May 2026

Show that f(f(x)) = x Show that \(f(f(x))=x\) Question: If $$ f(x)=\frac{x+1}{x-1} $$ show that $$ f(f(x))=x $$ Solution Given: $$ f(x)=\frac{x+1}{x-1} $$ Now, $$ f(f(x)) = \frac{\frac{x+1}{x-1}+1}{\frac{x+1}{x-1}-1} $$ Simplify the numerator: $$ \frac{x+1}{x-1}+1 = \frac{x+1+x-1}{x-1} = \frac{2x}{x-1} $$ Simplify the denominator: $$ \frac{x+1}{x-1}-1 = \frac{x+1-x+1}{x-1} = \frac{2}{x-1} $$ Therefore, $$ f(f(x)) = \frac{\frac{2x}{x-1}}{\frac{2}{x-1}} $$ […]

Show that f(f(f(x))) = x Show that \(f(f(f(x)))=x\) Question: If $$ f(x)=\frac{1}{1-x} $$ show that $$ f(f(f(x)))=x $$ Solution Given: $$ f(x)=\frac{1}{1-x} $$ First find \(f(f(x))\): $$ f(f(x)) = \frac{1}{1-\frac{1}{1-x}} $$ $$ = \frac{1}{\frac{(1-x)-1}{1-x}} $$ $$ = \frac{1-x}{-x} $$ $$ = \frac{x-1}{x} $$ Now find \(f(f(f(x)))\): $$ f\left(\frac{x-1}{x}\right) = \frac{1}{1-\frac{x-1}{x}} $$ $$ = \frac{1}{\frac{x-(x-1)}{x}} $$

Show that x = f(y) Show that \(x=f(y)\) Question: If $$ y=f(x)=\frac{ax-b}{bx-a} $$ show that $$ x=f(y) $$ Solution Given: $$ y=\frac{ax-b}{bx-a} $$ Now, $$ f(y)=\frac{ay-b}{by-a} $$ Substitute the value of \(y\): $$ f(y)= \frac{ a\left(\frac{ax-b}{bx-a}\right)-b }{ b\left(\frac{ax-b}{bx-a}\right)-a } $$ Taking LCM in numerator and denominator: $$ = \frac{ \frac{a(ax-b)-b(bx-a)}{bx-a} }{ \frac{b(ax-b)-a(bx-a)}{bx-a} } $$ $$

Find f(a+b) Find \(f(a+b)\) Question: If $$ f(x)=(x-a)^2(x-b)^2 $$ find $$ f(a+b) $$ Solution Given: $$ f(x)=(x-a)^2(x-b)^2 $$ Put \(x=a+b\): $$ f(a+b)=((a+b)-a)^2((a+b)-b)^2 $$ $$ =(b)^2(a)^2 $$ $$ =a^2b^2 $$ Hence, $$ \boxed{f(a+b)=a^2b^2} $$ “` Next Question / Full Exercise

Solve f(x)=f(2x+1) Solve \(f(x)=f(2x+1)\) Question: If $$ f(x)=x^2-3x+4 $$ then find the value of \(x\) satisfying $$ f(x)=f(2x+1) $$ Solution Given: $$ f(x)=x^2-3x+4 $$ First find \(f(2x+1)\): $$ f(2x+1)=(2x+1)^2-3(2x+1)+4 $$ $$ =4x^2+4x+1-6x-3+4 $$ $$ =4x^2-2x+2 $$ Now, $$ f(x)=f(2x+1) $$ $$ x^2-3x+4=4x^2-2x+2 $$ $$ 3x^2+x-2=0 $$ $$ 3x^2+3x-2x-2=0 $$ $$ 3x(x+1)-2(x+1)=0 $$ $$ (x+1)(3x-2)=0 $$

Express Function as Set of Ordered Pairs Express Function as Set of Ordered Pairs Question: Express the function $$ f:X\to\mathbb{R} $$ given by $$ f(x)=x^3+1 $$ as a set of ordered pairs, where $$ X=\{-1,0,3,9,7\} $$ Solution Given: $$ f(x)=x^3+1 $$ Find the value of \(f(x)\) for each element of \(X\). \(x\) \(f(x)=x^3+1\) \(-1\) \((-1)^3+1=0\)

Find Difference Quotient for f(x)=x² Find Difference Quotient for \(f(x)=x^2\) Question: If $$ f(x)=x^2 $$ find $$ \frac{f(1.1)-f(1)}{(1.1)-1} $$ Solution Given: $$ f(x)=x^2 $$ $$ f(1.1)=(1.1)^2=1.21 $$ $$ f(1)=1^2=1 $$ Therefore, $$ \frac{f(1.1)-f(1)}{(1.1)-1} = \frac{1.21-1}{0.1} $$ $$ = \frac{0.21}{0.1} =2.1 $$ Hence, $$ \boxed{2.1} $$ “` Next Question / Full Exercise

Show that f is a Function and g is not a Function Show that f is a Function and g is not a Function Question: The function \(f\) is defined by $$ f(x)= \begin{cases} x^2, & 0\le x\le 3 \\ 3x, & 3\le x\le 10 \end{cases} $$ The relation \(g\) is defined by $$ g(x)=

Find the Range of a Function Using Highest Prime Factor Find the Range of a Function Using Highest Prime Factor Question: Let $$ A=\{9,10,11,12,13\} $$ and let $$ f:A\to\mathbb{N} $$ be defined by $$ f(n)=\text{the highest prime factor of } n $$ Find the range of \(f\). Solution Find the highest prime factor of each

Which Relation is Not a Function? Which Relation is Not a Function? Question: Let $$ A=\{p,q,r,s\} $$ and $$ B=\{1,2,3\} $$ Which of the following relations from \(A\) to \(B\) is not a function? (i) $$ R_1=\{(p,1),(q,2),(r,1),(s,2)\} $$ (ii) $$ R_2=\{(p,1),(q,1),(r,1),(s,1)\} $$ (iii) $$ R_3=\{(p,1),(q,2),(p,2),(s,3)\} $$ (iv) $$ R_4=\{(p,2),(q,3),(r,2),(s,2)\} $$ Solution A relation is a