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Q.

If (2x1) is exactly divides 6x2+ax4 and bx211x+3, then the value of b2a2+ab is

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a

125

b

60

c

70

d

75

answer is A.

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Detailed Solution

f(x)=6x2+ax4;g(x)=bx211x+3

f(12)=2a10=0a=5

g(12)=b10=0b=10b2a2+ab=70

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If (2x−1) is exactly divides 6x2+ax−4 and bx2−11x+3, then the value of b2−a2+a−b is