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
- A \(92\)
- B \(118\)
- C \(102\)
- D \(108\)
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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