AP EAMCET · Maths · Indefinite Integration
If \(\frac{d}{d x}\left(\frac{x^2}{(x+2)(2 x+3)}\right)=\frac{A}{(x+2)^2}+\frac{B}{(2 x+3)^2}\) then \(\mathrm{A}+\mathrm{B}=\)
- A \(1 / 2\)
- B -5
- C \(-3 / 2\)
- D \(9 / 4\)
Answer & Solution
Correct Answer
(B) -5
Step-by-step Solution
Detailed explanation
\(\because \frac{d}{d x}\left(\frac{x^2}{(x+2)(2 x+3)}\right)=\frac{\mathrm{A}}{(x+2)^2}+\frac{\mathrm{B}}{(2 x+3)^2}\) \(\Rightarrow\left(\frac{7 x^2+12 x}{(x+2)(2 x+3)}\right)=\frac{\mathrm{A}}{(x+2)^2}+\frac{\mathrm{B}}{(2 x+3)^2}\) ...(i) The form of the partial fraction…
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\begin{aligned}
& \left(1+\cos \frac{\pi}{8}\right)\left(1+\cos \frac{2 \pi}{8}\right)\left(1+\cos \frac{3 \pi}{8}\right)\left(1+\cos \frac{4 \pi}{8}\right) \ldots \\
& \left(1+\cos \frac{7 \pi}{8}\right)=
\end{aligned}
\]AP EAMCET 2023 Hard