Question

The expression \(\dfrac{x^2 - 2x + a^2 + b^2}{x^2 + 2x + a^2 + b^2}\) lies between:

• A. \(\dfrac{\sqrt{a^2+b^2}+1}{\sqrt{a^2+b^2}-1}\) and \(\dfrac{\sqrt{a^2+b^2}-1}{\sqrt{a^2+b^2}+1}\)
• B. a and b
• C. \(\dfrac{\sqrt{a^2+b^2}+1}{\sqrt{a^2+b^2}-1}\) and 1
• D. \(\dfrac{\sqrt{a^2+b^2}-1}{\sqrt{a^2+b^2}+1}\)

Hint:

Treat the expression as y, cross multiply to get a quadratic in x, and use the real-roots (discriminant greater than or equal to 0) condition to find the range of y.

Answer 1

The Correct Option is A

Step 1: Set the expression equal to y and clear denominators.
Let \(k=\sqrt{a^2+b^2}\), so \(a^2+b^2=k^2\). Let
\[ y = \frac{x^2-2x+k^2}{x^2+2x+k^2} \]
Cross multiplying,
\[ y(x^2+2x+k^2) = x^2-2x+k^2 \]

Step 2: Collect everything into one quadratic in x.
\[ x^2(y-1) + 2x(y+1) + k^2(y-1) = 0 \]
This is a quadratic equation in \(x\) whose coefficients involve \(y\). Since \(x\) is a real number for any valid position, this quadratic must have real roots, so its discriminant cannot be negative.

Step 3: Apply the real-roots condition (discriminant \(\geq 0\)).
\[ [2(y+1)]^2 - 4(y-1)\cdot k^2(y-1) \geq 0 \]
\[ 4(y+1)^2 - 4k^2(y-1)^2 \geq 0 \]
\[ (y+1)^2 \geq k^2(y-1)^2 \]
Taking this as a difference of squares,
\[ \big[(y+1)-k(y-1)\big]\big[(y+1)+k(y-1)\big] \geq 0 \]

Step 4: Solve the boundary equations.
Setting each bracket to zero gives the two critical values of \(y\):
\[ (y+1)-k(y-1)=0 \implies y=\frac{k+1}{k-1} \]
\[ (y+1)+k(y-1)=0 \implies y=\frac{k-1}{k+1} \]
Since \(k=\sqrt{a^2+b^2}>1\) in this setting, \(\frac{k+1}{k-1}\) is the larger critical value and \(\frac{k-1}{k+1}\) is the smaller one. Because the product of the two brackets is a downward-opening quadratic in \(y\), the inequality holds for \(y\) between the two roots.

Step 5: State the range and check with a number.
So the expression lies between \(\dfrac{k-1}{k+1}\) and \(\dfrac{k+1}{k-1}\), that is, between \(\dfrac{\sqrt{a^2+b^2}-1}{\sqrt{a^2+b^2}+1}\) and \(\dfrac{\sqrt{a^2+b^2}+1}{\sqrt{a^2+b^2}-1}\). As a quick check, take \(a=2,b=0\) so \(k=2\): at \(x=-2\) the expression equals \(\frac{4+4+4}{4-4+4}=\frac{12}{4}=3=\frac{k+1}{k-1}\), and at \(x=2\) it equals \(\frac{4-4+4}{4+4+4}=\frac{4}{12}=\frac{1}{3}=\frac{k-1}{k+1}\), confirming both ends of the range exactly.

Final Answer:
The expression is bounded between the reciprocal pair \(\dfrac{\sqrt{a^2+b^2}-1}{\sqrt{a^2+b^2}+1}\) and \(\dfrac{\sqrt{a^2+b^2}+1}{\sqrt{a^2+b^2}-1}\). \[ \boxed{\dfrac{\sqrt{a^2+b^2}-1}{\sqrt{a^2+b^2}+1} \le y \le \dfrac{\sqrt{a^2+b^2}+1}{\sqrt{a^2+b^2}-1}} \]