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Question

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


A
0
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B
1
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C
2
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D
3
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Solution

The correct option is C 1
Number 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 $$1$$.

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