Question

for all real values of x, the maximum value of the expression \[\dfrac{x}{{{x^2} - 5x + 9}}\] is
$
  \left( a \right)\,\,1 \\
  \left( b \right)\,\,45 \\
  \left( c \right)\,\,90 \\
  \left( d \right)\,none\,of\,these \\
 $

Answer 1

Hint: For finding the maximum values of expression first let us consider the given expression as some constant term and form the quadratic expression and find the discriminant values. From the obtained values take out the maximum value. Here , use ${b^2} - 4ac \geqslant 0$ since x is real.

Complete step by step solution:
The objective of the problem is to find the maximum value of the expression.
Given expression , \[\dfrac{x}{{{x^2} - 5x + 9}}\]
Let us assume given expression is equal to y
That is $y = \dfrac{x}{{{x^2} - 5x + 9}}$
By cross multiplication we get $y\left( {{x^2} - 5x + 9} \right) = x$
Now expand the terms by multiplying right hand side term, we get
$ \Rightarrow y{x^2} - 5xy + 9y = x$
Taking the right hand side term to the left side to make it as a quadratic equation.
$ \Rightarrow y{x^2} - 5xy - x + 9y = 0$
Taking -x as a common from second and third terms.
$ \Rightarrow y{x^2} - x\left( {5y + 1} \right) + 9y = 0$
Comparing above equation with standard quadratic equation $a{x^2} + bx + c = 0$
Here $a = y,b = - \left( {5y + 1} \right),c = 9y$
Since x is real the discriminant value should be greater than or equal to zero..
That is ${b^2} - 4ac \geqslant 0$
On substituting the values of a, b,c we get,
${\left[ { - \left( {5y + 1} \right)} \right]^2} - 4\left( y \right)\left( {9y} \right) \geqslant 0$
On expanding the terms, we get
$
  {\left( {5y + 1} \right)^2} - 36{y^2} \geqslant 0 \\
   \Rightarrow {\left( {5y} \right)^2} + 2\left( {5y} \right)\left( 1 \right) + {1^2} - 36{y^2} \geqslant 0 \\
   \Rightarrow 25{y^2} + 10y + 1 - 36{y^2} \geqslant 0 \\
   \Rightarrow - 11{y^2} + 10y + 1 \geqslant 0 \\
 $
Here we used the formula ${\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab$ . Now multiplying above the equation with minus sign. By multiplying with negative sign greater than or equal changes to less than or equal.
$ \Rightarrow 11{y^2} - 10y - 1 \leqslant 0$
Multiply coefficient of first term with last term and factor the product such that the sum of factors is equal to coefficient of second term.
$ \Rightarrow 11{y^2} - 11y + y - 1 \leqslant 0$
Taking 11y as common from the first two terms and one from next two terms . we get
$
   \Rightarrow 11y\left( {y - 1} \right) + 1\left( {y - 1} \right) \leqslant 0 \\
   \Rightarrow \left( {y - 1} \right)\left( {11y + 1} \right) \leqslant 0 \\
 $
y lies between $\left[ { - \dfrac{{11}}{2},1} \right]$
Thus , the maximum value is 1.

Therefore , option a is correct.

Note:
The roots of the equation $a{x^2} + bx + c = 0$ can also be find by using the formula $\dfrac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$. If the discriminant value is greater than zero then the roots are real and distinct. And if the discriminant value is less than zero then the roots are real and imaginary, and if the discriminant value is equal to zero then the roots are equal.