Question

Question is in attachment​

Answer 1

Answer:

(b) 6 is the correct option

Step-by-step explanation:

Given :

The function

$\begin{aligned}f(x) & = \dfrac{sin3x}{x} , \ \ \ \ &x \neq0 \\& = \dfrac{k}{2}, \ \ \ \ &x = 0\end{aligned}$

is continuous at x = 0

To find :

Value of k

Solution :

As f(x) is continuous at x = 0,

$\boxed{\mathrm{\lim \limits_{x \mapsto 0} f(x) = f(0)}}$

Here, f(0) = k/2 . So, we get,

$\implies \mathrm{\lim \limits_{x\mapsto 0} \ \dfrac{sin3x}{x} = \dfrac{k}{2}}$

Multiply and divide by 3 on LHS

$\implies \mathrm{\lim \limits_{x\mapsto 0} \ \dfrac{3sin3x}{3x} = \dfrac{k}{2}}$

We know that,

$\boxed{\mathrm{\lim \limits_{a\mapsto 0} \ \dfrac{sina}{a}= 1}}$

Here, a = 3x. So,

$\implies \mathrm{3(1) = \dfrac{k}{2}}$

$\implies \underline{\boxed{\mathbf{k = 6}}}$

Hope it helps!!

Answer 2

Answer:

b) 6 is the answer

Step-by-step explanation:

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