Q41 / 120 · VITEEE 2010 · Mathematics
If $\mathrm{F}$ is function such that $\mathrm{F}(0)=2, \mathrm{~F}(1)=3$, $\mathrm{F}(\mathrm{x}+2)=2 \mathrm{~F}(\mathrm{x})-\mathrm{F}(\mathrm{x}+1)$ for $\mathrm{x} \geq 0$, then $\mathrm{F}(5)$ is equal to
- A$-7$
- B$-3$
- C17
- Correct answerD13
Solution
$F(x+2)=2 F(x)-F(x+1)$...(i)
Putting $x=0$, we get
$$
\begin{array}{l}
F(2)=2 F(0)-F(1) \\
\Rightarrow F(2)=2(2)-3 \\
\quad\{\because F(0)=2, F(1)=3\}
\end{array}
$$
$$
\Rightarrow F(2)=4-3 \Rightarrow F(2)=1
$$
Putting $x=1$, in eq. (i), we get
$$
\begin{aligned}
F(3) &=2 F(1)-F(2) \\
&=2(3)-1 \quad\{\because F(1)=3, F(2)=1\} \\
\Rightarrow F(3)=5 &
\end{aligned}
$$
Putting $x=2$, in eq. (i), we get
$$
\begin{array}{l}
F(4)=2 F(2)-F(3) \\
\quad=2(1)-5 \quad\{\because F(2)=1, F(3)=5\} \\
\Rightarrow F(4)=-3
\end{array}
$$
Putting $x=3$, in eq. (i), we get
$$
\begin{array}{l}
F(5)=2 F(3)-F(4) \\
\quad=2(5)+3 \quad\{: F(3)=5, F(4)=-3\} \\
\Rightarrow F(5)=13
\end{array}
$$
Hence, the correct answer is option: 'D'.