Consider the quadratic equation (c - 5)x² - 2cx + (c - 4) = 0, c ≠ 5. Let S be the set of all integr

Q: Wait, why do we use $f(0) \cdot f(2) 0$?

The discriminant $D > 0$ only guarantees that the roots are real and distinct. It does not tell us *where* they are. The Intermediate Value Theorem (IVT) is the standard tool for 'location of roots' problems because it directly constrains the roots to specific intervals by checking the sign of the function at the boundaries.

Q: Why does $f(2)$ appear in both inequalities?

Great observation! The point $x = 2$ is the shared boundary between our two intervals $(0, 2)$ and $(2, 3)$. Since one root is on one side of $2$ and the other is on the other side, $f(2)$ acts as the pivot point for both conditions.

Q: Why is the condition $c \neq 5$ given in the problem?

The equation is given as $(c - 5)x^2 - 2cx + (c - 4) = 0$. If $c = 5$, the $x^2$ term vanishes, and the equation becomes linear: $-10x + 1 = 0$, which has only one root. Since the problem specifies 'one root' and 'other root' (implying two roots), it must be a quadratic, so $c$ cannot be $5$.

Question

Consider the quadratic equation $(c - 5)x^2 - 2cx + (c - 4) = 0, c 
e 5$. Let S be the set of all integral values of c for which one root of the equation lies in the interval (0, 2) and its other root lies in the interval (2, 3). Then the number of elements in S is :

Options:

(A) 11

(B) 18

(C) 10

(D) 12

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the $x$-axis, looking at a parabola defined by the function $f(x) = (c - 5)x^2 - 2cx + (c - 4)$. This is not just an algebraic expression; it is a geometric entity.

We are told that this parabola crosses the $x$-axis at two distinct points. One root, $\alpha$, is trapped between $0$ and $2$, and the other root, $\beta$, is caught between $2$ and $3$.

### The Power of the Intermediate Value Theorem

To ensure a root exists in an interval $(a, b)$, we rely on the **Intermediate Value Theorem**. If a continuous function like our parabola is positive at one endpoint and negative at the other, it must have crossed the $x$-axis somewhere in between.

Mathematically, this is expressed as:
$$f(a) \cdot f(b) < 0$$

We have two intervals, $(0, 2)$ and $(2, 3)$. For $\alpha \in (0, 2)$, we must have $f(0) \cdot f(2) < 0$. For $\beta \in (2, 3)$, we must have $f(2) \cdot f(3) < 0$.

### The Algebraic Execution

Let us calculate the values of the function at these critical points:

$$f(0) = (c - 5)(0)^2 - 2c(0) + (c - 4) = c - 4$$

$$f(2) = (c - 5)(2)^2 - 2c(2) + (c - 4) = 4c - 20 - 4c + c - 4 = c - 24$$

$$f(3) = (c - 5)(3)^2 - 2c(3) + (c - 4) = 9c - 45 - 6c + c - 4 = 4c - 49$$

Now, we translate our geometric conditions into algebraic inequalities. The first condition, $f(0) \cdot f(2) < 0$, becomes:
$$(c - 4)(c - 24) < 0$$

Using the wavy curve method, we find that $c$ must lie in the interval $(4, 24)$.

The second condition, $f(2) \cdot f(3) < 0$, becomes:
$$(c - 24)(4c - 49) < 0$$

The critical points are $24$ and $\frac{49}{4} = 12.25$. Thus, $c$ must lie in the interval $(12.25, 24)$.

### The Final Intersection

We need both conditions to be true simultaneously. We are looking for the intersection of $(4, 24)$ and $(12.25, 24)$.

The tighter constraint is $c \in (12.25, 24)$. The question asks for the number of integral values of $c$ in this set.

The integers strictly between $12.25$ and $24$ are $13, 14, 15, 16, 17, 18, 19, 20, 21, 22,$ and $23$. Counting these, we find there are exactly **11** such values.

Select Answer:

Visualized Solution

Visualizing the Roots

The Geometric Setup

Location of Roots Condition

Setting Up the Inequalities

Calculating $f (0)$

Calculating $f (2)$

Calculating $f (3)$

Solving the First Inequality

Solving the Second Inequality

Combining the Conditions

Final Answer

The Sigma Insight: Location of Roots

Analyzing the Setup

The Power of the Intermediate Value Theorem

The Algebraic Execution

The Final Intersection

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