The number of real solutions of the equation | vedclass

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(C) The given equation is $(\sin x - x)(\cos x - x^2) = 0$. This implies $\sin x = x$ or $\cos x = x^2$. For the first part,$\sin x = x$,the only real

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$3$

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(C) The given equation is $(\sin x - x)(\cos x - x^2) = 0$. This implies $\sin x = x$ or $\cos x = x^2$. For the first part,$\sin x = x$,the only real solution is $x = 0$. For the second part,$\cos x = x^2$,we observe the graphs of $y = \cos x$ and $y = x^2$. The graph of $y = \cos x$ is a wave oscillating between $-1$ and $1$,and $y = x^2$ is a parabola opening upwards with its vertex at $(0, 1)$. At $x = 0$,$\cos(0) = 1$ and $0^2 = 0$,so $x=0$ is not a solution for this part. Since $\cos x$ decreases from $1$ to $-1$ as $x$ goes from $0$ to $\pi$,and $x^2$ increases from $0$ to $\pi^2$,there is exactly one intersection point in the interval $(0, 1)$. Due to symmetry,there is also exactly one intersection point in the interval $(-1, 0)$. Thus,$\cos x = x^2$ has $2$ real solutions. Including the solution $x = 0$ from the first part,the total number of real solutions is $1 + 2 = 3$.

The number of real solutions of the equation $(\sin x - x)(\cos x - x^2) = 0$ is

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