\displaystyle \textbf{Question 1: } \text{If }f(x)=x^2-3x+4,\text{ then find the values of }x\text{ satisfying}
\displaystyle \text{the equation }f(x)=f(2x+1).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=x^2-3x+4.
\displaystyle \therefore f(2x+1)=(2x+1)^2-3(2x+1)+4=4x^2-2x+2.
\displaystyle \text{Since }f(x)=f(2x+1),
\displaystyle x^2-3x+4=4x^2-2x+2.
\displaystyle \Rightarrow 3x^2+x-2=0.
\displaystyle \Rightarrow 3x^2+3x-2x-2=0.
\displaystyle \Rightarrow 3x(x+1)-2(x+1)=0.
\displaystyle \Rightarrow (x+1)(3x-2)=0.
\displaystyle \therefore x=-1\text{ or }x=\frac{2}{3}.
\\

\displaystyle \textbf{Question 2: } \text{If }f(x)=(x-a)^2(x-b)^2,\text{ find }f(a+b).
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=(x-a)^2(x-b)^2.
\displaystyle \therefore f(a+b)=(a+b-a)^2(a+b-b)^2.
\displaystyle =b^2a^2=a^2b^2.
\\

\displaystyle \textbf{Question 3: } \text{If }y=f(x)=\frac{ax-b}{bx-a},\text{ show that }x=f(y).
\displaystyle \text{Answer:}
\displaystyle \text{Given }y=f(x)=\frac{ax-b}{bx-a}.
\displaystyle \Rightarrow y=\frac{ax-b}{bx-a}.
\displaystyle \Rightarrow y(bx-a)=ax-b.
\displaystyle \Rightarrow bxy-ay=ax-b.
\displaystyle \Rightarrow bxy-ax=ay-b.
\displaystyle \Rightarrow x(by-a)=ay-b.
\displaystyle \Rightarrow x=\frac{ay-b}{by-a}.
\displaystyle \therefore x=f(y).\text{ Hence proved.}
\\

\displaystyle \textbf{Question 4: } \text{If }f(x)=\frac{1}{1-x},\text{ show that }f[f\{f(x)\}]=x.
\displaystyle \text{Answer:}
\displaystyle f(x)=\frac{1}{1-x}.
\displaystyle \therefore f(f(x))=\frac{1}{1-\frac{1}{1-x}}=\frac{1-x}{1-x-1}=\frac{x-1}{x}.
\displaystyle \therefore f(f(f(x)))=\frac{1}{1-\frac{x-1}{x}}=\frac{1}{\frac{x-(x-1)}{x}}=\frac{1}{\frac{1}{x}}=x.
\displaystyle \therefore f(f(f(x)))=x.\text{ Hence proved.}
\\

\displaystyle \textbf{Question 5: } \text{If }f(x)=\frac{x-1}{x+1},\text{ show that }f[f(x)]=x.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=\frac{x-1}{x+1}.
\displaystyle \therefore f(f(x))=f\!\left(\frac{x-1}{x+1}\right).
\displaystyle =\frac{\frac{x-1}{x+1}-1}{\frac{x-1}{x+1}+1}.
\displaystyle =\frac{\frac{x-1-(x+1)}{x+1}}{\frac{x-1+(x+1)}{x+1}}.
\displaystyle =\frac{\frac{-2}{x+1}}{\frac{2x}{x+1}}.
\displaystyle =\frac{-2}{x+1}\times\frac{x+1}{2x}=-\frac{1}{x}.

\displaystyle \textbf{Question 6: } \text{If }f(x)=x^3-\frac{1}{x^3},\text{ show that }f(x)+f\left(\frac{1}{x}\right)=0,\ x\neq0.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=x^3-\frac{1}{x^3},\quad x\neq0.
\displaystyle \therefore f\left(\frac{1}{x}\right)=\left(\frac{1}{x}\right)^3-\frac{1}{\left(\frac{1}{x}\right)^3}=\frac{1}{x^3}-x^3.
\displaystyle \therefore f(x)+f\left(\frac{1}{x}\right)=\left(x^3-\frac{1}{x^3}\right)+\left(\frac{1}{x^3}-x^3\right).
\displaystyle =x^3-\frac{1}{x^3}+\frac{1}{x^3}-x^3=0.
\displaystyle \therefore f(x)+f\left(\frac{1}{x}\right)=0.\text{ Hence proved.}
\\

\displaystyle \textbf{Question 7: } \text{If }f(x)=\frac{2x}{1+x^2},\text{ show that }f(\tan\theta)=\sin2\theta.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=\frac{2x}{1+x^2}.
\displaystyle \therefore f(\tan\theta)=\frac{2\tan\theta}{1+\tan^2\theta}.
\displaystyle =2\cdot\frac{\sin\theta}{\cos\theta}\cdot\frac{\cos^2\theta}{\cos^2\theta+\sin^2\theta}.
\displaystyle =2\sin\theta\cos\theta.
\displaystyle =\sin2\theta.
\displaystyle \therefore f(\tan\theta)=\sin2\theta.\text{ Hence proved.}
\\

\displaystyle \textbf{Question 8: } \text{If }f(x)=\frac{x-1}{x+1},\text{ then show that:}
\displaystyle \text{(i) }f\left(\frac{1}{x}\right)=-f(x)\qquad\text{(ii) }f\left(-\frac{1}{x}\right)=-\frac{1}{f(x)}.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=\frac{x-1}{x+1},\quad x\neq0,\pm1.
\displaystyle \text{(i) }f\left(\frac{1}{x}\right)=\frac{\frac{1}{x}-1}{\frac{1}{x}+1}=\frac{1-x}{1+x}=-\frac{x-1}{x+1}=-f(x).
\displaystyle \text{(ii) }f\left(-\frac{1}{x}\right)=\frac{-\frac{1}{x}-1}{-\frac{1}{x}+1}=\frac{-1-x}{-1+x}.
\displaystyle =-\frac{x+1}{x-1}=-\frac{1}{\frac{x-1}{x+1}}=-\frac{1}{f(x)}.
\displaystyle \therefore f\left(\frac{1}{x}\right)=-f(x)\text{ and }f\left(-\frac{1}{x}\right)=-\frac{1}{f(x)}.\text{ Hence proved.}
\\

\displaystyle \textbf{Question 9: } \text{If }f(x)=(a-x^n)^{1/n},\ a>0,\ n\in N,\text{ prove that }f(f(x))=x.
\displaystyle \text{Answer:}
\displaystyle \text{Given }f(x)=(a-x^n)^{1/n},\ a>0,\ n\in N.
\displaystyle \therefore f(f(x))=\left(a-\left[\left(a-x^n\right)^{1/n}\right]^n\right)^{1/n}.
\displaystyle =\left(a-(a-x^n)\right)^{1/n}.
\displaystyle =(x^n)^{1/n}.
\displaystyle =x.
\displaystyle \therefore f(f(x))=x.\text{ Hence proved.}
\\

\displaystyle \textbf{Question 11: }\text{If, for non-zero }x,\ af(x)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5,
\displaystyle \text{where }a\neq b\text{ and }a\neq-b,\text{ find }f(x).
\displaystyle \text{Answer:}
\displaystyle \text{Given }af(x)+bf\left(\frac{1}{x}\right)=\frac{1}{x}-5. \qquad\text{...(i)}
\displaystyle \text{Replacing }x\text{ by }\frac{1}{x}\text{ in (i),}
\displaystyle af\left(\frac{1}{x}\right)+bf(x)=x-5. \qquad\text{...(ii)}
\displaystyle \text{Adding (i) and (ii),}
\displaystyle (a+b)f(x)+(a+b)f\left(\frac{1}{x}\right)=x+\frac{1}{x}-10.
\displaystyle \therefore f(x)+f\left(\frac{1}{x}\right)=\frac{1}{a+b}\left(x+\frac{1}{x}-10\right). \qquad\text{...(iii)}
\displaystyle \text{Subtracting (ii) from (i),}
\displaystyle (a-b)f(x)-(a-b)f\left(\frac{1}{x}\right)=\frac{1}{x}-x.
\displaystyle \therefore f(x)-f\left(\frac{1}{x}\right)=\frac{1}{a-b}\left(\frac{1}{x}-x\right). \qquad\text{...(iv)}
\displaystyle \text{Adding (iii) and (iv),}
\displaystyle 2f(x)=\frac{1}{a+b}\left(x+\frac{1}{x}-10\right)+\frac{1}{a-b}\left(\frac{1}{x}-x\right).
\displaystyle =\frac{(a-b)\left(x+\frac{1}{x}-10\right)+(a+b)\left(\frac{1}{x}-x\right)}{a^2-b^2}.
\displaystyle =\frac{\frac{2a}{x}-2bx-10a+10b}{a^2-b^2}.
\displaystyle \therefore f(x)=\frac{\frac{a}{x}-bx-5a+5b}{a^2-b^2}.
\displaystyle =\frac{\frac{a}{x}-bx}{a^2-b^2}-\frac{5(a-b)}{(a-b)(a+b)}.
\displaystyle \therefore f(x)=\frac{\frac{a}{x}-bx}{a^2-b^2}-\frac{5}{a+b}.
\\

Discover more from ICSE / ISC / CBSE Mathematics Portal for K12 Students

Subscribe to get the latest posts sent to your email.