Answered

At Westonci.ca, we make it easy to get the answers you need from a community of informed and experienced contributors. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

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

Note:-

Spams will be deleted on the spot.
Answer with proper explanation​


Sagot :

Answer is in attachment.

k=6

View image Nepalieducation

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]

We appreciate your time. Please come back anytime for the latest information and answers to your questions. We hope our answers were useful. Return anytime for more information and answers to any other questions you have. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.