MHT CET · Maths · Trigonometric Ratios & Identities
\(\int \frac{d x}{32-2 x^2}=A \log (4-x)+B \log (4+x)+c\), then the value of \(\mathrm{A}\) and \(\mathrm{B}\) are respectively (where \(\mathrm{c}\) is a constant of integration)
- A \(\frac{-1}{8}, \frac{1}{8}\)
- B \(\frac{1}{8}, \frac{-1}{8}\)
- C \(\frac{-1}{16}, \frac{1}{16}\)
- D \(\frac{1}{8}, \frac{1}{8}\)
Answer & Solution
Correct Answer
(C) \(\frac{-1}{16}, \frac{1}{16}\)
Step-by-step Solution
Detailed explanation
Let \(I=\int \frac{d x}{32-2 x^2}\)
\(
=\frac{1}{2}\left[\frac{1}{2(4)} \log \left|\frac{4+\mathrm{x}}{4-\mathrm{x}}\right|\right]+\mathrm{c}=\frac{1}{16}[\log |4+\mathrm{x}|-\log |4-\mathrm{x}|]+\mathrm{c}
\)
Comparing with given data we get \(\mathrm{A}=\frac{-1}{16}, \mathrm{~B}=\frac{1}{16}\)
\(
=\frac{1}{2}\left[\frac{1}{2(4)} \log \left|\frac{4+\mathrm{x}}{4-\mathrm{x}}\right|\right]+\mathrm{c}=\frac{1}{16}[\log |4+\mathrm{x}|-\log |4-\mathrm{x}|]+\mathrm{c}
\)
Comparing with given data we get \(\mathrm{A}=\frac{-1}{16}, \mathrm{~B}=\frac{1}{16}\)
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