If f(x)= 1+x 1-x , show that f [f(x) ]=x - Vidyadip

Question

If $\text{f(x)}=\frac{1+\text{x}}{1-\text{x}},$ show that $\text{f}\big[\text{f}\text{(x)}\big]=\text{x}$

✓

Answer

We have,
$\text{f(x)}=\frac{1+\text{x}}{1-\text{x}}$
Now, $\text{f}\big[\text{f}\text{(x)}\big]=\text{f}\Big(\frac{\text{x}+1}{1-\text{x}}\Big)$
$=\frac{\big(\frac{\text{x}+1}{1-\text{x}}\big)+1}{\big(\frac{\text{x}+1}{1-\text{x}}\big)-1}$
$=\frac{\frac{\text{x}+1+{\text{x}}-1}{{\text{x}}-1}}{\frac{\text{x}+1-1(\text{x}-1)}{\text{x}-1}}$
$=\frac{\frac{2\text{x}}{\text{x}-1}}{\frac{\text{x}+1-\text{x}+1}{\text{x}-1}}$
$=\frac{2\text{x}}{2}$
$=\text{x}$
$\therefore\ \text{f}\big[\text{f}\text{(x)}\big]=\text{x}$ Hence, proved.

Need a full question paper?

Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.

Start Generating Free

Explore more

More "5 Marks Questions" from Functions→Functions — all question types→Maths STD 11 Science question papers→CBSE Board STD 11 Science question papers→CBSE Board question paper generator→

Similar questions

Find the sum of the series whose $n^{th}$ term is:$n(n + 1)(n + 4)$

Solve the following equation:
$\sin^{2}\text{x}-\cos\text{x}=\frac{1}{4}$

Evaluate the following
$\Big(\sqrt{\text{x}+1}+\sqrt{\text{x}-1}\Big)^6+\Big(\sqrt{\text{x}+1}-\sqrt{\text{x}-1}\Big)^6$

Evaluate:
$\lim\limits_{\text{x} \rightarrow{\text{a}}}\frac{\sin\text{x}-\sin\text{a}}{\sqrt{\text{x}}-\sqrt{\text{a}}}$

A visitor with sign board 'DO NOT LITTER' is moving on a circular path in an exhibition. During the movement he stops at points represented by (3, - 2) and (-2, 0). Also, centre of the circular path is on the line 2x - y = 3. What is the equation of the path? What message he wants to give to the public?

Find the equations of the circles passing through two points on $y-$axis at distances $3$ from the origin and having radius $5.$

$\text{a}(\cos\text{C}-\cos\text{B})=2(\text{b}-\text{c})\cos^2\frac{\text{A}}{2}.$

Prove that $\frac{1}{2}\tan\Big(\frac{\text{x}}{2}\Big)+\frac{1}{4}\tan\Big(\frac{\text{x}}{4}\Big)+...+\frac{1}{2^{\text{n}}}\tan\Big(\frac{\text{x}}{2^{\text{n}}}\Big)=\frac{1}{2^{\text{n}}}\cot\Big(\frac{\text{x}}{2^{\text{n}}}\Big)-\cot\text{x}$ for all $\text{n}\in\text{N}$ and $0<\text{x}<\frac{\pi}{2}$

$\text {A}$ speaks truth in $75 \%$ of the situations and $\text {B}$ in $80 \%$ of the situations. Find out in how many percent situations do they oppose each other?

If $x^p$ occurs in the expansion of $\Big(\text{x}^2+\frac{1}{\text{x}}\Big)^{2\text{n}},$ prove that its coefficient is $\frac{2\text{n}!}{\Big(\frac{4\text{n}-\text{p}}{3}\Big)!\Big(\frac{2\text{n}+\text{p}}{3}\Big)!}.$