MHT CET · Maths · Functions
Find a polynomial \(f(x)\) of degree 2 where \(f(0)=8, f(1)=12, f(2)=18\)
- A \(x^{2}+3 x-8\)
- B \(x^{2}-3 x+8\)
- C \(2 x^{2}-x+3\)
- D \(x^{2}+3 x+8\)
Answer & Solution
Correct Answer
(D) \(x^{2}+3 x+8\)
Step-by-step Solution
Detailed explanation
Let the polynomial is \(a x^{2}+b x+c\).
Now, \(f(0)=8\)
\(\Rightarrow\)
\(c=8\)
\(\therefore\) Equation is \(a x^{2}+b x+8\).
Again, \(f(1)=12 \Rightarrow a+b+8=12\)
\(\Rightarrow \quad a+b=4 \quad \ldots(\mathrm{i})\)
and \(f(2)=18\)
\(\Rightarrow \quad 4 a+2 b+8=18\)
\(\Rightarrow \quad 2 a+b=5 \quad \ldots\) (ii)
On solving Eqs. (i) and (ii), we get
\(a=1, b=3\)
\(\therefore\) Required equation is \(x^{2}+3 x+8\).
Now, \(f(0)=8\)
\(\Rightarrow\)
\(c=8\)
\(\therefore\) Equation is \(a x^{2}+b x+8\).
Again, \(f(1)=12 \Rightarrow a+b+8=12\)
\(\Rightarrow \quad a+b=4 \quad \ldots(\mathrm{i})\)
and \(f(2)=18\)
\(\Rightarrow \quad 4 a+2 b+8=18\)
\(\Rightarrow \quad 2 a+b=5 \quad \ldots\) (ii)
On solving Eqs. (i) and (ii), we get
\(a=1, b=3\)
\(\therefore\) Required equation is \(x^{2}+3 x+8\).
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