JEE MAIN - Mathematics (2021 - 25th February Evening Shift - No. 24)
If $$\mathop {\lim }\limits_{x \to 0} {{ax - ({e^{4x}} - 1)} \over {ax({e^{4x}} - 1)}}$$ exists and is equal to b, then the value of a $$-$$ 2b is __________.
Answer
5
Explanation
$$\mathop {\lim }\limits_{x \to 0} {{ax - \left( {{e^{4x}} - 1} \right)} \over {ax\left( {{e^{4x}} - 1} \right)}}$$
Applying L' Hospital Rule
$$\mathop {\lim }\limits_{x \to 0} {{a - 4{e^{4x}}} \over {a\left( {{e^{4x}} - 1} \right) + ax\left( {4{e^{4x}}} \right)}}$$
This is $${{a - 4} \over 0}$$.
limit exist only when $$a - 4 = 0$$ $$ \Rightarrow $$ a = 4
Applying L' Hospital Rule
$$\mathop {\lim }\limits_{x \to 0} {{ - 16{e^{4x}}} \over {a\left( {4{e^{4x}}} \right) + a\left( {4{e^{4x}}} \right) + ax\left( {16{e^{4x}}} \right)}}$$
= $${{ - 16} \over {4a + 4a}} = {{ - 16} \over {32}} = - {1 \over 2} = b$$
$$a - 2b = 4 - 2\left( {{{ - 1} \over 2}} \right) = 4 + 1 = 5$$
Applying L' Hospital Rule
$$\mathop {\lim }\limits_{x \to 0} {{a - 4{e^{4x}}} \over {a\left( {{e^{4x}} - 1} \right) + ax\left( {4{e^{4x}}} \right)}}$$
This is $${{a - 4} \over 0}$$.
limit exist only when $$a - 4 = 0$$ $$ \Rightarrow $$ a = 4
Applying L' Hospital Rule
$$\mathop {\lim }\limits_{x \to 0} {{ - 16{e^{4x}}} \over {a\left( {4{e^{4x}}} \right) + a\left( {4{e^{4x}}} \right) + ax\left( {16{e^{4x}}} \right)}}$$
= $${{ - 16} \over {4a + 4a}} = {{ - 16} \over {32}} = - {1 \over 2} = b$$
$$a - 2b = 4 - 2\left( {{{ - 1} \over 2}} \right) = 4 + 1 = 5$$
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