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Bunuel
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case 1: assume p/2 = a
then a^a = a => a=1
--> p/2 =1 or p=2 (sufficient)

case 2: assume 2/p = a
then a^a =a => a=1
--> 2/p = 1 or p=2 (sufficient)

Ans D
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assume P/2=X , then P=2X then option 1 can be converted to
x^x=x
x=1
then p=2
Do the same thing with option 2
Which will again give P=2.
Ans. D
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Bunuel
What is the value of p?

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


(2) \((\frac{2}{p}) ^ {(\frac{2}{p})} = \frac{2}{p}\)



statement 1: As p is in numerator P can be 0 or a definite value =2 if we equate the bases so insufficient
statement 2: P is in the denominator so p not equal 0 and therefore we get a unique value on equating the bases : sufficient

therefore B
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Bunuel
What is the value of p?

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


(2) \((\frac{2}{p}) ^ {(\frac{2}{p})} = \frac{2}{p}\)



statement 1: As p is in numerator P can be 0 or a definite value =2 if we equate the bases so insufficient
statement 2: P is in the denominator so p not equal 0 and therefore we get a unique value on equating the bases : sufficient

therefore B




0^0 is not defined if it is defined as in most cases 0^0 =1
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I am not sure but for option B (-2) also satisfies the value
I think it should b E
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+2 and -2 both satisfy both the statements. Hence, E
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0 and 2 satisfies both the statements.
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Given statements essentially say a^a=a
Only possible for values of 1 or -1.
St 1 - P can be 2 or -2
St 2 - P can be 2 or -2

Hence, Insufficient information given.
IMO E
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Bunuel
What is the value of p?

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


(2) \((\frac{2}{p}) ^ {(\frac{2}{p})} = \frac{2}{p}\)


Are You Up For the Challenge: 700 Level Questions


(1) \((\frac{p}{2}) ^ {(\frac{p}{2})} = \frac{p}{2}\)
Squaring both sides
(p/2)^p = (p/2)^2
p^(p-2) = 2^(p-2) - Eq 1

Here, we need to be careful with EVEN powers.
Eq 1 satisfies both +2, -2 as a solution.

(2) \((\frac{2}{p}) ^ {(\frac{2}{p})} = \frac{2}{p}\)
Just like (1)
2^(2-p) = p^(2-p)
Here as well, +2, -2 both satisfies.

Hence, can't be determined by both. (E)
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