Question

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

• A. 0
• B. 1
• C. 2
• D. 3

Answer 1

Answer: (B)

1

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

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

$f\prime \left(x\right)=3{\sec }^{2}x+3x^2$

$f\prime \left(x\right)=3[{\sec }^{2}x+x^2]>0\forall xϵ(0,\frac{\pi }{4})$

$[\because {\sec }^{2}x+x^2\ne 0]$, there it is always increasing

$\therefore f\left(x\right)$ is strictly increasing function.

Now $f\left(0\right)=-2<0$ and $f\left(\frac{\pi }{4}\right)=3\tan \frac{\pi }{4}+{\left(\frac{\pi }{4}\right)}^3-2$

$=3+{\left(\frac{\pi }{4}\right)}^3-2$

$=1+{\left(\frac{\pi }{4}\right)}^3>0$

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

Therefore, $f$ willl cross the $x-$axis at least once.

And since, in $[0,\frac{\pi }{4}],x$ is an increasing graph, there number of roots of $f\left(x\right)$ will be $1$.