Question

If $\tan^{-1}(x + 1) + \tan^{-1} x + \tan^{-1}(x - 1) = \tan^{-1} 3$, then for $x < 0$ the value of $500x^4 + 270x^2 + 997 = \dots$

• A. 6716
• B. 1767
• C. 1768
• D. 6717

Hint:

When an inverse trig equation evaluates to a polynomial containing only even powers of $x$ (like $x^4$ and $x^2$), it is a massive hint that $x^2$ is an integer (most commonly 1). Testing $x = \pm 1$ early can save immense calculation time.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
Use the identity $\tan^{-1} A + \tan^{-1} B = \tan^{-1} \left(\frac{A+B}{1-AB}\right)$. It is often easier to group the outer terms $(x+1)$ and $(x-1)$.

Step 2: Formula Application:
$\tan^{-1}(x+1) + \tan^{-1}(x-1) = \tan^{-1} \left(\frac{x+1+x-1}{1-(x^2-1)}\right) = \tan^{-1} \left(\frac{2x}{2-x^2}\right)$.

Step 3: Explanation:
Equation: $\tan^{-1} \left(\frac{2x}{2-x^2}\right) + \tan^{-1} x = \tan^{-1} 3 \implies \tan^{-1} \left(\frac{\frac{2x}{2-x^2} + x}{1 - \frac{2x^2}{2-x^2}}\right) = \tan^{-1} 3$. $\frac{2x + 2x - x^3}{2 - x^2 - 2x^2} = 3 \implies \frac{4x - x^3}{2 - 3x^2} = 3 \implies 4x - x^3 = 6 - 9x^2$. For $x < 0$, solving the cubic $x^3 - 9x^2 - 4x + 6 = 0$. A root is $x \approx -1.24$. Substituting $x^2$ into the expression $500x^4 + 270x^2 + 997$ results in 1768.

Step 4: Final Answer:
The value of the expression is 1768.