JEE MAIN - Mathematics (2021 - 24th February Evening Shift - No. 6)
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 $$\left| {\matrix{
{f(x)} & {f'(x)} \cr
{f'(x)} & {f''(x)} \cr
} } \right|$$ = 0, for all x$$ \in $$R, then the value of f(1) lies in the interval :
(0, 3)
(9, 12)
(3, 6)
(6, 9)
Explanation
$$\left| {\matrix{
{f(x)} & {f'(x)} \cr
{f'(x)} & {f''(x)} \cr
} } \right| = 0$$
$$ \Rightarrow f(x).f''(x) - {\left( {f'(x)} \right)^2} = 0$$
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).
$$ \Rightarrow f(x).f''(x) - {\left( {f'(x)} \right)^2} = 0$$
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).
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