The number of real solutions of the equation | vedclass

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(D) Let $f(x) = -x^2 + 5x - 10$ and $g(x) = (\frac{3}{2})^x$.For the quadratic function $f(x) = -x^2 + 5x - 10$,the discriminant $D = b^2 - 4ac = 5^2

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(D) Let $f(x) = -x^2 + 5x - 10$ and $g(x) = (\frac{3}{2})^x$.For the quadratic function $f(x) = -x^2 + 5x - 10$,the discriminant $D = b^2 - 4ac = 5^2 - 4(-1)(-10) = 25 - 40 = -15$.The maximum value of $f(x)$ is given by $\frac{-D}{4a} = \frac{-(-15)}{4(-1)} = -\frac{15}{4} = -3.75$.Since the coefficient of $x^2$ is negative,the parabola opens downwards,meaning $f(x) \leq -3.75$ for all real $x$.For the exponential function $g(x) = (\frac{3}{2})^x$,we know that $g(x) > 0$ for all real $x$.Since $f(x)$ is always negative $(f(x) \leq -3.75)$ and $g(x)$ is always positive $(g(x) > 0)$,there is no value of $x$ for which $f(x) = g(x)$.Therefore,the equation has no real solutions.

The number of real solutions of the equation $(\frac{3}{2})^x = -x^2 + 5x - 10$ is:

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