CBSE 2026 · Region 2 · Set 1 · Q28 · 3 marks
If $\displaystyle \frac{\mathrm{d}}{\mathrm{d} x}(\mathrm{~F}(x))=\frac{1}{\mathrm{e}^{x}+1}$, then find $\displaystyle \mathrm{F}(x)$ given that $\displaystyle \mathrm{F}(0)=\log \frac{1}{2}$.
Marking-scheme solution
$\displaystyle \dfrac{d}{dx}(F(x))=\dfrac{1}{\mathrm{e}^{x}+1}$
$\displaystyle \Rightarrow F(x)=\int \dfrac{1}{\mathrm{e}^{x}+1}\, dx$
$\displaystyle =\int \dfrac{\mathrm{e}^{-x}}{1+\mathrm{e}^{-x}}\, dx$
$\displaystyle =-\log\left(1+\mathrm{e}^{-x}\right)+C$
Now, $\displaystyle F(0)=-\log 2+C \Rightarrow \log \dfrac{1}{2}+\log 2=C \Rightarrow C=0$
$\displaystyle \therefore F(x)=-\log\left(1+\mathrm{e}^{-x}\right)$
Note: The above expression is equivalent to $\displaystyle F(x)=x-\log\left(\mathrm{e}^{x}+1\right)$IntegralsIntegration as an Inverse Process of DifferentiationApplyshort_answermedium
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CBSE Class 12 Mathematics past-paper question from the 2026board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.