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Inverse Trigonometric Functions question

2023 · 1 Feb · Shift 2 · Q37
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Inverse Trigonometric Functions question

2023 · 1 Feb · Shift 2 · Q37

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
Let S={x∈R:0<x<1 and 2tan⁡−1(1−x1+x)=cos⁡−1(1−x21+x2)}S = \left\{ x \in R: 0 \lt x \lt 1 \text{ and } 2\tan^{-1}\left(\frac{1-x}{1+x}\right) = \cos^{-1}\left(\frac{1-x^2}{1+x^2}\right) \right\}S={x∈R:0<x<1 and 2tan−1(1+x1−x​)=cos−1(1+x21−x2​)}. If n(S)\mathrm{n(S)}n(S) denotes the number of elements in S\mathrm{S}S then :
  1. A
    n(S)=0\mathrm{n}(\mathrm{S})=0n(S)=0
  2. B
    n(S)=1\mathrm{n}(\mathrm{S})=1n(S)=1 and only one element in S\mathrm{S}S is less than 12\frac{1}{2}21​.
  3. C
    n(S)=1\mathrm{n}(\mathrm{S})=1n(S)=1 and the elements in S\mathrm{S}S is more than 12\frac{1}{2}21​.
  4. D
    n(S)=1\mathrm{n}(\mathrm{S})=1n(S)=1 and the element in S\mathrm{S}S is less than 12\frac{1}{2}21​.
View written solutionFree

Correct answer: D

  1. We need to solve 2tan⁡−1(1−x1+x)=cos⁡−1(1−x21+x2),0<x<1.2\tan^{-1}\left(\frac{1-x}{1+x}\right)=\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right),\qquad 0<x<1.2tan−1(1+x1−x​)=cos−1(1+x21−x2​),0<x<1.

We must find how many such xxx exist, and whether that value is less than or greater than 12\frac1221​.

  1. Simplify the left-hand side using the identity tan⁡(A−B2)=sin⁡A−sin⁡Bcos⁡A+cos⁡B,\tan\left(\frac{A-B}{2}\right)=\frac{\sin A-\sin B}{\cos A+\cos B},tan(2A−B​)=cosA+cosBsinA−sinB​, but here the standard substitution is more direct: tan⁡(π4−tan⁡−1x)=1−x1+x.\tan\left(\frac{\pi}{4}-\tan^{-1}x\right)=\frac{1-x}{1+x}.tan(4π​−tan−1x)=1+x1−x​. Hence, tan⁡−1(1−x1+x)=π4−tan⁡−1x,\tan^{-1}\left(\frac{1-x}{1+x}\right)=\frac{\pi}{4}-\tan^{-1}x,tan−1(1+x1−x​)=4π​−tan−1x, since 0<x<10<x<10<x<1 implies 1−x1+x>0\frac{1-x}{1+x}>01+x1−x​>0 and both sides lie in the principal range.

Therefore, 2tan⁡−1(1−x1+x)=2(π4−tan⁡−1x)=π2−2tan⁡−1x.2\tan^{-1}\left(\frac{1-x}{1+x}\right)=2\left(\frac{\pi}{4}-\tan^{-1}x\right)=\frac{\pi}{2}-2\tan^{-1}x.2tan−1(1+x1−x​)=2(4π​−tan−1x)=2π​−2tan−1x.

  1. Simplify the right-hand side.

Recall the identity cos⁡(2θ)=1−tan⁡2θ1+tan⁡2θ.\cos(2\theta)=\frac{1-\tan^2\theta}{1+\tan^2\theta}.cos(2θ)=1+tan2θ1−tan2θ​. If we put θ=tan⁡−1x\theta=\tan^{-1}xθ=tan−1x, then cos⁡(2θ)=1−x21+x2.\cos(2\theta)=\frac{1-x^2}{1+x^2}.cos(2θ)=1+x21−x2​. So, cos⁡−1(1−x21+x2)=cos⁡−1(cos⁡2θ).\cos^{-1}\left(\frac{1-x^2}{1+x^2}\right)=\cos^{-1}(\cos 2\theta).cos−1(1+x21−x2​)=cos−1(cos2θ).

Now since 0<x<10<x<10<x<1, we have

\implies 0<2\theta<\frac{\pi}{2}.$$ Because $2\theta\in[0,\pi]$, the principal value gives $$\cos^{-1}(\cos 2\theta)=2\theta=2\tan^{-1}x.$$ Thus the equation becomes $$\frac{\pi}{2}-2\tan^{-1}x=2\tan^{-1}x.$$ --- 4. Solve for $x$: $$\frac{\pi}{2}=4\tan^{-1}x$$ $$\tan^{-1}x=\frac{\pi}{8}$$ $$x=\tan\frac{\pi}{8}.$$ So there is exactly one solution in $(0,1)$. --- 5. Check whether this solution is less than $\frac12$. Use the standard value: $$\tan\frac{\pi}{8}=\sqrt{2}-1.$$ Since $$\sqrt{2}-1\approx 1.414-1=0.414<\frac12,$$ the unique element of $S$ is less than $\frac12$. --- 6. Therefore, $$n(S)=1,$$ and the only element of $S$ is less than $\frac12$. So the correct option is: $$\boxed{\text{D}}.$$
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