If a differentiable function $f(x)$ satisfies a functional rule $f(x) + f(x+2) + f(x+4) = 0$ for all $x$ belonging to real numbers,

Then find the value of :

$$\lim \limits_{x \to 0} \frac{\bigl(f(x+12)\bigr)^2- f(x) f(0) - f(x+6) f(18) + \bigl(f(18)\bigr)^2}{x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)} $$

what's the method?

The answer is $32$.

  • 1
    $\begingroup$ If the limit goes to 1 the limit is just $(\text{something})/\frac\pi4$. Are you sure the equation is copied correctly? Also, $f(x)=0$ is a differentiable function satisfying the functional equation but the limit is obviously 0 not 32. $\endgroup$ – kennytm May 15 '16 at 14:11
  • $\begingroup$ There is obviously a mistake in question (perhaps the book has typo). Because from the given hypotheses we can't conclude that limit is $32$. See the answer by user A s. Also the comment by @kennytm is valid. $\endgroup$ – Paramanand Singh May 16 '16 at 2:56
  • $\begingroup$ I have corrected the question it is limit goes to 0 not 1. Please note. $\endgroup$ – Harry Karwasra May 16 '16 at 8:11
  • $\begingroup$ Even if you change $x \to 1$ to $x \to 0$ the answer is not guaranteed to be $32$ rather the answer is equal to $2\{f'(0)\}^{2}$. Also as noted by @kennytm the function $f(x) = 0$ satisfies all the requirements of the question and then the limit is $0$. $\endgroup$ – Paramanand Singh May 16 '16 at 9:53

If dont know if the following can help you \begin{align} &f(x) + f(x+2) + f(x+4) = 0\hspace{1cm} (I)\\ &f(x+2) + f(x+4) + f(x+6) = 0\hspace{1cm} (II)\\ &f(x+6) + f(x+8) + f(x+10) = 0\hspace{1cm} (III)\\ &f(x+8) + f(x+10) + f(x+12) = 0\hspace{1cm} (IV)\\ &f(x+12) + f(x+14) + f(x+16) = 0\hspace{1cm} (V)\\ &f(x+14) + f(x+16) + f(x+18) = 0\hspace{1cm} (VI)\\ \end{align} First $(I)-(II)$ gives $f(x)=f(x+6)$ In particular $f(0)=f(6)$

also $(I)-(II)+(III)-(IV)$ gives $f(x)=f(x+12)$ In particular $f(0)=f(12)$

also also $(I)-(II)+(III)-(IV)+(V)-(VI)$ gives $f(x)=f(x+18)$

etc ... etc .. Noting that $\lim \limits_{x \to 1} x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)=\frac{\pi}{4}$. So $$\lim \limits_{x \to 1} \frac{\bigl(f(x+12)\bigr)^2- f(x) f(0) - f(x+6) f(18) + \bigl(f(18)\bigr)^2}{x\left({\frac{\pi}{4} - \tan^{-1}{(1-x)}}\right)}= \frac{4}{\pi} \lim \limits_{x \to 1} (f(x)+f(0))^{2}$$

  • $\begingroup$ I am sorry for the mistake but it is limit x goes to 0 not 1. Please note that. Also from your question it is evident that the period of function is 6. $\endgroup$ – Harry Karwasra May 16 '16 at 8:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.