If f x - left begin matrix k- -when - X -0lendmatrix right is continuous at x -0- then the value of k will beA- 1 B- 2-5C- -2-5D- None of these

The correct option is B 2/5lim_x → 0f(x)=lim_x → 02 sin 2x/2x.5=2/5=k ...

237

You visited us 237 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Mathematics

Expansions to Remove Indeterminate Form

If f x = left...

Question

If $f (x) = {\begin{matrix} \frac{s i n 2 x}{5 x}, & w h e n x \neq 0 k, & w h e n x = 0 \end{matrix}$ is continuous at x = 0, then the value of k will be

1

No worries! We‘ve got your back. Try BYJU‘S free classes today!

\frac{2}{5}

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

- \frac{2}{5}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

N o n e o f t h e s e

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is B $\frac{2}{5}$
${lim}_{x \to 0} f (x) = {lim}_{x \to 0} \frac{2 s i n 2 x}{2 x .5} = \frac{2}{5} = k$

Suggest Corrections

Similar questions

Q. Let

f (x) = ⎧ ⎨ ⎩ \begin{matrix} \frac{x^{3} + x^{2} - 16 x + 20}{(x - 2)^{2}}, & i f x \neq 2 k, i f x = 2 \end{matrix}

f f(x)be continuous for all x, then k =

Q. If the function

f (x) = {\begin{matrix} \frac{k c o s x}{π - 2 x}, & w h e n & x \neq \frac{π}{2} 3, & w h e n & x = \frac{π}{2} \end{matrix}

be continuous at

x - \frac{π}{2}

, then k=

If function be $f (x) = {\begin{matrix} \frac{x^{2} - 1}{x - 1} & when x \neq 1 k & when x = 1 \end{matrix}$ is continuous at $x = 1$ , then the value of $k$ is?

Q. If

f (x) = {\begin{matrix} \frac{36^{x} - 9^{x} - 4^{x} + 1}{\sqrt{2} - \sqrt{1 + c o s x}} & , x \neq 0 k & , x = 0 \end{matrix}

Is continuous at x =0, then k equals:

Q. If

f (x) = ⎧ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎩ \frac{1 - cos x}{\begin{matrix} x k \end{matrix}} \begin{matrix} x \neq 0 x = 0 \end{matrix} ⎫ ⎪ ⎪ ⎪ ⎪ ⎬ ⎪ ⎪ ⎪ ⎪ ⎭

is continuous at

x = 0

, then the value of

k

is:

Join BYJU'S Learning Program

If f(x)={sin2x5x,when x≠0k,when x=0 is continuous at x = 0, then the value of k will be

The correct option is B 25limx→0f(x)=limx→02 sin2x2x.5=25=k

If $f (x) = {\begin{matrix} \frac{s i n 2 x}{5 x}, & w h e n x \neq 0 k, & w h e n x = 0 \end{matrix}$ is continuous at x = 0, then the value of k will be

The correct option is B $\frac{2}{5}$
${lim}_{x \to 0} f (x) = {lim}_{x \to 0} \frac{2 s i n 2 x}{2 x .5} = \frac{2}{5} = k$