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  • Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 22 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 22 Maths Textbook Solution.

Answers (1)

Answer:

                k=\frac{2}{5}

Hint:

f\left ( x \right )  must be defined. The limit of f\left ( x \right )the  approaches the value x must exist.

Given:

                f(x)=\left\{\begin{array}{l} \frac{\sin 2 x}{5 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.  is continuous at  x=0

Solution:

                f(x)=\left\{\begin{array}{l} \frac{\sin 2 x}{5 x}, \text { if } x \neq 0 \\\\ k, \text { if } x=0 \end{array}\right.

If f\left ( x \right )  is continuous at x=0 , then

                \begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\\\ &\lim _{x \rightarrow 0} \frac{\sin 2 x}{5 x}=k \end{aligned}

                \lim _{x \rightarrow 0} \frac{2 \sin 2 x}{5 \times 2 x}=k    [Multiplying and dividing by 2 ]

                \frac{2}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}=k                                                                                                             \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

                \frac{2}{5}\times 1=k

                k=\frac{2}{5}

 

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