91Ó°ÊÓ

We can check if there is a change in the sign of the equation within the interval. If there is a change in sign, we can be sure that there is at least one solution. For \(x=0\): \(0^3+2(0)-3=-3\) (Negative) For \(x=1\): \(1^3+2(1)-3=0\) (Zero) The function changes its sign from negative to zero within the interval \([0,1]\). This means that there is at least one solution. Since the function is a cubic polynomial, it can have at most three solutions. However, within the given interval, it has exactly one solution.

Let us define a new function \(f(x)=\mathrm{e}^{-x}-x+1\). In that case, finding a solution for \(\mathrm{e}^{-x}=\mathrm{x}-1\) is equivalent to finding the roots of the equation \(f(x)=0\). Now let's check the signs of the function at \(x=1\) and \(x=2\): For \(x=1\): \(f(1)=\mathrm{e}^{-1}-1+1>0\) (Positive) For \(x=2\): \(f(2)=\mathrm{e}^{-2}-2+1<0\) (Negative) The function changes its sign from positive to negative within the interval \([1,2]\). This means that there is at least one solution in the interval. \(f'(x)=-\mathrm{e}^{-x}-1<0\) for all \(x\ge1\). This means that the function is strictly decreasing, implying that there can be no more than one solution in the given interval. So, there is exactly one solution for this equation.

Like before, let's define a new function \(g(x)=x \ln x - 3\). We want to find the roots of the equation \(g(x)=0\). Check the signs of the function at \(x=2\) and \(x=4\): For \(x=2\): \(g(2)=2 \ln 2 - 3<0\) (Negative) For \(x=4\): \(g(4)=4 \ln 4 - 3>0\) (Positive) The function changes its sign within the interval \([2,4]\). This means that there is at least one solution. \(g'(x)=\ln x + 1 > 0\) for all \(x\ge 2\). The function is strictly increasing, so there can be no more than one solution in the given interval. So, there is exactly one solution for this equation.

Once again, let's define a new function \(h(x)=\sin x-3 x+1\). We want to find the roots of the equation \(h(x)=0\). Check the signs of the function at \(x=-1\) and \(x=1\): For \(x=-1\): \(h(-1)=\sin(-1)-3(-1)+1>0\) (Positive) For \(x=1\): \(h(1)=\sin(1)-3(1)+1<0\) (Negative) The function changes its sign within the interval \([-1,1]\). This means that there is at least one solution. However, we cannot argue that there is only one solution in this case, because the sine function is periodic and oscillates infinitely. Comparing all of the four given equations, we find that #tag_bold# Equations A, B, and C have exactly one solution #tag_bold_end# in their respective intervals. Equation D cannot be guaranteed to have only one solution due to the periodicity of the sine function.