    Question

# If f(x) is polynomial of degree 3 with leading coefficient 2 and f(1)=4,f(2)=7,f(3)=12 and f(5)=11α−1, then α=

A
11
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B
10
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C
7
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D
5
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Solution

## The correct option is C 7f(1)=4,f(2)=7,f(3)=12 By the given values, we can assume f(x)=x2+3 for x=1,2,3 Now, let g(x)=f(x)−x2−3, then x=1,2,3 are roots of g(x) As f(x) is a polynomial of degree 3, so degree of g(x) is also 3 Therefore, g(x)=a(x−1)(x−2)(x−3)⇒f(x)=a(x−1)(x−2)(x−3)+x2+3 As the leading coefficient of f(x) is 2, so f(x)=2(x−1)(x−2)(x−3)+x2+3⇒f(5)=2(4)(3)(2)+25+3⇒11α−1=76∴α=7  Suggest Corrections  2      Similar questions
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