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Provide Solutio for RD Sharma Class 12 Chapter Continuity Exercise 8.1 Subquestion (v) Question 10

Answers (1)

Answer:

                Discontinuous

Hint:

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

Solution:

Given,

Given,

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

Here,  f\left ( 1 \right )=n-1

\begin{aligned} &\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1} \frac{1-x^{n}}{1-x} \\\\ &\lim _{x \rightarrow 1} f(x)=\lim _{x \rightarrow 1}\left[(1-x)^{n-1}+n C_{1}(1-x)^{n-2} x+n C_{2}(1-x)^{n-3} x^{2}+\ldots .+n C_{n-1}(1-x)^{0} x^{n-1}\right] \end{aligned}  

                                                                                               \left[\because(a+b)^{n}={ }^{n} C_{0} a^{n} b^{0}+{ }^{n} C_{1} a^{n-1} b^{1}+{ }^{n} C_{2} d^{n-2} b^{2}+\ldots+{ }^{n} C_{n} a^{0} b^{n}\right]

\lim _{x \rightarrow 1} f(x)=0+0+\ldots+(1)^{n-1}=1 \neq f(1)

Thus, f\left ( x \right ) is discontinuous at  x=1.

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