For any y ∈R, let cot^-1 (y ) ∈ (0 , ) and tan^-1 (y ) ∈(- 2 , 2). Then, the sum of all solutions of the equation tan^-1 6y9-y^2+ cot^-1 9-y^26y= 2 3 for 0

Question

For any y ∈R, let cot^-1 (y ) ∈ (0 , ) and tan^-1 (y ) ∈(- 2 , 2). Then, the sum of all solutions of the equation tan^-1 6y9-y^2+ cot^-1 9-y^26y= 2 3 for 0< |y |<3, is equal to [2023 Adv.]

Accepted Answer

Correct answer: C. (c) Given, tan^-1 ( 6y9-y^2 )+ cot^-1 ( 9-y^26y )= 2 3 … . . (i ) and 0< |y |<3 ⇒-3<y<0 or 0<y<3 So, there is two cases arise to solve Eq. (i). Case I When y ∈ (-3 ,0 ), i.e. -3<y<0 ⇒ tan^-1 ( 6y9-y^2 )+( + tan^-1 ( 6y9-y^2 ))= 2 3 ⇒2 tan^-1 ( 6y9-y^2 )=- 3 ⇒ tan^-1 ( 6y9-y^2 )=- 6 ⇒ 6y9-y^2=- 1 3 ⇒6 3y=-9+y^2 ⇒y^2-6 3y-9=0 ⇒y=3 3 6 but y ∈ (-3 ,0 ) i.e. y is negative. So ,y=3 3-6 … . . ( ii ) Case II When y ∈ (0 ,3 ), i.e. 0<y<3 ⇒ tan^-1 ( 6y9-y^2 )+ tan^-1 ( 6y9-y^2 )= 2 3 ⇒2 tan^-1 ( 6y9-y^2 )= 2 3 ⇒ 6y9-y^2= tan ( 3 )= 3 ⇒6y=9 3- 3y^2 ⇒ 3y^2+6y-9 3=0 ⇒y= 3 ,-3 3 buty ∈ (0 ,3 ) i.e. y is positive. So, y= 3 … . ( iii ) Hence, sum of Eqs. (ii) and (iii) gives us ( 3 )+ (3 3-6 )=4 3-6

$For any y \in R, let cot^{- 1} (y) \in (0, π) and tan^{- 1} (y) \in (- \frac{π}{2}, \frac{π}{2}) . Then, the sum of all solutions of the equation tan^{- 1} \frac{6 y}{9 - y ^{2}} + cot^{- 1} \frac{9 - y ^{2}}{6 y} = \frac{2 π}{3} for 0 < | y | < 3, is equal to [2023 Adv.]$

Solution

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