JEE PYQ: Differential Equation Question 9
Question 9 - 2021 (24 Feb Shift 2)
Let $f$ be a twice differentiable function defined on $\mathbb{R}$ such that $f(0) = 1$, $f’(0) = 2$ and $f’(x) \neq 0$ for all $x \in \mathbb{R}$. If $\begin{vmatrix} f(x) & f’(x) \ f’(x) & f’’(x) \end{vmatrix} = 0$, for all $x \in \mathbb{R}$, then the value of $f(1)$ lies in the interval:
(1) $(9, 12)$
(2) $(6, 9)$
(3) $(3, 6)$
(4) $(0, 3)$
Show Answer
Answer: (2)
Solution
$f(x)f’’(x) - (f’(x))^2 = 0 \Rightarrow \frac{d}{dx}\left(\frac{f’(x)}{f(x)}\right) = 0 \Rightarrow \frac{f’(x)}{f(x)} = k$. So $f(x) = f(0)e^{kx} = e^{2x}$ (since $k = \frac{f’(0)}{f(0)} = 2$). $f(1) = e^2 \approx 7.39 \in (6,9)$.
BETA
