Q49 / 90 · KVPY 2017 (19 Nov SB/SX) · Mathematics
Let $\mathrm{f}(\mathrm{x})$ be a polynomial with integer coefficients satisfying $\mathrm{f}(1)=5$ and $\mathrm{f}(2)=7$. The smallest possible positive value of $\mathrm{f}(12)$ is
- A5
- B7
- Correct answerC27
- D15
Solution
$\mathrm{f}(\mathrm{x})=\mathrm{ax}+\mathrm{b}$
$5=\mathrm{a}+\mathrm{b}$
$7=2 \mathrm{a}+\mathrm{b}$
solve $\quad \mathrm{a}=2$
$\quad \mathrm{~b}=3$
$\quad \mathrm{f}(\mathrm{x})=2 \mathrm{x}+3$
$\mathrm{f}(12)=24+3$
$\mathrm{f}(12)=27$
Hence, the correct answer is option: 'C'.