Let a, b ∈ R, b ∈ 0, Define a function
f(x) = { a sin π/2(x - 1),, for x ≤ 0; (tan 2x - sin 2x)/bx³,, for x > 0 .
If f is continuous at x = 0, then 10 - ab is equal to ________________.

Solution

Maths sir · 4 steps

Get the value from the left piece and the limit from the right piece, then set them equal.

1. Value at zero. For x ≤ 0 the sine piece applies, and it is continuous, so the left-hand limit equals f(0). This gives f(0) = a sin(-π/2) = -a.
2. Factor the numerator. Write tan 2x - sin 2x = tan 2x (1 - cos 2x). Near 0, tan 2x ≈ 2x and 1 - cos 2x ≈ (2x)²/2 = 2x², so the numerator behaves like 4x³.

   lim(x → 0⁺) (tan 2x - sin 2x)/(b x³) = lim(x → 0⁺) (tan 2x)/2x · (1 - cos 2x)/4x² · 8/b = 1 × 1/2 × 8/b = 4/b

3. Match both sides. Continuity at 0 needs -a = 4/b, so ab = -4.
4. So. 10 - ab = 10 - (-4) = 14.

The common slip: 6. Dropping the sign of sin(-π/2) = -1 gives ab = 4 and an answer of 6, not 14.

The idea to keep. For tan t - sin t, factor it as tan t (1 - cos t). Near zero it behaves like t³/2, and that settles every limit of this kind.

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