GATE CSE 2024 | Set 1 | GA: 5

Question

For positive non-zero real variables $p$ and $q$, if

$\log \left(p^2+q^2\right)=\log p+\log q+2 \log 3,$

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

• A. $79$
• B. $81$
• C. $9$
• D. $83$

Answer 1

Given that, $\log(p^2 + q^2) = \log p + \log q + 2 \log 3$

We can assume it's a $\log$ base $10$

$\log_{10}(p^2 + q^2) = \log_{10} p + \log_{10} q + 2 \log_{10} 3$

$\log_{10}(p^2 + q^2) = \log_{10} (pq) + \log_{10} 3^2$

$\log_{10}(p^2 + q^2) = \log_{10} (9pq) $

$p^2+q^2 = 9pq$

We can take a square on both sides.

$(p^2+q^2)^2 = (9pq)^2$

$p^4 + q^4 + 2p^2q^2 = 81p^2q^2$

$p^4 + q^4  = 81p^2q^2 - 2p^2q^2$

$p^4 + q^4  = 79p^2q^2$

$\dfrac{p^4 + q^4}{p^2q^2} = 79$

Correct Answer: A

Comments

Once you have substituted $p=p^2$ and $q=q^2,$ you have restricted the values of $p$ and $q$ because $p=p^2$ and $q=q^2$ implies non-zero $p=1$ and $q=1$ and it does not satisfy $\log(p^2+q^2)=\log p+\log q+2\log3.$

You can find $\frac{p^4+q^4}{p^2q^2}$ if some $p,q$ would satisfy

$\log(p^2+q^2)=\log p+\log q+2\log3$

$p^2=p$

$q^2=q$

$\log(p^4+q^4)=\log p^2+\log q^2+2\log3$  

but you will not get any $(p,q)$ which satisfy all these 4 equations simultaneously.

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@ankitgupta.1729

I feel $p^2$ and $q^2$ doesn't need to be the alternative solutions only when $p = p^2$ and $q = q^2$.

In other words, if $p = p^2$ and $q = q^2$, then obviously $p^2$ and $q^2$ will also satisfy the given equation.

However, the equation may have both pairs of solutions $(p,q)$ and $(p^2, q^2)$ even though $p^2 \neq p$ and $q^2 \neq q$.

Please correct me.

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@Sachin Mittal 1

of course if $p=p^2$ and $q=q^2$ then $(p^2,q^2)$ will also satisfy the equation because $(p,q)$ satisfies the equation but in this way you will be imposing the constraint as $p=1$ and $q=1.$ 

Suppose in the real domain you have $f(x,y)=0$

and you have only one point $(x_0,y_0)$ satisfies $f(x,y)=0$ i.e. $f(x_0,y_0)=0$ 

Now, if you say $x=x^2$ and $y=y^2$ for all the points $(x,y)$ it means $x_0^2=x_0$ and $y_0^2=y_0$ are true and that implies $x_0=0,1$ and $y_0=0,1$.

So, in this way you are imposing some constraints on the points that satisfy the given equation.  

Answer 2

Easy and Simple