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  • Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 45 maths textbook solution

Please solve RD Sharma class 12 chapter 8 Continuity exercise multiple choice question 45 maths textbook solution

Answers (1)

Answer:

 The correct option is (a)

Given:

 The given function f is

f(x)=x^{2} \sin \frac{1}{x}, \text { where } x \neq 0

It is evident that f is defined at all points of the real line.

 let C be a real number.

\begin{aligned} &\text { Case } 1 \text { : If } c \neq 0 \text {, then } f(c)=c^{2} \sin \frac{1}{c}\\ &\lim _{x \rightarrow c} f(x)=\lim _{x \rightarrow c}\left(x^{2} \sin \frac{1}{x}\right)=\left(\lim _{x \rightarrow c} x^{2}\right)\left(\lim _{x \rightarrow c} \sin \frac{1}{x}\right)=c^{2} \sin \frac{1}{c} \end{aligned}

\therefore \lim _{x \rightarrow c} f(x)=f(c)

Therefore, f is continuous at all points x≠0

Case 2: If c=0, then f(0)=0

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

It is known that

\begin{aligned} &-1 \leq \sin \frac{1}{x} \leq 1, x \neq 0 \\ &=-x^{2} \leq \sin \frac{1}{x} \leq x^{2} \end{aligned}

\begin{aligned} &=\lim _{x \rightarrow 0}\left(-x^{2}\right) \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq \lim _{x \rightarrow 0} x^{2} \\ &=0 \leq \lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right) \leq 0 \end{aligned}

\begin{aligned} &=\lim _{x \rightarrow 0}\left(x^{2} \sin \frac{1}{x}\right)=0 \\ &\therefore \lim _{x \rightarrow 0^{-}} f(x)=0 \end{aligned}

Similarly,

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

\therefore \lim _{x \rightarrow 0^{-}} f(x)=f(0)=\lim _{x \rightarrow 0^{+}} f(x)

There fore f is continuous at x = 0

So, the correct option is (a)

Posted by

Gurleen Kaur

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