Let f(x)={k cosxπ−2x when x≠π2,3when x=π2 , and if limx→π2f(x)=f(π2) , find the value of k

This is a Short Answer Type Questions as classified in NCERT Exemplar Sol: Given f(x)={kcosxπ−2x when x≠π2,3when x=π2 LHL f(x)=limx→π2−kcosxπ−2x=limh→0kcos(π2−h)π−2(π2−h) =limh→0ksinhπ−π+2h=limh→0ksinh2h=k2.1=k2 [?limx→0sinxx=1] RHL f(x)=limx→π2+kcosxπ−2x=limh→0kcos(π2+h)π−2(π2+h) =limh→0−ksinhπ−π−2h=limh→0−ksinh−2h=k2 [?limx→0sinxx=1]We are given that limx→π2f(x)=3So, k2=3 ⇒k=6

Let $f (x) = {\begin{matrix} \frac{k c o s x}{π - 2 x} & when x \neq \frac{π}{2}, \\ 3 & when x = \frac{π}{2} \end{matrix}$ , and if $l i m_{x \to \frac{π}{2}} f (x) = f (\frac{π}{2})$ , find the value of $k$

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Alok Kumar Singh | Contributor-Level 10

7 months ago

This is a Short Answer Type Questions as classified in NCERT Exemplar

Sol:

$\begin{array}{l} G i v e n f (x) = {\begin{matrix} \frac{k c o s x}{π - 2 x} & w h e n x \neq \frac{π}{2}, \\ 3 & w h e n x = \frac{π}{2} \end{matrix} \\ L H L f (x) = \underset{x \to {\frac{π}{2}}^{-}}{l i m} \frac{k c o s x}{π - 2 x} = \underset{h \to 0}{l i m} \frac{k c o s (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} \\ = \underset{h \to 0}{l i m} \frac{k s i n h}{π - π + 2 h} = \underset{h \to 0}{l i m} \frac{k s i n h}{2 h} = \frac{k}{2} . 1 = \frac{k}{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ R H L f (x) = \underset{x \to {\frac{π}{2}}^{+}}{l i m} \frac{k c o s x}{π - 2 x} = \underset{h \to 0}{l i m} \frac{k c o s (\frac{π}{2} + h)}{π - 2 (\frac{π}{2} + h)} \\ = \underset{h \to 0}{l i m} \frac{- k s i n h}{π - π - 2 h} = \underset{h \to 0}{l i m} \frac{- k s i n h}{- 2 h} = \frac{k}{2} [? \underset{x \to 0}{l i m} \frac{s i n x}{x} = 1] \\ W e a r e g i v e n t h a t \underset{x \to \frac{π}{2}}{l i m} f (x) = 3 \\ S o, \frac{k}{2} = 3 \Rightarrow k = 6 \end{array}$

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