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

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