Answered

At Westonci.ca, we provide reliable answers to your questions from a community of experts. Start exploring today! Get immediate and reliable solutions to your questions from a community of experienced professionals on our platform. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.

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 :

Nepalieducation

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 hope you found what you were looking for. Feel free to revisit us for more answers and updated information. We appreciate your visit. Our platform is always here to offer accurate and reliable answers. Return anytime. Your questions are important to us at Westonci.ca. Visit again for expert answers and reliable information.