Question

Between any two real roots of e^x sinx + 1 =0 there is ________ real roots of the equation tanx +1

Answer 1

We will use bisection method to solve this problem.

f(x)$= e^x sinx + 1$

The roots of  $e^x sinx + 1 =0$ will lie between , $x=\frac{-\pi}{4} =\frac{-11}{14}$ and x=0.

As, $f(\frac{-\pi}{4})={\text{-ve Real number}},$

and f(0)= Positive Real number,

So, one root lies between $\frac{-11}{14},$ and 0.

Which is equal to , $(\frac{\frac{-11}{14}+0}{2},0)=(\frac{-11}{28},0)$.

Now coming to equation, g(x) = tan x +1

Using bisection method we can see that , it's root will lie between, $(\frac{-5\pi }{18},0)$.

As, $g(\frac{-5\pi }{18})={\text{-ve Real number}}$ and g(0)=Positive Real number

The one root will lie between  $\frac{-55}{63},$ and 0.

Which is equal to , $(\frac{\frac{-55}{63}+0}{2},0)=(\frac{-55}{126},0)$.

As, one end of both the equation f(x)$= e^x sinx + 1$ andg(x) = tan x +1 in which their root lies is 0.

→→So, we can say that , Between any two real roots of e^x sinx + 1 =0 there is Infinite real roots of the equation tanx +1.