JEE MAIN - Mathematics (2019 - 10th January Evening Slot - No. 2)
Let f be a differentiable function such that f '(x) = 7 - $${3 \over 4}{{f\left( x \right)} \over x},$$ (x > 0) and f(1) $$ \ne $$ 4. Then $$\mathop {\lim }\limits_{x \to 0'} \,$$ xf$$\left( {{1 \over x}} \right)$$ :
does not exist
exists and equals $${4 \over 7}$$
exists and equals 4
exists and equals 0
Explanation
$$f'(x) = 7 - {3 \over 4}{{f\left( x \right)} \over x}\,\,\,\left( {x > 0} \right)$$
Given f(1) $$ \ne $$ 4 $$\mathop {\lim }\limits_{x \to {0^ + }} \,xf\left( {{1 \over x}} \right)\, = ?$$
$${{dy} \over {dx}} + {3 \over 4}{y \over x} = 7$$ (This is LDE)
IF $$ = {e^{\int {{3 \over {4x}}dx} }} = {e^{{3 \over 4}\ln \left| x \right|}} = {x^{{3 \over 4}}}$$
$$y.{x^{{3 \over 4}}} = \int {7.{x^{{3 \over 4}}}} dx$$
$$y.{x^{{3 \over 4}}} = 7.{{{x^{{7 \over 4}}}} \over {{7 \over 4}}} + C$$
$$f(x) = 4x + C.{x^{ - {3 \over 4}}}$$
$$f\left( {{1 \over 4}} \right) = {4 \over x} + C.{x^{{3 \over 4}}}$$
$$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {4 + C.{x^{{7 \over 4}}}} \right) = 4$$
Given f(1) $$ \ne $$ 4 $$\mathop {\lim }\limits_{x \to {0^ + }} \,xf\left( {{1 \over x}} \right)\, = ?$$
$${{dy} \over {dx}} + {3 \over 4}{y \over x} = 7$$ (This is LDE)
IF $$ = {e^{\int {{3 \over {4x}}dx} }} = {e^{{3 \over 4}\ln \left| x \right|}} = {x^{{3 \over 4}}}$$
$$y.{x^{{3 \over 4}}} = \int {7.{x^{{3 \over 4}}}} dx$$
$$y.{x^{{3 \over 4}}} = 7.{{{x^{{7 \over 4}}}} \over {{7 \over 4}}} + C$$
$$f(x) = 4x + C.{x^{ - {3 \over 4}}}$$
$$f\left( {{1 \over 4}} \right) = {4 \over x} + C.{x^{{3 \over 4}}}$$
$$\mathop {\lim }\limits_{x \to {0^ + }} xf\left( {{1 \over x}} \right) = \mathop {\lim }\limits_{x \to {0^ + }} \left( {4 + C.{x^{{7 \over 4}}}} \right) = 4$$
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