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