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Question

x2 is a factor of f(x)=x3ax2bx+16. When f(x) is divided by (x3), the remainder is 5. The value of a and b is:

A
a=4,b=4
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B
a=5,b=2
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C
a=9,b=8
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D
a=7,b=7
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Solution

The correct option is A a=4,b=4
Since (x2) is a factor of f(x),using factor theorem, f(2)=0.

23a×(2)2b×2+16=084a2b+16=02a+b=12...(i)

When f(x) is divided by (x - 3), theremainder is f(3) by remainder theorem. The remainder is given tobe (-5).f(3)=533a×32b×3+16=5279a3b+16=53a+b=16...(ii)

Subtracting equation (i) from equation (ii), we get:(3a+b)(2a+b)=1612a=4

Substiuting value of a in (i), we get:2×4+b=12b=4

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