CBSE 2022 · Region 4 · Set 1 · Q8 · 3 marks
Find : \[\int \frac{1}{\mathrm{e}^{\mathrm{x}}+1} \mathrm{dx} \]Evaluate : \[\int_{1}^{4}\{|\mathrm{x}|+|3-\mathrm{x}|\} d \mathrm{x} \]
Find : \[\int \frac{1}{\mathrm{e}^{\mathrm{x}}+1} \mathrm{dx} \]
Evaluate : \[\int_{1}^{4}\{|\mathrm{x}|+|3-\mathrm{x}|\} d \mathrm{x} \]
Marking-scheme solution
(a)
Let, \(\displaystyle e^{x}=t, e^{x} d x=d t\)
\[\begin{aligned}
\int \frac{1}{e^{x}+1} d x & =\int \frac{1}{t(t+1)} d t \\
& =\int \frac{d t}{t}-\int \frac{d t}{t+1} \\
& =\log |t|-\log |t+1|+C \\
& =\log e^{x}-\log \left(e^{x}+1\right)+C \text { or } \log \left|\frac{e^{x}}{1+e^{x}}\right|+C
\end{aligned}
\]
(b)
\(\displaystyle I=\int_{1}^{4}\{|x|+|3-x|\} d x\)
\[\begin{aligned}
& =\int_{1}^{3}\{|x|+|3-x|\} d x+\int_{3}^{4}\{|x|+|3-x|\} d x \\
& =\int_{1}^{3} 3 d x+\int_{3}^{4}(2 x-3) d x \\
& =|3 x|_{1}^{3}+\left|x^{2}-3 x\right|_{3}^{4} \\
& =6+4=10
\end{aligned}
\]
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CBSE Class 12 Mathematics past-paper question from the 2022board exam, with the answer as CBSE’s own marking scheme gives it. Where our answers come from.