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Limits — JEE Maths practice question

JEE Maths question with a full step-by-step solution.

Question

\displaystyle\lim_{x\to 0}\dfrac{x(1+a\cos x)-b\sin x}{(f(x))^{3}}=1

and

\displaystyle\lim_{x\to 0}\dfrac{f(x)}{x}=1

then

a=-\dfrac{5}{2}

correct

b=\dfrac{27}{2}

a=-\dfrac{5}{3}

b=-\dfrac{3}{2}

correct

Solution

Step 1: Since

\dfrac{f(x)}{x}\to 1

, we can replace

(f(x))^{3}

x^{3}

in the limit. So

\lim_{x\to 0}\dfrac{x(1+a\cos x)-b\sin x}{x^{3}}=1

Step 2: Use the Taylor series expansions:

\cos x=1-\dfrac{x^{2}}{2!}+\dfrac{x^{4}}{4!}-\dfrac{x^{6}}{6!}+\ldots

\sin x=x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\ldots

Step 3: Substitute these expansions in the numerator:

x\left(1+a\left(1-\dfrac{x^{2}}{2!}+\dfrac{x^{4}}{4!}-\ldots\right)\right)-b\left(x-\dfrac{x^{3}}{3!}+\dfrac{x^{5}}{5!}-\ldots\right)

Step 4: Simplify and divide by

x^{3}

=\dfrac{(1+a-b)x+\left(-\dfrac{a}{2!}+\dfrac{b}{3!}\right)x^{3}+\left(\dfrac{a}{4!}-\dfrac{b}{5!}\right)x^{5}+\ldots}{x^{3}}

=\dfrac{1+a-b}{x^{2}}+\left(-\dfrac{a}{2}+\dfrac{b}{6}\right)+x^{2}\left(\dfrac{a}{24}-\dfrac{b}{120}\right)+\ldots

Step 5: For the limit to exist and equal

1

: The coefficient of

\dfrac{1}{x^{2}}

must vanish:

1+a-b=0

. The constant term must equal

1

-\dfrac{a}{2}+\dfrac{b}{6}=1

. Step 6: Solve the two equations: From the first:

b=1+a

. Substituting:

-\dfrac{a}{2}+\dfrac{1+a}{6}=1\;\Rightarrow\;-3a+1+a=6\;\Rightarrow\;-2a=5\;\Rightarrow\;a=-\dfrac{5}{2}

. Then

b=1+a=1-\dfrac{5}{2}=-\dfrac{3}{2}

. Correct answers: (1)

a=-\dfrac{5}{2}

and (4)

b=-\dfrac{3}{2}

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