QJEEM19Ma5c1b: JEE Main 2019 Mathematics - Class 12 JEEM19Ma5c1b Solution: Let ƒ : R R be a differentiable function satisfying ƒ'(3) +…

Question

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Let ƒ : R

\to

R be a differentiable function satisfying ƒ'(3) + ƒ'(2) = 0.
Then

\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}

is equal to

Accepted Answer

The general formula for indeterminate form 1^$\infty$ is

\mathop {\lim }\limits_{x \to a} {\left( {f\left( x \right)} \right)^{g\left( x \right)}} = {e^{\mathop {\lim }\limits_{x \to a} g\left( x \right)\left( {f\left( x \right) - 1} \right)}}

I =

\mathop {\lim }\limits_{x \to 0} {\left( {{{1 + f(3 + x) - f(3)} \over {1 + f(2 - x) - f(2)}}} \right)^{{1 \over x}}}

Here I is in 1^$\infty$ form.

\therefore

I =

{e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}} - 1} \right){1 \over x}}}

=

{e^{\mathop {\lim }\limits_{x \to 0} \left( {{{1 + f\left( {3 + x} \right) - f\left( 3 \right) - 1 - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}

=

{e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right){1 \over x}}}

Here

{{{f\left( {3 + x} \right) - f\left( 3 \right) - f\left( {2 - x} \right) + f\left( 2 \right)} \over x}}

is in

{0 \over 0}

form.

So using L'Hopital rule we get

=

{e^{\mathop {\lim }\limits_{x \to 0} \left( {{{f'\left( {3 + x} \right) + f'\left( {2 - x} \right)} \over 1}} \right).\mathop {\lim }\limits_{x \to 0} \left( {{1 \over {1 + f\left( {2 - x} \right) - f\left( 2 \right)}}} \right)}}

=

{e^{\left( {f'\left( 3 \right) + f'\left( 2 \right)} \right).1}}

= e⁰ [ as given ƒ'(3) + ƒ'(2) = 0 ]

= 1

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