
We call ‘p’ a good number if the inequality \[\dfrac{{2{x^2} + 2x + 3}}{{{x^2} + x + 1}} \leqslant p\] is satisfied for any real x. Find the smallest integral good number.
Answer
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Hint: We will start solving this question, by simplifying the left-hand side of the given term in the question. For simplifying the left-hand side, we will divide the numerator by the denominator of the given term. After simplifying the term, we will solve the inequality and find the smallest integral value of p.
Complete step by step solution:
Now, the inequality is \[\dfrac{{2{x^2} + 2x + 3}}{{{x^2} + x + 1}} \leqslant p\]. We will simplify the left-hand side by dividing the numerator by the denominator. The given term can be written as,
\[\dfrac{{2{x^2} + 2x + 2 + 1}}{{{x^2} + x + 1}} \leqslant p\]
\[\dfrac{{2({x^2} + x + 1) + 1}}{{{x^2} + x + 1}} \leqslant p\]
Now, on dividing, we get
$2 + \dfrac{1}{{{x^2} + x + 1}} \leqslant p$
Now, we have to get the smallest value of p. So, we have to find the maximum value of the left-hand side term. The maximum value of the left-hand side term depends on the term $\dfrac{1}{{{x^2} + x + 1}}$. The value of $\dfrac{1}{{{x^2} + x + 1}}$ is maximum, when the value of ${x^2} + x + 1$ is minimum.
For getting the minimum value of ${x^2} + x + 1$, we will make a perfect square in the term ${x^2} + x + 1$.
So, we get
${x^2} + x + 1$ = ${x^2} + x + 1 + \dfrac{1}{4} - \dfrac{1}{4}$ = ${(x + \dfrac{1}{2})^2} + \dfrac{3}{4}$
Now, the minimum square of a number = 0. So, by putting x = $ - \dfrac{1}{2}$, we get
${x^2} + x + 1$ = ${( - \dfrac{1}{2} + \dfrac{1}{2})^2} + \dfrac{3}{4}$ = $\dfrac{3}{4}$
So, the minimum value of ${x^2} + x + 1$ is $\dfrac{3}{4}$.
So, the maximum value of left-hand side term = $2 + \dfrac{1}{{\dfrac{3}{4}}}$ = $2 + \dfrac{4}{3}$ = $\dfrac{{10}}{3}$ = 3.33
So, $p \geqslant 3.33$
So, p = 4. Therefore, the smallest integral value of good number p is 4.
Note: Whenever we come up with such types of questions, we will first simplify the given term. After simplification, we will solve the inequality. This question can be done by solving inequality and by using the theory of equations, which is another easiest way to solve this type of question. When we solve inequality, we should be careful in performing operations on inequalities, because in some operations, inequality changes.
Complete step by step solution:
Now, the inequality is \[\dfrac{{2{x^2} + 2x + 3}}{{{x^2} + x + 1}} \leqslant p\]. We will simplify the left-hand side by dividing the numerator by the denominator. The given term can be written as,
\[\dfrac{{2{x^2} + 2x + 2 + 1}}{{{x^2} + x + 1}} \leqslant p\]
\[\dfrac{{2({x^2} + x + 1) + 1}}{{{x^2} + x + 1}} \leqslant p\]
Now, on dividing, we get
$2 + \dfrac{1}{{{x^2} + x + 1}} \leqslant p$
Now, we have to get the smallest value of p. So, we have to find the maximum value of the left-hand side term. The maximum value of the left-hand side term depends on the term $\dfrac{1}{{{x^2} + x + 1}}$. The value of $\dfrac{1}{{{x^2} + x + 1}}$ is maximum, when the value of ${x^2} + x + 1$ is minimum.
For getting the minimum value of ${x^2} + x + 1$, we will make a perfect square in the term ${x^2} + x + 1$.
So, we get
${x^2} + x + 1$ = ${x^2} + x + 1 + \dfrac{1}{4} - \dfrac{1}{4}$ = ${(x + \dfrac{1}{2})^2} + \dfrac{3}{4}$
Now, the minimum square of a number = 0. So, by putting x = $ - \dfrac{1}{2}$, we get
${x^2} + x + 1$ = ${( - \dfrac{1}{2} + \dfrac{1}{2})^2} + \dfrac{3}{4}$ = $\dfrac{3}{4}$
So, the minimum value of ${x^2} + x + 1$ is $\dfrac{3}{4}$.
So, the maximum value of left-hand side term = $2 + \dfrac{1}{{\dfrac{3}{4}}}$ = $2 + \dfrac{4}{3}$ = $\dfrac{{10}}{3}$ = 3.33
So, $p \geqslant 3.33$
So, p = 4. Therefore, the smallest integral value of good number p is 4.
Note: Whenever we come up with such types of questions, we will first simplify the given term. After simplification, we will solve the inequality. This question can be done by solving inequality and by using the theory of equations, which is another easiest way to solve this type of question. When we solve inequality, we should be careful in performing operations on inequalities, because in some operations, inequality changes.
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