MHT CET · Maths · Application of Derivatives
If \(f(x)=x^2+a x+b\) has minima at \(x=3\) whose value is 5 , then the values of \(a\) and \(b\) are respectively.
- A -6,-14
- B -6,14
- C 14,-6
- D 6,14
Answer & Solution
Correct Answer
(B) -6,14
Step-by-step Solution
Detailed explanation
\(
f(x)=x^2+a x+b
\)
\(\therefore \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+\mathrm{a}\) and when \(\mathrm{f}^{\prime}(\mathrm{x})=0\), we get \(\mathrm{x}=\frac{-\mathrm{a}}{2}\)
Now \(\mathrm{f}^{\prime}(\mathrm{x})=2\) and \(2>0\)
\(\therefore \mathrm{f}(\mathrm{x})\) has minima at \(\mathrm{x}=\frac{-\mathrm{a}}{2}=3 \quad \ldots\) [as per given data]
\(
\therefore \mathrm{a}=-6
\)
Since Minimum value of \(f(x)\) is 5 at \(x=3\), we write
\(
5=(3)^2+(-6)(3)+b \quad \Rightarrow b=14
\)
f(x)=x^2+a x+b
\)
\(\therefore \mathrm{f}^{\prime}(\mathrm{x})=2 \mathrm{x}+\mathrm{a}\) and when \(\mathrm{f}^{\prime}(\mathrm{x})=0\), we get \(\mathrm{x}=\frac{-\mathrm{a}}{2}\)
Now \(\mathrm{f}^{\prime}(\mathrm{x})=2\) and \(2>0\)
\(\therefore \mathrm{f}(\mathrm{x})\) has minima at \(\mathrm{x}=\frac{-\mathrm{a}}{2}=3 \quad \ldots\) [as per given data]
\(
\therefore \mathrm{a}=-6
\)
Since Minimum value of \(f(x)\) is 5 at \(x=3\), we write
\(
5=(3)^2+(-6)(3)+b \quad \Rightarrow b=14
\)
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