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  • Please Solve RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 7 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter Continuity Exercise 8.1 Question 7 Maths Textbook Solution.

Answers (1)

Answer:

                x=0 (Discontinuous)

Hint:

For a function to be continuous at a point, its LHL RHL and value at that point should be equal.

Solution:

Given,

f(x)=\left(\begin{array}{c} \frac{1-\cos x}{x^{2}}, \text { if } x \neq 0 \\\\ 1, \text { if } x=0 \end{array}\right)   

Consider,

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

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

\begin{aligned} &=\lim _{x \rightarrow 0}\left(\frac{2 \sin \frac{x}{2}}{4\left(\frac{x^{2}}{4}\right)}\right) \\\\ &=\lim _{x \rightarrow 0}\left(\frac{2\left(\sin \frac{x}{2}\right)^{2}}{4\left(\frac{x}{2}\right)^{2}}\right) \end{aligned}

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

\lim _{x \rightarrow 0} f(x)=\frac{1}{2} \times 1^{2}=\frac{1}{2}

Given

f\left ( 0 \right )=1

\lim _{x \rightarrow 0} f(x) \neq f(0)

Thus,  f\left ( x \right ) is discontinuous at x=0.

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