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?

& A.\dfrac{1}{2} \\\ & B.1 \\\ & C.\dfrac{1}{\sqrt{2}} \\\ & D.2 \\\ \end{aligned}$$

Answer 1

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

\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{aligned} & I=\dfrac{\sqrt{2}\left( \dfrac{1}{\sqrt{2}} \right)-1}{1-1} \\\ & I=\dfrac{0}{0} \\\ \end{aligned}$$ Hence, we have obtained $\dfrac{0}{0}\text{ form}$ Applying L.Hospital rule in equation (ii) we get: $$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{aligned} & 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{aligned}$$ Applying limit and using $\sin \dfrac{\pi }{4}=\sqrt{2}\text{ and cose}{{\text{c}}^{\text{2}}}\dfrac{\pi }{4}=2$ we get: $$\begin{aligned} & 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{aligned}$$ From equation (i) we have; $$\begin{aligned} & \dfrac{1}{2}=k \\\ & \Rightarrow k=\dfrac{1}{2} \\\ \end{aligned}$$ Hence, the value of $k=\dfrac{1}{2}$ **So, the correct answer is “Option A”.** **Note:** A possible confusion for students is how $\text{cose}{{\text{c}}^{\text{2}}}\dfrac{\pi }{4}$ became equal to 2. We have trigonometric identity as $$\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.