Question

Let $f$ be a twice differentiable function defined on $R$ such that $f\left(0\right)=1$, $f\text{'}\left(0\right)=2$and $f\text{'}\left(x\right)\ne 0$ for all $x\in R$. If $\left|\begin{array}{cc}f\left(x\right)& f\text{'}\left(x\right)\\ f\text{'}\left(x\right)& f"\left(x\right)\end{array}\right|=0$, for all $x\in R$, then the value of $f\left(1\right)$ lies in the interval:

• A. $\left(9,12\right)$
• B. $\left(6,9\right)$
• C. $\left(3,6\right)$
• D. $\left(0,3\right)$

Answer 1

The correct option is B

$\left(6,9\right)$

Explanation for correct option(s)

Option B: $\left(6,9\right)$

Given data

$f\left(0\right)=1$

$f\text{'}\left(0\right)=2$

$f\text{'}\left(x\right)\ne 0$ for all $x\in R$

$\left|\begin{array}{cc}f\left(x\right)& f\text{'}\left(x\right)\\ f\text{'}\left(x\right)& f"\left(x\right)\end{array}\right|=0$

Consider the given equation as,

$\left|\begin{array}{cc}f\left(x\right)& f\text{'}\left(x\right)\\ f\text{'}\left(x\right)& f"\left(x\right)\end{array}\right|=0$

$f\left(x\right)f"\left(x\right)-{\left(f\text{'}\left(x\right)\right)}^2=0......\left(1\right)$

Divide by ${\left(f\text{'}\left(x\right)\right)}^2$ in Equation (1)

$\frac{f\left(x\right)f"\left(x\right)-{\left(f\text{'}\left(x\right)\right)}^2}{{\left(f\text{'}\left(x\right)\right)}^2}=0$

Rewrite the above Equation as,

$\frac{d}{dx}\left(\frac{f\text{'}\left(x\right)}{f\left(x\right)}\right)=0$

Integrating both side

$$\begin{gathered} \int \frac{d}{dx}\left(\frac{f\text{'}\left(x\right)}{f\left(x\right)}\right)=\int 0dx \\ \frac{f\text{'}\left(x\right)}{f\left(x\right)}=c \end{gathered}$$

Integrating the above Equation with respect to $x$.

$\int \frac{f\text{'}\left(x\right)}{f\left(x\right)}dx=\int cdx......\left(2\right)$

We know that

$\log f\left(x\right)=\frac{1}{f\left(x\right)}f\text{'}\left(x\right)$

Then the Equation (2) becomes

$$\begin{gathered} \log \left(f\left(x\right)\right)=cx+d \\ f\left(x\right)=e^{cx+d} \\ f\left(x\right)=A.e^{cx} \end{gathered}$$

If $x=0$, from the given data $f\left(0\right)=1$

$$\begin{gathered} f\left(0\right)=A.e^0 \\ 1=A \end{gathered}$$

$$\begin{gathered} f\left(x\right)=1.e^{cx} \\ f\left(x\right)=e^{cx} \\ f\text{'}\left(x\right)=e^{cx}.c \end{gathered}$$

From the given data

$f\text{'}\left(0\right)=2$

$$\begin{gathered} f\text{'}\left(0\right)=e^0.c \\ 2=1.c \\ 2=c \end{gathered}$$

$$\begin{gathered} f\left(x\right)=1.e^{cx} \\ f\left(x\right)=1.e^{2x} \\ f\left(x\right)=e^{2x} \end{gathered}$$

If $x=1$,

$$\begin{gathered} f\left(1\right)=e^{2\times 1} \\ f\left(1\right)=e^2 \end{gathered}$$

The Approximate value of $e=2.7$

$$\begin{gathered} f\left(1\right)={\left(2.7\right)}^2 \\ f\left(1\right)=7.29 \end{gathered}$$

Since, the value of $f\left(1\right)$ is lies between the interval $\left(6,9\right)$

Hence, the correct answer is Option B.