Question

Let f(x)=\(x-\frac{x^3}{3^2.2!}+\frac{x^5}{5^2.4!}-\frac{x^7}{7^2.6!}+…\infty\). Find a point nearest to c such that f'(c) = 1

(a) 1

(b) 0

(c) 2.3445 * 10^-9

(d) 458328.33 * 10^-3

I have been asked this question in a national level competition.

I'd like to ask this question from Lagrange’s Mean Value Theorem topic in chapter Differential Calculus of Engineering Mathematics

Answer 1

Correct option is (d) 458328.33 * 10^-3

The explanation: First find f'(x)

f'(x)=1-\(\frac{x^2}{3!}+\frac{x^4}{5!}-\frac{x^6}{7!}+…\infty\)

Multiplying and dividing by x We have the well known series expansion of \(\frac{sin(x)}{x}\)  

We get

f'(x)=\(\frac{1}{x}\times (x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+…\infty)\)

f'(x)=\(\frac{sin(x)}{x}\)

Equating this to 1 We have

 \(\frac{sin(x)}{x}\) = 1

We know the well known limit   \(lim_{x \rightarrow 0} \frac{sin(x)}{x}\) = 1

Thus we have to choose a point nearer to 0 as our answer which is,

458328.33 * 10^-3.