माना समीकरण \(x^2-\sqrt{2} x+\sqrt{6}=0\) के मूल \(\alpha, \beta\) हैं तथा समीकरण \(x^2+a x+b=0\) के मूल \(\frac{1}{\alpha^2}+1, \frac{1}{\beta^2}+1\) हैं। तब समीकरण \(x ^2-( a + b -2) x +( a + b +2)=0\) के मूल

A अवास्तविक संमिश्र संख्याएँ हैं
B वास्तविक तथा दोनों ऋणात्मक है।
C वास्तविक तथा दोनों धनात्मक हैं।
D वास्तविक तथा उनमें से ठीक एक धनात्मक है

Verified Solution

Answer & Solution

Correct Answer

(B) वास्तविक तथा दोनों ऋणात्मक है।

Step-by-step Solution

Detailed explanation

\(a=\frac{-1}{\alpha^{2}}-\frac{1}{\beta^{2}}-2\) \(b =\frac{1}{\alpha^{2}}+\frac{1}{\beta^{2}}+1+\frac{1}{\alpha^{2} \beta^{2}}\) \(a + b =\frac{1}{(\alpha \beta)^{2}}-1=\frac{1}{6}-1=-\frac{5}{6}\) \(x^{2}-\left(-\frac{5}{6}-2\right) x+\left(2-\frac{5}{6}\right)=0\)…

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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