If the solve the problem

Question:

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

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

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

(c) 1

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

Solution:

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

Maximum value of $\frac{1}{4 x^{2}+2 x+1}=$ Minimum value of $4 x^{2}+2 x+1$

Now,

$f(x)=4 x^{2}+2 x+1$

$\Rightarrow f^{\prime}(x)=8 x+2$

For a local maxima or a local minima, we must have

$f^{\prime}(x)=0$

$\Rightarrow 8 x+2=0$

$\Rightarrow 8 x=-2$

$\Rightarrow x=\frac{-1}{4}$

Now,

$f^{\prime \prime}(x)=8$

$\Rightarrow f^{\prime \prime}(1)=8>0$

So, $x=\frac{-1}{4}$ is a local minima.

Thus, $\frac{1}{4 x^{2}+2 x+1}$ is maximum at $x=\frac{-1}{4}$.

$\Rightarrow$ Maximum value of $\frac{1}{4 x^{2}+2 x+1}=\frac{1}{4\left(\frac{-1}{4}\right)^{2}+2\left(\frac{-1}{4}\right)+1}$

$=\frac{1}{\frac{4}{16}-\frac{1}{2}+1}=\frac{16}{12}=\frac{4}{3}$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now