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Inverse Trigonometric Functions question

2023 · 31 Jan · Shift 2 · Q26
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  5. /2023 · 31 Jan · Shift 2 · Q26

Inverse Trigonometric Functions question

2023 · 31 Jan · Shift 2 · Q26

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
Let (a, b) ⊂(0,2π)\subset(0,2 \pi)⊂(0,2π) be the largest interval for which sin⁡−1(sin⁡θ)−cos⁡−1(sin⁡θ)>0,θ∈(0,2π)\sin ^{-1}(\sin \theta)-\cos ^{-1}(\sin \theta)\gt 0, \theta \in(0,2 \pi)sin−1(sinθ)−cos−1(sinθ)>0,θ∈(0,2π), holds. If αx2+βx+sin⁡−1(x2−6x+10)+cos⁡−1(x2−6x+10)=0\alpha x^{2}+\beta x+\sin ^{-1}\left(x^{2}-6 x+10\right)+\cos ^{-1}\left(x^{2}-6 x+10\right)=0αx2+βx+sin−1(x2−6x+10)+cos−1(x2−6x+10)=0 and α−β=b−a\alpha-\beta=b-aα−β=b−a, then α\alphaα is equal to :
  1. A
    π16\frac{\pi}{16}16π​
  2. B
    π48\frac{\pi}{48}48π​
  3. C
    π8\frac{\pi}{8}8π​
  4. D
    π12\frac{\pi}{12}12π​
View written solutionFree

Correct answer: D

  1. Simplify the inequality

We need the largest interval (a,b)⊂(0,2π)(a,b)\subset(0,2\pi)(a,b)⊂(0,2π) such that

sin⁡−1(sin⁡θ)−cos⁡−1(sin⁡θ)>0.\sin^{-1}(\sin\theta)-\cos^{-1}(\sin\theta)>0.sin−1(sinθ)−cos−1(sinθ)>0.

Use the identity

cos⁡−1x=π2−sin⁡−1x(x∈[−1,1]).\cos^{-1}x=\frac{\pi}{2}-\sin^{-1}x \qquad (x\in[-1,1]).cos−1x=2π​−sin−1x(x∈[−1,1]).

So the inequality becomes

sin⁡−1(sin⁡θ)−(π2−sin⁡−1(sin⁡θ))>0\sin^{-1}(\sin\theta)-\left(\frac{\pi}{2}-\sin^{-1}(\sin\theta)\right)>0sin−1(sinθ)−(2π​−sin−1(sinθ))>0 2sin⁡−1(sin⁡θ)−π2>02\sin^{-1}(\sin\theta)-\frac{\pi}{2}>02sin−1(sinθ)−2π​>0 sin⁡−1(sin⁡θ)>π4.\sin^{-1}(\sin\theta)>\frac{\pi}{4}.sin−1(sinθ)>4π​.
  1. Write sin⁡−1(sin⁡θ)\sin^{-1}(\sin\theta)sin−1(sinθ) piecewise on (0,2π)(0,2\pi)(0,2π)

Recall:

sin⁡−1(sin⁡θ)={θ,θ∈(0,π2),π−θ,θ∈(π2,3π2),θ−2π,θ∈(3π2,2π).\sin^{-1}(\sin\theta)= \begin{cases} \theta, & \theta\in\left(0,\frac{\pi}{2}\right),\\[4pt] \pi-\theta, & \theta\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right),\\[4pt] \theta-2\pi, & \theta\in\left(\frac{3\pi}{2},2\pi\right). \end{cases}sin−1(sinθ)=⎩⎨⎧​θ,π−θ,θ−2π,​θ∈(0,2π​),θ∈(2π​,23π​),θ∈(23π​,2π).​

Now solve

sin⁡−1(sin⁡θ)>π4.\sin^{-1}(\sin\theta)>\frac{\pi}{4}.sin−1(sinθ)>4π​.

  • For θ∈(0,π2)\theta\in\left(0,\frac{\pi}{2}\right)θ∈(0,2π​): θ>π4  ⟹  θ∈(π4,π2).\theta>\frac{\pi}{4} \implies \theta\in\left(\frac{\pi}{4},\frac{\pi}{2}\right).θ>4π​⟹θ∈(4π​,2π​).

  • For θ∈(π2,3π2)\theta\in\left(\frac{\pi}{2},\frac{3\pi}{2}\right)θ∈(2π​,23π​): π−θ>π4  ⟹  θ<3π4.\pi-\theta>\frac{\pi}{4} \implies \theta<\frac{3\pi}{4}.π−θ>4π​⟹θ<43π​. Hence θ∈(π2,3π4).\theta\in\left(\frac{\pi}{2},\frac{3\pi}{4}\right).θ∈(2π​,43π​).

  • For θ∈(3π2,2π)\theta\in\left(\frac{3\pi}{2},2\pi\right)θ∈(23π​,2π): θ−2π>π4\theta-2\pi>\frac{\pi}{4}θ−2π>4π​ is impossible since θ−2π<0\theta-2\pi<0θ−2π<0.

Combining,

θ∈(π4,3π4).\theta\in\left(\frac{\pi}{4},\frac{3\pi}{4}\right).θ∈(4π​,43π​).

Thus the largest interval is

(a,b)=(π4,3π4).(a,b)=\left(\frac{\pi}{4},\frac{3\pi}{4}\right).(a,b)=(4π​,43π​).

So

b−a=3π4−π4=π2.b-a=\frac{3\pi}{4}-\frac{\pi}{4}=\frac{\pi}{2}.b−a=43π​−4π​=2π​.
  1. Use the second equation

Given

αx2+βx+sin⁡−1(x2−6x+10)+cos⁡−1(x2−6x+10)=0.\alpha x^2+\beta x+\sin^{-1}(x^2-6x+10)+\cos^{-1}(x^2-6x+10)=0.αx2+βx+sin−1(x2−6x+10)+cos−1(x2−6x+10)=0.

Since for any t∈[−1,1]t\in[-1,1]t∈[−1,1],

sin⁡−1t+cos⁡−1t=π2,\sin^{-1}t+\cos^{-1}t=\frac{\pi}{2},sin−1t+cos−1t=2π​,

we get

αx2+βx+π2=0.\alpha x^2+\beta x+\frac{\pi}{2}=0.αx2+βx+2π​=0.

For this to hold identically in xxx, coefficients must be zero appropriately. Equivalently,

αx2+βx=−π2.\alpha x^2+\beta x=-\frac{\pi}{2}.αx2+βx=−2π​.

This can only happen as a constant polynomial if

α=0,β=0,\alpha=0,\quad \beta=0,α=0,β=0,

which is impossible because then π2=0\frac{\pi}{2}=02π​=0.

So the intended meaning is that the equation holds for all admissible xxx after comparing coefficients with the constant term, hence we interpret the condition together with

α−β=b−a=π2.\alpha-\beta=b-a=\frac{\pi}{2}.α−β=b−a=2π​.

Now, because

x2−6x+10=(x−3)2+1,x^2-6x+10=(x-3)^2+1,x2−6x+10=(x−3)2+1,

and for inverse trigonometric functions to be defined,

−1≤x2−6x+10≤1.-1\le x^2-6x+10\le 1.−1≤x2−6x+10≤1.

But (x−3)2+1≥1(x-3)^2+1\ge 1(x−3)2+1≥1, so this forces

x2−6x+10=1  ⟹  (x−3)2=0  ⟹  x=3.x^2-6x+10=1 \implies (x-3)^2=0 \implies x=3.x2−6x+10=1⟹(x−3)2=0⟹x=3.

Substitute x=3x=3x=3 into the equation:

9α+3β+sin⁡−1(1)+cos⁡−1(1)=0.9\alpha+3\beta+\sin^{-1}(1)+\cos^{-1}(1)=0.9α+3β+sin−1(1)+cos−1(1)=0.

Now

sin⁡−1(1)=π2,cos⁡−1(1)=0.\sin^{-1}(1)=\frac{\pi}{2},\qquad \cos^{-1}(1)=0.sin−1(1)=2π​,cos−1(1)=0.

Hence

9α+3β+π2=0.9\alpha+3\beta+\frac{\pi}{2}=0.9α+3β+2π​=0.

Divide by 333:

3α+β=−π6.(1)3\alpha+\beta=-\frac{\pi}{6}.\tag{1}3α+β=−6π​.(1)

Also

α−β=π2.(2)\alpha-\beta=\frac{\pi}{2}.\tag{2}α−β=2π​.(2)
  1. Solve for α\alphaα and β\betaβ

From (2),

β=α−π2.\beta=\alpha-\frac{\pi}{2}.β=α−2π​.

Substitute into (1):

3α+α−π2=−π63\alpha+\alpha-\frac{\pi}{2}=-\frac{\pi}{6}3α+α−2π​=−6π​ 4α=π2−π6=π34\alpha=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}4α=2π​−6π​=3π​ α=π12.\alpha=\frac{\pi}{12}.α=12π​.
  1. Check options
α=π12\alpha=\frac{\pi}{12}α=12π​

which is Option D.

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