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Let

[tex]\boxed{\sf f(x)=\begin{cases}\sf \dfrac{kcosx}{\pi -2x},x\neq \dfrac{\pi}{2} \\ \sf 3,x=\dfrac{\pi}{2}\end{cases}}[/tex]
If

[tex]\\ \rm\hookrightarrow {\displaystyle{\lim_{x\to \dfrac{\pi}{2}}}}f(x)=f\left(\dfrac{\pi}{2}\right)[/tex]

Find k

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Sagot :

Nepalieducationidn

Answer is in attachment.

k=6

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We are provided with ;

[tex]{:\implies \quad \bf f(x)=\displaystyle \begin{cases}\bf \dfrac{k\cos (x)}{\pi -2x}\:\:,\:\: x\neq \dfrac{\pi}{2}\\ \\ \bf 3\:\:,\:\: x=\dfrac{\pi}{2}\end{cases}}[/tex]

Also we are given with ;

[tex]{:\implies \quad \displaystyle \bf \lim_{x\to \footnotesize \dfrac{\pi}{2}}f(x)=f\left(\dfrac{\pi}{2}\right)}[/tex]

At first , let's define the function at x = π/2 . Now , as given that f(x) = 3 , x = π/2. Implies , f(π/2) = 3

Now , we have ;

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to \footnotesize \dfrac{\pi}{2}}f(x)=3}[/tex]

Now , As in RHS , x is approaching π/2 , means that x is in neighbourhood of π/2 , x is coming towards π/2 , but it's not π/2 , implies f(x) for the limit in LHS is defined for x ≠ π/2 or we don't have to take value of x as π/2 , means x ≠ π/2 in that case , means we have to take f(x) = {kcos(x)}/π-2x , x ≠ π/2 for the limit given in LHS ,

[tex]{:\implies \quad \displaystyle \sf \lim_{x\to \footnotesize \dfrac{\pi}{2}}\dfrac{k\cos (x)}{\pi -2x}=3}[/tex]

Now , As k is constant , so take it out of the limit

[tex]{:\implies \quad \displaystyle \sf k \lim_{x\to \footnotesize \dfrac{\pi}{2}}\dfrac{\cos (x)}{\pi -2x}=3}[/tex]

For , further evaluation of the limit , we will use substitution , putting ;

[tex]{:\implies \quad \sf x=\dfrac{\pi}{2}-y\:\: , as\:\: x\to \dfrac{\pi}{2}\:\:,\: So\:\: y\to0}[/tex]

Putting ;

[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\cos \left(\dfrac{\pi}{2}-y\right)}{\pi -2\left(\dfrac{\pi}{2}-y\right)}=3}[/tex]

Now , we knows that

  • [tex]{\boxed{\bf{\cos \left(\dfrac{\pi}{2}-\theta \right)=\sin (\theta)}}}[/tex]

Using this , we have :

[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\sin (y)}{\pi -\bigg\{2\left(\dfrac{\pi}{2}\right)-2y\bigg\}}=3}[/tex]

[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\sin (y)}{\pi -(\pi -2y)}=3}[/tex]

[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\sin (y)}{\cancel{\pi}-\cancel{\pi} +2y}=3}[/tex]

[tex]{:\implies \quad \displaystyle \sf k \lim_{y\to0}\dfrac{\sin (y)}{2y}=3}[/tex]

Take ½ out of the limit as it's too constant ;

[tex]{:\implies \quad \displaystyle \sf \dfrac{k}{2} \lim_{y\to0}\dfrac{\sin (y)}{y}=3}[/tex]

Now , we also knows that ;

  • [tex]{\boxed{\displaystyle \bf \lim_{h\to0}\dfrac{\sin (h)}{h}=1}}[/tex]

Using this we have ;

[tex]{:\implies \quad \sf \dfrac{k}{2}=3}[/tex]

[tex]{:\implies \quad \bf \therefore \quad \underline{\underline{k=6}}}[/tex]

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