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Exercise 4.5 · Q10

Q.Find the number of solutions of the equation tan⁡−1(x−1)+tan⁡−1x+tan⁡−1(x+1)=tan⁡−1(3x)\tan^{-1}(x-1) + \tan^{-1}x + \tan^{-1}(x+1) = \tan^{-1}(3x).

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Combining tan⁡−1(x−1)+tan⁡−1(x+1)\tan^{-1}(x-1)+\tan^{-1}(x+1) first (their sum is symmetric) and then adding tan⁡−1x\tan^{-1}x, matching against tan⁡−1(3x)\tan^{-1}(3x), reduces the equation to a cubic with three real roots — and all three check out directly in the original equation.

Step 1. Combine the outer two terms. For ∣x∣<2|x|<\sqrt2 (so that (x−1)(x+1)=x2−1<1(x-1)(x+1)=x^2-1<1), tan⁡−1(x−1)+tan⁡−1(x+1)=tan⁡−1(x−1)+(x+1)1−(x−1)(x+1)=tan⁡−12x2−x2\tan^{-1}(x-1)+\tan^{-1}(x+1)=\tan^{-1}\dfrac{(x-1)+(x+1)}{1-(x-1)(x+1)}=\tan^{-1}\dfrac{2x}{2-x^2}.

Step 2. Add the middle term. tan⁡−12x2−x2+tan⁡−1x=tan⁡−12x2−x2+x1−2x2−x2⋅x=tan⁡−1x(4−x2)2−3x2\tan^{-1}\dfrac{2x}{2-x^2}+\tan^{-1}x=\tan^{-1}\dfrac{\frac{2x}{2-x^2}+x}{1-\frac{2x}{2-x^2}\cdot x}=\tan^{-1}\dfrac{x(4-x^2)}{2-3x^2} (multiplying numerator and denominator by 2−x22-x^2 and simplifying).

Step 3. Equate with the right-hand side. Matching tan⁡−1x(4−x2)2−3x2=tan⁡−1(3x)\tan^{-1}\dfrac{x(4-x^2)}{2-3x^2}=\tan^{-1}(3x) gives x(4−x2)2−3x2=3x\dfrac{x(4-x^2)}{2-3x^2}=3x (for 2−3x2≠02-3x^2\ne0).

Step 4. Clear the denominator and simplify. x(4−x2)=3x(2−3x2)⇒4x−x3=6x−9x3⇒8x3−2x=0⇒2x(4x2−1)=0x(4-x^2)=3x(2-3x^2)\Rightarrow4x-x^3=6x-9x^3\Rightarrow8x^3-2x=0\Rightarrow2x(4x^2-1)=0.

Step 5. Solve the cubic. x=0x=0 or x2=14⇒x=±12x^2=\dfrac14\Rightarrow x=\pm\dfrac12 — three real roots. …

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