AP EAMCET · Maths · Quadratic Equation
If \(\mathrm{P}(x)=x^5+a x^4+b x^3+c x^2+d x+e\) is a polynomial such that \(\mathrm{P}(0)=1, \mathrm{P}(1)=2, \mathrm{P}(2)=5, \mathrm{P}(3)=10\) and \(\mathrm{P}(4)=17\), then \(\mathrm{P}(5)=\)
- A \(26\)
- B \(146\)
- C \(126\)
- D \(76\)
Answer & Solution
Correct Answer
(B) \(146\)
Step-by-step Solution
Detailed explanation
\(P(x)=x^5+a x^4+6 x^3+c x^2+d x+e\) \(P(x)-x^2-1\) is a polynomial of degree 5 and has 5 roots \(0,1,2,3,4\) \(P(x)-x^2-1=(x-0)(x-1)(x-2)(x-3)(x-4)\) \(P(x)=x(x-1)(x-2)(x-3)(x-4)+x^2+1\) \(P(5)=5!+25+1=146\)
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