TS EAMCET · Maths · Quadratic Equation
If a polynomial \(\mathrm{P}(x)\) given by \(P(x)=2 x^4+a x^3+b x^2+c x+d\) is such that \(\mathrm{P}(1)=4, \mathrm{P}(2)=7, \mathrm{P}(3)=12\) and \(\mathrm{P}(4)=19\), then \(\mathrm{P}(5)=\)
- A 28
- B 76
- C 26
- D 72
Answer & Solution
Correct Answer
(B) 76
Step-by-step Solution
Detailed explanation
\(\mathrm{P}(x)=2 x^4+a x^3+b x^2+c x+d\) \(\begin{aligned} & \mathrm{P}(1)=4=1^2+3 \\ & \mathrm{P}(2)=7=2^2+3 \\ & \mathrm{P}(3)=12=3^2+3 \\ & \mathrm{P}(4)=19=4^2+3 \\ & \therefore \mathrm{P}(x)=x^2+3\end{aligned}\) But to make it degree 4 polynomial with leading coefficient…
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