Question

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Answer 1

$\large\underline{\sf{Given \:Question - }}$

$\sf \: If \: f(x) = \dfrac{ \sqrt{2}cosx - 1 }{cotx - 1}, \: x \: \ne \: \dfrac{\pi}{4}, \: find \: f\bigg(\dfrac{\pi}{4}\bigg) \: if$

$\sf \: is \: continuous \: at \: x \: = \: \dfrac{\pi}{4}$

$\large\underline{\sf{Solution-}}$

Given that

$\rm :\longmapsto\:\sf \: \: f(x) = \dfrac{ \sqrt{2}cosx - 1 }{cotx - 1}, \: x \: \ne \: \dfrac{\pi}{4}\:$

Let assume that

$\rm :\longmapsto\:f\bigg[\dfrac{\pi}{4} \bigg] = k$

We know,

A function f(x) is said to be continuous at x = a, iff

$\boxed{ \rm \:\displaystyle\lim_{x \to a}f(x) = f(a)}$

Since, it is given that,

$\rm :\longmapsto\:\sf \: f(x) \: is \: continuous \: at \: x \: = \: \dfrac{\pi}{4}$

Therefore,

$\rm :\longmapsto\:\displaystyle\lim_{x \to \dfrac{\pi}{4}} \: f(x) = f\bigg(\dfrac{\pi}{4}\bigg)$

$\rm :\longmapsto\:k = \displaystyle\lim_{x \to \dfrac{\pi}{4}} \: \frac{ \sqrt{2}cosx - 1}{cotx - 1}$

$\rm :\longmapsto\:k = \displaystyle\lim_{x \to \dfrac{\pi}{4}} \: \frac{ \sqrt{2}cosx - 1}{\dfrac{cosx}{sinx} - 1}$

$\rm :\longmapsto\:k = \displaystyle\lim_{x \to \dfrac{\pi}{4}} \: \frac{ \sqrt{2}cosx - 1}{\dfrac{cosx - sinx}{sinx}}$

$\rm :\longmapsto\:k = \displaystyle\lim_{x \to \dfrac{\pi}{4}} \: \frac{sinx(\sqrt{2}cosx - 1)}{cosx - sinx}$

On rationalizing both numerator and denominator, we get

$\rm :\longmapsto\:k = \displaystyle\lim_{x \to \dfrac{\pi}{4}} \: \frac{sinx(\sqrt{2}cosx - 1)}{cosx - sinx} \times \dfrac{ \sqrt{2} cosx + 1}{ \sqrt{2} cosx + 1} \times \dfrac{cosx + sinx}{cosx + sinx}$

We know,

$\boxed{ \rm \:(x + y)(x - y) = {x}^{2} - {y}^{2}}$

So, using this, we get

$\rm :\longmapsto\:k = \displaystyle\lim_{x \to \dfrac{\pi}{4}} \: \frac{sinx( {2cos}^{2} x - 1)(cosx + sinx)}{ ({cos}^{2}x - {sin}^{2}x)( \sqrt{2}cosx + 1) }$

We know,

$\boxed{ \rm \:cos2x = {2cos}^{2}x - 1}$

and

$\boxed{ \rm \:cos2x = {cos}^{2}x - {sin}^{2} x}$

So, using this, we get

$\rm :\longmapsto\:k = \displaystyle\lim_{x \to \dfrac{\pi}{4}} \: \frac{sinx( cos2x)(cosx + sinx)}{ cos2x( \sqrt{2}cosx + 1) }$

$\rm :\longmapsto\:k = \displaystyle\lim_{x \to \dfrac{\pi}{4}} \: \frac{sinx(cosx + sin)}{\sqrt{2}cosx + 1}$

$\rm :\longmapsto\:k = sin\dfrac{\pi}{4}\bigg(cos\dfrac{\pi}{4} + sin\dfrac{\pi}{4} \bigg) \div \bigg( \sqrt{2}cos\dfrac{\pi}{4} + 1 \bigg)$

$\rm :\longmapsto\:k = \dfrac{1}{ \sqrt{2} } \times \bigg(\dfrac{1}{ \sqrt{2} } + \dfrac{1}{ \sqrt{2} } \bigg) \div \bigg( \sqrt{2} \times \dfrac{1}{ \sqrt{2} } + 1\bigg)$

$\rm :\longmapsto\:k = \dfrac{1}{ \sqrt{2} } \times \bigg(\dfrac{2}{ \sqrt{2} } \bigg) \div \bigg(1 + 1 \bigg)$

$\bf\implies \:k \: = \: \dfrac{1}{2}$

Hence,

$\rm :\longmapsto\:\boxed{ \bf \: \: \: \:f\bigg[\dfrac{\pi}{4} \bigg] = \dfrac{1}{2} \: \: \: \: }$