How do I solve such type of questions when time limit is about 2–3 minutes?

Question

If $f(x)= \ln\left ( \frac{1+x}{1-x} \right )$ then

$A.\,f(x_1)\cdot f(x_2)= f(x_1+x_2)\\ B.\,f(x+2)-2f(x+1)+f(x)=0\\ C.\,f(x)+f(x+1)=f(x^2+x)\\ D.\,f(x_1)+f(x_2)=f\left(\frac{x_1+x_2}{1+x_1 x_2}\right) $

As you can see, there four options and in the exam we get only 2–3 minutes to solve and I can't keep checking each and every option. Is there any other option?

In questions like this I find it easier and faster to give $x_1,x_2$ values and checking the results — The Integrator, Jun 01 '18 at 18:11
If $x=0$ then B and C are out. Two options left. Took me 20 seconds. Can you do the rest? — imranfat, Jun 01 '18 at 18:12
+1) This is actually a very nice problem. I got to take a pic of this for reference! — imranfat, Jun 01 '18 at 18:16
If you were very good you might notice that $D$ has some resemblance to the $\tan A+ B$ rule, and then you might look to see that $f(x) = 2\tanh^{-1} (x)$ — Doug M, Jun 01 '18 at 18:32
Can anyone give me step by step solution for the substitution method? — The ReBel, Jun 01 '18 at 18:57

Rhys Hughes · Accepted Answer · 2018-06-01T18:23:45.403

1

Given your constraints, $-1<x<1$

Set $x_1=x_2=0$ and try them.

Then: $$f(x)=\ln(\frac 11)=0$$ $f(x+1)$ is undefined.

$f(x+2)$ is also undefined.

This eliminates $B$ and $C$.

Now to check $A$ and $D$.

$A$ works, and so does $D$

Now let $x_1=\frac 12, x_2=0$

We have that $f(x_1)=\ln 3, f(x_2)=0, f(x_1+x_2)=\ln3$, and so $A$ is false, leaving only $D$

edited Jun 01 '18 at 18:23

answered Jun 01 '18 at 18:18

Rhys Hughes

12,842

mistake made allow me to edit – Rhys Hughes Jun 01 '18 at 18:19
Answer is D. D is the answer given in the book – The ReBel Jun 01 '18 at 18:20
Why didn't you take x_{1}=1 , it's easier to solve? @Rhys Hughes – The ReBel Jun 01 '18 at 18:48
@TheReBel, $\frac{1+(1)}{1-(1)}=\frac 20$ which is undefined – Rhys Hughes Jun 01 '18 at 18:50
But , in the 2nd option when you put x=0 then f(x+1) becomes f(1), then it's also undefined so doesn't that mean that we can't keep x=0. Sorry for so much reasoning, but you know, I can't help. – The ReBel Jun 01 '18 at 18:54
1

There are values of $x$ for which $f(x)$ is defined but $f(x+1)$ is not. Hence equations which use $f(x+1)$ are invalid – Rhys Hughes Jun 01 '18 at 22:01

How do I solve such type of questions when time limit is about 2–3 minutes?

1 Answers1