Question

If the function $f$ defined as $f(x) = \dfrac{1}{x} - \dfrac{{k - 1}}{{{e^{2x}} - 1}},{\text{ }}x \ne 0$, is continuous at $x = 0$, then the ordered pair $(k,f(0))$ is equal to?
A) $\left( {3,1} \right)$
B) $\left( {3,2} \right)$
C) $\left( {\dfrac{1}{3},2} \right)$
D) $\left( {2,1} \right)$

Answer 1

Hint:In the above question we are given a function continuous at $x = 0$ and we have to find value of $(k,f(0))$, that is in short we have to find value of $k{\text{ and }}f(0)$. Now in order to solve the question use the fact given in question that function continuous at $x = 0$, by using definition of continuity of any function at a point.Let a function $g(x)$ be continuous at $x = a$, if
$g(a + ) = g(a) = g(a - )$, where $g(a + ) = \mathop {\lim }\limits_{h \to 0} g(a + h){\text{ and }}g(a - ) = \mathop {\lim }\limits_{h \to 0} g(a - h)$.
So, using this we can easily find the required values.

Complete step-by-step answer:
We are given a function $f(x) = \dfrac{1}{x} - \dfrac{{k - 1}}{{{e^{2x}} - 1}},{\text{ }}x \ne 0$ …………….(1)
which is continuous at $x = 0$.
So here the question is giving us an idea about continuity of the function at $x = 0$.
So now we will consider the definition of continuity.
Let a function $g(x)$ be continuous at $x = a$, if
$g(a + ) = g(a) = g(a - )$, where $g(a + ) = \mathop {\lim }\limits_{h \to 0} g(a + h){\text{ and }}g(a - ) = \mathop {\lim }\limits_{h \to 0} g(a - h)$ …………….(2)
Now we are given function $f(x)$ which is continuous at $x = 0$ so now using definition of continuity from (2), we get,
$f(0 + ) = f(0) = f(0 - )$, where $f(0 + ) = \mathop {\lim }\limits_{h \to 0} f(0 + h){\text{ and }}f(0 - ) = \mathop {\lim }\limits_{h \to 0} f(0 - h)$ …………….(3)
Firstly, lets find the value of $f(0 - )$ using (3), we get,
Therefore, $f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{{0 + h}} - \dfrac{{k - 1}}{{{e^{2(0 + h)}} - 1}}$
Now further simplifying, we get,
$f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{1}{h} - \dfrac{{k - 1}}{{{e^{2h}} - 1}}$
Now taking L.C.M, we get,
$f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{2h}} - 1 - (k - 1)(h)}}{{(h)({e^{2h}} - 1)}}$
$f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{2h}} - 1 - kh + h}}{{(h)({e^{2h}} - 1)}}$ …………….(4)
Now we how the series expansion for ${e^x}$ is,
${e^x} = 1 + x + \dfrac{{{x^2}}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{{x^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{{x^4}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_$, where $\left| \!{\underline {\,
  n \,}} \right. = 1 \times 2 \times 3 \times 4 \times \_\_\_ \times n$
Hence using the above formula, we can get value of ${e^h}$, so we get
${e^h} = 1 + x + \dfrac{{{h^2}}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{{h^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{{h^4}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_$
Now according to (4), we need ${e^{2h}}$, so we will get,
 ${e^{2h}} = 1 + 2h + \dfrac{{4{h^2}}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^4}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_$ …………….(5)
Now, substituting value of ${e^{ - 2h}}$ from (5) in (4), we get,
$
  f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {1 + 2h + \dfrac{{4{h^2}}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^4}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right) - 1 - kh + h}}{{(h)\left\{ {\left( {1 + 2h + \dfrac{{4{h^2}}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^4}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right) - 1} \right\}}} \\
  f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {2h + \dfrac{{4{h^2}}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^4}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right) - kh + h}}{{(h)\left\{ {\left( {2h + \dfrac{{4{h^2}}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^4}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)} \right\}}} \\
 $
Now taking $h$ common from numerator.
$f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{h\left\{ {\left( {2 + \dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right) - k + 1} \right\}}}{{(h)\left\{ {\left( {2h + \dfrac{{4{h^2}}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^4}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)} \right\}}}$
Now cancelling $h$ from numerator and denominator we get,
$f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {3 + \dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right) - k}}{{\left( {2h + \dfrac{{4{h^2}}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^3}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^4}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)}}$
Now taking $h$ common from denominator we get
$f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {3 + \dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right) - k}}{{(h)\left( {2 + \dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)}}$ …………….(6)
Now for this limit to exist $h$ present in denominator must be cancelled, and to cancel it, we know $3 - k = 0$ and then we can take $h$ common from numerator and cancel it with $h$ present in denominator.
So, from this we can get value of $k$-
$
  3 - k = 0 \\
   \Rightarrow k = 3 \\
 $ …………….(7)
Now considering (6) and substituting value of $k$ from (7) in it, we get,
 $
  f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {3 + \dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right) - 3}}{{(h)\left( {2 + \dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)}} \\
  f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)}}{{(h)\left( {2 + \dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)}} \\
 $
Now taking $h$ common from numerator and cancelling it with $h$ in denominator, we get,
$
  f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{(h)\left( {\dfrac{4}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8h}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^2}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)}}{{(h)\left( {2 + \dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)}} \\
  f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{\left( {\dfrac{4}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8h}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^2}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)}}{{\left( {2 + \dfrac{{4h}}{{\left| \!{\underline {\,
  2 \,}} \right. }} + \dfrac{{8{h^2}}}{{\left| \!{\underline {\,
  3 \,}} \right. }} + \dfrac{{16{h^3}}}{{\left| \!{\underline {\,
  4 \,}} \right. }} + \_\_\_} \right)}} \\
 $
Now simply solving the limit $h \to 0$, we get,
$f(0 + ) = \dfrac{{\dfrac{4}{{\left| \!{\underline {\,
  2 \,}} \right. }}}}{2}$
Now using $\left| \!{\underline {\,
  2 \,}} \right. = 1 \times 2 = 2$, we get,
$
  f(0 + ) = \dfrac{{\dfrac{4}{2}}}{2} \\
  f(0 + ) = \dfrac{4}{4} \\
  f(0 + ) = 1 \\
 $ …………….(8)
Now using definition of continuity for $f$, we have
$f(0 + ) = f(0) = f(0 - )$
So, we can write that using (8),
$f(0 + ) = f(0) = 1$
Hence $f(0) = 1$ …………….(9)
Now from (7) and (9), we have,
$k = 3{\text{ and }}f(0) = 1$ …………….(10)
Now according to question, we have to find the ordered pair $\left( {k,f(0)} \right)$
Now substituting values of $k{\text{ and }}f(0)$ from (10) in ordered pair we get-
$\left( {k,f(0)} \right) \equiv \left( {3,1} \right)$

So, the correct answer is “Option A”.

Note:In the above question to find the values of $k{\text{ and }}f(0)$ we used $f(0 + )$, but if we use $f(0 - )$ there will be no change in the answers. Therefore, we can use $f(0 + ){\text{ as well as }}f(0 - )$.
We can also use another method to solve limit in (4)
$f(0 + ) = \mathop {\lim }\limits_{h \to 0} \dfrac{{{e^{2h}} - 1 - kh + h}}{{(h)({e^{2h}} - 1)}}$
As we can clearly see for $x \to 0,\dfrac{{{e^{2h}} - 1 - kh + h}}{{(h)({e^{2h}} - 1)}} \to \dfrac{0}{0}$
So, it is indeterminate form $\dfrac{0}{0}$, and we can use L hospital rule.
L hospital rule states that
$\mathop {\lim }\limits_{x \to 0} \dfrac{{m(x)}}{{n(x)}} = \mathop {\lim }\limits_{x \to 0} \dfrac{{\dfrac{{dm(x)}}{{dx}}}}{{\dfrac{{dn(x)}}{{dx}}}}$, where $m(x) \to 0,n(x) \to 0{\text{ for }}x \to 0$
Therefore, we can now easily find $k{\text{ and }}f(0)$, using L hospital rule for (4).