JEE Mathematics 2023 Question with Solution

Given: The quadratic equation is $2x^2 - 8x + k = 0$.

Find: The number of integral values of $k$ such that one root lies in $(1, 2)$ and the other lies in $(2, 3)$.

Let

$$f(x) = 2x^2 - 8x + k$$

For one root in $(1, 2)$, we need

$$f(1) \cdot f(2) < 0$$

For the other root in $(2, 3)$, we need

$$f(2) \cdot f(3) < 0$$

Now,

$$f(1) = 2(1)^2 - 8(1) + k = k - 6$$ $$f(2) = 2(2)^2 - 8(2) + k = k - 8$$ $$f(3) = 2(3)^2 - 8(3) + k = k - 6$$

Using the first condition,

$$(k - 6)(k - 8) < 0$$

which gives

$$k \in (6, 8)$$

Using the second condition,

$$(k - 8)(k - 6) < 0$$

which also gives

$$k \in (6, 8)$$

So the required interval is

$$k \in (6, 8)$$

The only integer in this interval is $7$.

Therefore, the number of integral values of $k$ is $1$. The correct option is C.

Given: $f(x) = 2x^2 - 8x + k$.

Find: Why the interval condition leads to the inequalities used above.

A polynomial is continuous. Therefore, if a root lies in the open interval $(1, 2)$, then the function values at the endpoints must have opposite signs, so

$$f(1)f(2) < 0$$

Similarly, if another root lies in $(2, 3)$, then

$$f(2)f(3) < 0$$

Here,

$$f(1) = k - 6, \quad f(2) = k - 8, \quad f(3) = k - 6$$

Hence both interval conditions reduce to the same inequality:

$$(k - 6)(k - 8) < 0$$

A product of two linear factors is negative between its distinct zeros, so

$$6 < k < 8$$

Thus the only integral value is $k = 7$, and the count is $1$.