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: Understanding the Concept.
Instead of jumping straight to a discriminant, notice a neat symmetry in this expression first. Write $k=\sqrt{a^2+b^2}$ and $f(x)=\dfrac{x^2-2x+k^2}{x^2+2x+k^2}$. Replacing $x$ by $-x$ swaps the numerator and denominator of $f$, so $f(-x) = \dfrac{1}{f(x)}$. This tells us the set of values $f$ can take must be symmetric under taking the reciprocal, which already hints that the two boundary values of the range are reciprocals of each other.

Step 2: Key Formula or Approach.
To pin down the actual boundary values, treat $y=f(x)$ and rewrite the equation as a quadratic in $x$:
\[ (y-1)x^2 + 2(y+1)x + (y-1)k^2 = 0 \]
For a real $x$ to exist, the discriminant of this quadratic (in $x$) must be non-negative.

Step 3: Detailed Explanation.
Discriminant condition:
\[ 4(y+1)^2 - 4k^2(y-1)^2 \ge 0 \]
\[ (y+1)^2 \ge k^2(y-1)^2 \]
This reduces cleanly to the factored inequality
\[ [y(1-k)+(1+k)]\,[y(1+k)+(1-k)] \ge 0 \]
Solving $y(1-k)+(1+k)=0$ gives $y=\dfrac{k+1}{k-1}$.
Solving $y(1+k)+(1-k)=0$ gives $y=\dfrac{k-1}{k+1}$.
These two values are reciprocals of each other, exactly matching the symmetry noticed in Step 1. Since the leading coefficient of the quadratic in $y$ is negative (because $k>1$ makes $1-k^2<0$), the expression is non-negative between its two roots, so $y$ is confined between $\dfrac{k-1}{k+1}$ and $\dfrac{k+1}{k-1}$.

Step 4: Final Answer.
Substituting back $k=\sqrt{a^2+b^2}$, the expression always lies between $\dfrac{\sqrt{a^2+b^2}-1}{\sqrt{a^2+b^2}+1}$ and $\dfrac{\sqrt{a^2+b^2}+1}{\sqrt{a^2+b^2}-1}$, a reciprocal pair of bounds, confirming the symmetry argument from the start. \[ \boxed{\dfrac{\sqrt{a^2+b^2}-1}{\sqrt{a^2+b^2}+1} \le y \le \dfrac{\sqrt{a^2+b^2}+1}{\sqrt{a^2+b^2}-1}} \]