Answer
Verified
108.3k+ 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.
Recently Updated Pages
If x is real then the maximum and minimum values of class 10 maths JEE_Main
If one of the roots of equation x2+ax+30 is 3 and one class 10 maths JEE_Main
The HCF of two numbers is 96 and their LCM is 1296 class 10 maths JEE_Main
The height of a cone is 21 cm Find the area of the class 10 maths JEE_Main
In a family each daughter has the same number of brothers class 10 maths JEE_Main
If the vertices of a triangle are ab cc b0 and b0c class 10 maths JEE_Main
Other Pages
Electric field due to uniformly charged sphere class 12 physics JEE_Main
Lattice energy of an ionic compound depends upon A class 11 chemistry JEE_Main
As a result of isobaric heating Delta T 72K one mole class 11 physics JEE_Main
The graph of current versus time in a wire is given class 12 physics JEE_Main
If a wire of resistance R is stretched to double of class 12 physics JEE_Main
A 5m long pole of 3kg mass is placed against a smooth class 11 physics JEE_Main