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JEE Advanced2021MathematicsQuadratic EquationActual

For x ∈ R , the number of real roots of the equation 3 x 2 - 4 x 2 - 1 + x - 1 = 0 is

Correct answer

4

Step-by-step solution

x 2 - 1 = x 2 - 1 ,                     x 2 ≥ 1 - x 2 - 1           x 2 < 1 Case 1 x 2 - 1 < 0 ⇒ x ∈ ( - 1 , 1 ) 3 x 2 + 4 x 2 - 1 + x - 1 = 0 7 x 2 + x - 5 = 0 x = - 1 ± 1 + 140 2 × 7 = - 1 + 141 14 , - 1 - 141 14 We know, 141 ≈ 11 . 5 So, both roots lie in the interval - 1 , 1 Case 2 x 2 - 1 ≥ 0 ⇒ x ∈ ( - ∞ , - 1 ] ∪ [ 1 , ∞ ) 3 x 2 - 4 x 2 - 1 + x - 1 = 0 - x 2

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