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Inverse Trigonometric Functions question

2021 · 16 Mar · Shift 2 · Q36
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Inverse Trigonometric Functions question

2021 · 16 Mar · Shift 2 · Q36

JEE MainMathematicsInverse Trigonometric FunctionsMCQ+4 / −1
Given that the inverse trigonometric functions take principal values only. Then, the number of real values of x which satisfy sin⁡−1(3x5)+sin⁡−1(4x5)=sin⁡−1x{\sin ^{ - 1}}\left( {{{3x} \over 5}} \right) + {\sin ^{ - 1}}\left( {{{4x} \over 5}} \right) = {\sin ^{ - 1}}xsin−1(53x​)+sin−1(54x​)=sin−1x is equal to :
  1. A
    2
  2. B
    0
  3. C
    3
  4. D
    1
View written solutionFree

Correct answer: C

  1. Domain of the equation

We need all inverse sine terms to be defined:

∣3x5∣≤1,∣4x5∣≤1,∣x∣≤1.\left|\frac{3x}{5}\right| \le 1, \quad \left|\frac{4x}{5}\right| \le 1, \quad |x|\le 1.​53x​​≤1,​54x​​≤1,∣x∣≤1.

The strongest condition is

∣x∣≤1.|x|\le 1.∣x∣≤1.

So we solve the equation for x∈[−1,1]x\in[-1,1]x∈[−1,1]:

sin⁡−1(3x5)+sin⁡−1(4x5)=sin⁡−1(x).\sin^{-1}\left(\frac{3x}{5}\right)+\sin^{-1}\left(\frac{4x}{5}\right)=\sin^{-1}(x).sin−1(53x​)+sin−1(54x​)=sin−1(x).

  1. Introduce angles

Let

A=sin⁡−1(3x5),B=sin⁡−1(4x5).A=\sin^{-1}\left(\frac{3x}{5}\right), \qquad B=\sin^{-1}\left(\frac{4x}{5}\right).A=sin−1(53x​),B=sin−1(54x​).

Then the equation becomes

A+B=sin⁡−1(x).A+B=\sin^{-1}(x).A+B=sin−1(x).

Since inverse sine gives principal values, we have

A,B,sin⁡−1(x)∈[−π2,π2].A,B,\sin^{-1}(x)\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right].A,B,sin−1(x)∈[−2π​,2π​].

Taking sine on both sides is valid if we also check principal-value consistency later.

  1. Use the sine addition formula

From A+B=sin⁡−1(x)A+B=\sin^{-1}(x)A+B=sin−1(x),

sin⁡(A+B)=x.\sin(A+B)=x.sin(A+B)=x.

Now,

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B.\sin(A+B)=\sin A\cos B+\cos A\sin B.sin(A+B)=sinAcosB+cosAsinB.

We know

sin⁡A=3x5,sin⁡B=4x5.\sin A=\frac{3x}{5}, \qquad \sin B=\frac{4x}{5}.sinA=53x​,sinB=54x​.

Because A,B∈[−π2,π2]A,B\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right]A,B∈[−2π​,2π​], their cosines are nonnegative:

cos⁡A=1−9x225=25−9x25,\cos A=\sqrt{1-\frac{9x^2}{25}}=\frac{\sqrt{25-9x^2}}{5},cosA=1−259x2​​=525−9x2​​, cos⁡B=1−16x225=25−16x25.\cos B=\sqrt{1-\frac{16x^2}{25}}=\frac{\sqrt{25-16x^2}}{5}.cosB=1−2516x2​​=525−16x2​​.

Hence

x=3x5⋅25−16x25+4x5⋅25−9x25.x=\frac{3x}{5}\cdot \frac{\sqrt{25-16x^2}}{5}+\frac{4x}{5}\cdot \frac{\sqrt{25-9x^2}}{5}.x=53x​⋅525−16x2​​+54x​⋅525−9x2​​.

So,

x=x25(325−16x2+425−9x2).x=\frac{x}{25}\left(3\sqrt{25-16x^2}+4\sqrt{25-9x^2}\right).x=25x​(325−16x2​+425−9x2​).

Thus

x[25−325−16x2−425−9x2]=0.x\left[25-3\sqrt{25-16x^2}-4\sqrt{25-9x^2}\right]=0.x[25−325−16x2​−425−9x2​]=0.

Therefore either

Case 1: x=0x=0x=0

This is clearly a solution.

  1. Solve the nonzero case

For x≠0x\neq 0x=0,

25=325−16x2+425−9x2.25=3\sqrt{25-16x^2}+4\sqrt{25-9x^2}.25=325−16x2​+425−9x2​.

Let

u=25−16x2,v=25−9x2.u=\sqrt{25-16x^2}, \qquad v=\sqrt{25-9x^2}.u=25−16x2​,v=25−9x2​.

Then

3u+4v=25.3u+4v=25.3u+4v=25.

Notice that if ∣x∣≤1|x|\le 1∣x∣≤1, then

25−16x2≥9  ⟹  u≥3,25-16x^2\ge 9 \implies u\ge 3,25−16x2≥9⟹u≥3, 25−9x2≥16  ⟹  v≥4.25-9x^2\ge 16 \implies v\ge 4.25−9x2≥16⟹v≥4.

Therefore

3u+4v≥3⋅3+4⋅4=25.3u+4v \ge 3\cdot 3+4\cdot 4=25.3u+4v≥3⋅3+4⋅4=25.

Equality holds only when

u=3andv=4.u=3 \quad \text{and} \quad v=4.u=3andv=4.

So,

25−16x2=3  ⟹  25−16x2=9  ⟹  16x2=16  ⟹  x2=1,\sqrt{25-16x^2}=3 \implies 25-16x^2=9 \implies 16x^2=16 \implies x^2=1,25−16x2​=3⟹25−16x2=9⟹16x2=16⟹x2=1,

and

25−9x2=4  ⟹  25−9x2=16  ⟹  9x2=9  ⟹  x2=1.\sqrt{25-9x^2}=4 \implies 25-9x^2=16 \implies 9x^2=9 \implies x^2=1.25−9x2​=4⟹25−9x2=16⟹9x2=9⟹x2=1.

Thus the nonzero solutions are

x=1, −1.x=1,\,-1.x=1,−1.

  1. Verify all candidates in the original equation

We have candidates:

x=0, 1, −1.x=0,\ 1,\ -1.x=0, 1, −1.

  • For x=0x=0x=0: sin⁡−1(0)+sin⁡−1(0)=0=sin⁡−1(0).\sin^{-1}(0)+\sin^{-1}(0)=0=\sin^{-1}(0).sin−1(0)+sin−1(0)=0=sin−1(0). True.

  • For x=1x=1x=1: sin⁡−1(35)+sin⁡−1(45).\sin^{-1}\left(\frac35\right)+\sin^{-1}\left(\frac45\right).sin−1(53​)+sin−1(54​). Since these are acute angles of a 333-444-555 triangle, let sin⁡A=35,sin⁡B=45,\sin A=\frac35,\quad \sin B=\frac45,sinA=53​,sinB=54​, then cos⁡A=45,cos⁡B=35,\cos A=\frac45,\quad \cos B=\frac35,cosA=54​,cosB=53​, so sin⁡(A+B)=35⋅35+45⋅45=1.\sin(A+B)=\frac35\cdot\frac35+\frac45\cdot\frac45=1.sin(A+B)=53​⋅53​+54​⋅54​=1. Since A,B∈(0,π2)A,B\in\left(0,\frac\pi2\right)A,B∈(0,2π​), we get A+B=π2=sin⁡−1(1).A+B=\frac\pi2=\sin^{-1}(1).A+B=2π​=sin−1(1). True.

  • For x=−1x=-1x=−1: sin⁡−1(−35)+sin⁡−1(−45)=−sin⁡−1(35)−sin⁡−1(45)=−π2=sin⁡−1(−1).\sin^{-1}\left(-\frac35\right)+\sin^{-1}\left(-\frac45\right)=-\sin^{-1}\left(\frac35\right)-\sin^{-1}\left(\frac45\right)=-\frac\pi2=\sin^{-1}(-1).sin−1(−53​)+sin−1(−54​)=−sin−1(53​)−sin−1(54​)=−2π​=sin−1(−1). True.

So all three work.

  1. Number of real solutions

3\boxed{3}3​

Hence the correct option is C.

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