Question
If $\text{f}(\text{x})=\frac{1}{4\text{x}^{2}+2\text{x}+1}$, then its maximum value is :
- $\frac{4}{3}$
- $\frac{2}{3}$
- $1$
- $\frac{3}{4}$
$\frac{4}{3}$
Solution:
Maximum value of $\frac{1}{4\text{x}^{2}+2\text{x}+1}$ = Minimum value of $4\text{x}^{2}+2\text{x}+1$
Now, $\text{f}(\text{x})=\text4\text{x}^{2}+2\text{x}+1$
lmplies that $\text{f}'(\text{x})=8\text{x}+2$
For a local maxima or a local minima, We must have f'(x) = 0
lmplies that $8\text{x}+2 =0$
lmplies that $8\text{x}=-2$
lmplies that $\text{x}=-14$
Now, $\text{f}''(\text{x})=8$
lmplies that $\text{f}''(\text{1})=8 >0$
Therefore, $\text{x}=\frac{-1}{4}$ is a local minima.
Thus, $\frac{1}{4\text{x}^{2}+2\text{x}+1}$ is maximum at $\text{x}=\frac{-1}{4}$
lmplies that maximum value of $\frac{1}{4\text{x}^{2}+2\text{x}+1}=\frac{1}{4\Big(\frac{-1}{4}\Big)^{2}+2\Big(\frac{-1}{4}\Big)+1}$
$=\frac{16}{12}=\frac{4}{3}$
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