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Greatest Value of x in a Rational Product Equation = 42 | IOQM

JEE Maths question with a full step-by-step solution.

Question
Find the greatest value of xx for which
(13x−x2x+1)(x+13−xx+1)=42.\left(\frac{13x-x^2}{x+1}\right)\left(x+\frac{13-x}{x+1}\right)=42.
Solution
Answer: 6
Step 1: Simplify the second factor. Writing 13−x=14−(x+1)13-x=14-(x+1),
x+13−xx+1=x+14−(x+1)x+1=x−1+14x+1.x+\frac{13-x}{x+1}=x+\frac{14-(x+1)}{x+1}=x-1+\frac{14}{x+1}.
 Let t=x−1+14x+1t=x-1+\dfrac{14}{x+1}, so the second factor equals tt. Step 2: Express the first factor through tt. Since 13x−x2=13(x+1)−(x2−1)−1413x-x^2=13(x+1)-(x^2-1)-14,
13x−x2x+1=13−x2−1x+1−14x+1=13−(x−1)−14x+1=13−(x−1+14x+1)=13−t.\frac{13x-x^2}{x+1}=13-\frac{x^2-1}{x+1}-\frac{14}{x+1}=13-(x-1)-\frac{14}{x+1}=13-\left(x-1+\frac{14}{x+1}\right)=13-t.
 So the first factor equals 13−t13-t. Step 3: The equation becomes the product (13−t) t=42(13-t)\,t=42, that is
t2−13t+42=0 ⇒ (t−6)(t−7)=0 ⇒ t=6 or t=7.t^2-13t+42=0\ \Rightarrow\ (t-6)(t-7)=0\ \Rightarrow\ t=6\ \text{or}\ t=7.
 Step 4: Case t=6t=6. From x−1+14x+1=6x-1+\dfrac{14}{x+1}=6, i.e. x+14x+1=7x+\dfrac{14}{x+1}=7,
x(x+1)+14=7(x+1) ⇒ x2−6x+7=0 ⇒ (x−3)2=2 ⇒ x=3±2.x(x+1)+14=7(x+1)\ \Rightarrow\ x^2-6x+7=0\ \Rightarrow\ (x-3)^2=2\ \Rightarrow\ x=3\pm\sqrt2.
 Step 5: Case t=7t=7. From x−1+14x+1=7x-1+\dfrac{14}{x+1}=7, i.e. x+14x+1=8x+\dfrac{14}{x+1}=8,
x(x+1)+14=8(x+1) ⇒ x2−7x+6=0 ⇒ (x−1)(x−6)=0 ⇒ x=1 or x=6.x(x+1)+14=8(x+1)\ \Rightarrow\ x^2-7x+6=0\ \Rightarrow\ (x-1)(x-6)=0\ \Rightarrow\ x=1\ \text{or}\ x=6.
 Step 6: The four real solutions are x=3+2, 3−2, 1, 6x=3+\sqrt2,\ 3-\sqrt2,\ 1,\ 6. Since 3+2≈4.41<63+\sqrt2\approx4.41<6, the greatest value is
x=6.x=6.
 Answer: 6
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