Question

The interval of \[a\] for which the expression \[{x^2} - ax + 1 - 2{a^2}\] is always positive is
A) \[\left( { - 1,1} \right)\]
B) \[\left( { - 2,2} \right)\]
C) \[\left( { - \infty , - 1} \right) \cup \left( {1,\infty } \right)\]
D) \[\left( { - \dfrac{2}{3},\dfrac{2}{3}} \right)\]

Answer 1

Hint:
Here, we will use the formula of discriminant that is calculated \[{b^2} - 4ac\] of the standard form of quadratic equation is \[a{x^2} + bx + c\]. Then we will compare the given expression with the standard form of the quadratic equation to find the value of \[a\], \[b\] and \[c\]. Then we will substitute the above value of \[a\], \[b\] and \[c\] in the formula of the discriminant. Since we know that when \[y\] is always positive, the discriminant \[D\] is less than 0 and then simplifies to find the required interval.

Complete step by step solution:
We are given that \[y = {x^2} - ax + 1 - 2{a^2}\] is always positive.
So, we have \[y > 0\].
We know that the discriminant is calculated using the formula \[{b^2} - 4ac\] of the standard form of quadratic equation is \[a{x^2} + bx + c\].
Comparing the given expression with the standard form of a quadratic equation to find the value of \[a\], \[b\] and \[c\], we get
\[a = 1\]
\[b = - a\]
\[c = 1 - 2{a^2}\]
Substituting the above value of \[a\], \[b\] and \[c\] in the formula of discriminant, we get
\[
   \Rightarrow {\left( { - a} \right)^2} - 4 \cdot 1 \cdot \left( {1 - 2{a^2}} \right) \\
   \Rightarrow {a^2} - 4 + 8{a^2} \\
   \Rightarrow 9{a^2} - 4 \\
 \]
Since we know that when \[y\] is always positive, the discriminant \[D\] is less than 0.
\[
   \Rightarrow 9{a^2} - 4 < 0 \\
   \Rightarrow {\left( {3a} \right)^2} - {2^2} < 0 \\
 \]
Using the rule, \[{a^2} - {b^2} = \left( {a - b} \right)\left( {a + b} \right)\] in the above equation, we get
\[ \Rightarrow \left( {3a + 2} \right)\left( {3a - 2} \right) < 0\]
Taking \[3a + 2 < 0\] and then subtracting the equation by 2 on both sides, we get
\[
   \Rightarrow 3a + 2 - 2 < 0 - 2 \\
   \Rightarrow 3a < - 2 \\
 \]
Dividing the above equation by 3 on both sides, we get
\[ \Rightarrow a < - \dfrac{2}{3}{\text{ ......eq.(1)}}\]
Taking \[3a - 2 < 0\] and then adding the equation by 2 on both sides, we get
\[
   \Rightarrow 3a - 2 + 2 < 0 + 2 \\
   \Rightarrow 3a < 2 \\
 \]
Dividing the above equation by 3 on both sides, we get
\[ \Rightarrow a < \dfrac{2}{3}{\text{ ......eq.(2)}}\]
Using equation (1) and equation (2) to form a interval, we get
\[\left( { - \dfrac{2}{3},\dfrac{2}{3}} \right)\]

Hence, option D is correct.

Note:
A quadratic equation represents a parabola graphically. when we talk about roots of a quadratic equation then it’s nothing but the intersection of the parabola with the x-axis. A quadratic equation with a positive coefficient of ${x}^{2}$ will be negative in between in the roots and will be positive outside.