Let f(x)= 2^ x+2 +16 2^ 2 x+1 +2^ x+4 +32 . Then the value of 8 (f ((1)/(15))+f((2)/(15))+ +f((59)/(15))) is equal to

Correct answer: B. 118 aligned & f(x)= 42^x+16 2.2^ 2 x +16.2^x+32 & f(x)= 2(2^x+4) 2^ 2 x +8.2^x+16 & f(x)=(2)/(2^x+4) & f(4-x)=(2^x)/(2(2^x+4)) & f(x)+f(4-x)=(1)/(2) aligned So, f((1)/(15))+f((59)/(15))=(1)/(2) aligned & Similarly =f((29)/(15))+f((31)/(15))=(1)/(2) & f((30)/(15))=f(2)=(2)/(2^2+4)=(2)/(8)=(1)/(4) & 8(29 × (1)/(2)+(1)/(4)) aligned Ans. 118

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Functions

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Functions

Let f(x)= 2^ x+2 +16 2^ 2 x+1 +2^ x+4 +32 . Then the value of 8 (f...

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

Options

A $92$
B $118$
C $102$
D $108$

Correct answer

B. $118$

Step-by-step solution

\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,

\quad \mathrm{f}\left(\frac{1}{15}\right)+\mathrm{f}\left(\frac{59}{15}\right)=\frac{1}{2}

\begin{aligned} & \text{ Similarly }=f\left(\frac{29}{15}\right)+f\left(\frac{31}{15}\right)=\frac{1}{2} \\ & f\left(\frac{30}{15}\right)=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) \end{aligned}

$$ Ans. 118

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