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Q.Find the maximum of the expression 2x+5−3x22x+5-3x^2 as xx varies over RR.

Telangana TsbieTelangana Board of Intermediate Education 2020Subjective· 2mImportance★★★★★
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2x+5−3x22x+5-3x^2 is a downward-opening quadratic in xx; its maximum occurs at the vertex x=−b/2ax=-b/2a.

Write the expression as f(x)=−3x2+2x+5f(x) = -3x^2+2x+5, a quadratic with a=−3, b=2, c=5a=-3,\ b=2,\ c=5.

Since a=−3<0a=-3<0, the parabola opens downward, so ff has a maximum (not minimum) at

x=−b2a=−22(−3)=13x = -\frac{b}{2a} = -\frac{2}{2(-3)} = \frac{1}{3}

Substitute back:

f(13)=2(13)+5−3(13)2=23+5−3⋅19=23+5−13f\left(\frac13\right) = 2\left(\frac13\right)+5-3\left(\frac13\right)^2 = \frac23+5-3\cdot\frac19 = \frac23+5-\frac13 …

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