Question

If the function f defined on $\left( \dfrac{\pi }{6},\dfrac{\pi }{3} \right)$ by \[f\left( x \right)=\left\{ \begin{matrix}
   \dfrac{\sqrt{2}\cos x-1}{\cot \text{ }x-1}, & x\ne \dfrac{\pi }{4} \\
   k, & x\ne \dfrac{\pi }{4} \\
\end{matrix} \right.\] is continuous then k is equal to?
\[\begin{align}
  & A.\dfrac{1}{2} \\
 & B.1 \\
 & C.\dfrac{1}{\sqrt{2}} \\
 & D.2 \\
\end{align}\]

Answer 1

Hint: To solve this question, we will first understand what are continuous functions. A function $g:A\to B$ is continuous as $a\in A$ if g (a) is well defined and \[\underset{x\to a}{\mathop{\lim }}\,\text{ }g\left( x \right)=g\left( a \right)\]
Now, in our question, as we are already given f (x) is continuous so, we only need to compare and equate $\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ f}\left( x \right)$ to $f\left( \dfrac{\pi }{4} \right)$ to get value of k. And while calculating $\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ f}\left( x \right)$ we will use L.Hospital Rule stated as
L.Hospital Rule is a method which is applicable where the obtained value is of type $\dfrac{0}{0}\Rightarrow \dfrac{\infty }{\infty }$
To apply this rule after we get $\dfrac{0}{0}\Rightarrow \dfrac{\infty }{\infty }$ form, we just differentiate both numerator and denominator separately with respect to the given function.

Complete step-by-step answer:
We are going to define f on $\left( \dfrac{\pi }{6},\dfrac{\pi }{3} \right)$ as
\[f\left( x \right)=\left\{ \begin{matrix}
   \dfrac{\sqrt{2}\cos x-1}{\cot \text{ }x-1}, & x\ne \dfrac{\pi }{4} \\
   k, & x\ne \dfrac{\pi }{4} \\
\end{matrix} \right.\]
Given that, f is continuous. So, first of all we will define when a function is called continuous in a given domain.
A function defined as $g:A\to B$ is called continuous for $a\in A$ every 'a' in domain if \[\underset{x\to a}{\mathop{\lim }}\,\text{ }g\left( x \right)=g\left( a \right)\] and g (a) is defined.
Or for more elaborative definition we have
\[\underset{x\to {{a}^{-}}}{\mathop{\lim }}\,\text{ }g\left( x \right)=g\left( a \right)=\underset{x\to {{a}^{+}}}{\mathop{\lim }}\,\text{ }g\left( x \right)\]
So, for given f (x) in our question we will check for both values ${{\dfrac{\pi }{4}}^{+}}\text{ and }{{\dfrac{\pi }{4}}^{-}}$ than
\[\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,\text{ f}\left( x \right)=f\left( \dfrac{\pi }{4} \right)=\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\text{ f}\left( x \right)\]
Now, here as we are given
\[f\left( x \right)=\left\{ \begin{matrix}
   \dfrac{\sqrt{2}\cos x-1}{\cot \text{ }x-1}, & x\ne \dfrac{\pi }{4} \\
   k, & x\ne \dfrac{\pi }{4} \\
\end{matrix} \right.\]
That is, for both ${{\dfrac{\pi }{4}}^{+}}\text{ and }{{\dfrac{\pi }{4}}^{-}}$ we have $f\left( x \right)=\dfrac{\sqrt{2}\cos x-1}{\cot \text{ }x-1}$ so we can calculate just;
\[\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ f}\left( x \right)=f\left( \dfrac{\pi }{4} \right)\]
Then, if this holds, function is continuous.
Now, as we are given f (x) is continuous
\[\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ f}\left( x \right)=f\left( \dfrac{\pi }{4} \right)\text{ holds }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (i)}\]
Now $f\left( \dfrac{\pi }{4} \right)=k$ and consider \[\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ f}\left( x \right)=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ }\dfrac{\sqrt{2}\cos x-1}{\cot \text{ }x-1}\]
We have \[\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ }\dfrac{\sqrt{2}\cos x-1}{\cot \text{ }x-1}\] let it be I.
We will first try to obtain answer of the question that if after applying limit value are we getting $\dfrac{0}{0}\text{ form}\Rightarrow \dfrac{\infty }{\infty }\text{ form}$
If so then, we will apply L.Hospital Rule.
Let us define L.Hospital Rule:
L.Hospital Rule is a method which is applicable when the obtained value is of type $\dfrac{0}{0}\Rightarrow \dfrac{\infty }{\infty }$
To apply this rule after, we get $\dfrac{0}{0}\Rightarrow \dfrac{\infty }{\infty }\text{ form}$ we first differentiate both numerator and denominator separately with respect to the given function.
Here, we have \[\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ }\dfrac{\sqrt{2}\cos x-1}{\cot \text{ }x-1}=I\text{ }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. }\text{. (ii)}\]
Applying \[x\to \dfrac{\pi }{4}\Rightarrow I=\dfrac{\sqrt{2}\cos \dfrac{\pi }{4}-1}{\cot \dfrac{\pi }{4}-1}\]
As value of \[\cos \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\] and we have $\cot \dfrac{\pi }{4}=1$
\[\begin{align}
  & I=\dfrac{\sqrt{2}\left( \dfrac{1}{\sqrt{2}} \right)-1}{1-1} \\
 & I=\dfrac{0}{0} \\
\end{align}\]
Hence, we have obtained $\dfrac{0}{0}\text{ form}$
Applying L.Hospital rule in equation (ii) we get:
\[I=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ }\dfrac{\dfrac{d}{dx}\left( \sqrt{2}\cos x-1 \right)}{\dfrac{d}{dx}\left( \cot \text{ }x-1 \right)}\]
Now, value of $\dfrac{d}{dx}\cos x=-\sin x,\dfrac{d}{dx}\cot x=-\text{cose}{{\text{c}}^{\text{2}}}x\text{ and }\dfrac{d}{dx}\left( 1 \right)=0$
Using this all in above, we get:
\[\begin{align}
  & I=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ }\dfrac{-\sqrt{2}\sin x}{\text{cose}{{\text{c}}^{\text{2}}}x} \\
 & I=\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ }\dfrac{\sqrt{2}\sin x}{\text{cose}{{\text{c}}^{\text{2}}}x} \\
\end{align}\]
Applying limit and using $\sin \dfrac{\pi }{4}=\sqrt{2}\text{ and cose}{{\text{c}}^{\text{2}}}\dfrac{\pi }{4}=2$ we get:
\[\begin{align}
  & I=\dfrac{\sqrt{2}\sin \dfrac{\pi }{4}}{\text{cose}{{\text{c}}^{\text{2}}}\dfrac{\pi }{4}} \\
 & I=\dfrac{\sqrt{2}\left( \dfrac{1}{\sqrt{2}} \right)}{2} \\
 & I=\dfrac{1}{2} \\
\end{align}\]
From equation (i) we have;
\[\begin{align}
  & \dfrac{1}{2}=k \\
 & \Rightarrow k=\dfrac{1}{2} \\
\end{align}\]
Hence, the value of $k=\dfrac{1}{2}$

So, the correct answer is “Option A”.

Note: A possible confusion for students is how $\text{cose}{{\text{c}}^{\text{2}}}\dfrac{\pi }{4}$ became equal to 2.
We have trigonometric identity as \[\sec \theta =\dfrac{1}{\text{cosec}\theta }\Rightarrow \sin \dfrac{\pi }{4}=\dfrac{1}{\text{cose}{{\text{c}}^{\text{2}}}\dfrac{\pi }{4}}\]
Now, value of \[\sin \dfrac{\pi }{4}=\dfrac{1}{\sqrt{2}}\Rightarrow \text{cosec}\dfrac{\pi }{4}\Rightarrow \dfrac{1}{\sqrt{2}}=\dfrac{1}{\text{cosec}\dfrac{\pi }{4}}\Rightarrow \text{cosec}\dfrac{\pi }{4}=\sqrt{2}\text{ and cose}{{\text{c}}^{\text{2}}}\dfrac{\pi }{4}=\sqrt{2}-\sqrt{2}=2\]
Also, a key point to note in this question is that, we only calculated $\underset{x\to \dfrac{\pi }{4}}{\mathop{\lim }}\,\text{ f}\left( x \right)$ and not \[\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\text{ f}\left( x \right)\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,\text{ f}\left( x \right)\]
This was so because value of f (x) at ${{\dfrac{\pi }{4}}^{+}}\text{ and }{{\dfrac{\pi }{4}}^{-}}$ was same. If it is different in any other question, then we would calculate \[\underset{x\to {{\dfrac{\pi }{4}}^{-}}}{\mathop{\lim }}\,\text{ f}\left( x \right)\text{ and }\underset{x\to {{\dfrac{\pi }{4}}^{+}}}{\mathop{\lim }}\,\text{ f}\left( x \right)\] separately.