A) 1
B) 2
C) 3
D) Infinite
Correct Answer: A
Solution :
[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.You need to login to perform this action.
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