Question

solve it fast everyone...!!!!​

Answer 1

Given Question :-

The value of k, if f(x) is continuous at x = 0

$\begin{gathered}\begin{gathered}\bf\: f(x) \: = \: \begin{cases} &\sf{ {\bigg(cosx\bigg) }^{\dfrac{1}{x}} \: \: when \: x \: \ne \: 0} \\ &\sf{k \: \: when \: x \: = \: 0} \end{cases}\end{gathered}\end{gathered} \: \: is$

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

Given function is

$\begin{gathered}\begin{gathered}\bf\:\rm :\longmapsto\: f(x) \: = \: \begin{cases} &\sf{ {\bigg(cosx\bigg) }^{\dfrac{1}{x}} \: \: when \: x \: \ne \: 0} \\ &\sf{k \: \: when \: x \: = \: 0} \end{cases}\end{gathered}\end{gathered}$

Its given that f(x) is continuous at x = 0.

We know, A function f(x) is continuous at x = 0, iff

$\boxed{ \tt{ \: f(0) = \displaystyle\lim _{x \to 0}f(x) \: }}$

So, using this

$\rm :\longmapsto\:k = \displaystyle\lim _{x \to 0} {\bigg(cosx\bigg) }^{\dfrac{1}{x} }$

To evaluate this limit, Taking log on both sides, we get

$\rm :\longmapsto\:logk = \displaystyle\lim _{x \to 0} log{\bigg(cosx\bigg) }^{\dfrac{1}{x} }$

$\rm :\longmapsto\:logk = \displaystyle\lim _{x \to 0}\dfrac{1}{x}log(cosx)$

$\rm :\longmapsto\:logk = \displaystyle\lim _{x \to 0}\dfrac{log(cosx)}{x}$

On applying L - Hospital Rule, we get

$\rm :\longmapsto\:logk = \displaystyle\lim _{x \to 0}\dfrac{\dfrac{d}{dx}log(cosx)}{\dfrac{d}{dx}x}$

We know

$\boxed{ \tt{ \: \dfrac{d}{dx}logx \: = \: \frac{1}{x} \: }} \\ \\ and \\ \\ \boxed{ \tt{ \: \dfrac{d}{dx}x \: = \: 1 \: }}$

So, using this, we get

$\rm :\longmapsto\:logk = \displaystyle\lim _{x \to 0}\dfrac{\dfrac{1}{cosx}\dfrac{d}{dx}cosx}{1}$

$\rm :\longmapsto\:logk = \displaystyle\lim _{x \to 0}\dfrac{1}{cosx} \times ( - sinx)$

$\rm :\longmapsto\:logk = - \displaystyle\lim _{x \to 0} \: tanx$

$\rm :\longmapsto\:logk = - \: tan0$

$\rm :\longmapsto\:logk = -0$

$\rm :\longmapsto\:logk =0$

$\rm :\longmapsto\:k = {e}^{0}$

$\bf\implies \:k \: = \: 1$

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Formula Used :-

$\boxed{ \tt{ \: log_{x}(y) = z \: \rm \implies\:y = {x}^{z} \: }}$

$\boxed{ \tt{ \: \dfrac{d}{dx}sinx \: = \: - \: cosx \: }}$

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Additional Information :-

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$