JEE Mains · Maths · STD 12 - 6. Application of derivatives
The curve \(y(x)=a x^{3}+b x^{2}+c x+5\) touches the \(x\)-axis at the point \(P (-2,0)\) and cuts the \(y\)-axis at the point \(Q\), where \(y ^{\prime}\) is equal to \(3\) . Then the local maximum value of \(y ( x )\) is.
- A \(\frac{27}{4}\)
- B \(\frac{29}{4}\)
- C \(\frac{37}{4}\)
- D \(\frac{9}{2}\)
Answer & Solution
Correct Answer
(A) \(\frac{27}{4}\)
Step-by-step Solution
Detailed explanation
\(y(x)=a x^{3}+b x^{2}+c x+5\) is passing through \((-2,0)\) then \(8 a-4 b+2 c=5 \ldots \ldots(1)\) \(y^{\prime}(x)=3 a x^{2}+2 b x+c\) touches \(x\)-axis at \((-2,0)\) \(12 a-4 b+c=0\) again, for \(x=0, y^{\prime}(x)=3\) \(c=3\) Solving eq. \((1), (2)\) and \((3)\)…
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