AP EAMCET · Maths · Application of Derivatives

Let \(P(x)=x^4+a x^3+b x^2+c x+d\) be such that \(x=0\) is the only real root of \(\mathrm{P}^1(\mathrm{x})=0\). If \(\mathrm{P}(-1) < \mathrm{P}(1)\), then in the interval \([-1,1]\)

A \(\mathrm{P}(-1)\) is not minimum of \(\mathrm{P}(\mathrm{x})\), but \(\mathrm{P}(1)\) is the maximum of \(\mathrm{P}(\mathrm{x})\)
B \(\mathrm{P}(-1)\) is minimum of \(\mathrm{P}(\mathrm{x})\), but \(\mathrm{P}(1)\) is not the maximum of \(\mathrm{P}(\mathrm{x})\)
C Neither \(\mathrm{P}(-1)\) is the minimum nor \(\mathrm{P}(1)\) is the maximum of \(\mathrm{P}(\mathrm{x})\)
D \(\mathrm{P}(-1)\) is the minimum and \(\mathrm{P}(1)\) is the maximum of \(\mathrm{P}(\mathrm{x})\)

Verified Solution

Answer & Solution

Correct Answer

(A) \(\mathrm{P}(-1)\) is not minimum of \(\mathrm{P}(\mathrm{x})\), but \(\mathrm{P}(1)\) is the maximum of \(\mathrm{P}(\mathrm{x})\)

Step-by-step Solution

Detailed explanation

Let \(P'(x) = 4x^3 + 3ax^2 + 2bx + c\). \(x=0\) is the only real root of \(P'(x)=0 \implies P'(0)=0 \implies c=0\). So, \(P'(x) = 4x^3 + 3ax^2 + 2bx = x(4x^2 + 3ax + 2b)\). For \(x=0\) to be the only real root, the quadratic \(4x^2 + 3ax + 2b\) must have no real roots or a…

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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