AP EAMCET · Maths · Basic of Mathematics
If \(\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3}=a+\frac{b}{x+p}+\frac{c}{q x+3}+\frac{d}{3 x+p}\) then \(\frac{a+b}{p+q}=\)
- A 2
- B 3
- C \(-\frac{2}{5}\)
- D \(\frac{2}{3}\)
Answer & Solution
Correct Answer
(A) 2
Step-by-step Solution
Detailed explanation
Given that: \(\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3}=a+\frac{b}{x+p}+\frac{c}{q x+3}+\frac{d}{3 x+p}\) ...(i) \(=\frac{6 x^3+7 x^2-14 x+11}{6 x^3+x^2-10 x+3}=1+\frac{6 x^2-4 x+8}{6 x^3+x^2-10 x+3}\) \(=1+\frac{6 x^2-4 x+8}{(x-1)(3 x-1)(2 x+3)}\) ...(ii) Now let…
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