question
\(\large \color{black}{\begin{align}&\{a,b\}\in \mathbb{R}\ \ \normalsize \text{if} \ 0<a<1\ \text{and } \ 0<b<1 \ \text{and }\ a>b \hspace{.33em}\\~\\ &\normalsize \text{which of the expressions will take highest value ? } \ \hspace{.33em}\\~\\ &a.)\ a-b \hspace{.33em}\\~\\ &b.)\ a+b \hspace{.33em}\\~\\ &c.)\ \dfrac{a}{b} \hspace{.33em}\\~\\ &d.)\ \normalsize \text{cannot be determined} \hspace{.33em}\\~\\ \end{align}}\)
D, both b and are okay
0.9/0.0000000000000001 = ?
Both B and C
so its c.) ?
yes it is
but book has given option d.) cannot be determined
.99/.98=? .99+.98=?
your book better have some explanation for why it cannot be determined hmm
yea wait
actually it depends on how you interpret the question, i hate these questions
cannot be determined
a+b > a/b and a/b > a+b find the intersection
Since no restrictions on a,b. My example and @ganeshie8 both worked, then it's D
\(\large \color{black}{\begin{align}\normalsize \text{Book: any of}\ (a+b) \ \text{or }\ \dfrac{a}{b} \text{could be greater thus we cannot determine this} \hspace{.33em}\\~\\ \end{align}}\)
yeah i get that, let me ask a side question : consider two functions f(x) = x^2, g(x) = -x^2+1 which function takes highest value ?
f(x)>g(x)
you can find the exact solutions too if u want
y1=a+b y2=a/b solve y1>y2 , 0<a,b<1, a>b
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3D solutions
z=x+y, z=x/y
you could say "clearly it depends on x" you could also say "clearly f(x) takes highest values because it is shooting up" depends on what the author thinks i guess
a=0.9; b=0.1: a/b > a+b a=0.99; b=0.98: a+b > a/b
here is complete solution set
my mind says its option c.)
|dw:1435174861889:dw|
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