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Difference between revisions of "2007 Cyprus MO/Lyceum/Problems"

(→‎Problem 3)
(→‎Problem 4)
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== Problem 4 ==
 
== Problem 4 ==
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We define the operation <math>a*b = \frac{1+a}{1+b^2}</math>, <math>\forall a,b \in \real</math>.
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 +
The value of <math>(2*0)*1</math> is
 +
 +
A. 2
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 +
B. 1
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 +
C. 0
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 +
D. \frac{1}{2}
 +
 +
E. \frac{5}{2}
 +
  
 
[[2007 Cyprus MO/Lyceum/Problem 4|Solution]]
 
[[2007 Cyprus MO/Lyceum/Problem 4|Solution]]

Revision as of 11:10, 6 May 2007

Problem 1

If $x-y=1$,then the value of the expression $K=x^2+x-2xy+y^2-y$ is

A. $2$

B. $-2$

C. $1$

D. $-1$

E. $0$

Solution

Problem 2

Given the formula $f(x) = 4^x$, then $f(x+1)-f(x)$ equals to

A. $4$

B. $4^x$

C. $2$•$4^x$

D. $4^{x+1}$

E. $3$•$4^x$

Solution

Problem 3

A cyclist drives form town A to town B with velocity 40$\frac{km}{h}$ and comes back with velocity 60$\frac{km}{h}$. The mean valocity in $\frac{km}{h}$ for the total distance is

A. 45

B. 48

C. 50

D. 55

E. 100

Solution

Problem 4

We define the operation $a*b = \frac{1+a}{1+b^2}$, $\forall a,b \in \real$.

The value of $(2*0)*1$ is

A. 2

B. 1

C. 0

D. \frac{1}{2}

E. \frac{5}{2}


Solution

Problem 5

Solution

Problem 6

Solution

Problem 7

Solution

Problem 8

Solution

Problem 9

Solution

Problem 10

Solution

Problem 11

Solution

Problem 12

Solution

Problem 13

Solution

Problem 14

Solution

Problem 15

Solution

See also

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