f(x) is the function such that limx→0f(x)x=1, if limx→0x(1+acosx)−bsinx(f(x))3=56, then

Given,limx→0 f(x)x=1∴limx→0 x(1+acos⁡x)−bsin⁡xx3⋅f(x)x3=56limx→0 (1+acos⁡x)−bxsin⁡xx2=56⇒limx→0 1+a1−x22!+x44!+……−bxx−x33!+x55!−……..x2=56 For finite limit, 1+a−b=0⇒b=a+1 And −a2!+b3!=56⇒−a2+b6=56⇒−3a+b=5⇒−3a+a+1=5⇒ −2a=4⇒a=−2 From Eq. (i), b=−2+1=−1

f(x) is the function such that limx→0f(x)x=1, if limx→0x(1+acosx)−bsinx(f(x))3=56, then

Given,limx→0 f(x)x=1∴limx→0 x(1+acos⁡x)−bsin⁡xx3⋅f(x)x3=56limx→0 (1+acos⁡x)−bxsin⁡xx2=56⇒limx→0 1+a1−x22!+x44!+……−bxx−x33!+x55!−……..x2=56 For finite limit, 1+a−b=0⇒b=a+1 And −a2!+b3!=56⇒−a2+b6=56⇒−3a+b=5⇒−3a+a+1=5⇒ −2a=4⇒a=−2 From Eq. (i), b=−2+1=−1

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$f (x)$ is the function such that $\lim_{x \to 0} \frac{f (x)}{x} = 1,$ if $\lim_{x \to 0} \frac{x (1 + a \cos x) - b \sin x}{{(f (x))}^{3}} = \frac{5}{6},$ then

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$a = - 1$ and $b = - 2$

$a = - 2$ and $b = - 1$

$a = 1$ and $b = - 2$

$a = - 1$ and $b = + 2$

answer is B.

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Detailed Solution

$\begin{array}{l} G i v e n, lim_{x \to 0} \frac{f (x)}{x} = 1 \\ ∴ lim_{x \to 0} \frac{x (1 + a \cos x) - b \sin x}{x^{3} \cdot {(\frac{f (x)}{x})}^{3}} = \frac{5}{6} \\ lim_{x \to 0} \frac{(1 + a \cos x) - \frac{b}{x} \sin x}{x^{2}} = \frac{5}{6} \end{array}$
$\Rightarrow lim_{x \to 0} \frac{(1 + a (1 - \frac{x^{2}}{2!} + \frac{x^{4}}{4!} + \dots \dots) - \frac{b}{x} (x - \frac{x^{3}}{3!} + \frac{x^{5}}{5!} - \dots \dots . .))}{x^{2}} = \frac{5}{6}$
$For finite limit, 1 + a - b = 0 \Rightarrow b = a + 1$
$And \frac{- a}{2!} + \frac{b}{3!} = \frac{5}{6} \Rightarrow \frac{- a}{2} + \frac{b}{6} = \frac{5}{6} \Rightarrow - 3 a + b = 5 \Rightarrow - 3 a + a + 1 = 5$
$\Rightarrow - 2 a = 4 \Rightarrow a = - 2$
$From Eq. (i), b = - 2 + 1 = - 1$