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

The Correct Option is A

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