is
- A.6
- B.2
- C.3
- D.1
Show answer and solution
Correct option: B
Solution
At x→0, both numerator and denominator→0, so rationalize the denominator by multiplying by the conjugate.
Numerator after rationalization: (x + 2sinx)(√(x²+2sinx+1) + √(sin²x−x+1))
Denominator after rationalization: (x²+2sinx+1) − (sin²x−x+1) = x² + 2sinx − sin²x + x
Factor denominator: x² + x + 2sinx − sin²x. As x→0, use sinx ≈ x, so this ≈ x² + x + 2x − x² = 3x.
Numerator: (x + 2sinx) ≈ (x + 2x) = 3x, and the square root factor → (1 + 1) = 2.
So the limit = (3x · 2)/(3x) = 2.
Numerator after rationalization: (x + 2sinx)(√(x²+2sinx+1) + √(sin²x−x+1))
Denominator after rationalization: (x²+2sinx+1) − (sin²x−x+1) = x² + 2sinx − sin²x + x
Factor denominator: x² + x + 2sinx − sin²x. As x→0, use sinx ≈ x, so this ≈ x² + x + 2x − x² = 3x.
Numerator: (x + 2sinx) ≈ (x + 2x) = 3x, and the square root factor → (1 + 1) = 2.
So the limit = (3x · 2)/(3x) = 2.
[JEE Main 2019 · 12 Apr · Shift II]