JEE Main · 2021 · 24 Feb 2021 · Shift 2 · Differential Equations

The determinant condition gives:

$f(x)\cdot f^{\prime \prime }(x)-[f^{\prime }(x){]}^2=0$

This means $\frac{f^{\prime \prime }(x)}{f(x)}=\frac{[f^{\prime }(x){]}^2}{[f(x){]}^2}$

Let $g(x)=\ln |f(x)|$. Then $g^{\prime }(x)=\frac{f^{\prime }(x)}{f(x)}$ and $g^{\prime \prime }(x)=\frac{f^{\prime \prime }(x)f(x)-[f^{\prime }(x){]}^2}{[f(x){]}^2}=0$

Since $g^{\prime \prime }(x)=0$, we have $g(x)=Ax+B$ for constants $A,B$.

From initial conditions: $g(0)=\ln |f(0)|=\ln 1=0$, so $B=0$.

$g^{\prime }(0)=\frac{f^{\prime }(0)}{f(0)}=\frac{2}{1}=2$, so $A=2$.

Thus $g(x)=2x$, which gives $\ln |f(x)|=2x$, so $f(x)=e^{2x}$ (since $f(x)>0$).

Therefore, $f(1)=e^2\approx 7.39$, which lies in the interval $(6,9)$.