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Inverse Trigonometric Functions question

2023 · 25 Jan · Shift 1 · Q46
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  5. /2023 · 25 Jan · Shift 1 · Q46

Inverse Trigonometric Functions question

2023 · 25 Jan · Shift 1 · Q46

JEE MainMathematicsInverse Trigonometric FunctionsNumerical+4 / −1
If the sum of all the solutions of tan⁡−1(2x1−x2)+cot⁡−1(1−x22x)=π3,−1<x<1,xe0{\tan ^{ - 1}}\left( {{{2x} \over {1 - {x^2}}}} \right) + {\cot ^{ - 1}}\left( {{{1 - {x^2}} \over {2x}}} \right) = {\pi \over 3}, - 1 \lt x \lt 1,x e 0tan−1(1−x22x​)+cot−1(2x1−x2​)=3π​,−1<x<1,xe0, is α−43\alpha - {4 \over {\sqrt 3 }}α−3​4​, then α\alphaα is equal to ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 2

  1. Given equation

We need to solve

tan⁡−1 ⁣(2x1−x2)+cot⁡−1 ⁣(1−x22x)=π3,\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)+\cot^{-1}\!\left(\frac{1-x^2}{2x}\right)=\frac{\pi}{3},tan−1(1−x22x​)+cot−1(2x1−x2​)=3π​,

with −1<x<1,x≠0.-1<x<1,\quad x\ne 0.−1<x<1,x=0.

We are then asked to find the sum of all solutions, and compare it with the form α−43.\alpha-\frac{4}{\sqrt{3}}.α−3​4​.

  1. Relating the two inverse expressions

Let y=2x1−x2.y=\frac{2x}{1-x^2}.y=1−x22x​. Then the second term becomes cot⁡−1 ⁣(1−x22x)=cot⁡−1 ⁣(1y).\cot^{-1}\!\left(\frac{1-x^2}{2x}\right)=\cot^{-1}\!\left(\frac{1}{y}\right).cot−1(2x1−x2​)=cot−1(y1​).

Now use the identity based on principal values:

  • If y>0y>0y>0, then cot⁡−1 ⁣(1y)=tan⁡−1(y).\cot^{-1}\!\left(\frac1y\right)=\tan^{-1}(y).cot−1(y1​)=tan−1(y).
  • If y<0y<0y<0, then cot⁡−1 ⁣(1y)=π+tan⁡−1(y),\cot^{-1}\!\left(\frac1y\right)=\pi+\tan^{-1}(y),cot−1(y1​)=π+tan−1(y), since cot⁡−1\cot^{-1}cot−1 takes values in (0,π)(0,\pi)(0,π).

So we must split into cases.

  1. Sign of yyy in the interval −1<x<1-1<x<1−1<x<1

Since −1<x<1-1<x<1−1<x<1, we have 1−x2>0.1-x^2>0.1−x2>0. Therefore the sign of y=2x1−x2y=\frac{2x}{1-x^2}y=1−x22x​ is the same as the sign of xxx.

So:

  • if x>0x>0x>0, then y>0y>0y>0;
  • if x<0x<0x<0, then y<0y<0y<0.
  1. Case 1: x>0x>0x>0

Then cot⁡−1 ⁣(1−x22x)=tan⁡−1 ⁣(2x1−x2).\cot^{-1}\!\left(\frac{1-x^2}{2x}\right)=\tan^{-1}\!\left(\frac{2x}{1-x^2}\right).cot−1(2x1−x2​)=tan−1(1−x22x​). Hence the equation becomes 2tan⁡−1 ⁣(2x1−x2)=π3.2\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}.2tan−1(1−x22x​)=3π​. So tan⁡−1 ⁣(2x1−x2)=π6.\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{6}.tan−1(1−x22x​)=6π​. Taking tangent, 2x1−x2=13.\frac{2x}{1-x^2}=\frac{1}{\sqrt{3}}.1−x22x​=3​1​. Thus 23x=1−x2,2\sqrt{3}x=1-x^2,23​x=1−x2, so x2+23x−1=0.x^2+2\sqrt{3}x-1=0.x2+23​x−1=0.

Solving: x=−23±12+42=−23±42=−3±2.x=\frac{-2\sqrt{3}\pm\sqrt{12+4}}{2}=\frac{-2\sqrt{3}\pm 4}{2}=-\sqrt{3}\pm 2.x=2−23​±12+4​​=2−23​±4​=−3​±2. So the roots are x=2−3,x=−2−3.x=2-\sqrt{3},\quad x=-2-\sqrt{3}.x=2−3​,x=−2−3​. Only x=2−3x=2-\sqrt{3}x=2−3​ lies in (0,1)(0,1)(0,1).

  1. Case 2: x<0x<0x<0

Then cot⁡−1 ⁣(1−x22x)=π+tan⁡−1 ⁣(2x1−x2).\cot^{-1}\!\left(\frac{1-x^2}{2x}\right)=\pi+\tan^{-1}\!\left(\frac{2x}{1-x^2}\right).cot−1(2x1−x2​)=π+tan−1(1−x22x​). Hence tan⁡−1 ⁣(2x1−x2)+π+tan⁡−1 ⁣(2x1−x2)=π3.\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)+\pi+\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)=\frac{\pi}{3}.tan−1(1−x22x​)+π+tan−1(1−x22x​)=3π​. That is, 2tan⁡−1 ⁣(2x1−x2)+π=π3,2\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)+\pi=\frac{\pi}{3},2tan−1(1−x22x​)+π=3π​, which gives 2tan⁡−1 ⁣(2x1−x2)=−2π3.2\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)=-\frac{2\pi}{3}.2tan−1(1−x22x​)=−32π​. So tan⁡−1 ⁣(2x1−x2)=−π3.\tan^{-1}\!\left(\frac{2x}{1-x^2}\right)=-\frac{\pi}{3}.tan−1(1−x22x​)=−3π​. Taking tangent, 2x1−x2=−3.\frac{2x}{1-x^2}=-\sqrt{3}.1−x22x​=−3​. Thus 2x=−3(1−x2),2x=-\sqrt{3}(1-x^2),2x=−3​(1−x2), so 3x2−2x−3=0.\sqrt{3}x^2-2x-\sqrt{3}=0.3​x2−2x−3​=0.

Solving: x=2±4+1223=2±423.x=\frac{2\pm\sqrt{4+12}}{2\sqrt{3}}=\frac{2\pm 4}{2\sqrt{3}}.x=23​2±4+12​​=23​2±4​. Hence x=3,x=−13.x=\sqrt{3},\quad x=-\frac{1}{\sqrt{3}}.x=3​,x=−3​1​. Only x=−13x=-\frac{1}{\sqrt{3}}x=−3​1​ lies in (−1,0)(-1,0)(−1,0).

  1. Sum of all valid solutions

The two valid solutions are x1=2−3,x2=−13.x_1=2-\sqrt{3},\qquad x_2=-\frac{1}{\sqrt{3}}.x1​=2−3​,x2​=−3​1​. Their sum is

Now 3+13=3+13=43.\sqrt{3}+\frac{1}{\sqrt{3}}=\frac{3+1}{\sqrt{3}}=\frac{4}{\sqrt{3}}.3​+3​1​=3​3+1​=3​4​. Therefore S=2−43.S=2-\frac{4}{\sqrt{3}}.S=2−3​4​.

Comparing with α−43,\alpha-\frac{4}{\sqrt{3}},α−3​4​, we get α=2.\alpha=2.α=2.

  1. Comparison with stored answer

Stored correct answer = 222.

Our derived answer is also 222, so they agree.

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