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  • Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 36 subquestion (ii)

Explain solution RD Sharma class 12 Chapter 8 Continuity exercise 8.1 question 36 subquestion (ii)

Answers (1)

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Answer:

k=\frac{-2}{\pi}

Hint:

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

Given:

                f(x)=\left\{\begin{array}{c} (x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right.

Solution:

                f(x)=\left\{\begin{array}{c} (x-1) \tan \frac{\pi x}{2}, \text { if } x \neq 1 \\\\ k, \text { if } x=1 \end{array}\right.

If  f\left ( x \right ) is continuous at x=1 , then

                \begin{aligned} &\lim _{x \rightarrow 1} f(x)=f(1) \\\\ &\lim _{x \rightarrow 1}(x-1) \tan \frac{\pi x}{2}=k \end{aligned}

Putting x-1=y , we get

                \begin{aligned} &\lim _{y \rightarrow 0} y \tan \frac{\pi(y+1)}{2}=k \\\\ &\lim _{y \rightarrow 0} y \tan \left(\frac{\pi y}{2}+\frac{\pi}{2}\right)=k \end{aligned}

                \lim _{y \rightarrow 0} y \tan \left(\frac{\pi}{2}+\frac{\pi y}{2}\right)=k

                -\lim _{y \rightarrow 0} y \cot \left(\frac{\pi y}{2}\right)=k                                                                                   [\because \tan (90+\theta)=-\cot \theta]

                -\frac{2}{\pi} \lim _{y \rightarrow 0} \frac{\frac{\pi y}{2} \cos \left(\frac{\pi y}{2}\right)}{\sin \left(\frac{\pi y}{2}\right)}=k                                                                   \left[\because \cot x=\frac{\cos x}{\sin x}\right]

                -\frac{2}{\pi} \frac{\lim _{y \rightarrow 0} \cos \left(\frac{\pi y}{2}\right)}{\lim _{y \rightarrow 0}\left(\frac{\sin \left(\frac{\pi y}{2}\right)}{\frac{\pi y}{2}}\right)}=k

                \begin{aligned} & \\\\ &\frac{-2}{\pi} \times \frac{1}{1}=k \\\\ &k=\frac{-2}{\pi} \end{aligned}                                                                                         \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=1\right]

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