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Inverse Trigonometric Functions question

2023 · Shift 2 · Q20
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  5. /2023 · Shift 2 · Q20

Inverse Trigonometric Functions question

2023 · Shift 2 · Q20

JEE AdvancedMathematicsInverse Trigonometric FunctionsMCQ+3 / −1
For any y∈Ry \in \mathbb{R}y∈R, let cot⁡−1(y)∈(0,π)\cot ^{-1}(y) \in(0, \pi)cot−1(y)∈(0,π) and tan⁡−1(y)∈(−π2,π2)\tan ^{-1}(y) \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)tan−1(y)∈(−2π​,2π​). Then the sum of all the solutions of the equation tan⁡−1(6y9−y2)+cot⁡−1(9−y26y)=2π3\tan ^{-1}\left(\frac{6 y}{9-y^2}\right)+\cot ^{-1}\left(\frac{9-y^2}{6 y}\right)=\frac{2 \pi}{3}tan−1(9−y26y​)+cot−1(6y9−y2​)=32π​ for 0<∣y∣<30\lt |y|\lt 30<∣y∣<3, is equal to :
  1. A
    23−32 \sqrt{3}-323​−3
  2. B
    3−233-2 \sqrt{3}3−23​
  3. C
    43−64 \sqrt{3}-643​−6
  4. D
    6−436-4 \sqrt{3}6−43​
View written solutionFree

Correct answer: C

  1. Given equation

We need to solve, for 0<∣y∣<30<|y|<30<∣y∣<3, an−1(6y9−y2)+cot⁡−1(9−y26y)=2π3. an^{-1}\left(\frac{6y}{9-y^2}\right)+\cot^{-1}\left(\frac{9-y^2}{6y}\right)=\frac{2\pi}{3}.an−1(9−y26y​)+cot−1(6y9−y2​)=32π​.

The principal value ranges are: tan⁡−1(t)∈(−π2,π2),cot⁡−1(t)∈(0,π).\tan^{-1}(t)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right), \qquad \cot^{-1}(t)\in(0,\pi).tan−1(t)∈(−2π​,2π​),cot−1(t)∈(0,π).

  1. Introduce a variable

Let u=6y9−y2.u=\frac{6y}{9-y^2}.u=9−y26y​. Then 9−y26y=1u.\frac{9-y^2}{6y}=\frac{1}{u}.6y9−y2​=u1​. So the equation becomes tan⁡−1(u)+cot⁡−1(1u)=2π3.\tan^{-1}(u)+\cot^{-1}\left(\frac{1}{u}\right)=\frac{2\pi}{3}.tan−1(u)+cot−1(u1​)=32π​.

  1. Use the relation between cot⁡−1(1/u)\cot^{-1}(1/u)cot−1(1/u) and tan⁡−1(u)\tan^{-1}(u)tan−1(u)

Let α=tan⁡−1(u).\alpha=\tan^{-1}(u).α=tan−1(u). Then α∈(−π2,π2)\alpha\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right)α∈(−2π​,2π​) and tan⁡α=u\tan\alpha=utanα=u.

Now cot⁡−1(1/u)\cot^{-1}(1/u)cot−1(1/u) is the angle β∈(0,π)\beta\in(0,\pi)β∈(0,π) such that cot⁡β=1u  ⟹  tan⁡β=u.\cot\beta=\frac{1}{u} \implies \tan\beta=u.cotβ=u1​⟹tanβ=u. Because of the principal range of cot⁡−1\cot^{-1}cot−1:

  • if u>0u>0u>0, then β=α\beta=\alphaβ=α;
  • if u<0u<0u<0, then β=π+α\beta=\pi+\alphaβ=π+α (since α<0\alpha<0α<0 and π+α∈(π2,π)\pi+\alpha\in(\frac\pi2,\pi)π+α∈(2π​,π)).

So we consider cases.

  1. Case 1: u>0u>0u>0

Then tan⁡−1(u)+cot⁡−1(1/u)=α+α=2α.\tan^{-1}(u)+\cot^{-1}(1/u)=\alpha+\alpha=2\alpha.tan−1(u)+cot−1(1/u)=α+α=2α. Hence 2α=2π3  ⟹  α=π3.2\alpha=\frac{2\pi}{3} \implies \alpha=\frac{\pi}{3}.2α=32π​⟹α=3π​. Therefore u=tan⁡π3=3.u=\tan\frac{\pi}{3}=\sqrt{3}.u=tan3π​=3​. So 6y9−y2=3.\frac{6y}{9-y^2}=\sqrt{3}.9−y26y​=3​.

Solve: 6y=3(9−y2)6y=\sqrt{3}(9-y^2)6y=3​(9−y2) 3y2+6y−93=0.\sqrt{3}y^2+6y-9\sqrt{3}=0.3​y2+6y−93​=0. Divide by 3\sqrt{3}3​: y2+23 y−9=0.y^2+2\sqrt{3}\,y-9=0.y2+23​y−9=0. Thus y=−23±12+362=−23±432.y=\frac{-2\sqrt{3}\pm\sqrt{12+36}}{2}=\frac{-2\sqrt{3}\pm4\sqrt{3}}{2}.y=2−23​±12+36​​=2−23​±43​​. So

\qquad y=-3\sqrt{3}.$$ But $0<|y|<3$, so only $$y=\sqrt{3}$$ is allowed. --- 5. **Case 2: $u<0$** Then $$\tan^{-1}(u)+\cot^{-1}(1/u)=\alpha+(\pi+\alpha)=\pi+2\alpha.$$ Hence $$\pi+2\alpha=\frac{2\pi}{3} \implies 2\alpha=-\frac{\pi}{3} \implies \alpha=-\frac{\pi}{6}.$$ Therefore $$u=\tan\left(-\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}}.$$ So $$\frac{6y}{9-y^2}=-\frac{1}{\sqrt{3}}.$$ Solve: $$6\sqrt{3}y=-(9-y^2)$$ $$y^2-6\sqrt{3}\,y-9=0.$$ Thus $$y=\frac{6\sqrt{3}\pm\sqrt{108+36}}{2}=\frac{6\sqrt{3}\pm12}{2}=3\sqrt{3}\pm6.$$ So $$y=3\sqrt{3}+6, \qquad y=3\sqrt{3}-6.$$ Now check $0<|y|<3$: - $3\sqrt{3}+6>3$, rejected. - $3\sqrt{3}-6\approx -0.804$, accepted. So the second valid solution is $$y=3\sqrt{3}-6.$$ --- 6. **Sum of all valid solutions** The solutions are $$y=\sqrt{3}, \qquad y=3\sqrt{3}-6.$$ Their sum is $$\sqrt{3}+(3\sqrt{3}-6)=4\sqrt{3}-6.$$ --- 7. **Compare with options** $$4\sqrt{3}-6$$ which is **Option C**. --- 8. **Verification with stored answer** Stored correct answer: **C**. Our derived answer also gives **C**. Hence they agree.
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