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  • Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 6

Provide solution for RD Sharma maths class 12 chapter 8 Continuity exercise multiple choice question 6

Answers (1)

Answer:

 The correct option is (C)

Hint:

 Use the given formula:

\text { (i) } \lim _{x \rightarrow 0} \frac{\log (1-x)}{x}=1 \text { and } \lim _{x \rightarrow 0} \frac{\sin x}{x}=1

(ii) A function f(x) is said to be continuous at a point x=a of its domain, if

\begin{aligned} &\lim _{x \rightarrow a^{+}} f(a+h)=\lim _{x \rightarrow a^{-}} f(a-h)=f(a) \\ &(\text { iii }) \lim _{x \rightarrow a}\{f(x) g(x)\}=1 m, \text { where } \lim _{x \rightarrow a} f(x)=1, \lim _{x \rightarrow a} g(x)=m \end{aligned}

Given:

 f(x)=\left\{\begin{array}{l} \frac{1-\sin x}{(\pi-2 x)^{2}} \frac{\log \sin x}{\left(\log \left(1+\pi^{2}-4 \pi x+4 x^{2}\right)\right)} \\ \\ k \end{array}\right.        \begin{aligned} , \quad &x \neq \frac{\pi}{2} \\ , \quad &x=\frac{\pi}{2} \end{aligned}

Solution:

Function f(x) is continuous at

x = \frac{\pi }{2}

\lim _{x \rightarrow \frac{\pi}{2}} f(x)=f\left(\frac{\pi}{2}\right)

Using substitution method

\begin{aligned} &\text { Substitute } \frac{\pi}{2}-x=t \\ &\Rightarrow \lim _{t \rightarrow 0}\left(\frac{\pi}{2}-t\right)=f\left(\frac{\pi}{2}\right) \end{aligned}

\Rightarrow \lim _{t \rightarrow 0}\left\{\frac{1-\sin \left(\frac{\pi}{2}-t\right)}{4 t^{2}} \frac{\log \sin \left(\frac{\pi}{2}-t\right)}{\log \left(1+\pi^{2}-4 \pi\left(\frac{\pi}{2}-t\right)+4\left(\frac{\pi}{2}-t\right)^{2}\right.}\right\}=k

\Rightarrow \lim _{t \rightarrow 0} \frac{\left(2 \sin ^{2} t / 2\right) \log \cos 1}{4 t^{2} \cdot \log \left(1+\pi^{2}-2 \pi^{2}+4 \pi t+\pi^{2}+4 t^{2}-4 \pi t\right)}=k

\begin{aligned} &\Rightarrow \lim _{t \rightarrow 0} \frac{2 \sin ^{2} t / 2 \log \cos t}{4 t^{2} \log \left(1+4 t^{2}\right)}=k \\ &\Rightarrow \lim _{t \rightarrow 0} \frac{2 \sin ^{2} t / 2 \log \sqrt{1-\sin ^{2} t}}{4 \times 4 \frac{t^{2}} {4}} \frac{1}{4 t^{2} \cdot \frac{\log (1+4 t^{2})}{4 t^{2}}}=k \quad\left[\because \cos ^{2} x=1-\sin ^{2} x\right] \end{aligned}

\Rightarrow \lim _{t \rightarrow 0} \frac{2}{16}\left(\frac{\sin t / 2}{t / 2}\right)^{2} \cdot \frac{\frac{1}{2} \log (1-\sin 2 t)}{4 t^{2} \cdot \frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}}=k

\Rightarrow \frac{1}{8} \lim _{t \rightarrow 0}\left(\frac{\sin t / 2}{t / 2}\right)^{2} \cdot \frac{-\sin ^{2} f}{2} \cdot \frac{\log \left(1-\sin ^{2} t\right)}{4 t^{2}\left(\sin ^{2} t\right)} \times \frac{1}{\log \left(1+4 t^{2}\right)}=k

\Rightarrow \frac{-1}{8} \lim _{t \rightarrow 0}\left(\frac{\sin t / 2}{t / 2}\right)^{2} \cdot \frac{\sin ^{2} f}{8 t^{2}} \cdot \frac{\log (1-\sin 2 t)}{\left(-\sin ^{2} t-1\right)} \times \frac{1}{\frac{\log (1+4+2)}{4 t^{2}}}=k

\Rightarrow-\frac{1}{8 \times 8} \lim _{t \rightarrow 0}\left(\frac{\sin t / 2}{t / 2}\right)^{2} \cdot \lim _{t \rightarrow 0}\left(\frac{\sin t}{t}\right)^{2} \cdot \lim _{t \rightarrow 0} \frac{\log \left(1-\sin ^{2} t\right)}{-\sin 2 t} \cdot \lim _{d \rightarrow 0} \frac{1}{\frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}}=k

Using formula (iii)

\Rightarrow \frac{1}{64} \lim _{t \rightarrow 0}\left\{\frac{\sin ^{2} \frac{t}{2} \frac{\log \left(1-\sin ^{2} t\right)}{8 t^{2}}}{\left(\frac{t}{2}\right)^{2}} \frac{\log \left(1+4 t^{2}\right)}{4 t^{2}}\right\}=k

\Rightarrow-\frac{1}{64} \lim _{t \rightarrow 0}\left(\frac{\sin t}{t}\right)^{2} \lim _{t \rightarrow 0} \log \frac{1-\sin ^{2} t}{-\sin ^{2} t}=k        \left[\because \lim _{x \rightarrow 0} \frac{\sin x}{x}=4, \lim _{x \rightarrow 0} \frac{\log (1+x)}{x}=1\right]

Using standard limit formula (i)

\Rightarrow k=-\frac{1}{64}

So, option (c) is correct.

Posted by

Gurleen Kaur

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