Question

# Number of solution of the equation $$\displaystyle 3\tan x+x^{3}=2$$ in $$\left ( 0,\dfrac{\pi}{4} \right )$$ is

A
0
B
1
C
2
D
3

Solution

## The correct option is C 1Number of solutions of the equation $$3\tan x+x^3=2$$ in $$\left(0, \cfrac{\pi}{4}\right)$$:Let,$$f\left(x\right)=3\tan x+x^3-2$$$$f\prime\left(x\right)= 3\sec^2x+3x^2$$$$f\prime \left( x \right) =3\left[ \sec ^{ 2 } x+x^{ 2 } \right] >0\quad \forall \quad x\quad \epsilon \quad \left( 0,\cfrac { \pi }{ 4 } \right)$$$$\left[\because \sec^{2}x+x^2\neq 0\right]$$,  there it is always increasing$$\therefore f\left(x\right)$$ is strictly increasing function.Now $$f\left(0\right)=-2<0$$ and $$f\left(\cfrac{\pi}{4}\right)=3\tan \cfrac{\pi}{4}+\left(\cfrac{\pi}{4}\right)^{3}-2$$$$=3+\left(\cfrac{\pi}{4}\right)^{3}-2$$$$=1+\left(\cfrac{\pi}{4}\right)^{3}>0$$$$\therefore f\left(0\right)<-ve$$ and $$f\left(\cfrac{\pi}{4}\right)>+ve$$Therefore, $$f$$ willl cross the $$x-$$axis at least once.And since, in $$\left[0,\cfrac{\pi}{4}\right],x$$ is an increasing graph, there number of roots of $$f\left(x\right)$$ will be $$1$$.Maths

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