NIMCET 2011 — Mathematics PYQ

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Question

NIMCET 2011 — Mathematics PYQ

NIMCET | Mathematics | 2011

The minimum value of $px + qy$ , when $xy = r^2$ is:

Choose the correct answer:

Explanation

Step 1: Apply AM-GM to the terms $px$ and $qy$

Let $a = px$ and $b = qy$ .

\frac{px + qy}{2} \geq \sqrt{(px)(qy)}

Step 2: Simplify the expression

Multiply both sides by 2:

px + qy \geq 2\sqrt{pqxy}

Step 3: Substitute the given constraint

We are given that $xy = r^2$ . Substitute this into the inequality:

px + qy \geq 2\sqrt{pq(r^2)}

px + qy \geq 2r\sqrt{pq}

Conclusion:

The minimum value of the expression is $2r\sqrt{pq}$ .

Correct Option:

(a) $2r\sqrt{pq}$