Question

If In=e^nxTan(x), and \(\frac{I_{n+2}-2nI_{n+1}+n^2 I_n}{nI_{n-I}}\) = c (1+x^2)\(\frac{d^2}{dx^2} (\frac{1}{1+x^2})\),  Then value of ‘c’ equals to

(a) 1

(b) 2

(c) 3

(d) 4

I got this question in an online quiz.

The doubt is from The nth Derivative of Some Elementary Functions in section Differential Calculus of Engineering Mathematics

Answer 1

Right answer is (a) 1

For explanation: In=enxTan(x)

⇨In+1=ne^xnTan(x) + e^xn/(1+x2)

⇨In+1=nIn + e^xn/(1+x2)

⇨In+2=nIn+1 + \(\frac{ne^nx}{1+x^2} – \frac{2xe^nx}{(1+x^2)^2}\)

⇨In+2=nIn+1 + n[In+1-nIn] – \(\frac{2xe^nx}{(1+x^2)^2}\)

⇨\(\frac{I_{n+2}-2nI_{n+1}+n^2 I_n}{nI_{n-I}}=(1+x^2)\frac{d^2}{dx^2} (\frac{1}{1+x^2})\), Hence c=1.