Q97 / 150 · MHT CET 2026 (15 April Shift 1) · Mathematics
If the differential equation $\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$ for all $x$ and $f(0) = 1, f'(0) = 2$, then...
- A$f'(x) = -f(x)$
- B$f'(x) = f(x)$
- Correct answerC$f'(x) = 2f(x)$
- D$f'(x) = 0$
Solution
The given differential equation is $\begin{vmatrix} f(x) & f'(x) \\ f'(x) & f''(x) \end{vmatrix} = 0$
Expanding the determinant, we get $f(x)f''(x) - (f'(x))^2 = 0$
Dividing both sides by $(f(x))^2$, we get $\dfrac{f(x)f''(x) - (f'(x))^2}{(f(x))^2} = 0$
This can be written as $\dfrac{d}{dx} \left( \dfrac{f'(x)}{f(x)} \right) = 0$
Integrating both sides with respect to $x$, we get $\dfrac{f'(x)}{f(x)} = c$
Given $f(0) = 1$ and $f'(0) = 2$
Substituting $x = 0$, we get $c = \dfrac{f'(0)}{f(0)} = \dfrac{2}{1} = 2$
Therefore, $\dfrac{f'(x)}{f(x)} = 2 \Rightarrow f'(x) = 2f(x)$
Answer: $f'(x) = 2f(x)$
Expanding the determinant, we get $f(x)f''(x) - (f'(x))^2 = 0$
Dividing both sides by $(f(x))^2$, we get $\dfrac{f(x)f''(x) - (f'(x))^2}{(f(x))^2} = 0$
This can be written as $\dfrac{d}{dx} \left( \dfrac{f'(x)}{f(x)} \right) = 0$
Integrating both sides with respect to $x$, we get $\dfrac{f'(x)}{f(x)} = c$
Given $f(0) = 1$ and $f'(0) = 2$
Substituting $x = 0$, we get $c = \dfrac{f'(0)}{f(0)} = \dfrac{2}{1} = 2$
Therefore, $\dfrac{f'(x)}{f(x)} = 2 \Rightarrow f'(x) = 2f(x)$
Answer: $f'(x) = 2f(x)$
Hence, the correct answer is option: 'C'.