\(a >0\) के लिए, माना वक्र \(C _{1}: y ^{2}= ax\) तथा \(C _{2}: x ^{2}=\) \(ay\), मूलबिंदु \(O\) तथा एक बिंदु \(P\) पर काटते हैं। माना रेखा \(x\) \(= b ,(0< b < a )\), जीवा \(OP\) तथा \(x\)-अक्ष को क्रमशः बिंदुओं \(Q\) तथा \(R\) पर काटती है। यदि रेखा \(x = b\), वक्रों \(C _{1}\) तथा \(C _{2}\) द्वारा परिबद्ध क्षेत्र को समद्विभाजित करती है तथा \(\triangle OQR\) का क्षेत्रफल \(=\frac{1}{2}\) है, तो ' \(a\) ' जिस समीकरण को संतुष्ट करता है, वह है

A \(x^{6}-12 x^{3}+4=0\)
B \(x^{6}-12 x^{3}-4=0\)
C \(x^{6}+6 x^{3}-4=0\)
D \(x^{6}-6 x^{3}+4=0\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(x^{6}-12 x^{3}+4=0\)

Step-by-step Solution

Detailed explanation

\(\int_{0}^{b}\left(\sqrt{a x}-\frac{x^{2}}{a}\right) d x=\frac{1}{2} \times \frac{16\left(\frac{a}{4}\right)\left(\frac{a}{4}\right)}{3}\) \(\Rightarrow\left[\frac{2 \sqrt{a}}{3} x^{3 / 2}-\frac{x^{3}}{3 a}\right]_{0}^{b}=\frac{a^{2}}{6}\)…

Apply the relevant identity from the chapter and substitute the values established in the previous step, keeping the original constraint in view.

Use the simplified expression to isolate the unknown, then plug the result back into the question setup to confirm the working is internally consistent.

Cross-check against a boundary or limiting case. Both the dimensional balance and the qualitative behaviour at the extreme should agree with the candidate answer.

Combine the steps above to identify the correct option. The remaining choices each fail at least one of the consistency, dimensional, or boundary checks.

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