Search any question & find its solution
Question:
Answered & Verified by Expert
It is given that at $x=1$, the function $x^4-62 x^2+a x+9$ attains its maximum value, on the interval $[0,2]$. Find the value of a.
Solution:
1106 Upvotes
Verified Answer
$\because \mathrm{f}(\mathrm{x})=\mathrm{x}^4-62 \mathrm{x}^2+\mathrm{ax}+9 \therefore \mathrm{f}^{\prime}(\mathrm{x})=4 \mathrm{x}^3-124 \mathrm{x}+\mathrm{a}$
Now $\mathrm{f}^{\prime}(\mathrm{x})=0$ at $\mathrm{x}=1 \Rightarrow 4-124+\mathrm{a}=0 \Rightarrow \mathrm{a}=120$
$\operatorname{Nowf}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}^2-124$ :
At $\mathrm{x}=1 \mathrm{f}^{\prime \prime}(1)=12-124=-112 < 0$
$\Rightarrow \mathrm{f}(\mathrm{x})$ has a maximum at $\mathrm{x}=1$ when $\mathrm{a}=120$.
Now $\mathrm{f}^{\prime}(\mathrm{x})=0$ at $\mathrm{x}=1 \Rightarrow 4-124+\mathrm{a}=0 \Rightarrow \mathrm{a}=120$
$\operatorname{Nowf}^{\prime \prime}(\mathrm{x})=12 \mathrm{x}^2-124$ :
At $\mathrm{x}=1 \mathrm{f}^{\prime \prime}(1)=12-124=-112 < 0$
$\Rightarrow \mathrm{f}(\mathrm{x})$ has a maximum at $\mathrm{x}=1$ when $\mathrm{a}=120$.
Looking for more such questions to practice?
Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.