The function f ( x ) = { k cos x - 2 x , if x 2 3 , if x = 2 is continuous at x = 2 , when k equals:

Question

The function $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & if x \neq \frac{π}{2} \\ 3, & if x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , when $k$ equals:

Smart Achievers · Accepted Answer

(2)

Ans. 6

Explanation:

We have, $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & if x \neq \frac{π}{2} \\ 3, & if x = \frac{π}{2} \end{matrix}$

$f (x)$ is continuous at $x = \frac{π}{2}$

$∴$ $\lim_{x \to \frac{π^{-}}{2}} \frac{k \cos x}{π - 2 x} = 3$

$\Rightarrow \lim_{h \to 0} \frac{k \cos (\frac{π}{2} - h)}{π - 2 (\frac{π}{2} - h)} = 3$

$\Rightarrow \lim_{h \to 0} \frac{k \sin h}{2 h} = 3$

$∴$ $\frac{k}{2} = 3 \Rightarrow k = 6$

Solution

Q.
The function $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & if x \neq \frac{π}{2} \\ 3, & if x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , when $k$ equals:

1 - 6

2 6

3 5

4 - 5

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Solution

Q. The function f(x)={kcosxπ-2x,if x≠π23,if x=π2 is continuous at x=π2, when k equals:

1 - 6

2 6

3 5

4 - 5

Ans. (2) Ans. 6 Explanation: We have, f(x)={kcosxπ-2x,if x≠π23,if x=π2 f(x) is continuous at x=π2 ∴ limx→π-2kcosxπ-2x=3 ⇒limh→0kcos(π2-h)π-2(π2-h)=3 ⇒limh→0ksinh2h=3 ∴ k2=3⇒k=6

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Q.
The function $f (x) =$ ${\begin{matrix} \frac{k \cos x}{π - 2 x}, & if x \neq \frac{π}{2} \\ 3, & if x = \frac{π}{2} \end{matrix}$ is continuous at $x = \frac{π}{2}$ , when $k$ equals: