Let f be a twice differentiable function defined on R such that f(0) = 1, f'(0) = 2 and f'…

Question

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

Accepted Answer

$$\left| {\begin{pmatrix} {f(x)} & {f'(x)}  \
   {f'(x)} & {f''(x)}  \

\end{pmatrix} } \right| = 0$$

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

Dividing by $${\left( {f(x)} \right)^2}$$, we get

$$ \Rightarrow {{f(x).f''(x) - {{\left( {f'(x)} \right)}^2}} \over {{{\left( {f(x)} \right)}^2}}} = 0$$

$$ \Rightarrow {d \over {dx}}\left( {{{f'(x)} \over {f(x)}}} \right) = 0$$

Integrating both side,

$${{f'(x)} \over {f(x)}} = c$$ (constant)

At, $x = 0$, $${{f'(0)} \over {f(0)}} = c$$

$$ \Rightarrow {2 \over 1} = c$$

$$ \Rightarrow c = 2$$

$$ \therefore $$ $${{f'(x)} \over {f(x)}} = 2$$

$$ \Rightarrow \int {{{f'(x)} \over {f(x)}}} dx = 2\int {dx} $$

$$ \Rightarrow \ln |f(x)|\, = 2x + c'$$

at x = 0,

$$ln|f(0)|\, = 0 + c'$$

$$ \Rightarrow 0 = 0 + c'$$

$$ \Rightarrow c' = 0$$

$$ \therefore $$ $n|f(x)| = 2x$

$$ \Rightarrow f(x) = {e^{2x}}$$

$f(1) = {e^2} = {(2.71)^2} = 7.34$

So it lie between (6, 9).

Solutions · 1