Q.Solve for $x$: $\dfrac{4}{x+1} \le 3 \le \dfrac{6}{x+1}$, $(x > 0)$.

Concept and Intuition

When solving rational inequalities like $\frac{A}{B} > \frac{C}{D}$, the common mistake is to cross-multiply without considering the signs of the denominators. Instead, we bring everything to one side, combine into a single fraction, and then analyze where that fraction is positive (or negative). The key idea: a rational expression changes sign only at points where its numerator or denominator is zero. These "critical points" split the number line into intervals; we test a point in each interval to determine the sign.

Step-by-step solution

1. Bring all terms to one side

2. Combine into a single fraction Common denominator: $(x-3)(3x+1)$.

3. Expand and simplify the numerator First product: $(2x-1)(3x+1) = 6x^2 + 2x - 3x - 1 = 6x^2 - x - 1$ Second product: $(x+2)(x-3) = x^2 - 3x + 2x - 6 = x^2 - x - 6$ Subtract:

So the inequality becomes:

4. Simplify further Since $x^2+1 > 0$ for all real $x$, and $5 > 0$, the sign of the whole expression depends only on the denominator:

5. Find critical points

   Denominator zero at $x = 3$ and $x = -\frac{1}{3}$. These are the points where the expression is undefined (and where sign can change).

6. Build a sign chart

   The real line is divided into three intervals:

   • $(-\infty, -\frac{1}{3})$
   • $(-\frac{1}{3}, 3)$
   • $(3, \infty)$

   Test a point in each interval:

   • For $x = -1$ (in $(-\infty, -\frac{1}{3})$):

     $(x-3)(3x+1) = (-4)(-2) = 8 > 0$ → expression positive.

   • For $x = 0$ (in $(-\frac{1}{3}, 3)$):

     $(0-3)(0+1) = (-3)(1) = -3 < 0$ → expression negative.

   • For $x = 4$ (in $(3, \infty)$):

     $(4-3)(12+1) = (1)(13) = 13 > 0$ → expression positive.

7. Select intervals where inequality holds

   We need $> 0$, so we take intervals where the expression is positive:

Concept and Intuition

With an inequality like $\dfrac{P(x)}{Q(x)}\geq1$, you cannot simply cross-multiply by $Q(x)$ unless you know its sign, because multiplying by a negative quantity flips the inequality. The safe method is to bring everything to one side, combine into a single fraction, and then do a sign analysis (a number-line test) using all the zeros of numerator and denominator as critical points.

Step-by-Step Solution

1. Rewrite: $\dfrac{x^2-1}{(x-4)(x-3)} - 1 \geq 0$.
2. Combine over a common denominator: $\dfrac{x^2-1 - (x-4)(x-3)}{(x-4)(x-3)} \geq 0$.
3. Expand $(x-4)(x-3) = x^2-7x+12$, so the numerator becomes $x^2-1-x^2+7x-12 = 7x-13$.
4. The inequality is now $\dfrac{7x-13}{(x-4)(x-3)}\geq0$, with critical points $x=\tfrac{13}{7}\;(\approx1.857)$, $x=3$, $x=4$ (the latter two excluded since the expression is undefined there).
5. Sign chart:
   • $x<\tfrac{13}{7}$: numerator $<0$, denominator $>0$ (both factors negative-negative giving positive... check: $x<3$ so $(x-4)<0,(x-3)<0$, product $>0$) → fraction $<0$. Excluded.
   • $\tfrac{13}{7}\leq x<3$: numerator $\geq0$, denominator $>0$ → fraction $\geq0$. Included.
   • $3<x<4$: numerator $>0$, denominator: $(x-4)<0,(x-3)>0$, product $<0$ → fraction $<0$. Excluded.
   • $x>4$: numerator $>0$, denominator $>0$ → fraction $>0$. Included.
6. So the solution set is $\left[\tfrac{13}{7},3\right)\cup(4,\infty)$.

Common Mistakes

• Cross-multiplying by $(x-4)(x-3)$ without tracking its sign, which flips inequalities incorrectly in some intervals.
• Forgetting to exclude $x=3,4$ where the expression is undefined.

ANSWER: B

Concept and Intuition

Solve each quadratic inequality by factoring and sign analysis, then intersect the resulting intervals, finally counting the integers within the intersection.

Step-by-Step Solution

1. $x^2-7x+6\le0 \Rightarrow (x-1)(x-6)\le0 \Rightarrow 1\le x\le6$.
2. $x^2-3x>0 \Rightarrow x(x-3)>0 \Rightarrow x<0$ or $x>3$.
3. Intersect with $[1,6]$: the $x<0$ branch has no overlap; the $x>3$ branch overlaps as $3<x\le6$.
4. Integers satisfying $3<x\le6$: $x=4,5,6$.
5. Count $=3$.

Common Mistakes

• Including $x=3$ (it fails the strict inequality $x^2-3x>0$, since at $x=3$, $x^2-3x=0$, not $>0$).

ANSWER: C