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Trigonometric Ratio and Identites question

2024 · 31 Jan · Shift 2 · Q37
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  5. /2024 · 31 Jan · Shift 2 · Q37

Trigonometric Ratio and Identites question

2024 · 31 Jan · Shift 2 · Q37

JEE MainMathematicsTrigonometric Ratio and IdentitesMCQ+4 / −1
The number of solutions, of the equation esin⁡x−2e−sin⁡x=2e^{\sin x}-2 e^{-\sin x}=2esinx−2e−sinx=2, is :
  1. A
    0
  2. B
    1
  3. C
    2
  4. D
    more than 2
View written solutionFree

Correct answer: A

  1. Let y=sin⁡x.y=\sin x.y=sinx. Since sin⁡x∈[−1,1]\sin x\in[-1,1]sinx∈[−1,1], we must solve ey−2e−y=2.e^{y}-2e^{-y}=2.ey−2e−y=2.

  2. Multiply both sides by eye^yey: e2y−2=2ey.e^{2y}-2=2e^y.e2y−2=2ey. Rearrange: e2y−2ey−2=0.e^{2y}-2e^y-2=0.e2y−2ey−2=0.

  3. Put t=ey.t=e^y.t=ey. Since ey>0e^y>0ey>0, we get the quadratic t2−2t−2=0.t^2-2t-2=0.t2−2t−2=0.

  4. Solve the quadratic: t=2±4+82=1±3.t=\frac{2\pm\sqrt{4+8}}{2}=1\pm\sqrt{3}.t=22±4+8​​=1±3​. Now, t>0t>0t>0, so t=1+3t=1+\sqrt{3}t=1+3​ only, because 1−3<01-\sqrt{3}<01−3​<0 is not possible for eye^yey.

  5. Hence ey=1+3⇒y=ln⁡(1+3).e^y=1+\sqrt{3}\quad\Rightarrow\quad y=\ln(1+\sqrt{3}).ey=1+3​⇒y=ln(1+3​). So sin⁡x=ln⁡(1+3).\sin x=\ln(1+\sqrt{3}).sinx=ln(1+3​).

  6. Check whether this is possible. Since 1+3≈2.732,1+\sqrt{3}\approx 2.732,1+3​≈2.732, we have ln⁡(1+3)≈ln⁡(2.732)≈1.005.\ln(1+\sqrt{3})\approx \ln(2.732)\approx 1.005.ln(1+3​)≈ln(2.732)≈1.005. But for real xxx, −1≤sin⁡x≤1.-1\le \sin x\le 1.−1≤sinx≤1. Since ln⁡(1+3)>1,\ln(1+\sqrt{3})>1,ln(1+3​)>1, this value cannot equal sin⁡x\sin xsinx.

Therefore, there is no real solution.

  1. Hence the number of solutions is 0.0.0. So the correct option is A.
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