JEE Advanced 2020 · Paper 2 · Mathematics Q16

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MathematicsNumerical+4Question 52 of 54

Let the function $f:[0,1]\to\mathbb R$ be defined by

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

Then the value of

$f\left(\frac1{40}\right) + f\left(\frac2{40}\right) + f\left(\frac3{40}\right) + \cdots + f\left(\frac{39}{40}\right) - f\left(\frac12\right)$

is ______

Your answer

Answer and worked solution

Official answer: 19

Pairing identity.
$f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}=\frac{4}{4+2\cdot4^x}=\frac2{4^x+2},\qquad\text{so}\qquad f(x)+f(1-x)=1.$

Summing. The terms $k=1,\dots,39$ pair as $(k,40-k)$ for $k=1,\dots,19$ , giving 19, with $f\left(\frac{20}{40}\right)=f\left(\frac12\right)$ left over. That leftover cancels the subtracted $f\left(\frac12\right)$ .

Answer: 19

Topics:Special series sums Types of functions

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JEE Advanced 2020 · Paper 2, Mathematics question 16