UPSC 2023 Maths Optional Paper 1 Q1c — Step-by-Step Solution

10 marks · Section A

Taylor's theorem with remainders · Calculus · asked 2× in 13 yrs · Read the full method →

Question

Find the values of $p$ and $q$ for which $\displaystyle\lim_{x\to 0}\frac{x(1+p\cos x)-q\sin x}{x^3}$ exists and equals $1$ .

Taylor expansion of $\sin,\cos$ to enough terms; equate coefficients of $x^{-2}$ and $x^0$ to get a linear system in $p,q$ .

Strategy. Expand $\cos x$ and $\sin x$ as Taylor series and equate coefficients.

Step 1 — Taylor expansions.

$\cos x=1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots$ $\sin x=x-\dfrac{x^3}{6}+\dfrac{x^5}{120}-\cdots$

Step 2 — Numerator.

x(1+p\cos x)=x+px\Bigl(1-\dfrac{x^2}{2}+\dfrac{x^4}{24}-\cdots\Bigr)=(1+p)x-\dfrac{p}{2}x^3+\dfrac{p}{24}x^5-\cdots

q\sin x=qx-\dfrac{q}{6}x^3+\dfrac{q}{120}x^5-\cdots

Numerator:

x(1+p\cos x)-q\sin x=(1+p-q)x+\Bigl(-\dfrac{p}{2}+\dfrac{q}{6}\Bigr)x^3+O(x^5).

Step 3 — Divide by $x^3$ and take $x\to 0$ .

\frac{(1+p-q)x+\bigl(-p/2+q/6\bigr)x^3+O(x^5)}{x^3}=\dfrac{1+p-q}{x^2}+\Bigl(-\dfrac{p}{2}+\dfrac{q}{6}\Bigr)+O(x^2).

For the limit to exist (be finite), the $1/x^2$ term must vanish:

1+p-q=0\quad\Rightarrow\quad q=1+p.

For the limit to equal $1$ :

-\dfrac{p}{2}+\dfrac{q}{6}=1.

Substitute $q=1+p$ :

-\dfrac{p}{2}+\dfrac{1+p}{6}=1\Rightarrow\dfrac{-3p+1+p}{6}=1\Rightarrow\dfrac{1-2p}{6}=1\Rightarrow 1-2p=6\Rightarrow p=-\dfrac{5}{2}.

Then $q=1+p=-\dfrac{3}{2}$ .

\boxed{\;p=-\tfrac{5}{2},\quad q=-\tfrac{3}{2}.\;}