If _ x → 0 x(1 + acos x) - bsin x x ^ 3 = 1 , then the value of ab...

Using the expansion, we get,
$\lim_{x \to 0}\dfrac{x + ax\left(1 - \dfrac{{x}^{2}}{2 !} + \dfrac{{x}^{4}}{4 !} - . . . . . .\right) - b\left(x - \dfrac{{x}^{3}}{3 !} + . . . . . .\right)}{{x}^{3}} = 1$
$\lim_{x \to 0}\dfrac{\left(1 + a - b\right)x + \left(\dfrac{- a}{2} + \dfrac{b}{6}\right){x}^{3} + . . . . .}{{x}^{3}} = 1$
For limit to exist numerator & denominator must be of the same degree
$∴ 1 + a - b = 0$ … \left(1\right)$<br />Also,$\dfrac{- a}{2} + \dfrac{b}{6} = 1 ⇒ 3a - b + 6 = 0&hellip; \left(2\right)
By equations \left(1\right)& \left(2\right), we get,
$a = - \dfrac{5}{2}$ and $b = - \dfrac{3}{2}$

$⇒ ab = \dfrac{15}{4} = 3 .75$