Let the function f(x)=2 x^2-log _ e x, x>0, be decreasing in (0, a) and increasing in ( a, 4). A tangent to the parabola y^2=4 a x at a point P on it passes through the point (8 a, 8 a-1) but does not pass through the point (-(1 / a), 0). If the equation of the normal at P is : (x / alpha)+(y / beta)=1, then alpha+beta is equal to ________________.

f'(x) = (4x^2 - 1 / x) \ So f(x) is decreasing in ( 0, (1 / 2) ) and increasing in ( (1 / 2), infinity ) => a = (1 / 2) [1em] Tangent to y^2 = 2x => y = ,x + (1 / 2m) \ It passes through (4,3): \ 3 = 4m + (1 / 2m) => 8m^2 - 6m + 1 = 0 \ => m = (1 / 2) or (1 / 4) [1em] So the tangent may be: \ y = (1 / 2)x + 1 or y = (1 / 4)x + 2 [1em] But y = (1 / 2)x + 1 passes through (-2,0), so it is rejected. [1em] Equation of normal: \ y = -4x - 2( (1 / 2) )(-4) - (1 / 2)(-4)^3 \ => y = -4x + 4 + 32 \ => y = -4x + 36 \ or (x / 9) + (y / 36) = 1 \

Let the function fx2 x2log e x xandgt0 be decreasing in... | null (Mathematics)

\begin{aligned} & f'(x) = \frac{4x^2 - 1}{x} \\ & \text{So } f(x) \text{ is decreasing in } \left( 0, \frac{1}{2} \right) \text{ and increasing in } \left( \frac{1}{2}, \infty \right) \Rightarrow a = \frac{1}{2} \\[1em] & \text{Tangent to } y^2 = 2x \Rightarrow y = ,x + \frac{1}{2m} \\ & \text{It passes through } (4,3): \\ & 3 = 4m + \frac{1}{2m} \Rightarrow 8m^2 - 6m + 1 = 0 \\ & \Rightarrow m = \frac{1}{2} \text{ or } \frac{1}{4} \\[1em] & \text{So the tangent may be:} \\ & y = \frac{1}{2}x + 1 \quad \text{or} \quad y = \frac{1}{4}x + 2 \\[1em] & \text{But } y = \frac{1}{2}x + 1 \text{ passes through } (-2,0), \text{ so it is rejected.} \\[1em] & \text{Equation of normal:} \\ & y = -4x - 2\left( \frac{1}{2} \right)(-4) - \frac{1}{2}(-4)^3 \\ & \Rightarrow y = -4x + 4 + 32 \\ & \Rightarrow y = -4x + 36 \\ & \text{or } \frac{x}{9} + \frac{y}{36} = 1 \end{aligned}

Explanation

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