JEE Main 2021 (Online) 24th February Evening ShiftQuestion 1
Let $f$ be a twice differentiable function defined on $\mathbf{R}$ such that $f(0)=1$, $f'(0)=2$ and $f'(x)\neq0$ for all $x\in\mathbf{R}$. If $\begin{vmatrix}f(x)&f'(x)\\f'(x)&f''(x)\end{vmatrix}=0$, for all $x\in\mathbf{R}$, then the value of $f(1)$ lies in the interval :
Differential Equation
Solution
Concept. $ff''-f'^2=0$ says the derivative of $\frac{f'}{f}$ is zero. Formulas used. (1) $\left(\dfrac{f'}{f}\right)'=\dfrac{ff''-f'^2}{f^2}$Step 1 - Expand. The determinant is $ff''-(f')^2=0$. Step 2 - Use (1). $\left(\frac{f'}{f}\right)'=0$, so $\frac{f'}{f}=k$, a constant, and $f=Ae^{kx}$. Step 3 - Conditions. $f(0)=1$ gives $A=1$; $f'(0)=2$ gives $k=2$. So $f(x)=e^{2x}$. Step 4 - Value. $f(1)=e^2\approx7.39$, which lies in $(6,9)$. Answer: (C)