JEE Mains · Maths · STD 12 - 6. Application of derivatives
The function \(f(x)=x^{3}-6 x^{2}+a x+b\) is such that \(f(2)=f(4)=0\). Consider two statements. \((S_1)\) there exists \(\mathrm{x}_{1}, \mathrm{x}_{2} \in(2,4), \mathrm{x}_{1}<\mathrm{x}_{2}\), such that \(f^{\prime}\left(x_{1}\right)=-1\) and \(f^{\prime}\left(x_{2}\right)=0\) \((S_2)\) there exists \(\mathrm{x}_{3}, \mathrm{x}_{4} \in(2,4), \mathrm{x}_{3}<\mathrm{x}_{4}\), such that \(f\) is decreasing in \(\left(2, x_{4}\right)\), increasing in \(\left(x_{4}, 4\right)\) and \(2 f^{\prime}\left(x_{3}\right)=\sqrt{3} f\left(x_{4}\right)\). Then
- A both \((S_1)\) and \((S_2)\) are true
- B \((S_1)\) is false and \((S_2)\) is true
- C both \((S_1)\) and \((S_2)\) are false
- D \((S_1)\) is true and \((S_2)\) is false
Answer & Solution
Correct Answer
(A) both \((S_1)\) and \((S_2)\) are true
Step-by-step Solution
Detailed explanation
\(f(x)=x^{3}-6 x^{2}+a x+b\) \(f(2)=8-24+2 a+b=0\) \(2 a+b=16 \ldots(1)\) \(f(4)=64-96+4 a+b=0\) \(4 a+b=32 \ldots .(2)\) Solving \((1)\) and \((2)\) \(a=8, b=0\) \(f(x)=x^{3}-6 x^{2}+8 x\) \(f(x)=x^{3}-6 x^{2}+8 x\) \(f^{\prime}(x)=3 x^{2}-12 x+8\)…
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