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# 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}

Last updated date: 08th Sep 2024
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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.

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.