MHB STEP Paper 1 Q3 1998: True False Justification

[CaptainBlack]

Just to show that not everything in a STEP paper in difficult, this is an easy question:

Which of the following are true and which false? Justify your answers


(i) \(a^{\ln(b)}=b^{\ln(a)}\), for all \(a,b \gt 0\).


(ii) \(\cos(\sin(\theta))=\sin(\cos(\theta))\), for all real \(\theta\).


(iii) There exists a polynomial \(P\) such that \(|P(\theta)-\cos(\theta)| \lt 10^{ -6 } \) for all real \(\theta\)


(iv) \(x^4+3+x^{-4} \ge 5\) for all \(x\gt 0\).

 

[Amer]

1) True take the ln for both sides

2) False take theta = 0

3) that true using Taylor expansion of cos(theta)

4) How to solve it ?

 

[Evgeny.Makarov]

3) that true using Taylor expansion of cos(theta)

Really?

4) How to solve it ?

I assume z should be replaced by x. One way is to express $x^4+x^{-4}$ through $x+x^{-1}$. One needs to know that $x+x^{-1}\ge2$ for x > 0.

 

[chisigma]

3) that true using Taylor expansion of cos(theta)

The question is that is true for all $\theta$... the function $\cos \theta$ is bounded in $\theta \in \mathbb{R}$, any polinomial $P(\theta)$ which is not a constant is unbounded in $\theta \in \mathbb{R}$...

Kind regards

$\chi$ $\sigma$

 

[CaptainBlack]

1) True take the ln for both sides

It is true, but that is not as it stands a valid explanation, you are assuming it true and deriving a truth, which is invalid logic. You need to start with a known truth and from that derive the equality you are seeking to justify.

2) False take theta = 0

Yes.

3) that true using Taylor expansion of cos(theta)

No, a Taylor expansion is not a polynomial, and a Taylor polynomial does not satisfy what is to be demonstrated for all \(\theta\)

4) How to solve it ?

\(f(x)=x^4+3+x^{-4}\) is continuous and differentiable for \(x\gt 0\), it goes to \(+\infty\) at \(x=0\) and as \(x\to \infty\). It has one stationary point in \( (0,\infty)\) at \(x=1\), which therefore must be a minimum and \(f(1)=5\)

CB

 

[Amer]

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How about
$\ln(a)\ln(b) = \ln(b)\ln(a) $
$\ln(b^{\ln(a)}) = \ln(a^{\ln(b)})$

 

[CaptainBlack]

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How about
$\ln(a)\ln(b) = \ln(b)\ln(a) $
$\ln(b^{\ln(a)}) = \ln(a^{\ln(b)})$

Basically yes, though I would put in some words explaining what you are doing.

CB