What is the value of y^0 + y^(–1) when y = 1/2 ?

Question

What is the value of  y^0 + y^{(–1)}  when  y = \frac{1}{2}  ?

(A) 0.5
(B) 1.50
(C) 3
(D) 4
(E) 5

imeanup · Answer

y^0 + y^{-1} = 1+ \frac{1}{y} = 1+ 2 = 3 
 Ans C

Nipungupta9081 · Answer

IMO C

Anything raised to the power of 0 is 1

= 1 + 2 = 3

BrentGMATPrepNow · Answer

Useful property:  (\frac{a}{b})^{-n}=(\frac{b}{a})^{n}   
Example:  (\frac{5}{2})^{-3}=(\frac{2}{5})^{3}

So, we get:  y^0 + y^{(–1)}=(\frac{1}{2})^0 + (\frac{1}{2})^{(–1)}

=1 + (\frac{2}{1})^{1}

=1 + 2

=3

Answer: C

Abhishek009 · Answer

y^0 = 1   &  y^{(–1)}  given  y = \frac{1}{2}  , thus  y = 2

Hence,  y^0 + y^{(–1)} = 1 + 2 = 3 , Answer will be (C)

MathRevolution · Answer

y^0 = 1

y^{-1}  =  \frac{1}{y}

For y =  \frac{1}{2} =   \frac{1}{\frac{1}{2}}

=> 1 + 2 = 3

Answer C

ScottTargetTestPrep · Answer

Solution:

(1/2)^0 + (1/2)^(-1) = 1 + 2 = 3

Answer: C

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