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Inverse Trigonometric Functions question

2024 · 27 Jan · Shift 2 · Q31
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  5. /2024 · 27 Jan · Shift 2 · Q31

Inverse Trigonometric Functions question

2024 · 27 Jan · Shift 2 · Q31

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
Considering only the principal values of inverse trigonometric functions, the number of positive real values of xxx satisfying tan⁡−1(x)+tan⁡−1(2x)=π4\tan ^{-1}(x)+\tan ^{-1}(2 x)=\frac{\pi}{4}tan−1(x)+tan−1(2x)=4π​ is :
  1. A
    more than 2
  2. B
    2
  3. C
    0
  4. D
    1
View written solutionFree

Correct answer: D

  1. Let tan⁡−1(x)+tan⁡−1(2x)=π4.\tan^{-1}(x)+\tan^{-1}(2x)=\frac{\pi}{4}.tan−1(x)+tan−1(2x)=4π​. We need the number of positive real values of xxx satisfying this, using only principal values of inverse tangent.

  2. Since x>0x>0x>0, we have: tan⁡−1(x)∈(0,π2),tan⁡−1(2x)∈(0,π2).\tan^{-1}(x)\in \left(0,\frac{\pi}{2}\right), \qquad \tan^{-1}(2x)\in \left(0,\frac{\pi}{2}\right).tan−1(x)∈(0,2π​),tan−1(2x)∈(0,2π​). So both angles are principal values and their sum is valid.

  3. Apply the tangent addition formula: If A=tan⁡−1(x),B=tan⁡−1(2x),A=\tan^{-1}(x), \qquad B=\tan^{-1}(2x),A=tan−1(x),B=tan−1(2x), then

=\frac{x+2x}{1-2x^2}=\frac{3x}{1-2x^2}.$$ Given $$A+B=\frac{\pi}{4},$$ so $$\tan(A+B)=\tan\frac{\pi}{4}=1.$$ Hence, $$\frac{3x}{1-2x^2}=1.$$ 4. Solve: $$3x=1-2x^2$$ $$2x^2+3x-1=0.$$ Using the quadratic formula, $$x=\frac{-3\pm\sqrt{9+8}}{4}=\frac{-3\pm\sqrt{17}}{4}.$$ 5. Check positivity: - $$x=\frac{-3+\sqrt{17}}{4}>0$$ since $\sqrt{17}>3$ - $$x=\frac{-3-\sqrt{17}}{4}<0$$ So there is exactly **one positive real** solution. 6. Also verify no extraneous issue arises: For the positive root, both inverse tangents are positive and their sum can indeed equal $\pi/4$. Hence this solution is valid. Therefore, the number of positive real values of $x$ is $$\boxed{1}.$$ So the correct option is **D**.
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