Q2. f(x) = x3+2x2+5x+2cosx
f'(x) = 3x2+4x+5-2sinx
Now min value of 3x2+4x+5 is (4.5.3-16)/12 = 11/3 = 3.67
So f'(x) is always positive
So f(x) is an increasing function
So in the interval [0,2∩] its min value will be at x=0
At x=0 f(x) = 2
So in the given interval for no value of x is f(x) zero so no solution