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QJEEM21Ma038c: JEE Main 2021 Mathematics - Class 12 JEEM21Ma038c Solution: If exists and is equal to b, then the value of a 2b is…

If lim⁡x→0ax−(e4x−1)ax(e4x−1)\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 __________.

Step-by-Step Verified Solution

lim⁡x→0ax−(e4x−1)ax(e4x−1)\mathop {\lim }\limits_{x \to 0} {{ax - \left( {{e^{4x}} - 1} \right)} \over {ax\left( {{e^{4x}} - 1} \right)}}

Applying L' Hospital Rule

lim⁡x→0a−4e4xa(e4x−1)+ax(4e4x)\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−40{{a - 4} \over 0} .

limit exist only when a−4=0a - 4 = 0 ⇒\Rightarrow a = 4

Applying L' Hospital Rule

lim⁡x→0−16e4xa(4e4x)+a(4e4x)+ax(16e4x)\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)}}

= −164a+4a=−1632=−12=b{{ - 16} \over {4a + 4a}} = {{ - 16} \over {32}} = - {1 \over 2} = b

a−2b=4−2(−12)=4+1=5a - 2b = 4 - 2\left( {{{ - 1} \over 2}} \right) = 4 + 1 = 5

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