lim x → 0 cosec x(√(2 cos ² x+3 cos x)-√(cos ² x+sin x+4)) is:

Solution

Maths sir · 4 steps

Both roots tend to √5, so rationalise, then only the -sin x term survives against sin x.

1. Spot the form. At x = 0 both roots equal √5, so the bracket tends to 0 while cosec x blows up. Written as a fraction over sin x, this is the 0/0 form.
2. Rationalise. Multiply and divide by the sum of the two roots. The numerator becomes (2 cos² x + 3 cos x) - (cos² x + sin x + 4) = cos² x + 3 cos x - 4 - sin x.

   L = lim(x → 0) (cos² x + 3 cos x - 4 - sin x)/(sin x) · 1/(√(2 cos² x + 3 cos x) + √(cos² x + sin x + 4))

3. Split the numerator. Factorise cos² x + 3 cos x - 4 = (cos x + 4)(cos x - 1). Since 1 - cos x behaves like x²/2, the ratio (cos x - 1)/(sin x) → 0, while (-sin x)/(sin x) = -1. So the first fraction tends to 5 × 0 - 1 = -1.
4. So. The sum of roots tends to √5 + √5 = 2√5, giving L = -1/2√5. The answer is (d).

The idea to keep. When two square roots tend to the same value, rationalise first. Then compare each numerator term with the denominator: 1 - cos x is of order x² and dies, while sin x is of order x and survives.

Answer from the NTA final answer key. Solution by PrepWiser.