Question

No.of solutions xcube+2xsquare+5x+2cosx =0 in interval 0t360degrees

Answer 1

Answer:

1 Solution

your equation can be written as

$x ({x}^{2} + 2x + 5) = - 2 \cos(x)$

Draw graph only root of rhs is there x = 0

Now at x = π/2 value of cubic is > 2 and - cos 2x is from [ -2 , 2 ] and cubic is monotonic increasing so we won't found any root for x > 0.

There must be a root for x < 0 ( draw graph )

It's easiest way to solve . For each statement you can use signs of first and second derivative to prove . But no need if it's just objective question . Value of roots and function at boundaries is suffice