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Functions question

2022 · 27 Jun · Shift 1 · Q36
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  5. /2022 · 27 Jun · Shift 1 · Q36

Functions question

2022 · 27 Jun · Shift 1 · Q36

JEE MainMathematicsFunctionsNumerical+4 / −1
Let f : R →\to→ R be a function defined by f(x)=2e2xe2x+ef(x) = {{2{e^{2x}}} \over {{e^{2x}} + e}}f(x)=e2x+e2e2x​. Then f(1100)+f(2100)+f(3100)+   .....   +   f(99100)f\left( {{1 \over {100}}} \right) + f\left( {{2 \over {100}}} \right) + f\left( {{3 \over {100}}} \right) + \,\,\,.....\,\,\, + \,\,\,f\left( {{{99} \over {100}}} \right)f(1001​)+f(1002​)+f(1003​)+.....+f(10099​) is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
View written solutionFree

Correct answer: 99

  1. We are given f(x)=2e2xe2x+e.f(x)=\frac{2e^{2x}}{e^{2x}+e}.f(x)=e2x+e2e2x​. We need to find S=f(1100)+f(2100)+⋯+f(99100).S=f\left(\frac1{100}\right)+f\left(\frac2{100}\right)+\cdots+f\left(\frac{99}{100}\right).S=f(1001​)+f(1002​)+⋯+f(10099​).

  2. The key idea is to relate f(x)f(x)f(x) and f(1−x)f(1-x)f(1−x).

Let us compute f(1−x)f(1-x)f(1−x):

=\frac{2e^{2-2x}}{e^{2-2x}+e}.$$ Now simplify by dividing numerator and denominator by $e^{1-2x}$: $$f(1-x)=\frac{2e^{1}}{e^{1}+e^{2x}}= rac{2e}{e+e^{2x}}.$$ Also, $$f(x)=\frac{2e^{2x}}{e^{2x}+e}.$$ Therefore, $$f(x)+f(1-x)=\frac{2e^{2x}}{e^{2x}+e}+\frac{2e}{e^{2x}+e} =\frac{2(e^{2x}+e)}{e^{2x}+e}=2.$$ So we have the identity $$\boxed{f(x)+f(1-x)=2.}$$ 3. Now look at the required sum: $$S=f\left(\frac1{100}\right)+f\left(\frac2{100}\right)+\cdots+f\left(\frac{99}{100}\right).$$ Pair terms symmetrically: $$f\left(\frac1{100}\right)+f\left(\frac{99}{100}\right)=2,$$ $$f\left(\frac2{100}\right)+f\left(\frac{98}{100}\right)=2,$$ and so on. In general, $$f\left(\frac{k}{100}\right)+f\left(1-\frac{k}{100}\right)=f\left(\frac{k}{100}\right)+f\left(\frac{100-k}{100}\right)=2.$$ 4. The numbers from $1$ to $99$ give $99$ terms total. Among these, we get $49$ pairs: $$(1,99),(2,98),\dots,(49,51),$$ each contributing $2$. There is one middle term left: $$f\left(\frac{50}{100}\right)=f\left(\frac12\right).$$ Compute it: $$f\left(\frac12\right)=\frac{2e^{2\cdot(1/2)}}{e^{2\cdot(1/2)}+e} =\frac{2e}{e+e}=1.$$ 5. Hence, $$S=49\cdot 2 + 1 = 98+1=99.$$ Therefore, $$\boxed{99}.$$
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