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JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Inverse Trigonometric Functions

Let

x = sin (2 tan^{- 1} α)

and

y = sin (\frac{1}{2} tan^{- 1} \frac{4}{3})

. If

S = {α \in R : y^{2} = 1 - x}

, then

\sum_{α \in S} 16 α^{3}

is equal to

Answer: 130

Solution

First, compute

y

. Let

θ = tan^{- 1} \frac{4}{3}

, so

tan θ = \frac{4}{3}

. Then

sin θ = \frac{4}{5}

,

cos θ = \frac{3}{5}

.

y = sin \frac{θ}{2} = \frac{1 - c o s θ}{2} = \frac{1 - 3/5}{2} = \frac{1}{5} = \frac{1}{5}

.

So

y^{2} = \frac{1}{5}

.

Next,

x = sin (2 tan^{- 1} α)

. Let

ϕ = tan^{- 1} α

, so

tan ϕ = α

. Then:

x = sin 2 ϕ = \frac{2 t a n ϕ}{1 + t a n ^{2} ϕ} = \frac{2 α}{1 + α ^{2}}

.

Condition:

y^{2} = 1 - x

, so

\frac{1}{5} = 1 - \frac{2 α}{1 + α ^{2}}

.

\frac{2 α}{1 + α ^{2}} = \frac{4}{5}

, giving

10 α = 4 (1 + α^{2})

, i.e.,

4 α^{2} - 10 α + 4 = 0

, or

2 α^{2} - 5 α + 2 = 0

.

Solutions:

α = \frac{5 \pm 25 - 16}{4} = \frac{5 \pm 3}{4}

, so

α = 2

or

α = \frac{1}{2}

.

\sum_{α \in S} 16 α^{3} = 16 (2^{3} + (\frac{1}{2})^{3}) = 16 (8 + \frac{1}{8}) = 16 \cdot \frac{65}{8} = 130

.

JEE Main · 2022 · 25 Jul 2022 · Shift 2 · Integer answer

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