If the range of f(x) = frac{2x^4-14x^2-8x+49} | vedclass

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(C) Given $f(x) = \frac{2x^4-14x^2-8x+49}{x^4-7x^2-4x+23}$.We can rewrite the numerator as:$2(x^4-7x^2-4x+23) + 3 = 2x^4-14x^2-8x+46+3 = 2x^4-14x^2-8x

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$5$

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(C) Given $f(x) = \frac{2x^4-14x^2-8x+49}{x^4-7x^2-4x+23}$.We can rewrite the numerator as:$2(x^4-7x^2-4x+23) + 3 = 2x^4-14x^2-8x+46+3 = 2x^4-14x^2-8x+49$.So,$f(x) = \frac{2(x^4-7x^2-4x+23) + 3}{x^4-7x^2-4x+23} = 2 + \frac{3}{x^4-7x^2-4x+23}$.Let $h(x) = x^4-7x^2-4x+23$.We can express $h(x)$ as:$h(x) = (x^2-4)^2 + (x-2)^2 + 3$.Since $(x^2-4)^2 \geq 0$ and $(x-2)^2 \geq 0$,the minimum value of $h(x)$ is $3$ (at $x=2$).Thus,the range of $h(x)$ is $[3, \infty)$.Now,$f(x) = 2 + \frac{3}{h(x)}$.As $h(x) \in [3, \infty)$,$\frac{3}{h(x)} \in (0, 1]$.Therefore,$f(x) \in (2+0, 2+1] = (2, 3]$.Comparing $(2, 3]$ with $(a, b]$,we get $a=2$ and $b=3$.Thus,$a+b = 2+3 = 5$.

If the range of $f(x) = \frac{2x^4-14x^2-8x+49}{x^4-7x^2-4x+23}$ is $(a, b]$,then $(a + b)$ is

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