If f(x)=\frac2^x2^x+\sqrt2, x \in \mathbbR , then \sum_k=1^81 f(\frack82) is equal to

Functions PYQ — JEE Main 2025 · 28 Jan · Shift 1 · Q27 with Solution

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Correct answer: D

We are given $f(x)=\frac{2^x}{2^x+\sqrt{2}}.$ We need to find $S=\sum_{k=1}^{81} f\left(\frac{k}{82}\right).$
First, rewrite the function in a useful symmetric form.

Since $\sqrt{2}=2^{1/2}$ , $f(x)=\frac{2^x}{2^x+2^{1/2}}.$ Now consider $f\left(\frac12-x\right)$ : $f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}+2^{1/2}}.$ Factor $2^{1/2-x}$ from the denominator: $f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}(1+2^x)}=\frac{1}{1+2^x}.$ Also, $f(x)=\frac{2^x}{2^x+2^{1/2}}= rac{2^{x-1/2}}{2^{x-1/2}+1}.$ But a cleaner way is to directly add the two forms:

\qquad f\left(\frac12-x\right)=\frac{1}{1+2^x}.$$ Now rewrite $f(x)$ by dividing numerator and denominator by $2^x$: $$f(x)=\frac{1}{1+2^{1/2-x}}.$$ Then $$f\left(\frac12-x\right)=\frac{1}{1+2^x}.$$ Let $t=2^x$. Then $$f(x)=\frac{t}{t+\sqrt2}, \qquad f\left(\frac12-x\right)=\frac{1}{1+t}.$$ An even simpler direct verification is: $$f\left(\frac{k}{82}\right)+f\left(\frac{82-k}{82}\right) =f\left(\frac{k}{82}\right)+f\left(1-\frac{k}{82}\right),$$ but that is not the right symmetry here. The correct pairing comes from noticing that $$\frac{k}{82}+\frac{82-k}{82}=1,$$ which is not immediately useful for this function. So instead define $$x_k=\frac{k}{82}.$$ Then $$x_{81-k}=\frac{81-k}{82}=\frac12-\frac{2k-40}{82},$$ and the natural pairing is actually $$x_k+x_{82-k}=1.$$ Let us test the intended symmetry directly: For any $x$, $$f(x)+f\left(1-x\right)=\frac{2^x}{2^x+\sqrt2}+\frac{2^{1-x}}{2^{1-x}+\sqrt2}.$$ This is not obviously constant. So let us instead use the substitution $$g(t)=\frac{t}{t+\sqrt2}, \qquad t=2^x.$$ Then for the complementary exponent $\frac12-x$, we get $2^{1/2-x}=\frac{\sqrt2}{2^x}=\frac{\sqrt2}{t}$, hence $$f\left(\frac12-x\right)=\frac{\sqrt2/t}{\sqrt2/t+\sqrt2} =\frac{1/t}{1/t+1}=\frac{1}{1+t}.$$ And $$f(x)=\frac{t}{t+\sqrt2}.$$ This suggests checking whether the terms in our sum are paired by $$\frac{k}{82}+\frac{41-k}{82}=\frac12.$$ Indeed, $$\frac{k}{82}+\frac{41-k}{82}=\frac{41}{82}=\frac12.$$ So pair $$f\left(\frac{k}{82}\right) \text{ with } f\left(\frac{41-k}{82}\right).$$ But our sum runs from $k=1$ to $81$, so the cleaner pairing is $$\frac{k}{82} + \frac{41-k}{82} = \frac12$$ for $k=1,2,\dots,40$. 3. Now prove the key identity: $$f(x)+f\left(\frac12-x\right)=1.$$ Let $t=2^x$. Then $2^{1/2-x}=\frac{\sqrt2}{t}$. So $$f\left(\frac12-x\right)=\frac{\sqrt2/t}{\sqrt2/t+\sqrt2} =\frac{\sqrt2/t}{\sqrt2(1/t+1)} =\frac{1/t}{1/t+1} =\frac{1}{1+t}.$$ Also, $$f(x)=\frac{t}{t+\sqrt2}.$$ This does not yet look like $1-\frac{1}{1+t}$, so let us compute carefully again: Since $$f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}+\sqrt2},$$ and $\sqrt2=2^{1/2}$, $$f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}+2^{1/2}}.$$ Factor out $2^{1/2-x}$ from denominator: $$f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}(1+2^x)}=\frac{1}{1+2^x}.$$ Now for $f(x)$, divide numerator and denominator by $2^x$: $$f(x)=\frac{1}{1+2^{1/2-x}}.$$ Let $u=2^{1/2-x}$. Then $2^x=\frac{\sqrt2}{u}$, but more directly if $u=2^{1/2-x}$, then $$f(x)=\frac{1}{1+u}, \qquad f\left(\frac12-x\right)=\frac{1}{1+2^x}.$$ This route is messy. Let us instead verify with common denominator: $$f(x)=\frac{2^x}{2^x+\sqrt2},$$ $$f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}+\sqrt2}.$$ Now put $a=2^x$, so $2^{1/2-x}=\frac{\sqrt2}{a}$. Hence $$f\left(\frac12-x\right)=\frac{\sqrt2/a}{\sqrt2/a+\sqrt2} =\frac{\sqrt2/a}{\sqrt2(1/a+1)} =\frac{1/a}{1/a+1} =\frac{1}{1+a}.$$ And since $\sqrt2=2^{1/2}$, if we let $a=2^{x-1/2}$ instead, then $$f(x)=\frac{2^{x-1/2}}{2^{x-1/2}+1}.$$ Set $b=2^{x-1/2}$. Then $$f(x)=\frac{b}{b+1},$$ and $$f\left(\frac12-x\right)=\frac{1}{b+1}.$$ Therefore, $$f(x)+f\left(\frac12-x\right)=\frac{b}{b+1}+\frac{1}{b+1}=1.$$ So the identity is proved: $$\boxed{f(x)+f\left(\frac12-x\right)=1.}$$ 4. Now apply this to the sum. For $k=1,2,\dots,40$, $$\frac{k}{82}+\left(\frac12-\frac{k}{82}\right)=\frac12,$$ and $$\frac12-\frac{k}{82}=\frac{41-k}{82}.$$ Thus, $$f\left(\frac{k}{82}\right)+f\left(\frac{41-k}{82}\right)=1.$$ So the terms from $k=1$ to $40$ pair up to give 20 pairs? Let us count carefully. The set $$\left\{\frac{1}{82},\frac{2}{82},\dots,\frac{40}{82}\right\}$$ pairs internally as $$\frac{k}{82} \leftrightarrow \frac{41-k}{82}.$$ These form $20$ pairs, each summing to $1$. Hence, $$\sum_{k=1}^{40} f\left(\frac{k}{82}\right)=20.$$ This is incorrect because pairing only inside the first 40 terms gives total of 20, but we still need the remaining terms $k=41$ to $81$. So let us pair the full set properly. Observe that for $k=1,2,\dots,81$, $$\frac{k}{82} + \frac{41-k}{82} = \frac12,$$ but $41-k$ is not always in $1$ to $81$. A better indexing is: for each $k=1,2,\dots,40$, pair $$\frac{k}{82} \quad \text{with} \quad \frac{41-k}{82},$$ which covers $1$ to $40$ only. Similarly, for $k=42$ to $81$, those are of the form $$\frac12 + \frac{m}{82}, \quad m=1,2,\dots,40,$$ and can be paired with $$\frac12 - \frac{m}{82}.$$ Thus for each $m=1,2,\dots,40$, $$f\left(\frac12-\frac{m}{82}\right)+f\left(\frac12+\frac{m}{82}\right)=1.$$ That is, $$f\left(\frac{41-m}{82}\right)+f\left(\frac{41+m}{82}\right)=1.$$ These are exactly the pairs $$(1,81),(2,80),\dots,(40,42),$$ with the middle term $k=41$ left over. So, $$S=\sum_{m=1}^{40}\left[f\left(\frac{41-m}{82}\right)+f\left(\frac{41+m}{82}\right)\right]+f\left(\frac{41}{82}\right).$$ Each bracket equals $1$, so $$S=40+f\left(\frac{41}{82}\right).$$ But $$\frac{41}{82}=\frac12,$$ so $$f\left(\frac12\right)=\frac{2^{1/2}}{2^{1/2}+\sqrt2}= rac{\sqrt2}{\sqrt2+\sqrt2}= rac12.$$ Therefore, $$S=40+\frac12=\frac{81}{2}.$$ 5. Final answer: $$\boxed{\frac{81}{2}}$$ which corresponds to option **D**.

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Correct answer: D

We are given $f(x)=\frac{2^x}{2^x+\sqrt{2}}.$ We need to find $S=\sum_{k=1}^{81} f\left(\frac{k}{82}\right).$
First, rewrite the function in a useful symmetric form.

\qquad f\left(\frac12-x\right)=\frac{1}{1+2^x}.$$ Now rewrite $f(x)$ by dividing numerator and denominator by $2^x$: $$f(x)=\frac{1}{1+2^{1/2-x}}.$$ Then $$f\left(\frac12-x\right)=\frac{1}{1+2^x}.$$ Let $t=2^x$. Then $$f(x)=\frac{t}{t+\sqrt2}, \qquad f\left(\frac12-x\right)=\frac{1}{1+t}.$$ An even simpler direct verification is: $$f\left(\frac{k}{82}\right)+f\left(\frac{82-k}{82}\right) =f\left(\frac{k}{82}\right)+f\left(1-\frac{k}{82}\right),$$ but that is not the right symmetry here. The correct pairing comes from noticing that $$\frac{k}{82}+\frac{82-k}{82}=1,$$ which is not immediately useful for this function. So instead define $$x_k=\frac{k}{82}.$$ Then $$x_{81-k}=\frac{81-k}{82}=\frac12-\frac{2k-40}{82},$$ and the natural pairing is actually $$x_k+x_{82-k}=1.$$ Let us test the intended symmetry directly: For any $x$, $$f(x)+f\left(1-x\right)=\frac{2^x}{2^x+\sqrt2}+\frac{2^{1-x}}{2^{1-x}+\sqrt2}.$$ This is not obviously constant. So let us instead use the substitution $$g(t)=\frac{t}{t+\sqrt2}, \qquad t=2^x.$$ Then for the complementary exponent $\frac12-x$, we get $2^{1/2-x}=\frac{\sqrt2}{2^x}=\frac{\sqrt2}{t}$, hence $$f\left(\frac12-x\right)=\frac{\sqrt2/t}{\sqrt2/t+\sqrt2} =\frac{1/t}{1/t+1}=\frac{1}{1+t}.$$ And $$f(x)=\frac{t}{t+\sqrt2}.$$ This suggests checking whether the terms in our sum are paired by $$\frac{k}{82}+\frac{41-k}{82}=\frac12.$$ Indeed, $$\frac{k}{82}+\frac{41-k}{82}=\frac{41}{82}=\frac12.$$ So pair $$f\left(\frac{k}{82}\right) \text{ with } f\left(\frac{41-k}{82}\right).$$ But our sum runs from $k=1$ to $81$, so the cleaner pairing is $$\frac{k}{82} + \frac{41-k}{82} = \frac12$$ for $k=1,2,\dots,40$. 3. Now prove the key identity: $$f(x)+f\left(\frac12-x\right)=1.$$ Let $t=2^x$. Then $2^{1/2-x}=\frac{\sqrt2}{t}$. So $$f\left(\frac12-x\right)=\frac{\sqrt2/t}{\sqrt2/t+\sqrt2} =\frac{\sqrt2/t}{\sqrt2(1/t+1)} =\frac{1/t}{1/t+1} =\frac{1}{1+t}.$$ Also, $$f(x)=\frac{t}{t+\sqrt2}.$$ This does not yet look like $1-\frac{1}{1+t}$, so let us compute carefully again: Since $$f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}+\sqrt2},$$ and $\sqrt2=2^{1/2}$, $$f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}+2^{1/2}}.$$ Factor out $2^{1/2-x}$ from denominator: $$f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}(1+2^x)}=\frac{1}{1+2^x}.$$ Now for $f(x)$, divide numerator and denominator by $2^x$: $$f(x)=\frac{1}{1+2^{1/2-x}}.$$ Let $u=2^{1/2-x}$. Then $2^x=\frac{\sqrt2}{u}$, but more directly if $u=2^{1/2-x}$, then $$f(x)=\frac{1}{1+u}, \qquad f\left(\frac12-x\right)=\frac{1}{1+2^x}.$$ This route is messy. Let us instead verify with common denominator: $$f(x)=\frac{2^x}{2^x+\sqrt2},$$ $$f\left(\frac12-x\right)=\frac{2^{1/2-x}}{2^{1/2-x}+\sqrt2}.$$ Now put $a=2^x$, so $2^{1/2-x}=\frac{\sqrt2}{a}$. Hence $$f\left(\frac12-x\right)=\frac{\sqrt2/a}{\sqrt2/a+\sqrt2} =\frac{\sqrt2/a}{\sqrt2(1/a+1)} =\frac{1/a}{1/a+1} =\frac{1}{1+a}.$$ And since $\sqrt2=2^{1/2}$, if we let $a=2^{x-1/2}$ instead, then $$f(x)=\frac{2^{x-1/2}}{2^{x-1/2}+1}.$$ Set $b=2^{x-1/2}$. Then $$f(x)=\frac{b}{b+1},$$ and $$f\left(\frac12-x\right)=\frac{1}{b+1}.$$ Therefore, $$f(x)+f\left(\frac12-x\right)=\frac{b}{b+1}+\frac{1}{b+1}=1.$$ So the identity is proved: $$\boxed{f(x)+f\left(\frac12-x\right)=1.}$$ 4. Now apply this to the sum. For $k=1,2,\dots,40$, $$\frac{k}{82}+\left(\frac12-\frac{k}{82}\right)=\frac12,$$ and $$\frac12-\frac{k}{82}=\frac{41-k}{82}.$$ Thus, $$f\left(\frac{k}{82}\right)+f\left(\frac{41-k}{82}\right)=1.$$ So the terms from $k=1$ to $40$ pair up to give 20 pairs? Let us count carefully. The set $$\left\{\frac{1}{82},\frac{2}{82},\dots,\frac{40}{82}\right\}$$ pairs internally as $$\frac{k}{82} \leftrightarrow \frac{41-k}{82}.$$ These form $20$ pairs, each summing to $1$. Hence, $$\sum_{k=1}^{40} f\left(\frac{k}{82}\right)=20.$$ This is incorrect because pairing only inside the first 40 terms gives total of 20, but we still need the remaining terms $k=41$ to $81$. So let us pair the full set properly. Observe that for $k=1,2,\dots,81$, $$\frac{k}{82} + \frac{41-k}{82} = \frac12,$$ but $41-k$ is not always in $1$ to $81$. A better indexing is: for each $k=1,2,\dots,40$, pair $$\frac{k}{82} \quad \text{with} \quad \frac{41-k}{82},$$ which covers $1$ to $40$ only. Similarly, for $k=42$ to $81$, those are of the form $$\frac12 + \frac{m}{82}, \quad m=1,2,\dots,40,$$ and can be paired with $$\frac12 - \frac{m}{82}.$$ Thus for each $m=1,2,\dots,40$, $$f\left(\frac12-\frac{m}{82}\right)+f\left(\frac12+\frac{m}{82}\right)=1.$$ That is, $$f\left(\frac{41-m}{82}\right)+f\left(\frac{41+m}{82}\right)=1.$$ These are exactly the pairs $$(1,81),(2,80),\dots,(40,42),$$ with the middle term $k=41$ left over. So, $$S=\sum_{m=1}^{40}\left[f\left(\frac{41-m}{82}\right)+f\left(\frac{41+m}{82}\right)\right]+f\left(\frac{41}{82}\right).$$ Each bracket equals $1$, so $$S=40+f\left(\frac{41}{82}\right).$$ But $$\frac{41}{82}=\frac12,$$ so $$f\left(\frac12\right)=\frac{2^{1/2}}{2^{1/2}+\sqrt2}= rac{\sqrt2}{\sqrt2+\sqrt2}= rac12.$$ Therefore, $$S=40+\frac12=\frac{81}{2}.$$ 5. Final answer: $$\boxed{\frac{81}{2}}$$ which corresponds to option **D**.

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