If p and q are real number such that p+q=3, p^4+q^4=369, then the value of 1p+1q^(-2) is equal to

Correct answer: 4. Given, p^4+q^4=369⇒p2+q2^2-2pq^2=369 ⇒p+q^2-2pq2-2pq^2=369 ⇒9-2pq^2-2pq^2=369 ⇒pq^2-18pq-144=0 ⇒pq=24(rejected),-6 Now, 1p+1q^(-2)=pq^2p+q^2=pq^29=4

Q52 / 90 · JEE Main 2022 (26 Jun Shift 2) · Mathematics

If $p$ and $q$ are real number such that $p + q = 3, p^{4} + q^{4} = 369$ , then the value of ${(\frac{1}{p} + \frac{1}{q})}^{- 2}$ is equal to

Correct answer: 4

Solution

Given,

$p^{4} + q^{4} = 369 \Rightarrow {(p^{2} + q^{2})}^{2} - 2 {(p q)}^{2} = 369$

$\Rightarrow {({(p + q)}^{2} - 2 p q)}^{2} - 2 {(p q)}^{2} = 369$

$\Rightarrow {(9 - 2 p q)}^{2} - 2 {(p q)}^{2} = 369$

$\Rightarrow {(p q)}^{2} - 18 (p q) - 144 = 0$ $\Rightarrow p q = 24 (rejected), - 6$

Now, ${(\frac{1}{p} + \frac{1}{q})}^{- 2} = \frac{{(p q)}^{2}}{{(p + q)}^{2}} = \frac{{(p q)}^{2}}{9} = 4$

Hence, the correct answer is 4.

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If p and q are real number such that p+q=3, p4+q4=369, then the value of 1p+1q-2 is equal to

Solution

Relevant Questions

If $p$ and $q$ are real number such that $p + q = 3, p^{4} + q^{4} = 369$ , then the value of ${(\frac{1}{p} + \frac{1}{q})}^{- 2}$ is equal to