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  • Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 35 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 35 Maths Textbook Solution.

Answers (1)

Answer:

                k=1

Hint:

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

Given:

                f(x)=\left(\begin{array}{c} \frac{1-\cos 4 x}{8 x^{2}}, \text { when } x \neq 0 \\\\ k, w h e n\: x=0 \end{array}\right)

Solution:

                f(x)=\left(\begin{array}{c} \frac{1-\cos 4 x}{8 x^{2}}, \text { when } x \neq 0 \\\\ k, w h e n\: 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{1-\cos 4 x}{8 x^{2}}=f(0) \end{aligned}

                \lim _{x \rightarrow 0} \frac{2 \sin ^{2} 2 x}{8 x^{2}}=f(0)                                                               \left[\because 1-\cos 2 x=2 \sin ^{2} x\right]

                \frac{2}{2} \lim _{x \rightarrow 0} \frac{\sin ^{2} 2 x}{4 x^{2}}=f(0)                                                             \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

                \begin{aligned} &1 \times 1=f(0) \\ &k=1 \end{aligned}                                                                            [\because f(0)=k]

 

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