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JEE Mains · Maths · STD 11- 2. Relation and Function

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

  1. A \(92\)
  2. B \(118\)
  3. C \(102\)
  4. D \(108\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(118\)

Step-by-step Solution

Detailed explanation

\(\begin{aligned} & f(x)=\frac{42^x+16}{2.2^{2 x}+16.2^x+32} \\ & f(x)=\frac{2\left(2^x+4\right)}{2^{2 x}+8.2^x+16} \\ & f(x)=\frac{2}{2^x+4} \\ & f(4-x)=\frac{2^x}{2\left(2^x+4\right)} \\ & f(x)+f(4-x)=\frac{1}{2} \end{aligned}\) So,…
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