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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CGPET 2014 Solved Paper

Section: Mathematics

Question : 131 of 150

Marks: +1, -0

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

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}

\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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