Let the function f(x) = 2x^{2} - log_{e} x,x | vedclass

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(A) Given $f(x) = 2x^{2} - \log_{e} x$. The derivative is $f'(x) = 4x - \frac{1}{x} = \frac{4x^{2} - 1}{x}$.For $f(x)$ to be decreasing in $(0, a)$ an

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$45$

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(A) Given $f(x) = 2x^{2} - \log_{e} x$. The derivative is $f'(x) = 4x - \frac{1}{x} = \frac{4x^{2} - 1}{x}$.For $f(x)$ to be decreasing in $(0, a)$ and increasing in $(a, 4)$,we set $f'(x) = 0$,which gives $4x^{2} = 1$,so $x = \frac{1}{2}$ (since $x > 0$). Thus,$a = \frac{1}{2}$.The parabola is $y^{2} = 4(\frac{1}{2})x = 2x$.$A$ point $P$ on $y^{2} = 2x$ is $(2t^{2}, 2t)$. The tangent at $P$ is $ty = x + 2t^{2}$.This tangent passes through $(8a, 8a - 1) = (4, 3)$.So,$t(3) = 4 + 2t^{2} \Rightarrow 2t^{2} - 3t + 4 = 0$. This has no real roots. Re-evaluating the parabola equation: $y^{2} = 4ax$. If $a = 1/2$,$y^{2} = 2x$. The point is $(8a, 8a-1) = (4, 3)$.Using the standard form $y^{2} = 4ax$,the tangent at $(at^{2}, 2at)$ is $ty = x + at^{2}$.Substituting $(8a, 8a-1)$: $t(8a-1) = 8a + at^{2} \Rightarrow at^{2} - t(8a-1) + 8a = 0$.Given $a = 1/2$,$\frac{1}{2}t^{2} - t(4-1) + 4 = 0 \Rightarrow \frac{1}{2}t^{2} - 3t + 4 = 0 \Rightarrow t^{2} - 6t + 8 = 0$.Roots are $t = 2, 4$. If $t=2$,$P = (at^{2}, 2at) = (2, 2)$. If $t=4$,$P = (8, 4)$.The normal at $P(at^{2}, 2at)$ is $y = -tx + 2at + at^{3}$.For $t=4, a=1/2$: $y = -4x + 2(1/2)(4) + (1/2)(64) = -4x + 4 + 32 = -4x + 36$.$4x + y = 36 \Rightarrow \frac{x}{9} + \frac{y}{36} = 1$. Thus $\alpha = 9, \beta = 36$.$\alpha + \beta = 9 + 36 = 45$.

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