CBSE Boardहिन्दी माध्यमकक्षा 12 साइन्सगणितसमाकलन1 Mark
MCQ
$\int_0^1 \log \left(\frac{1}{x}-1\right) d x$ का मान है$-$
✓
$0$
B
$1$
C
$\log 2$
D
$\log \frac{3}{2}$
✓
Answer
Correct option: A.
$0$
माना $I =\int_0^1 \log \left(\frac{1}{x}-1\right) d x=\int_0^1 \log \left(\frac{1-x}{x}\right) d x \ldots(1)$
$I=\int_0^1 \log \left(\frac{1-x}{x}\right) d x$
$=\int_0^1 \log \left(\frac{1-(1-x)}{(1-x)}\right) d x$ गुणधर्म $P _5$ से
$I=\int_0^1 \log \left(\frac{1-1+x}{(1-x)}\right) d x=\int_0^1 \log \left(\frac{x}{(1-x)}\right) d x$
$I=-\int_0^1 \log \left(\frac{1-x}{x}\right) d x$
$\begin{aligned} I & =-I \\ 2 I & =0 \therefore I=0\end{aligned}$
अत: सही विकल्प $(A)$ है।
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