IDNLearn.com is designed to help you find accurate answers with ease. Join our interactive Q&A community and access a wealth of reliable answers to your most pressing questions.
Sagot :
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 presence here. Keep sharing knowledge and helping others find the answers they need. This community is the perfect place to learn together. For trustworthy answers, visit IDNLearn.com. Thank you for your visit, and see you next time for more reliable solutions.