Question

If(x) = $\frac{1}{4x2+2x+1}$, then its maximum value is

(a) $\frac{4}{3}$

(b) $\frac{2}{3}$

(c) 1

(d) $\frac{3}{4}$

Answer 1

(a) $\frac{4}{3}$

$\text{Maximum value of}\frac{1}{4x^2+2x+1}\text{= Minimum value of}4x^2+2x+1$
$$\begin{gathered} \mathrm{Now}, \\ f\left(x\right)=4x^2+2x+1 \\ \Rightarrow f\text{'}\left(x\right)=8x+2 \\ \mathrm{For}\mathrm{a}\mathrm{local}\mathrm{maxima}\mathrm{or}\mathrm{a}\mathrm{local}\mathrm{minima},\mathrm{we}\mathrm{must}\mathrm{have} \\ f\text{'}\left(x\right)=0 \\ \Rightarrow 8x+2=0 \\ \Rightarrow 8x=-2 \\ \Rightarrow x=\frac{-1}{4} \\ \mathrm{Now}, \\ f\text{'}\text{'}\left(x\right)=8 \\ \Rightarrow f\text{'}\text{'}\left(1\right)=8>0 \\ \mathrm{So},x=\frac{-1}{4}\mathrm{is}\mathrm{a}\mathrm{local}\text{minima}. \\ \text{Thus,}\frac{1}{4x^2+2x+1}\text{is maximum at}\mathit{\text{x}}=\frac{-1}{4}. \\ \Rightarrow \mathrm{Maximum}\mathrm{value}\mathrm{of}\frac{1}{4x^2+2x+1}=\frac{1}{4{\left({\displaystyle \frac{-1}{4}}\right)}^2+2\left({\displaystyle \frac{-1}{4}}\right)+1} \\ =\frac{1}{{\displaystyle \frac{4}{16}}-{\displaystyle \frac{1}{2}}+1}=\frac{16}{12}=\frac{4}{3} \end{gathered}$$