Solve this Ques of jee main exam February session 2021
Answers
EXPLANATION.
⇒ sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x².
As we know that,
Range of sinθ = [-1,1].
There are three digit exists between [-1,1].
⇒ - 1, 0, 1.
⇒ [x² + 1/3].
in this particular equation, we get.
Only two possibility exists = 0,1.
⇒ [x² - 2/3].
In this particular case, we get.
Three possibility exists = -1, 0, 1.
Total we can say that,
There are six [6] possibility exists.
⇒ [x² + 1/3] [x² - 2/3].
⇒ 0 -1.
⇒ 0 0.
⇒ 0 1.
⇒ 1 -1.
⇒ 1 0.
⇒ 1 1.
Now, we can equation as,
⇒ sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x².
Put the value (0, -1) in the equation, we get.
⇒ sin⁻¹[(0)² + 1/3] + cos⁻¹[(-1)² - 2/3] = x².
⇒ sin⁻¹(0) + cos⁻¹(1) = x².
⇒ 0 + π = x². [L.H.S.]
Now again put the value of x² = π in the equation, we get.
⇒ sin⁻¹[π + 1/3] + cos⁻¹[π - 2/3] = π.
⇒ sin⁻¹[3.14 + 0.33] + cos⁻¹[3.14 - 0.6667] = π. [R.H.S.]
As we can see that,
⇒ L.H.S ≠ R.H.S.
So 1st possibility is rejected.
Put the value (0,0) in the equation, we get.
⇒ sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x².
⇒ sin⁻¹[(0)² + 1/3] + cos⁻¹[(0)² - 2/3] = x².
⇒ sin⁻¹(0) + cos⁻¹[0] = x².
⇒ 0 + π/2 = x². [L.H.S.]
Put the value of π/2 in the equation, we get.
⇒ sin⁻¹[π/2 + 1/3] + cos⁻¹[π/2 - 2/3] = π/2.
⇒ sin⁻¹[1.57 + 0.33] + cos⁻¹[1.57 - 0.667] = 1.57. [R.H.S.]
As we can see that,
⇒ L.H.S. ≠ R.H.S.
So 2nd possibility is also rejected.
Put the value (0,-1) in the equation, we get.
⇒ sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x².
⇒ sin⁻¹[(0)² + 1/3] + cos⁻¹[(-1)² - 2/3] = x².
⇒ sin⁻¹(0) + cos⁻¹(1) = x².
⇒ 0 + 0 = x². [L.H.S.].
Put the value of x² = 0 in the equation, we get.
⇒ sin⁻¹[0 + 1/3] + cos⁻¹[0 - 2/3] = 0.
⇒ sin⁻¹(0) + cos⁻¹(-1) = 0. [R.H.S].
As we can see that,
⇒ L.H.S. ≠ R.H.S.
So 3rd possibility is also rejected.
Put the value (1, -1) in the equation, we get.
⇒ sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x².
⇒ sin⁻¹[(1)² + 1/3] + cos⁻¹[(-1)² - 2/3] = x².
⇒ sin⁻¹(1) + cos⁻¹(-1) = x².
⇒ π/2 + π = x².
⇒ 3π/2 = x². [L.H.S.].
Put the value of x² = 3π/2 in the equation, we get.
⇒ sin⁻¹[3π/2 + 1/3] + cos⁻¹[3π/2 - 2/3] = 3π/2.
⇒ sin⁻¹[9π + 2/6] + cos⁻¹[9π - 4/6] = 3π/2. [R.H.S.].
As we can see that,
⇒ L.H.S. ≠ R.H.S.
So, 4th possibility is also rejected.
Put the value (1,0) in the equation, we get.
⇒ sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x².
⇒ sin⁻¹[(1)² + 1/3] + cos⁻¹[(0)² - 2/3] = x².
⇒ sin⁻¹(1) + cos⁻¹(0) = x².
⇒ π/2 + π/2 = x².
⇒ x² = π. [L.H.S.].
Put the value of x² = π in the equation, we get.
⇒ sin⁻¹[π + 1/3] + cos⁻¹[π - 2/3] = π.
⇒ sin⁻¹[3.14 + 0.33] + cos⁻¹[3.14 - 0.667] = π. [R.H.S.].
As we can see that,
⇒ L.H.S. ≠ R.H.S.
So 5th possibility is also rejected.
Put the value (1,1) in the equation, we get.
⇒ sin⁻¹[x² + 1/3] + cos⁻¹[x² - 2/3] = x².
⇒ sin⁻¹[(1)² + 1/3] + cos⁻¹[(1)² - 2/3] = x².
⇒ sin⁻¹(1) + cos⁻¹(1) = x².
⇒ π/2 = x². [L.H.S.].
Put the value of x² = π/2 in the equation, we get.
⇒ sin⁻¹[π/2 + 1/3] + cos⁻¹[π/2 - 2/3] = π/2.
⇒ sin⁻¹[1.57 + 1.33] + cos⁻¹[1.57 - 0.667] = 1.57. [R.H.S.].
As we can see that,
⇒ L.H.S. ≠ R.H.S.
So 6th possibility is also rejected.
As we can observe all the possibility and see that no one possibility is matched.
Hence, Number of solutions = 0.
Option [A] is correct answer.