\displaystyle \textbf{Question 1. }\text{Write the range of the real function }f(x)=|x|.
\displaystyle \text{Answer:}
\displaystyle |x|\geq0
\displaystyle \therefore \text{Range}=[0,\infty)
\\

\displaystyle \textbf{Question 2. }\text{If }f\text{ is a real function satisfying }f\left(x+\frac{1}{x}\right)=x^2+\frac{1}{x^2}\text{ for all } \\ x\in R-\{0\},\text{ then write the expression for }f(x).
\displaystyle \text{Answer:}
\displaystyle \left(x+\frac{1}{x}\right)^2=x^2+\frac{1}{x^2}+2
\displaystyle x^2+\frac{1}{x^2}=\left(x+\frac{1}{x}\right)^2-2
\displaystyle \therefore f\left(x+\frac{1}{x}\right)=\left(x+\frac{1}{x}\right)^2-2
\displaystyle \therefore f(t)=t^2-2
\displaystyle \therefore f(x)=x^2-2

\displaystyle \textbf{Question 3. }\text{Write the range of the function }f(x)=\sin[x], \\ \text{ where }-\frac{\pi}{4}\leq x\leq\frac{\pi}{4}.
\displaystyle \text{Answer:}
\displaystyle -\frac{\pi}{4}\leq x\leq\frac{\pi}{4}
\displaystyle [x]\in\{-1,0\}
\displaystyle \therefore f(x)=\sin[x]\in\{\sin(-1),\sin0\}
\displaystyle \therefore \text{Range}=\{-\sin1,0\}
\\

\displaystyle \textbf{Question 4. }\text{If }f(x)=\cos[\pi^2]x+\cos[-\pi^2]x,\text{ where }[x]\text{ denotes the greatest} \\ \text{integer less than or equal to }x,\text{ then write the value of }f(\pi).
\displaystyle \text{Answer:}
\displaystyle [\pi^2]=9,\quad[-\pi^2]=-10
\displaystyle f(x)=\cos9x+\cos(-10x)
\displaystyle f(\pi)=\cos9\pi+\cos(-10\pi)
\displaystyle =-1+1=0
\\

\displaystyle \textbf{Question 5. }\text{Write the range of the function }f(x)=\cos[x], \\ \text{ where }-\frac{\pi}{2}<x<\frac{\pi}{2}.
\displaystyle \text{Answer:}
\displaystyle -\frac{\pi}{2}<x<\frac{\pi}{2}
\displaystyle [x]\in\{-2,-1,0,1\}
\displaystyle f(x)=\cos[x]\in\{\cos2,\cos1,1\}
\displaystyle \therefore \text{Range}=\{\cos2,\cos1,1\}
\\

\displaystyle \textbf{Question 6. }\text{Write the range of the function }f(x)=e^{x-[x]},\ x\in R.
\displaystyle \text{Answer:}
\displaystyle 0\leq x-[x]<1
\displaystyle 1\leq e^{x-[x]}<e
\displaystyle \therefore \text{Range}=[1,e)
\\

\displaystyle \textbf{Question 7. }\text{Let }f(x)=\frac{\alpha x}{x+1},\ x\neq-1.\text{ Then write the value of }\alpha\text{ satisfying } \\ f(f(x))=x\text{ for all }x\neq-1.
\displaystyle \text{Answer:}
\displaystyle f(f(x))=\frac{\alpha\cdot\frac{\alpha x}{x+1}}{\frac{\alpha x}{x+1}+1}
\displaystyle =\frac{\alpha^2x}{(\alpha+1)x+1}
\displaystyle \frac{\alpha^2x}{(\alpha+1)x+1}=x
\displaystyle \alpha^2=(\alpha+1)x+1
\displaystyle \text{For all }x,\text{ we must have }\alpha+1=0\text{ and }\alpha^2=1
\displaystyle \therefore \alpha=-1
\\

\displaystyle \textbf{Question 8. }\text{If }f(x)=1-\frac{1}{x},\text{ then write the value of }f\left(f\left(\frac{1}{x}\right)\right).
\displaystyle \text{Answer:}
\displaystyle f\left(\frac{1}{x}\right)=1-x
\displaystyle f\left(f\left(\frac{1}{x}\right)\right)=f(1-x)
\displaystyle =1-\frac{1}{1-x}
\displaystyle =\frac{-x}{1-x}=\frac{x}{x-1}
\\

\displaystyle \textbf{Question 9. }\text{Write the domain and range of the function }f(x)=\frac{x-2}{2-x}.
\displaystyle \text{Answer:}
\displaystyle 2-x\neq0\Rightarrow x\neq2
\displaystyle f(x)=\frac{x-2}{2-x}=-1
\displaystyle \therefore \text{Domain}=R-\{2\}
\displaystyle \text{Range}=\{-1\}

\displaystyle \textbf{Question 10. }\text{If }f(x)=4x-x^2,\ x\in R,\text{ then write the value of } \\ f(a+1)-f(a-1).
\displaystyle \text{Answer:}
\displaystyle f(a+1)=4(a+1)-(a+1)^2
\displaystyle =4a+4-a^2-2a-1
\displaystyle =-a^2+2a+3
\displaystyle f(a-1)=4(a-1)-(a-1)^2
\displaystyle =4a-4-a^2+2a-1
\displaystyle =-a^2+6a-5
\displaystyle \therefore f(a+1)-f(a-1)=8-4a
\\

\displaystyle \textbf{Question 11. }\text{If }f,g,h\text{ are real functions given by }f(x)=x^2,\ g(x)=\tan x\text{ and } \\ h(x)=\log_e x,\text{ then write the value of }(h\circ g\circ f)\left(\sqrt{\frac{\pi}{4}}\right).
\displaystyle \text{Answer:}
\displaystyle f\left(\sqrt{\frac{\pi}{4}}\right)=\frac{\pi}{4}
\displaystyle g\left(\frac{\pi}{4}\right)=\tan\frac{\pi}{4}=1
\displaystyle h(1)=\log_e1=0
\displaystyle \therefore (h\circ g\circ f)\left(\sqrt{\frac{\pi}{4}}\right)=0
\\

\displaystyle \textbf{Question 12. }\text{Write the domain and range of function }f(x)\text{ given by } \\ f(x)=\frac{1}{\sqrt{x-|x|}}.
\displaystyle \text{Answer:}
\displaystyle x-|x|>0
\displaystyle \text{If }x\geq0,\ x-|x|=0
\displaystyle \text{If }x<0,\ x-|x|=2x<0
\displaystyle \therefore \text{No real value of }x\text{ is possible}
\displaystyle \text{Domain}=\phi
\displaystyle \text{Range}=\phi
\\

\displaystyle \textbf{Question 13. }\text{Write the domain and range of }f(x)=\sqrt{x-[x]}.
\displaystyle \text{Answer:}
\displaystyle 0\leq x-[x]<1
\displaystyle \therefore 0\leq\sqrt{x-[x]}<1
\displaystyle \text{Domain}=R
\displaystyle \text{Range}=[0,1)
\\

\displaystyle \textbf{Question 14. }\text{Write the domain and range of function }f(x)\text{ given by } \\ f(x)=\sqrt{[x]-x}.
\displaystyle \text{Answer:}
\displaystyle [x]-x\leq0
\displaystyle \text{For square root to exist, }[x]-x\geq0
\displaystyle \therefore [x]-x=0
\displaystyle x=[x]\Rightarrow x\in Z
\displaystyle \text{Domain}=Z
\displaystyle \text{Range}=\{0\}
\\

\displaystyle \textbf{Question 15. }\text{Let }A\text{ and }B\text{ be two sets such that }n(A)=p\text{ and }n(B)=q, \\ \text{ write the number of functions from }A\text{ to }B.
\displaystyle \text{Answer:}
\displaystyle \text{Number of functions from }A\text{ to }B=q^p
\\

\displaystyle \textbf{Question 16. }\text{Let }f\text{ and }g\text{ be two functions given by}
\displaystyle f=\{(2,4),(5,6),(8,-1),(10,-3)\}\text{ and }g=\{(2,5),(7,1),(8,4),(10,13),(11,-5)\}.
\displaystyle \text{Find the domain of }f+g.
\displaystyle \text{Answer:}
\displaystyle \text{Domain of }f=\{2,5,8,10\}
\displaystyle \text{Domain of }g=\{2,7,8,10,11\}
\displaystyle \text{Domain of }(f+g)=\text{Domain of }f\cap\text{Domain of }g
\displaystyle =\{2,8,10\}
\\

\displaystyle \textbf{Question 17. }\text{Find the set of values of }x\text{ for which the functions } \\ f(x)=3x^2-1\text{ and }g(x)=3+x\text{ are equal.}
\displaystyle \text{Answer:}
\displaystyle 3x^2-1=3+x
\displaystyle 3x^2-x-4=0
\displaystyle (3x-4)(x+1)=0
\displaystyle x=\frac{4}{3},-1
\displaystyle \therefore \text{Required set}=\left\{-1,\frac{4}{3}\right\}
\\

\displaystyle \textbf{Question 18. }\text{Let }f\text{ and }g\text{ be two real functions given by}
\displaystyle f=\{(0,1),(2,0),(3,-4),(4,2),(5,1)\}\text{ and }g=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\}.
\displaystyle \text{Find the domain of }fg.
\displaystyle \text{Answer:}
\displaystyle \text{Domain of }f=\{0,2,3,4,5\}
\displaystyle \text{Domain of }g=\{1,2,3,4,5\}
\displaystyle \text{Domain of }fg=\text{Domain of }f\cap\text{Domain of }g
\displaystyle =\{2,3,4,5\}

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