MHT CET · Maths · Definite Integration
\(\int_2^{\mathrm{e}}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] d x=a+\frac{b}{\log 2}\), then
- A \(\mathrm{a}=-\mathrm{e}, \mathrm{b}=2\)
- B \(\mathrm{a}=\mathrm{e}, \mathrm{b}=-2\)
- C \(\mathrm{a}=\mathrm{e}, \mathrm{b}=2\)
- D \(\mathrm{a}=-\mathrm{e}, \mathrm{b}=2\)
Answer & Solution
Correct Answer
(B) \(\mathrm{a}=\mathrm{e}, \mathrm{b}=-2\)
Step-by-step Solution
Detailed explanation
\(
\mathrm{a}+\frac{\mathrm{b}}{\log 2}=\int_2^{\mathrm{e}}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] \mathrm{dx}
\)
Put \(\log \mathrm{x}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt}\)
When \(\mathrm{x}=\mathrm{e}, \mathrm{t}=1\) and when \(\mathrm{x}=2, \mathrm{t}=\log 2\)
\(\therefore a+\frac{b}{\log 2}=\int_{\log 2}^1\left(\frac{1}{t}-\frac{1}{t^2}\right) e^t \cdot d t=\) \(\left[e^t \cdot \frac{1}{t}\right]_{\log 2}^1=e-\frac{e^{\log 2}}{\log 2}=e-\frac{2}{\log 2} \)
\( \therefore a=e, b=-2\)
\mathrm{a}+\frac{\mathrm{b}}{\log 2}=\int_2^{\mathrm{e}}\left[\frac{1}{\log x}-\frac{1}{(\log x)^2}\right] \mathrm{dx}
\)
Put \(\log \mathrm{x}=\mathrm{t} \Rightarrow \frac{1}{\mathrm{x}} \mathrm{dx}=\mathrm{dt} \Rightarrow \mathrm{dx}=\mathrm{e}^{\mathrm{t}} \mathrm{dt}\)
When \(\mathrm{x}=\mathrm{e}, \mathrm{t}=1\) and when \(\mathrm{x}=2, \mathrm{t}=\log 2\)
\(\therefore a+\frac{b}{\log 2}=\int_{\log 2}^1\left(\frac{1}{t}-\frac{1}{t^2}\right) e^t \cdot d t=\) \(\left[e^t \cdot \frac{1}{t}\right]_{\log 2}^1=e-\frac{e^{\log 2}}{\log 2}=e-\frac{2}{\log 2} \)
\( \therefore a=e, b=-2\)
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