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  • Explain solution RD Sharma class 12 chapter Continuity exercise 8.2 question 4 subquestion (i) maths

Explain solution RD Sharma class 12 chapter Continuity exercise 8.2 question 4 subquestion (i) maths

Answers (1)

Answer:

 \frac{2}{15}

Hint:

 f(x) is continuous so it is also continuous at end points. Put LHL= RHL

Given:

 f(x)= \begin{cases}\frac{\sin 2 x}{5 x} & x \neq 0 \\ 3 k & x=0\end{cases}

Explanation:

At x = 0

\begin{aligned} &\lim _{x \rightarrow 0} f(x)=\lim _{x \rightarrow 0} \frac{\sin 2 x}{5 x} \\ &=\frac{1}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x} \times 2 \\ &=\frac{2}{5} \lim _{x \rightarrow 0} \frac{\sin 2 x}{2 x}\left[\operatorname{as} \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right] \\ &=\frac{2}{5} \times 1=\frac{2}{5} \\ &\qquad f(0)=3 k \end{aligned}

As f(X) is continuous at x = 0 when

\begin{aligned} &\lim _{x \rightarrow 0} f(x)=f(0) \\ &\frac{2}{5}=3 k \\ &k=2 / 15 \end{aligned}

Posted by

Gurleen Kaur

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