Question

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Answer 1

Given :

$\sf\:f(x)=\dfrac{k\cos\:x}{\pi-2x},when\:x\neq\dfrac{\pi}{2}$

$\sf\:f(x)=3,when\:x=\dfrac{\pi}{2}$

To Find :

If $\sf\:\lim_{x\to\frac{\pi}{2}}\:f(x)=f(\frac{\pi}{2})$ ,Find the value of k.

Solution :

It is given that

$\sf\:f(\frac{\pi}{2})=\lim_{x\to\frac{\pi}{2}}$

Here ,$\sf\:f(\frac{\pi}{2})=3...(1)$

We have ,

$\sf\lim_{x\to\frac{\pi}{2}}=\dfrac{k\cos\:x}{\pi-2x}$

Let, $\sf\:\pi-2x=t$

$\sf\implies\:x=\dfrac{\pi}{2}-\dfrac{h}{2}$

$\sf\:\pi-2x=t$

If $\sf\:x\to\dfrac{\pi}{2}$

then,$\sf\:\pi-2\times\dfrac{\pi}{2}=t$

$\sf\implies\:t\to\:0$

Thus,

$\sf\lim_{x\to\frac{\pi}{2}}f(x)$

$\sf=\lim_{x\to\frac{\pi}{2}}\dfrac{k\cos\:x}{\pi-2x}$

$\sf=\lim_{t\to\:0}\dfrac{k\cos(\frac{\pi}{2}-\frac{t}{2})}{t}$

We know that, $\sf\:\cos(\frac{\pi}{2}-x)=\sin\:x$

$\sf=\lim_{t\to\:0}\dfrac{k\sin(\frac{t}{2})}{t}$

We know that:

$\bf\lim_{x\to\:0}\dfrac{\sin(x)}{x}=1$ ,thus :

$\sf=\lim_{t\to\:0}\dfrac{k\times\sin(\frac{t}{2})}{\frac{t}{2}\times2}$

$\sf=\lim_{t\to\:0}\dfrac{k}{2}\times\dfrac{\sin(\frac{t}{2})}{\frac{t}{2}}$

$\sf=\lim_{t\to\:0}\dfrac{k}{2}\times1$

$\sf=\dfrac{k}{2}$

Given :

$\sf\:f(\frac{\pi}{2})=\lim_{x\to\frac{\pi}{2}}$

$\sf\lim_{x\to\frac{\pi}{2}}=3$ [from equation (1)]

$\sf\implies\:\dfrac{k}{2}=3$

$\sf\implies\:k=6$

Therefore, The value of k = 6

Answer 2

Añswèr :

k = 6

Solution :

• Given ,f(x) = (kcosx)/π-2xwhen x≠ π/2

• f(x) = 3 , when x = π/2 => f(π/2) = 3

• Also , Lim x--> π/2 f(x) = f(π/2)--(1)

• Eqn(1)=>Lim x --> π/2 (kcosx)/π-2x = 3

________________________________L.H.S is in form of 0/0 (apply , L'hospital rule)

_______________________________

=>Limx-->π/2 d/dx(kcosx) / d/dx(π-2x) =3

• Limx-->π/2 (-ksinx)/(-2) = 3

Now substitute the limit

________________________________

• -ksin(π/2)/-2 = 3

• k(1) = 3×2
• k = 6

Note :

• L'hospital rule : When a function is in 0/0 (or) ∞/∞ model then both the numerator and denominator of the function can be differentiated to evaluate the limit
• d/dx(kcosx) = -ksinx
• d/dx(π-2x) = -2
• Sin(π/2) = 1