QJEEM18Mae530: JEE Main 2018 Mathematics - Class 12 JEEM18Mae530 Solution: If the function f defined as is continuous at x = 0, then the…

If the function is continuous at x = 0, then
$\mathop {\lim }\limits_{x \to 0}$ f(x) will exist and f(0) = $\mathop {\lim }\limits_{x \to 0}$ f(x)

Now, $\mathop {\lim }\limits_{x \to 0}$ f(x) = $\mathop {\lim }\limits_{x \to 0} \left( {{1 \over x} - {{k - 1} \over {{e^{2x}} - 1}}} \right)$

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

= $\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ....} \right) - 1 - kx + x} \over {\left( x \right)\left( {\left( {1 + 2x + {{{{\left( {2x} \right)}^2}} \over {2!}} + {{{{\left( {2x} \right)}^3}} \over {3!}} + ...} \right) - 1} \right)}}} \right]$

= $\mathop {\lim }\limits_{x \to 0} \left[ {{{\left( {3 - k} \right)x + {{4{x^2}} \over {2!}} + {{8{x^3}} \over {3!}} + ...} \over {\left( {2{x^2} + {{4{x^3}} \over {2!}} + {{8{x^3}} \over {3!}} + ....} \right)}}} \right]$

For the limit to exist, power of x in the numerator should be greater than or equal to the power of x in the denominator. Therefore, coefficient of x in numerator is equal to zero

$\Rightarrow$   3 $-$ k = 0

$\Rightarrow$   k = 3

So the limit reduces to

$\mathop {\lim }\limits_{x \to 0} {{\left( {{x^2}} \right)\left( {{4 \over {2!}} + {{8x} \over {3!}} + ...} \right)} \over {\left( {{x^2}} \right)\left( {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...} \right)}}$

= $\mathop {\lim }\limits_{x \to 0} {{{4 \over {2!}} + {{8x} \over {3!}} + ...} \over {2 + {{4x} \over {2!}} + {{8{x^2}} \over {3!}} + ...}}$ = 1

Hence, f(0) = 1