MCQ
The maximum value of $\text{f}(\text{x})=\frac{\text{x}}{4-\text{x}+\text{x}^{2}}$ on $[-1, 1]$ is:
- A$-\frac{1}{4}$
- B$-\frac{1}{3}$
- ✓$\frac{1}{6}$
- D$\frac{1}{5}$
✓
Answer
Correct option: C.
$\frac{1}{6}$
$\text{f}(\text{x})=\frac{\text{x}}{4-\text{x}+\text{x}^{2}}$
$\Rightarrow\text{f}\ '(\text{x})=\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}$
To find minimum or maximum value $f\ '(x) = 0$
$\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}=0$
$4-x^2=0$
$\text{x}=\pm2\notin[-1, 1]$
$\text{f}(-1)=\frac{-1}{6}$ and $\text{f}(1)=\frac{1}{4}$
Hence, maximum value is $\frac{1}{4}$.
Note: options in the book are not matching with the solution.
Above solution is based on the question given in the book.
$\Rightarrow\text{f}\ '(\text{x})=\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}$
To find minimum or maximum value $f\ '(x) = 0$
$\frac{4-\text{x}^{2}}{(4-\text{x}+\text{x}^{2})^{2}}=0$
$4-x^2=0$
$\text{x}=\pm2\notin[-1, 1]$
$\text{f}(-1)=\frac{-1}{6}$ and $\text{f}(1)=\frac{1}{4}$
Hence, maximum value is $\frac{1}{4}$.
Note: options in the book are not matching with the solution.
Above solution is based on the question given in the book.
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