ISRO EC 2023 Q.3: solution and answer

ISRO 2023 · ECQ.3 · 1 mark · MCQ

$\frac{d}{d x} (x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \frac{x ^{7}}{7 !} \dots)$ is equal to:

See the question in the paper

Answer and solution

ANSWER

The expression equals $cos x$ , option (c).

The series $x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \frac{x ^{7}}{7 !} + \dots$ is the Maclaurin series of $sin x$ , which converges for every real $x$ .
A power series may be differentiated term by term inside its radius of convergence, so $\frac{d}{d x} (x - \frac{x ^{3}}{3 !} + \frac{x ^{5}}{5 !} - \frac{x ^{7}}{7 !} + \dots) = 1 - \frac{3 x ^{2}}{3 !} + \frac{5 x ^{4}}{5 !} - \frac{7 x ^{6}}{7 !} + \dots$
Cancel the leading factor in each term using $\frac{n}{n !} = \frac{1}{( n - 1 )!}$ : $1 - \frac{x ^{2}}{2 !} + \frac{x ^{4}}{4 !} - \frac{x ^{6}}{6 !} + \dots$
This is the Maclaurin series of $cos x$ . It also agrees with $\frac{d}{d x} (sin x) = cos x$ .
The other options do not fit: $e^{x}$ has every term positive, $sin x$ is the original function and not its derivative, and $tan x$ is not given by a series of this factorial form.
Hence the result is $cos x$ , option (c).

KEY IDEA

The given series is the Maclaurin series of $sin x$ , and differentiating $sin x$ (or the series term by term) gives $cos x$ .

WATCH OUT FOR

Choosing $sin x$ because the series is recognised as sine, forgetting that the question asks for its derivative.