
If $x$ be real, then the maximum value of $5 + 4x - 4{x^2}$ will be equal to
A. $5$
B. $6$
C. $1$
D. $2$
Answer
160.8k+ views
Hint: By graphing the equation and locating the maximum point on the graph, you can find the maximum value graphically. Identifying whether your equation produces a maximum or minimum is the first step. Use the given quadratic equation to find the value of discriminant and further it depends on the value of discriminant which decides if the value is maximum or minimum.
Formula Used:
You can use the following equation to determine the maximum if your equation has the form $a{x^2} + bx + c$ :
${b^2} - 4ac \geqslant 0$ .
Complete step-by-step solution:
We can write the given equation in terms of $y$ or $f(x)$ as shown below:
$f(x) = 5 + 4x - 4{x^2} = y$ .
To find the maximum value of $f(x) = 5 + 4x - 4{x^2} = y$ or $f(x) = 5 + 4x - 4{x^2} - y = 0$
We use the formula ${b^2} - 4ac \geqslant 0$ as $x$ is real.
This equation is of the form $a{x^2} + bx + c$ . Comparing this equation with the given equation, we get
$a = - 4$ , $b = 4$ and $c = 5 - y$ .
Substituting value in the formula, we get
$16 - 4 \times ( - 4)(5 - y) \geqslant 0$
$ - 6 + y \leqslant 0$
We get
$y \leqslant 6$
So, $f(x)$ has the maximum value of $6$ .
Hence, the correct option is B.
Note: We also have an alternative method to solve and get the maximum value of the given quadratic equation. If we have, $a{x^2} + bx + c$ Therefore, the minimum or maximum is attained when: $x = - \dfrac{b}{{2a}}$ . Determine whether it is a minimum or maximum by looking at the sign of $a$: $a > 0$ $ \Rightarrow $ minimum and \[a < 0\] $ \Rightarrow $ maximum.
Formula Used:
You can use the following equation to determine the maximum if your equation has the form $a{x^2} + bx + c$ :
${b^2} - 4ac \geqslant 0$ .
Complete step-by-step solution:
We can write the given equation in terms of $y$ or $f(x)$ as shown below:
$f(x) = 5 + 4x - 4{x^2} = y$ .
To find the maximum value of $f(x) = 5 + 4x - 4{x^2} = y$ or $f(x) = 5 + 4x - 4{x^2} - y = 0$
We use the formula ${b^2} - 4ac \geqslant 0$ as $x$ is real.
This equation is of the form $a{x^2} + bx + c$ . Comparing this equation with the given equation, we get
$a = - 4$ , $b = 4$ and $c = 5 - y$ .
Substituting value in the formula, we get
$16 - 4 \times ( - 4)(5 - y) \geqslant 0$
$ - 6 + y \leqslant 0$
We get
$y \leqslant 6$
So, $f(x)$ has the maximum value of $6$ .
Hence, the correct option is B.
Note: We also have an alternative method to solve and get the maximum value of the given quadratic equation. If we have, $a{x^2} + bx + c$ Therefore, the minimum or maximum is attained when: $x = - \dfrac{b}{{2a}}$ . Determine whether it is a minimum or maximum by looking at the sign of $a$: $a > 0$ $ \Rightarrow $ minimum and \[a < 0\] $ \Rightarrow $ maximum.
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