A) \[\frac{9}{2}\]
B) \[\frac{2}{9}\]
C) 0
D) \[\frac{8}{9}\]
Correct Answer: B
Solution :
Given that\[f\left( \frac{9}{2} \right)=\frac{2}{9}\] \[\underset{x\to 0}{\mathop{\lim }}\,f\left( \frac{1-\cos 3x}{{{x}^{2}}} \right)=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{x}^{2}}}{1-\cos 3x} \right)\] \[=\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{{{x}^{2}}}{2{{\sin }^{2}}\frac{3x}{2}} \right)\] \[=\frac{1}{2}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{\frac{9}{4}..{{x}^{2}}.\frac{4}{9}}{{{\sin }^{2}}\frac{3x}{2}} \right)\] \[=\frac{4}{9\times 2}\underset{x\to 0}{\mathop{\lim }}\,\left( \frac{1}{\frac{{{\sin }^{2}}\frac{3x}{2}}{{{\left( \frac{3x}{2} \right)}^{2}}}} \right)\] \[=\frac{2}{9}\left[ \underset{x\to 0}{\mathop{\lim 1}}\,\frac{\frac{\underset{x\to 0}{\mathop{\lim 1}}\,}{{{\sin }^{2}}\frac{3x}{2}}}{{{\left( \frac{3x}{2} \right)}^{2}}} \right]\]\[=\frac{2}{9}.\left[ \frac{1}{1} \right]=\frac{2}{9}\]
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