Question

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

• A. $0$
• B. $1$
• C. $2$
• D. $3$

Answer 1

The correct option is B

$1$

Explanation for correct option

Given, $f\left(x\right)=3\tan \left(x\right)+x^3=2$

Now differentiating with respect to $x$ we get,

$\Rightarrow f\text{'}\left(x\right)=3{\sec }^2\left(x\right)+3x^2>0$

$f\left(x\right)$ is increasing for all $x$ belong to $\left(0,\frac{\pi }{4}\right)$

$$\begin{gathered} f\left(0\right)=-2 \\ f\left(\frac{\pi }{4}\right)=2+{\left(\frac{\pi }{4}\right)}^3-1 \\ \Rightarrow f\left(\frac{\pi }{4}\right)=1+{\left(\frac{\pi }{4}\right)}^3>0 \end{gathered}$$

$f\left(0\right)<0$ and $f\left(\frac{\pi }{4}\right)>0$

$\therefore f\left(x\right)$ has exactly $1$ root in between $\left(0,\frac{\pi }{4}\right)$ (From intermediate value theorem)

Hence, the correct option is $\left(\mathrm{B}\right)$.