X)=3\tan \,x+{{x}^{3}}-2\]. Then \[f'(x)=3\,\,{{\sec }^{2}}x+3{{x}^{2}}>0\]. Hence, f(x) increases. ALSO, \[f(0)=-2\] and \[f\left( \frac{\pi }{4} \RIGHT)>0\]. So, by intermediate value theorem, \[f(c)=2\] for some \[c\in \left( 0,\frac{\pi }{4} \right)\]. Hence, \[f(x)=0\] has only one root.

"> X)=3\tan \,x+{{x}^{3}}-2\]. Then \[f'(x)=3\,\,{{\sec }^{2}}x+3{{x}^{2}}>0\]. Hence, f(x) increases. ALSO, \[f(0)=-2\] and \[f\left( \frac{\pi }{4} \RIGHT)>0\]. So, by intermediate value theorem, \[f(c)=2\] for some \[c\in \left( 0,\frac{\pi }{4} \right)\]. Hence, \[f(x)=0\] has only one root.

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The number of solutions of the equation \[3\tan x+{{x}^{3}}=2\,\,in\left( 0,\frac{\pi }{4} \right).\] is

Joint Entrance Exam - Main (JEE Main) Mathematics in Joint Entrance Exam - Main (JEE Main) . 10 months ago

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[a] Let \[f(X)=3\tan \,x+{{x}^{3}}-2\]. Then \[f'(x)=3\,\,{{\sec }^{2}}x+3{{x}^{2}}>0\]. Hence, f(x) increases. ALSO, \[f(0)=-2\] and \[f\left( \frac{\pi }{4} \RIGHT)>0\]. So, by intermediate value theorem, \[f(c)=2\] for some \[c\in \left( 0,\frac{\pi }{4} \right)\]. Hence, \[f(x)=0\] has only one root.

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Posted on 20 Aug 2024, this text provides information on Joint Entrance Exam - Main (JEE Main) related to Mathematics in Joint Entrance Exam - Main (JEE Main). Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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