Question

Let


`\boxed{\sf f(x)=\begin{cases}\sf (kcosx)/(\pi -2x),x\\eq (\pi)/(2) \\ \sf 3,x=(\pi)/(2)\end{cases}}`
If


`\\ \rm\hookrightarrow {\displaystyle{\lim_{x\to (\pi)/(2)}}}f(x)=f\left((\pi)/(2)\right)`

Find k

Note:-

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Answer with proper explanation​

Answer 1

We are provided with ;

`{:\implies \quad \bf f(x)=\displaystyle \begin{cases}\bf (k\cos (x))/(\pi -2x)\:\:,\:\: x\\eq (\pi)/(2)\\ \\ \bf 3\:\:,\:\: x=(\pi)/(2)\end{cases}}`

Also we are given with ;

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

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 ;

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

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 ,

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

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

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

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

`{:\implies \quad \sf x=(\pi)/(2)-y\:\: , as\:\: x\to (\pi)/(2)\:\:,\: So\:\: y\to0}`

Putting ;

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

Now , we knows that

• 
  `{\boxed{\bf{\cos \left((\pi)/(2)-\theta \right)=\sin (\theta)}}}`

Using this , we have :

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

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

`{:\implies \quad \displaystyle \sf k \lim_(y\to0)\frac{\sin (y)}{\cancel{\pi}-\cancel{\pi} +2y}=3}`

`{:\implies \quad \displaystyle \sf k \lim_(y\to0)(\sin (y))/(2y)=3}`

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

`{:\implies \quad \displaystyle \sf (k)/(2) \lim_(y\to0)(\sin (y))/(y)=3}`

Now , we also knows that ;

• 
  `{\boxed{\displaystyle \bf \lim_(h\to0)(\sin (h))/(h)=1}}`

Using this we have ;

`{:\implies \quad \sf (k)/(2)=3}`

`{:\implies \quad \bf \therefore \quad \underline{\underline{k=6}}}`

Answer 2

Answer is in attachment.

k=6