If x is real,then the maximum value of 5 + 4x | vedclass

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(B) Let $f(x) = 5 + 4x - 4x^2 = y$.Rearranging the terms,we get $4x^2 - 4x + (y - 5) = 0$.Since $x$ is real,the discriminant $D$ must be greater than

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$6$

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(B) Let $f(x) = 5 + 4x - 4x^2 = y$.Rearranging the terms,we get $4x^2 - 4x + (y - 5) = 0$.Since $x$ is real,the discriminant $D$ must be greater than or equal to $0$ $(D \ge 0)$.$D = b^2 - 4ac = (-4)^2 - 4(4)(y - 5) \ge 0$.$16 - 16(y - 5) \ge 0$.Dividing by $16$,we get $1 - (y - 5) \ge 0$.$1 - y + 5 \ge 0$.$6 - y \ge 0$,which implies $y \le 6$.Alternatively,$f(x) = 5 - (4x^2 - 4x) = 5 - (4x^2 - 4x + 1 - 1) = 5 - ((2x - 1)^2 - 1) = 6 - (2x - 1)^2$.Since $(2x - 1)^2 \ge 0$,the maximum value of $f(x)$ is $6$ when $2x - 1 = 0$,i.e.,$x = 1/2$.

If $x$ is real,then the maximum value of $5 + 4x - 4x^2$ is equal to:

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