessian matrix at a particular point? Otherwise I don't know how you got det hessian matrix = 0 Also about having negative first partial derivative in each direction, I haven't done any calculation to check whether is negative or not but I take your word for it. That does...

That's a nice method. Jensen's inequality is powerful, it's a shame we don't learn it in 4unit