MHT CET · Maths · Functions
Missing term in the following table is
\(\begin{array}{lllllll}x & : & 0 & 1 & 2 & 3 & 4 \ y=f(x) & : & 1 & 3 & 9 & ? & 81\end{array}\)
- A 27
- B 30
- C 31
- D 34
Answer & Solution
Correct Answer
(C) 31
Step-by-step Solution
Detailed explanation
Here, 4 values are given, therefore
\(\Delta^{4} f(x)=0 \forall x \)
\( \Rightarrow (E-1)^{4} f(x)=0 \)
\( \Rightarrow \left(E^{4}-4 E^{3}+6 E^{2}-4 E+1\right) f(x)=0 \)
\( \Rightarrow E^{4} f(x)-4 E^{3} f(x)+6 E^{2} f(x) \)
\( -4 E f(x)+f(x)=0 \)
\( \Rightarrow f(x+4)-4 f(x+3)+6 f(x+2) \)
\( -4 f(x+1)+f(x)=0 \)
On putting \(x=0\), we get
\(
f(4)-4 f(3)+6 f(2)-4 f(1)+f(0)=0 \quad \ldots(\mathrm{i})
\)
On substituting the values of \(f(0), f(1), f(2), f(4)\) in Eq. (i), we get
\(
\begin{array}{ll}
81-4 f(3)+6 \times 9-4 \times 3+1=0 \\
\Rightarrow 4 f(3)=124
\end{array}
\)
\(\Rightarrow \quad f(3)=31\)
\(\Delta^{4} f(x)=0 \forall x \)
\( \Rightarrow (E-1)^{4} f(x)=0 \)
\( \Rightarrow \left(E^{4}-4 E^{3}+6 E^{2}-4 E+1\right) f(x)=0 \)
\( \Rightarrow E^{4} f(x)-4 E^{3} f(x)+6 E^{2} f(x) \)
\( -4 E f(x)+f(x)=0 \)
\( \Rightarrow f(x+4)-4 f(x+3)+6 f(x+2) \)
\( -4 f(x+1)+f(x)=0 \)
On putting \(x=0\), we get
\(
f(4)-4 f(3)+6 f(2)-4 f(1)+f(0)=0 \quad \ldots(\mathrm{i})
\)
On substituting the values of \(f(0), f(1), f(2), f(4)\) in Eq. (i), we get
\(
\begin{array}{ll}
81-4 f(3)+6 \times 9-4 \times 3+1=0 \\
\Rightarrow 4 f(3)=124
\end{array}
\)
\(\Rightarrow \quad f(3)=31\)
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