U TE 3x– 8x+9_ a + b + c then a+b+c...
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U TE 3x– 8x+9_ a + b + c then a+b+c= (x+1)3 x+1'(x+1)? "(x+1) ▼ (a) 26 (b) 5 (c) 18 (d) 9

JEE/Engineering Exams
Maths
Solution
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( frac{3 x^{2}-8 x+9}{(x+1)^{3}}=frac{9}{x+1}+frac{b}{(x+1)^{2}}+frac{c}{(n+1) 3} ) ( 3 x^{2}-8 x+9=frac{9}{-}(x+1)^{2}+b(x+1)+c ) ( 3 x^{2}-8 x+9=9 x^{2}+29 x+9+b n+b+0 ) contant ( Rightarrow a+b+c=9 )
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