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Inverse Trigonometric Functions question

2019 · 12 Jan · Shift 1 · Q43
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  5. /2019 · 12 Jan · Shift 1 · Q43

Inverse Trigonometric Functions question

2019 · 12 Jan · Shift 1 · Q43

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
Considering only the principal values of inverse functions, the set A = { x ≥\ge≥ 0: tan −-− 1(2x) + tan −-− 1(3x) = π4{\pi \over 4}4π​}
  1. A
    contains two elements
  2. B
    contains more than two elements
  3. C
    is an empty set
  4. D
    is a singleton
View written solutionFree

Correct answer: D

We need to solve

tan⁡−1(2x)+tan⁡−1(3x)=π4,x≥0\tan^{-1}(2x)+\tan^{-1}(3x)=\frac{\pi}{4}, \qquad x\ge 0tan−1(2x)+tan−1(3x)=4π​,x≥0

using only the principal values of inverse tangent.

1. Domain and principal value facts

For principal values,

tan⁡−1(t)∈(−π2,π2).\tan^{-1}(t)\in\left(-\frac{\pi}{2},\frac{\pi}{2}\right).tan−1(t)∈(−2π​,2π​).

Since x≥0x\ge 0x≥0, we have 2x≥02x\ge 02x≥0 and 3x≥03x\ge 03x≥0, so

tan⁡−1(2x)≥0,tan⁡−1(3x)≥0.\tan^{-1}(2x)\ge 0,\qquad \tan^{-1}(3x)\ge 0.tan−1(2x)≥0,tan−1(3x)≥0.

Hence their sum is well-defined and lies in (0,π)(0,\pi)(0,π), and here it is given to be π/4\pi/4π/4.

2. Use tangent of sum formula

Let

α=tan⁡−1(2x),β=tan⁡−1(3x).\alpha=\tan^{-1}(2x),\qquad \beta=\tan^{-1}(3x).α=tan−1(2x),β=tan−1(3x).

Then

α+β=π4.\alpha+\beta=\frac{\pi}{4}.α+β=4π​.

Taking tangent on both sides:

tan⁡(α+β)=tan⁡π4=1.\tan(\alpha+\beta)=\tan\frac{\pi}{4}=1.tan(α+β)=tan4π​=1.

Now,

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=2x+3x1−(2x)(3x).\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} =\frac{2x+3x}{1-(2x)(3x)}.tan(α+β)=1−tanαtanβtanα+tanβ​=1−(2x)(3x)2x+3x​.

So,

5x1−6x2=1.\frac{5x}{1-6x^2}=1.1−6x25x​=1.

Thus,

5x=1−6x25x=1-6x^25x=1−6x2 6x2+5x−1=0.6x^2+5x-1=0.6x2+5x−1=0.

3. Solve the quadratic

6x2+5x−1=0.6x^2+5x-1=0.6x2+5x−1=0.

Factorizing,

6x2+6x−x−1=06x^2+6x-x-1=06x2+6x−x−1=0 6x(x+1)−1(x+1)=06x(x+1)-1(x+1)=06x(x+1)−1(x+1)=0 (x+1)(6x−1)=0.(x+1)(6x-1)=0.(x+1)(6x−1)=0.

So,

x=−1orx=16.x=-1 \quad \text{or} \quad x=\frac16.x=−1orx=61​.

Given x≥0x\ge 0x≥0, only

x=16x=\frac16x=61​

is allowed.

4. Verify the solution

At x=16x=\frac16x=61​,

2x=13,3x=12.2x=\frac13,\qquad 3x=\frac12.2x=31​,3x=21​.

Then

tan⁡−1(13)+tan⁡−1(12).\tan^{-1}\left(\frac13\right)+\tan^{-1}\left(\frac12\right).tan−1(31​)+tan−1(21​).

Using tangent addition,

tan⁡(tan⁡−113+tan⁡−112)=13+121−13⋅12=561−16=5656=1.\tan\left(\tan^{-1}\frac13+\tan^{-1}\frac12\right) =\frac{\frac13+\frac12}{1-\frac13\cdot\frac12} =\frac{\frac56}{1-\frac16} =\frac{\frac56}{\frac56}=1.tan(tan−131​+tan−121​)=1−31​⋅21​31​+21​​=1−61​65​​=65​65​​=1.

Since both angles are positive and their sum lies in (0,π/2)(0,\pi/2)(0,π/2), the sum must be

π4.\frac{\pi}{4}.4π​.

So x=16x=\frac16x=61​ is indeed a valid solution.

5. Number of elements in the set

Thus,

A={16}.A=\left\{\frac16\right\}.A={61​}.

So the set has exactly one element, i.e. it is a singleton.

6. Option check

  • A: contains two elements — False
  • B: contains more than two elements — False
  • C: is an empty set — False
  • D: is a singleton — True

Therefore, the correct option is

D\boxed{D}D​
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