If λ is an integer and α, β are the roots of 4x 2– 16x + λ/4 = 0 such that 1 < α < 2 and 2 < β < 3, then how many values can λ take?
15
The given question involves a quadratic equation and conditions on its roots. We are given the equation \(4x^2 - 16x + \frac{\lambda}{4} = 0\), where \(\lambda\) is an integer. The roots of this equation, denoted by \(\alpha\) and \(\beta\), satisfy the conditions \(1 < \alpha < 2\) and \(2 < \beta < 3\). We need to determine how many possible integer values \(\lambda\) can take under these conditions.
First, let's simplify the quadratic equation by dividing by 4:
\(\frac{4x^2 - 16x + \frac{\lambda}{4}}{4} = \frac{0}{4}\)
\(x^2 - 4x + \frac{\lambda}{16} = 0\)
Let \(f(x) = x^2 - 4x + \frac{\lambda}{16}\). The roots of \(f(x)=0\) are \(\alpha\) and \(\beta\). Since the coefficient of \(x^2\) is positive (which is 1), the parabola representing \(f(x)\) opens upwards.
The conditions \(1 < \alpha < 2\) and \(2 < \beta < 3\) tell us that one root lies strictly between 1 and 2, and the other root lies strictly between 2 and 3. This means that \(x=2\) lies between the two roots \(\alpha\) and \(\beta\).
For a quadratic function \(f(x) = ax^2 + bx + c\) with \(a > 0\), if a value \(k\) is between the roots, then \(f(k)\) must be negative. If a value \(k\) is outside the roots, \(f(k)\) must be positive.
In our case, \(a=1 > 0\). The roots are \(\alpha\) and \(\beta\). The conditions are \(1 < \alpha < 2\) and \(2 < \beta < 3\). This implies:
Let's calculate \(f(1)\), \(f(2)\), and \(f(3)\) using the function \(f(x) = x^2 - 4x + \frac{\lambda}{16}\):
Now we use the conditions derived from the root locations:
Let's solve each inequality for \(\lambda\):
From inequality 1:
\(-3 + \frac{\lambda}{16} > 0\)
\(\frac{\lambda}{16} > 3\)
\(\lambda > 3 \times 16\)
\(\lambda > 48\)
From inequality 2:
\(-4 + \frac{\lambda}{16} < 0\)
\(\frac{\lambda}{16} < 4\)
\(\lambda < 4 \times 16\)
\(\lambda < 64\)
Inequality 3 is the same as inequality 1, so it also gives \(\lambda > 48\).
Combining the inequalities \(\lambda > 48\) and \(\lambda < 64\), we get the range for \(\lambda\) as \(48 < \lambda < 64\).
We are given that \(\lambda\) is an integer. The integers strictly between 48 and 64 are \(49, 50, 51, \dots, 63\).
To find the number of integers in this range, we can subtract the smallest integer from the largest integer and add 1:
Number of values = \(63 - 49 + 1 = 14 + 1 = 15\)
Therefore, \(\lambda\) can take 15 different integer values.
| Condition | Inequality | Result for \(\lambda\) |
|---|---|---|
| \(f(1) > 0\) | \(-3 + \frac{\lambda}{16} > 0\) | \(\lambda > 48\) |
| \(f(2) < 0\) | \(-4 + \frac{\lambda}{16} < 0\) | \(\lambda < 64\) |
| \(f(3) > 0\) | \(-3 + \frac{\lambda}{16} > 0\) | \(\lambda > 48\) |
The combined condition is \(48 < \lambda < 64\). Since \(\lambda\) must be an integer, the possible values are \(49, 50, \dots, 63\). There are 15 such integer values.
| Concept | Description | Application in Problem |
|---|---|---|
| Quadratic Equation Roots | Values of \(x\) that satisfy \(ax^2+bx+c=0\). | Given roots \(\alpha, \beta\) for \(4x^2 - 16x + \frac{\lambda}{4} = 0\). |
| Parabola Direction | If \(a>0\), parabola opens upwards. If \(a<0\), downwards. | \(x^2 - 4x + \frac{\lambda}{16} = 0\) has \(a=1>0\), opens upwards. |
| Root Location Theory | Relates the position of roots relative to a point \(k\) based on the sign of \(f(k)\). | Used \(f(1)>0\), \(f(2)<0\), \(f(3)>0\) based on \(1<\alpha<2\) and \(2<\beta<3\). |
| Solving Inequalities | Finding the range of values for a variable that satisfy an inequality. | Solved for \(\lambda\) from \(f(1)>0\), \(f(2)<0\), \(f(3)>0\). |
| Counting Integers in a Range | Number of integers between \(a\) and \(b\) (exclusive) is \(b-a-1\). (Here it's inclusive of \(a+1\) and \(b-1\)). Number of integers from \(a\) to \(b\) (inclusive) is \(b-a+1\). | Counted integers from 49 to 63. |
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