QJEEM21Ma67b3: JEE Main 2021 Mathematics - Class 12 JEEM21Ma67b3 Solution: Let f be a twice differentiable function defined on R such that …

Question

Verified Solution • 2025–26 Session moderate

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

| \begin{matrix} f (x) & f^{'} (x) \\ f^{'} (x) & f^{″} (x) \end{matrix} |

= 0, for all x

\in

R, then the value of f(1) lies in the interval :

Accepted Answer

| \begin{matrix} f (x) & f^{'} (x) \\ f^{'} (x) & f^{″} (x) \end{matrix} | = 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).

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