Question

Solve for the number of integral values of $x: \frac{(x^2-15x+47)(x-13)}{(x-8)}<0$.

• A. 5
• B. 6
• C. 7
• D. 8
• E. 3

Answer 1

Answer: (A)

5

Answer 2

$\frac{(x^2-15x+47)(x-13)}{(x-8)}<0$ .… (1)

$\frac{(x^2-15x+47)(x-13)(x-8)}{(x-8)^2}<0$

As $(x-8)^2≥0$

[The denominator is $0$ when $x=8$, which is not allowed.]

$(x^2-15x+47)(x-13)(x-8)<0$

Let us try to find the value of $x^2-15x+47$.

$⇒ x^2-15x+\frac{225}{4}+47-\frac{225}{4}$

$⇒ (x-\frac{15}{2})^2+47-\frac{225}{4}$

$⇒ (x-\frac{15}{2})^2-\frac{37}{4}$

It is always positive when the value of $x > 10$ and $x < 5$.

The critical points for $(x-13)(x-8)<0$ are $13$ and $8$.

Case I: For $x≥13$,

The value $(x-13)(x-8)$ is either zero or positive.

$(x-\frac{15}{2})^2-\frac{37}4$ is always positive.

So, $(x^2-15x+47)(x-13)(x-8)$ is always zero or positive.

Hence, there is no solution for this range.

Case II: For $8<x<13$,

The value of $(x-13)(x-8)$ is always negative.

But the value of $(x-\frac{15}{2})^2-\frac{37}4$ is always positive when the value of $x > 10$ and $x < 5$.

Thus, the values possible are $x = 11, 12$.

Case III: For $x≤8$,

The value of $(x-13)(x-8)$ is either zero or positive.

But the value of $(x-\frac{15}{2})^2-\frac{37}{4}$ is always negative when the values of $x$ are $5, 6$, and $7$.

The number of integral solutions is 5, and these values are $5, 6, 7, 11$, and $12$.

Hence, option A is the correct answer.