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Tutor profile: Kieran F.

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Kieran F.
Qualified math teacher with 10+ years experience.
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Questions

Subject: Pre-Calculus

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Question:

Solve the equation $$\sin(2\theta) - \cos^2(\theta)=0$$, giving values of $$\theta$$ in the interval $$0^{\circ} \le \theta \le 360^{\circ}$$.

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Kieran F.
Answer:

First use the identity $$\sin(2\theta) \equiv 2\sin(\theta)\cos(\theta)$$. $(\begin{align} \sin(2\theta) - \cos^2(\theta) &=0\\ 2\sin(\theta)\cos(\theta) - \cos^2(\theta) &=0\\ \cos(\theta)\left(2\sin(\theta)-\cos(\theta) \right) &=0 \end{align}$) Hence, $$\cos(\theta) = 0$$ or $$2\sin(\theta)-\cos(\theta)=0$$. For $$\cos(\theta) = 0$$ we have $$\theta = 90, 270$$. For $$2\sin(\theta)-\cos(\theta)=0$$ we need to solve further, $(\begin{align} 2\sin(\theta)-\cos(\theta)&=0\\ 2\sin(\theta)&=\cos(\theta)\\ \tan(\theta)&=\frac{1}{2} \end{align}$) Hence for $$\tan(\theta)=\frac{1}{2}$$ we have $$\theta = 26.6, 206.6$$ correct to 1 decimal place.

Subject: Calculus

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Question:

A curve is defined as $$y=\sqrt{x^2+x+5}$$. Find the equation of a tangent to the curve at $$x=p$$ in the form $$y-y_1=m(x-x_1)$$. Hence show that there is one point on this curve such that the tangent line passes through the origin and find its exact coordinates.

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Kieran F.
Answer:

To find a tangent equation we need the gradient of $$y$$. Given that $$y=\sqrt{x^2+x+5}$$, let $$u = x^2+x+5$$ then $$y = u^{\frac{1}{2}}$$. $(\begin{align*} \frac{du}{dx} &= 2x+1, \quad\quad \frac{dy}{du} = \frac{1}{2}u^{-\frac{1}{2}}\\ \frac{dy}{dx} &= \frac{2x+1}{2\sqrt{x^2+x+5}} \quad\quad \text{by the chain rule}. \end{align*}$) When $$x=p$$ the equation of a tangent will be $$y - \sqrt{p^2+p+5} = \frac{2p+1}{2\sqrt{p^2+p+5}}(x-p)$$. If the tangent passes through the origin we can substitute $$x=0$$ and $$y=0$$ into our general tangent equation and then solve for $$p$$. $(\begin{align*} - \sqrt{p^2+p+5} &= \frac{2p+1}{2\sqrt{p^2+p+5}}(-p)\\ \sqrt{p^2+p+5} &= \frac{2p^2+p}{2\sqrt{p^2+p+5}}\\ 2(p^2+p+5) &= 2p^2+p\\ 2p+10 &= p\\ p &= -10. \end{align*}$) Substituting to get the $$y$$-value we obtain $$y=\sqrt{95}$$, hence a tangent to the curve at $$(-10, \sqrt{95})$$ will pass through the origin.

Subject: Algebra

TutorMe
Question:

Find the values of $$k$$ such that $$y=5x+3$$ is tangent to the curve $$y=4x^2+kx+12$$.

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Kieran F.
Answer:

If $$y=5x+3$$ is tangent to the curve $$y=4x^2+kx+12$$ then these functions will intersect at a single point. Setting them equal and equating to zero gives us: $(\begin{align} 5x+3 &= 4x^2 + kx+12\\ 0 &= 4x^2 + (k-5)x+9 \end{align} $) Now we want to know which values of $$k$$ give one root for this equation. Since it is a quadratic, we can use the discriminant $$b^2 -4ac =0$$. $(\begin{align} 0 &= (k-5)^2-4\cdot(4)\cdot(9)\\ 0 &= (k-5)^2 - 144\\ 0 &= k^2 - 10k + 25 -144\\ 0 &= k^2 -10k -119\\ 0 &= (k+7)(k-17) \end{align} $) Hence $$k=-7$$ or $$k=17$$. The reader is encouraged to check these answers using a graphing calculator.

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