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  • Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 33 Maths Textbook Solution.

Please Solve RD Sharma Class 12 Chapter 8 Continuity Exercise 8.1 Question 33 Maths Textbook Solution.

Answers (1)

Answer:

                f(\pi)=\frac{49}{10}

Hint:

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

Given:

                f(x)=\frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}, x=\pi

Solution:

                f(x)=\frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}, x=\pi

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

                 \begin{aligned} &\lim _{x \rightarrow \pi} f(x)=f(\pi) \\ &\lim _{x \rightarrow \pi} \frac{1-\cos 7(x-\pi)}{5(x-\pi)^{2}}=f(\pi) \end{aligned}

                \frac{2}{5} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(x-\pi)}{2}\right)}{(x-\pi)^{2}}=f(\pi)                                                 \left[\because 1-\cos x=2 \sin ^{2} \frac{x}{2}\right]

                \frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(x-\pi)}{2}\right)}{\frac{49}{4}(x-\pi)^{2}}=f(\pi)                           [Multiplying and dividing by \frac{49}{4} ]

                \frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi} \frac{\sin ^{2}\left(\frac{7(x-\pi)}{2}\right)}{\left(\frac{7}{2}(x-\pi)\right)^{2}}=f(\pi)

                \frac{2}{5} \times \frac{49}{4} \lim _{x \rightarrow \pi}\left[\frac{\sin \left(\frac{7(x-\pi)}{2}\right)}{\left(\frac{7}{2}(x-\pi)\right)}\right]^{2}=f(\pi)

                \frac{2}{5} \times \frac{49}{4} \times 1=f(\pi)                                                                               \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

                \begin{aligned} &\frac{1}{5} \times \frac{49}{2} \times 1=f(\pi) \\ &f(\pi)=\frac{49}{10} \end{aligned}

               

 

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