If ₓ → 0} x(1+a x)-b x/x³=1, then

Correct Answer is : a=−52,b=−32a=-5/2, b=-3/2a=−25,b=−23

Correct Answer is : a=−52,b=−32a=-5/2, b=-3/2a=−25,b=−23

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CGPET 2012 Solved Paper

Section: Mathematics

Question : 110 of 150

Marks: +1, -0

\lim\limits_{x \to 0} \frac{x(1+a \cos x)-b \sin x}{x^3}=1

Solution:

\lim\limits_{x \to 0} \frac{x(1+a \cos x)-b \sin x}{x^3}=1

Using series of

\sin x

and

\cos x

x \left[1+a \left(1-\frac{x^2}{2!}+\dots\right)\right]

\Rightarrow \lim\limits_{x \to 0} \frac{-b \left(x-\frac{x^3}{3!}+\dots\right)}{x^3}=1

\Rightarrow \lim\limits_{x \to 0} \frac{x(1+a)-\frac{x^3}{2!}+\dots-b x+b\frac{x^3}{3!}+\dots}{x^3}=1

\Rightarrow \lim\limits_{x \to 0} \frac{x(1+a-b)+x^3 \left(\frac{b}{6}-\frac{a}{2}\right)+\dots}{x^3}=1

For existence of finite limit,

1+a-b=0

. . . (i)

Then,

\lim\limits_{x \to 0} \frac{x^3 \left(\frac{b}{6}-\frac{a}{2}\right)+\dots}{x^3}=1

\Rightarrow \frac{b}{6}-\frac{a}{2}=1

\Rightarrow b-3a=6

. . . (ii)

On solving Eqs. (i) and (ii), we get

a=\frac{-5}{2}

and

b=-\frac{3}{2}

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