The values of a and b such that limx→0 x(1+acos⁡x)−bsin⁡xx3=1, are

A
$\frac{5}{2}, \frac{3}{2}$
B
$\frac{5}{2}, - \frac{3}{2}$
C
$- \frac{5}{2}, - \frac{3}{2}$
D
none of these

Solution:

We have,

$\begin{array}{l} lim_{x \to 0} \frac{x (1 + a \cos x) - b \sin x}{x^{3}} = 1 \\ \Rightarrow lim_{x \to 0} \frac{x \{1 + a (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} - \frac{x^{6}}{6!} + \dots)\} - b \{x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} \dots\}}{x^{3}} = 1 \\ \Rightarrow lim_{x \to 0} \frac{(1 + a - b) + x^{2} (\frac{b}{3!} - \frac{a}{2!}) + x^{4} (\frac{a}{4!} - \frac{b}{5!}) + \dots}{x^{2}} = 1 \dots (i) \end{array}$

If $1 + a - b \neq 0,$ then $LHS \to \infty as x \to 0$ while $RHS =$ $1$

$∴ 1 + a - b = 0$

From (i), we have

$\begin{aligned} lim_{x \to 0} \frac{x^{2} (\frac{b}{3!} - \frac{a}{2!}) + x^{4} (\frac{a}{4!} - \frac{b}{5!}) + \dots}{x^{2}} = 1 \\ ∴ \frac{b}{3!} - \frac{a}{2!} = 1 \Rightarrow b - 3 a = 6 \end{aligned}$

Solving $1 + a - b = 0$ and $b - 3 a = 6$ , we get $a = - \frac{5}{2}, b = - \frac{3}{2}$

The values of $a$ and $b$ such that $lim_{x \to 0} \frac{x (1 + a \cos x) - b \sin x}{x^{3}} = 1,$ are

Solution:

Related content