The number of polynomials p(x) with integer coefficients such that the curve y=p(x) passes through (2,2) and (4,5) is
A
0.0
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B
1
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C
more than 1 but finite
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D
infinite
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Solution
The correct option is A 0.0 Let y=p(x)=a0+a1x+a2x2+......+anxn We know, p(2)=2=a0+2a1+22a2+.....+2nanp(4)=5=a0+4a1+42a2+.....+4nan Now, p(4)−p(2)=(4−2)a1+(42−22)a2+......+(4n−2n)an 3=(4−2)a1+(42−22)a2+......+(4n−2n)an From the above equation, we can see that the RHS will always remain even in nature, while the LHS is odd. So they will never be equal. Hence, there will be no polynomial which sastisfies the given condition.