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Official PYQ · 2024

GATE CS 2024 Set 1 Q5 — General Aptitude

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For positive non-zero real variables pp and qq, if

log⁡(p2+q2)=log⁡p+log⁡q+2log⁡3,\log \left(p^2+q^2\right)=\log p+\log q+2 \log 3,

then, the value of p4+q4p2q2\frac{p^4+q^4}{p^2 q^2} is

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Answer: A

Given the logarithmic equation for positive non-zero real variables pp and qq:

log⁡(p2+q2)=log⁡p+log⁡q+2log⁡3\log \left(p^2+q^2\right) = \log p + \log q + 2 \log 3

Using the properties of logarithms, we can simplify the right-hand side:

  1. 2log⁡3=log⁡(32)=log⁡92 \log 3 = \log(3^2) = \log 9
  2. log⁡p+log⁡q+log⁡9=log⁡(9pq)\log p + \log q + \log 9 = \log(9pq)

Substituting this back into the equation:

log⁡(p2+q2)=log⁡(9pq)\log \left(p^2+q^2\right) = \log(9pq)

Since the logarithm function is one-to-one, we can equate the arguments:

p2+q2=9pqp^2 + q^2 = 9pq

To find the value of p4+q4p2q2\frac{p^4+q^4}{p^2 q^2}, we square both sides of the equation:

(p2+q2)2=(9pq)2\left(p^2 + q^2\right)^2 = (9pq)^2 p4+q4+2p2q2=81p2q2p^4 + q^4 + 2p^2q^2 = 81p^2q^2

Subtracting 2p2q22p^2q^2 from both sides:

p4+q4=79p2q2p^4 + q^4 = 79p^2q^2

Dividing both sides by p2q2p^2q^2:

p4+q4p2q2=79\frac{p^4+q^4}{p^2 q^2} = 79

Thus, the correct option is A.

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