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Tutor profile: Bilal H.

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Bilal H.
Student who has co-taught GCSE Further Maths for the past two years
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Questions

Subject: Calculus

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Question:

Find a general function for $$y$$ given, $$\frac{dy}{dx}$$$$=x\sqrt{y²-4}$$

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Bilal H.
Answer:

Start by using a technique called "the separation of variables" where we isolate the $$x$$ and $$y$$ by multiplying both sides by $$dx$$ and dividing both sides by $$\sqrt{y²-4}$$ to give: $$\frac{1}{\sqrt{y²-4}}$$$$dy=xdx$$ We can then integrate both sides to remove the $$dy$$ and $$dx$$, giving: The integral of $$\frac{1}{\sqrt{y²-4}}$$$$dy=$$ the integral of $$xdx$$ these are both indefinite integrals. The LHS (integral with respect to $$y$$ is a bit more tricky and requires a hyperbolic substitution) by letting $$y=2cosh(u)$$ we find that $$dy=2sinh(u)du$$ this allows the integral to become $$\frac{1}{\sqrt{4cosh²(u)-4}}$$$$(2sinh(u))du$$ which simplifies to the integral of $$\frac{1}{2sinh(u)}$$$$2sinh(u)$$$$du$$ which is just the the integral of $$udu$$ which gives $$u+c$$ Integrating the RHS gives $$\frac{1}{2}$$$$x²+d$$ Thus finally yielding $$y+c=$$$$\frac{1}{2}$$$$x²+d$$ where $$c$$ and $$d$$ are arbirary constants. As both $$c$$ and $$d$$ are arbitrary constants this can be simplified to: $$y=$$$$\frac{1}{2}$$$$x²+A$$ where $$A$$ is an arbitrary constant.

Subject: Physics

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Question:

Explain, in molecular terms, why a pressure of a gas in a sealed container increases when the temperature of the container increases.

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Bilal H.
Answer:

As the temperature of a gas increases, the mean speed of each particle also increases. This increases the number of collisions made by particles of the gas on the walls of the container per second. Thereby, the force on the walls of the container increases, and therefore the pressure increases.

Subject: Algebra

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Question:

Solve for all values of $$x$$ such that $$f(x)=0$$ $$x^3 + 5x^2 + 3x + 15$$

Inactive
Bilal H.
Answer:

To begin, we want to find a factor of $$f(x)$$ in order to find a linear or quadratic term that we can factor out, allowing us to solve this more easily. Looking for terms in the form $$(x+a)$$ or $$(x²+bx+c)$$ Notice that in $$f(x)$$ the final (constant) term is 15. As it is easier I will start by assuming there exists some linear factor in the form $$(x+a)$$ that can be factored out of $$f(x)$$. In order to find this linear factor, I will use the factor theorem which states: if $$f(a)=0$$, therefore $$(x-a)$$ must be a factor of $$f(x)$$. The factors of 15 are: -15, -5, -3, -1, 1, 3, 5, 15. I find it easiest to work away from 0 and check $$f$$(each factor) to find where the output is 0. Check: $$f(1)=24$$ therefore $$(x-1)$$ is not a factor of $$f(x)$$ $$f(-1)=16$$ therefore $$(x+1)$$ is not a factor of $$f(x)$$ Eventually, I find that $$f(-5)=0$$ therefore $$(x+5)$$ must be a factor of $$f(x)$$ Next, I will divide $$f(x)$$ by $$(x+5)$$ in order to find a quadratic I can hopefully factor and then solve. $$\frac{f(x)}{(x+5)}$$=$$(x²+3)$$ which tells us that $$f(x) = (x+5)(x²+3)$$ As we are trying to find the values of $$x$$ such that $$f(x)=0$$ we can make $$(x+5)(x²+3)=0$$ This tells us that either $$(x+5)$$ or $$(x²+3)$$ are $$0$$ So we can consider them individually if $$(x+5)=0$$ therefore $$x=-5$$ If $$(x²+3)=0$$ then we get $$x²=-3$$ which is not solvable in the real world, however in the complex world we can solve this by taking the square root of both sides, leaving: $$x=$$$$ ±\sqrt{3}i $$ Therefore the answers to $$f(x)=0$$ are: $$x=-5$$, $$x=$$$$\sqrt{3}i $$ and $$x=$$$$ -\sqrt{3}i $$

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