Let f(x) = 2^(x+2)+16/2^(2x+1)+2^(x+4)+32.

Question

Let \(f(x)=\frac{2^{x+2}+16}{2^{2 x+1}+2^{x+4}+32}.\) Then the value of \(8\left(\mathrm{f}\left(\frac{1}{15}\right)+\mathrm{f}\left(\frac{2}{15}\right)+\ldots+\mathrm{f}\left(\frac{59}{15}\right)\right)\) is equal to

(1) 118

(2) 92

(3) 102

(4) 108
 

Answer 1

Correct option is (1) 118  

\(f(x)=\frac{42^{x}+16}{2.2^{2 x}+16.2^{x}+32}\)

\(f(x)=\frac{2\left(2^{x}+4\right)}{2^{2 x}+8.2^{x}+16}\)

\(f(x)=\frac{2}{2^{x}+4}\)

\(f(4-x)=\frac{2^{x}}{2\left(2^{x}+4\right)}\)

\(\mathrm{f}(\mathrm{x})+\mathrm{f}(4-\mathrm{x})=\frac{1}{2}\)

So, \( f\left(\frac{1}{15}\right)+f\left(\frac{59}{15}\right)=\frac{1}{2}\)

Similarly \(=f\left(\frac{29}{15}\right)+f\left(\frac{31}{15}\right)=\frac{1}{2}\) 

\(\mathrm{f}\left(\frac{30}{15}\right)=\mathrm{f}(2)=\frac{2}{2^{2}+4}=\frac{2}{8}=\frac{1}{4} \)

\( \Rightarrow 8\left(29 \times \frac{1}{2}+\frac{1}{4}\right)\)

\(\Rightarrow 118\)