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  • Let <img alt="\begin{gathered} \sin ^{-1}(1-\{x\}) \\ \\f(x)=\frac{\cdot \cos ^{-1}(1-\{x\})}{\sqrt{2\{x\}} \cdot(1-\{x\})}, \end{gathered}" src="https://entrancecorner.oncodecogs.com/gif.late

Let \begin{gathered} \sin ^{-1}(1-\{x\}) \\ \\f(x)=\frac{\cdot \cos ^{-1}(1-\{x\})}{\sqrt{2\{x\}} \cdot(1-\{x\})}, \end{gathered} then the value of \frac{2008 \sqrt{2}}{\pi}  \lim _{x \rightarrow 0^{-}} f(x) must be (where {x} denotes the fractional part of x ).

Option: 1

1010

 


Option: 2

1000

 


Option: 3

1004

 


Option: 4

625


Answers (1)

best_answer

\because \quad \lim _{x \rightarrow 0^{-}} f(x)=\lim _{h \rightarrow 0} f(0-h)

                 \begin{aligned} & =\lim _{h \rightarrow 0} \frac{\sin ^{-1}(1-\{0-h\}) \cdot \cos ^{-1}(1-\{0-h\})}{\sqrt{2\{0-h} \cdot(1-\{0-h\})} \\ \\& =\lim _{h \rightarrow 0} \frac{\sin ^{-1}(1-(1-h)) \cdot \cos ^{-1}(1-(1-h))}{\sqrt{2(1-h)} \cdot(1-i(1-h))} \\ \\& =\lim _{h \rightarrow 0} \frac{\sin ^{-1} h \cdot \cos ^{-1} h}{\sqrt{2(1-h)} \cdot h} \\ \\& =1 \cdot \frac{\pi / 2}{\sqrt{2}}=\frac{\pi}{2 \sqrt{2}} \end{aligned}

\therefore \frac{2008 \sqrt{2}}{\pi} \lim _{x \rightarrow 0^{-}} f(x)=\frac{2008 \sqrt{2}}{\pi} \cdot \frac{\pi}{2 \sqrt{2}}=1004

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seema garhwal

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