If 3x2 - 2x - 4x - 12 - A1x - 1 - A2x - 12 - A3x - 13 - A4x - 14 - A5x - 15 - A6x - 16- then A1 - A3 - A5- A2 - A4 - A6 -

If $a_{1}$ < $a_{2}$ < $a_{3}$ < $a_{4}$ < $a_{5}$ < $a_{6}$ , then the equation $(x - a_{1}) (x - a_{3}) (x - a_{5}) + 2 (x - a_{2}) (x - a_{4}) (x - a_{6})$ = 0 has

Q. If

a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0

then

\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}

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If 3x2−2x+4(x−1)2=A1x+1+A2(x+1)2+A3(x+1)3+A4(x+1)4+A5(x+1)5+A6(x+1)6, then (A1+A3+A5,A2+A4+A6)=

The correct option is D (−8,12)A1x+1+A2(x+1)2+A3(x+1)3+A4(x+1)4+A5(x+1)5+A6(x+1)6=3x2−2x+4(x−1)6 .... (1)Now, put x=0 in (1)A1+A2+A3+A4+A5+A6=4…(2)If A1+A3+A5=−8 and A2+A4+A6=12 satisfies the equationHence, only option D i.e. (−8,12) satisfies the solution.

If $\frac{3 x^{2} - 2 x + 4}{(x - 1)^{2}} = \frac{A_{1}}{x + 1} + \frac{A_{2}}{(x + 1)^{2}} + \frac{A_{3}}{(x + 1)^{3}} + \frac{A_{4}}{(x + 1)^{4}} + \frac{A_{5}}{(x + 1)^{5}} + \frac{A_{6}}{(x + 1)^{6}}$ , then $(A_{1} + A_{3} + A_{5}, A_{2} + A_{4} + A_{6}) =$