problem on differentiation

Question

show that $\cos x=x^2$ has exactly one solution in $\left(0,π/2\right)$ by the concept of differentiation let,$f\left(x\right)=\cos x-x^2$ first i assume that $f\left(x\right)$ has more than one solution.Suppose a&b are solution where $a<b$.Now,by rolle's theorem on $\left[a,b\right]$ i got a contradiction.Hence i can say that $f\left(x\right)$ has atmost one solution but how can i show that $f\left(x\right)$ has exactly one solution

Answer 1

Assume F(x) = cos(x)-x². Lets Analyse this function.

• Derivative of this function turns out to be : F '(x) = - sin(x) - 2*x
• Double Derivative of this function turns out to be : F ''(x) = - cos(x) - 2

Now, F ''(x) is always negative in interval [0, pie/2], since -cos(x) lies in [-1,0]. So F ''(x) is bound to lie in [-3, -2].

This implies, F '(x) is a strictly decreasing function in interval [0, pie/2], with max value at 0, which is F '(0) = 0.

So, we can say that F '(x) is always negative in interval (0, pie/2].

This implies, F (x) is strictly decreasing function in (0, pie/2]. Now F(0) = 1, So F(0⁺) is 1 since function is continuous.

And F(pie/2) = -(pie²/4). This implies F, being a strictly decreasing function, should cross 'X-axis' exactly once in [0, pie/2], since F(0⁺) > 0 and F(pie/2) < 0. Hence, unique root exists.

Therefore, cos(x) = x², only at a unique point in interval [0, pie/2].