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Question:
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Assertion (A)
$$
\int_2^e\left(\frac{1}{\log _e x}-\frac{1}{\left(\log _e x\right)^2}\right) d x=e-2 \log _2 e
$$
Reason (R)
$$
\int_a^b e^x\left(f(x)+f^{\prime}(x)\right) d x=e^b f(b)-e^a f(a)
$$
Options:
$$
\int_2^e\left(\frac{1}{\log _e x}-\frac{1}{\left(\log _e x\right)^2}\right) d x=e-2 \log _2 e
$$
Reason (R)
$$
\int_a^b e^x\left(f(x)+f^{\prime}(x)\right) d x=e^b f(b)-e^a f(a)
$$
Solution:
1899 Upvotes
Verified Answer
The correct answer is:
A and $R$ are true, $R$ is the correct explanation to $A$.
Assertion
Let $I=\int_2^e\left(\frac{1}{\log _e x}-\frac{1}{\left(\log _e x\right)^2}\right) d x$
Let $\log _e x=y$
$$
\begin{aligned}
\Rightarrow \quad x & =e^y \\
\Rightarrow \quad d x & =e^y d y \\
I & =\int_{\log _e 2}^1 e^y\left[\frac{1}{y}+\left(\frac{-1}{y^2}\right)\right] d y
\end{aligned}
$$
[Using the formula $\int_a^b e^x\left[f(x)+f^{\prime}(x)\right] d x=\left[e^x f(x)\right]_a^b$ ]
$$
\begin{aligned}
I & =\left[e^y \frac{1}{y}\right]_{-\log e^2}^1 \\
& =e-\frac{2}{\log _e 2}=e-2 \log _2 e
\end{aligned}
$$
Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.
Let $I=\int_2^e\left(\frac{1}{\log _e x}-\frac{1}{\left(\log _e x\right)^2}\right) d x$
Let $\log _e x=y$
$$
\begin{aligned}
\Rightarrow \quad x & =e^y \\
\Rightarrow \quad d x & =e^y d y \\
I & =\int_{\log _e 2}^1 e^y\left[\frac{1}{y}+\left(\frac{-1}{y^2}\right)\right] d y
\end{aligned}
$$
[Using the formula $\int_a^b e^x\left[f(x)+f^{\prime}(x)\right] d x=\left[e^x f(x)\right]_a^b$ ]
$$
\begin{aligned}
I & =\left[e^y \frac{1}{y}\right]_{-\log e^2}^1 \\
& =e-\frac{2}{\log _e 2}=e-2 \log _2 e
\end{aligned}
$$
Hence, Assertion and Reason both are true and Reason is the correct explanation of Assertion.
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