Let f be a twice differentiable function defi | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the determinant condition: $f(x)f^{\prime \prime}(x) - (f^{\prime}(x))^2 = 0$.This can be rewritten as $\frac{f^{\prime \prime}(x)}{f^{\prim

Vedclass · Accepted Answer

$(6, 9)$

Vedclass · Answer

(B) Given the determinant condition: $f(x)f^{\prime \prime}(x) - (f^{\prime}(x))^2 = 0$.This can be rewritten as $\frac{f^{\prime \prime}(x)}{f^{\prime}(x)} = \frac{f^{\prime}(x)}{f(x)}$.Integrating both sides with respect to $x$,we get $\ln(f^{\prime}(x)) = \ln(f(x)) + \ln(c)$,which implies $f^{\prime}(x) = c f(x)$.This is a first-order linear differential equation: $\frac{f^{\prime}(x)}{f(x)} = c$.Integrating again,$\ln(f(x)) = cx + k_1$,so $f(x) = k e^{cx}$.Using the initial conditions:$f(0) = 1 \implies k e^0 = 1 \implies k = 1$.$f^{\prime}(x) = c e^{cx}$,and $f^{\prime}(0) = 2 \implies c e^0 = 2 \implies c = 2$.Thus,$f(x) = e^{2x}$.Calculating $f(1) = e^2 \approx 7.389$.Since $6 < 7.389 < 9$,the value lies in the interval $(6, 9)$.

Similar Questions

The rank of the matrix $A = \begin{bmatrix} 2 & 1 & 2 \\ 1 & 0 & 1 \\ 4 & 1 & 4 \end{bmatrix}$ is

Consider a homogeneous system of three linear equations in three unknowns represented by $AX=O$. If $X=\left[\begin{array}{c}l \\ m \\ 0\end{array}\right]$, where $l \neq 0, m \neq 0, l, m \in \mathbb{R}$, represents an infinite number of solutions of this system, then the rank of $A$ is:

If the ranks of the matrices $A = \begin{bmatrix} 1 & 0 & 1 \\ 2 & 1 & 2 \\ 1 & 0 & -1 \end{bmatrix}$ and $B = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 2 & 4 & 6 & -8 \end{bmatrix}$ are $r_1$ and $r_2$ respectively,then $r_1 - r_2 =$

The rank of the matrix $\begin{bmatrix} 4 & 2 & 1-x \\ 5 & k & 1 \\ 6 & 3 & 1+x \end{bmatrix}$ is $1$,then

The number of distinct real roots of $\left|\begin{array}{lll}\sin x & \cos x & \cos x \\ \cos x & \sin x & \cos x \\ \cos x & \cos x & \sin x\end{array}\right|=0$ in the interval $-\frac{\pi}{4} \leq x \leq \frac{\pi}{4}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module