If ₓ arrow 0} \; { ( x) - x}{a x³ + b x^{5} + c} = \; -1/12, then

Correct Answer is : a=−2,b∈R,c=0a = -2, b R, c = 0a=−2,b∈R,c=0

Correct Answer is : a=−2,b∈R,c=0a = -2, b R, c = 0a=−2,b∈R,c=0

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Manipal Entrance Test 2020 Math Paper

Question : 21 of 70

Marks: +1, -0

\lim\limits_{x \rightarrow 0} \; \frac{\sin (\sin x) - \sin x}{a x^{3} + b x^{5} + c} = \; -\frac{1}{12}

, then

Solution:

Given,

\lim\limits_{x \rightarrow 0} \; \frac{\sin (\sin x) - \sin x}{a x^{3} + b x^{5} + c} = - \; \frac{1}{12}

.....(i)

\Rightarrow \; \; \; \frac{\sin \sin 0 - \sin 0}{a(0)^{3} + b(0)^{5} + c} = - \; \frac{1}{12}

\Rightarrow \; \; \; \frac{0}{c} = - \; \frac{1}{12} \Rightarrow c = 0

Applying L' Hospital's rule in Eq.(i), we get

\lim\limits_{x \rightarrow 0} \; \frac{\cos (\sin x) \cos x - \cos x}{3 a x^{2} + 5 b x^{4} + 0} = - \; \frac{1}{12}

\Rightarrow \lim\limits_{x \rightarrow 0} \; \frac{\cos x (\cos (\sin x) - 1)}{x^{2} (3 a + 5 b x)} = - \; \frac{1}{12}

\Rightarrow \lim\limits_{x \rightarrow 0} \cos x \; \frac{2 \sin ^{2} \left( \frac{\sin x}{2} \right)}{x^{2} (3 a + 5 b x)} = - \; \frac{1}{12}

\Rightarrow \; \frac{\cos 0}{3 a + 5 b \times 0} \times \lim\limits_{x \rightarrow 0} \; \frac{2 \sin ^{2} \left( \; \frac{\sin x}{2} \right)}{4 \left( \; \frac{\sin x}{2} \right)^{2} \times \left( \; \frac{x}{\sin x} \right)^{2}} = - \; \frac{1}{12}

\Rightarrow \; \frac{1}{3 a + 0} \times \; \frac{1}{2} \times \; \frac{1}{1} = - \; \frac{1}{12} \left( \because \lim\limits_{\theta \rightarrow 0} \frac{\sin \theta}{\theta} = 1 \right)

\Rightarrow \; \; \; \frac{1}{6 a} = - \; \frac{1}{12} \Rightarrow 6 a = -12

\therefore \; \; a = -2

Hence,

a = -2, b \in R

and

c = 0

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