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VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Mathematical Methods CAS => Topic started by: vea on October 28, 2010, 06:32:11 pm

Title: How to find the range of this without using a calculator.
Post by: vea on October 28, 2010, 06:32:11 pm: Hi guys

Was doing a practice exam and came across this question in Neap 2006 Exam 1.

What is the range of $1-9sin^2(2x)$ ?

I tried to guess and said [-8,10] but the answer is actually [-8,1]. Can someone please explain this?

Thanks in advanced.
Title: Re: How to find the range of this without using a calculator.
Post by: jasoN- on October 28, 2010, 06:36:16 pm: need to realise that $9sin^2(2x)$ has values from $0\leq y \leq9$ , as squaring all negative values gives positive
hence $-9sin^2(2x)$ has values from $-9\leq y \leq0$
+1 to everything -> $-8\leq y \leq1$
Title: Re: How to find the range of this without using a calculator.
Post by: vea on October 28, 2010, 06:39:28 pm: Thanks! That makes sense.

Seems like we're going for the same uni course hehe
Title: Re: How to find the range of this without using a calculator.
Post by: LachyMc on October 28, 2010, 06:40:17 pm: Wow that is tricky..
Title: Re: How to find the range of this without using a calculator.
Post by: jasoN- on October 28, 2010, 06:40:23 pm: awesome, but my chances are slim due to fail UMAT, but let's not worry about that :)
Title: Re: How to find the range of this without using a calculator.
Post by: 98.40_for_sure on October 28, 2010, 06:42:23 pm: I just remember DR. T
Dilation -> reflection -> translation in that order