Class 11: Question Bank & Solutions (ISC)

$\displaystyle \text{MULTIPLE CHOICE QUESTIONS (MCQs)}$

$\displaystyle \textbf{Question 1: }\text{If }A=\{1,2,4\},\ B=\{2,4,5\},\ C=\{2,5\},\text{ then}$
$\displaystyle (A-B)\times(B-C)\text{ is}$
$\displaystyle \text{(a) }\{(1,2),(1,5),(2,5)\}\qquad\text{(b) }\{(1,4)\}\qquad\text{(c) }(1,4)$
$\displaystyle \text{(d) none of these.}$
$\displaystyle \text{Answer:}$
$\displaystyle A-B=\{1\}$
$\displaystyle B-C=\{4\}$
$\displaystyle \therefore (A-B)\times(B-C)=\{1\}\times\{4\}=\{(1,4)\}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{If }R\text{ is a relation on the set }A=\{1,2,3,4,5,6,7,8,9\}\text{ given}$
$\displaystyle \text{by }xRy\Leftrightarrow y=3x,\text{ then }R=$
$\displaystyle \text{(a) }\{(3,1),(6,2),(8,2),(9,3)\}\qquad\text{(b) }\{(3,1),(6,2),(9,3)\}$
$\displaystyle \text{(c) }\{(3,1),(2,6),(3,9)\}\qquad\text{(d) none of these.}$
$\displaystyle \text{Answer:}$
$\displaystyle xRy\Leftrightarrow y=3x$
$\displaystyle \text{For }x=1,\ y=3;\qquad x=2,\ y=6;\qquad x=3,\ y=9.$
$\displaystyle \text{For }x\geq4,\ y=3x>9,\text{ so }y\notin A.$
$\displaystyle \therefore R=\{(1,3),(2,6),(3,9)\}$
$\displaystyle \text{This relation is not among options (a), (b) and (c).}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{Let }A=\{1,2,3\},\ B=\{1,3,5\}.\text{ If relation }R\text{ from }A\text{ to }B$
$\displaystyle \text{is given by }R=\{(1,3),(2,5),(3,3)\}.\text{ Then, }R^{-1}\text{ is}$
$\displaystyle \text{(a) }\{(3,3),(3,1),(5,2)\}\qquad\text{(b) }\{(1,3),(2,5),(3,3)\}$
$\displaystyle \text{(c) }\{(1,3),(5,2)\}\qquad\text{(d) none of these.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The inverse relation is obtained by interchanging the components of each ordered pair.}$
$\displaystyle R^{-1}=\{(3,1),(5,2),(3,3)\}$
$\displaystyle =\{(3,3),(3,1),(5,2)\}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If }A=\{1,2,3\},\ B=\{1,4,6,9\}\text{ and }R\text{ is a relation from }A\text{ to }B$
$\displaystyle \text{defined by `}x\text{ is greater than }y\text{'. The range of }R\text{ is}$
$\displaystyle \text{(a) }\{1,4,6,9\}\qquad\text{(b) }\{4,6,9\}\qquad\text{(c) }\{1\}$
$\displaystyle \text{(d) none of these.}$
$\displaystyle \text{Answer:}$
$\displaystyle R=\{(x,y):x\in A,\ y\in B\text{ and }x>y\}$
$\displaystyle \text{For }x=2,\ y=1,\text{ and for }x=3,\ y=1.$
$\displaystyle \therefore R=\{(2,1),(3,1)\}$
$\displaystyle \therefore \text{Range}(R)=\{1\}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }R=\{(x,y):x,y\in Z,\ x^2+y^2\leq4\}\text{ is a relation on }Z,$
$\displaystyle \text{then domain of }R\text{ is}$
$\displaystyle \text{(a) }\{0,1,2\}\qquad\text{(b) }\{0,-1,-2\}\qquad\text{(c) }\{-2,-1,0,1,2\}$
$\displaystyle \text{(d) none of these.}$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+y^2\leq4$
$\displaystyle \text{Since }y^2\geq0,\text{ we must have }x^2\leq4.$
$\displaystyle \therefore -2\leq x\leq2$
$\displaystyle \text{For each }x=-2,-1,0,1,2,\text{ there exists an integer }y\text{ satisfying the relation.}$
$\displaystyle \therefore \text{Domain}(R)=\{-2,-1,0,1,2\}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{A relation }R\text{ is defined from }\{2,3,4,5\}\text{ to }\{3,6,7,10\}\text{ by:}$
$\displaystyle xRy\Leftrightarrow x\text{ is relatively prime to }y.\text{ Then, domain of }R\text{ is}$
$\displaystyle \text{(a) }\{2,3,5\}\qquad\text{(b) }\{3,5\}\qquad\text{(c) }\{2,3,4\}$
$\displaystyle \text{(d) }\{2,3,4,5\}.$
$\displaystyle \text{Answer:}$
$\displaystyle \gcd(2,3)=1,\qquad \gcd(3,7)=1$
$\displaystyle \gcd(4,3)=1,\qquad \gcd(5,3)=1$
$\displaystyle \text{Thus, each element of }\{2,3,4,5\}\text{ is related to at least one element of the second set.}$
$\displaystyle \therefore \text{Domain}(R)=\{2,3,4,5\}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{A relation }\phi\text{ from }C\text{ to }R\text{ is defined by }x\phi y\Leftrightarrow |x|=y.$
$\displaystyle \text{Which one is correct?}$
$\displaystyle \text{(a) }(2+3i)\phi13\qquad\text{(b) }3\phi(-3)\qquad\text{(c) }(1+i)\phi2\qquad\text{(d) }i\phi1$
$\displaystyle \text{Answer:}$
$\displaystyle x\phi y\Leftrightarrow |x|=y$
$\displaystyle |2+3i|=\sqrt{2^2+3^2}=\sqrt{13}\ne13$
$\displaystyle |3|=3\ne-3$
$\displaystyle |1+i|=\sqrt{1^2+1^2}=\sqrt2\ne2$
$\displaystyle |i|=1$
$\displaystyle \therefore i\phi1$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{Let }R\text{ be a relation on }N\text{ defined by }x+2y=8.\text{ The domain of }R\text{ is}$
$\displaystyle \text{(a) }\{2,4,8\}\qquad\text{(b) }\{2,4,6,8\}\qquad\text{(c) }\{2,4,6\}\qquad\text{(d) }\{1,2,3,4\}.$
$\displaystyle \text{Answer:}$
$\displaystyle x+2y=8\Rightarrow x=8-2y$
$\displaystyle \text{Since }x,y\in N,\text{ take }y=1,2,3.$
$\displaystyle y=1\Rightarrow x=6,\qquad y=2\Rightarrow x=4,\qquad y=3\Rightarrow x=2$
$\displaystyle \therefore \text{Domain}(R)=\{2,4,6\}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{R is a relation from }\{11,12,13\}\text{ to }\{8,10,12\}\text{ defined by}$
$\displaystyle y=x-3.\text{ Then, }R^{-1}\text{ is}$
$\displaystyle \text{(a) }\{(8,11),(10,13)\}\qquad\text{(b) }\{(11,8),(13,10)\}$
$\displaystyle \text{(c) }\{(10,13),(8,11),(12,10)\}\qquad\text{(d) none of these.}$
$\displaystyle \text{Answer:}$
$\displaystyle y=x-3$
$\displaystyle x=11\Rightarrow y=8,\qquad x=12\Rightarrow y=9,\qquad x=13\Rightarrow y=10$
$\displaystyle \text{Since }9\notin\{8,10,12\},\ (12,9)\text{ does not belong to }R.$
$\displaystyle \therefore R=\{(11,8),(13,10)\}$
$\displaystyle \therefore R^{-1}=\{(8,11),(10,13)\}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If the set }A\text{ has }p\text{ elements, }B\text{ has }q\text{ elements, then the number}$
$\displaystyle \text{of elements in }A\times B\text{ is}$
$\displaystyle \text{(a) }p+q\qquad\text{(b) }p+q+1\qquad\text{(c) }pq\qquad\text{(d) }p^2$
$\displaystyle \text{Answer:}$
$\displaystyle n(A\times B)=n(A)\times n(B)$
$\displaystyle =p\times q=pq$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Let }R\text{ be a relation from a set }A\text{ to a set }B,\text{ then}$
$\displaystyle \text{(a) }R=A\cup B\qquad\text{(b) }R=A\cap B\qquad\text{(c) }R\subseteq A\times B$
$\displaystyle \text{(d) }R\subseteq B\times A.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{By definition, a relation from }A\text{ to }B\text{ is a subset of }A\times B.$
$\displaystyle \therefore R\subseteq A\times B$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{If }R\text{ is a relation from a finite set }A\text{ having }m\text{ elements to a}$
$\displaystyle \text{finite set }B\text{ having }n\text{ elements, then the number of relations from }A\text{ to }B\text{ is}$
$\displaystyle \text{(a) }2^{mn}\qquad\text{(b) }2^{mn}-1\qquad\text{(c) }2mn\qquad\text{(d) }m^n$
$\displaystyle \text{Answer:}$
$\displaystyle n(A\times B)=n(A)n(B)=mn$
$\displaystyle \text{Every subset of }A\times B\text{ defines a relation from }A\text{ to }B.$
$\displaystyle \therefore \text{Number of relations}=2^{mn}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{If }R\text{ is a relation on a finite set having }n\text{ elements, then the number}$
$\displaystyle \text{of relations on }A\text{ is}$
$\displaystyle \text{(a) }2^n\qquad\text{(b) }2^{n^2}\qquad\text{(c) }n^2\qquad\text{(d) }n^n$
$\displaystyle \text{Answer:}$
$\displaystyle n(A\times A)=n(A)n(A)=n^2$
$\displaystyle \text{A relation on }A\text{ is any subset of }A\times A.$
$\displaystyle \therefore \text{Number of relations on }A=2^{n^2}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{Let }A=\{1,2,3\},\ B=\{2,3,4\},\text{ then which of the following is a}$
$\displaystyle \text{function from }A\text{ to }B?$
$\displaystyle \text{(a) }\{(1,2),(1,3),(2,3),(3,3)\}$
$\displaystyle \text{(b) }\{(1,3),(2,4)\}$
$\displaystyle \text{(c) }\{(1,3),(2,2),(3,3)\}$
$\displaystyle \text{(d) }\{(1,2),(2,3),(3,2),(3,4)\}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For a function from }A\text{ to }B,\text{ every element of }A\text{ must have exactly one image in }B.$
$\displaystyle \text{In option (c), }1\mapsto3,\quad2\mapsto2,\quad3\mapsto3.$
$\displaystyle \text{Thus, every element of }A\text{ has exactly one image in }B.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{If }f:Q\to Q\text{ is defined as }f(x)=x^2,\text{ then }f^{-1}(9)\text{ is equal to}$
$\displaystyle \text{(a) }3\qquad\text{(b) }-3\qquad\text{(c) }\{-3,3\}\qquad\text{(d) }\phi$
$\displaystyle \text{Answer:}$
$\displaystyle f^{-1}(9)=\{x\in Q:f(x)=9\}$
$\displaystyle \therefore x^2=9$
$\displaystyle \therefore x=\pm3$
$\displaystyle \therefore f^{-1}(9)=\{-3,3\}$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{Which one of the following is not a function?}$
$\displaystyle \text{(a) }\{(x,y):x,y\in R,\ x^2=y\}$
$\displaystyle \text{(b) }\{(x,y):x,y\in R,\ y^2=x\}$
$\displaystyle \text{(c) }\{(x,y):x,y\in R,\ x=y^3\}$
$\displaystyle \text{(d) }\{(x,y):x,y\in R,\ y=x^3\}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For option (b), }y^2=x\Rightarrow y=\pm\sqrt{x}.$
$\displaystyle \text{Thus, for }x>0,\text{ one value of }x\text{ corresponds to two values of }y.$
$\displaystyle \therefore \text{Option (b) is not a function.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{If }f(x)=\cos(\log x),\text{ then }f(x^2)f(y^2)-\frac{1}{2}\left\{f\left(\frac{x^2}{y^2}\right)\right.$
$\displaystyle \left.+f(x^2y^2)\right\}\text{ has the value}$
$\displaystyle \text{(a) }-2\qquad\text{(b) }-1\qquad\text{(c) }\frac{1}{2}\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a=\log x\text{ and }b=\log y.$
$\displaystyle f(x^2)=\cos2a,\qquad f(y^2)=\cos2b$
$\displaystyle f\left(\frac{x^2}{y^2}\right)=\cos(2a-2b),\qquad f(x^2y^2)=\cos(2a+2b)$
$\displaystyle \text{Required value}=\cos2a\cos2b$
$\displaystyle \qquad-\frac{1}{2}\{\cos(2a-2b)+\cos(2a+2b)\}$
$\displaystyle \text{Using }\cos(A-B)+\cos(A+B)=2\cos A\cos B,$
$\displaystyle \text{Required value}=\cos2a\cos2b-\cos2a\cos2b=0.$
$\displaystyle \therefore \text{None of the given numerical values is correct.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{If }f(x)=\cos(\log x),\text{ then }f(x)f(y)-\frac{1}{2}\left\{f\left(\frac{x}{y}\right)+f(xy)\right\}$
$\displaystyle \text{has the value}$
$\displaystyle \text{(a) }-1\qquad\text{(b) }\frac{1}{2}\qquad\text{(c) }-2\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a=\log x\text{ and }b=\log y.$
$\displaystyle f(x)=\cos a,\qquad f(y)=\cos b$
$\displaystyle f\left(\frac{x}{y}\right)=\cos(a-b),\qquad f(xy)=\cos(a+b)$
$\displaystyle \text{Required value}=\cos a\cos b-\frac{1}{2}\{\cos(a-b)+\cos(a+b)\}$
$\displaystyle \text{Using }\cos(a-b)+\cos(a+b)=2\cos a\cos b,$
$\displaystyle \text{Required value}=\cos a\cos b-\cos a\cos b=0.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Let }f(x)=|x-1|.\text{ Then,}$
$\displaystyle \text{(a) }f(x^2)=[f(x)]^2$
$\displaystyle \text{(b) }f(x+y)=f(x)f(y)$
$\displaystyle \text{(c) }f(|x|)=|f(x)|$
$\displaystyle \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For (a), }f(x^2)=|x^2-1|,\quad [f(x)]^2=(x-1)^2.$
$\displaystyle \text{These are not equal for all }x.$
$\displaystyle \text{For (b), }f(x+y)=|x+y-1|,\quad f(x)f(y)=|x-1||y-1|.$
$\displaystyle \text{These are not equal for all }x,y.$
$\displaystyle \text{For (c), }f(|x|)=||x|-1|,\quad |f(x)|=|x-1|.$
$\displaystyle \text{For }x=-2,\quad f(|x|)=1\text{ but }|f(x)|=3.$
$\displaystyle \therefore \text{None of (a), (b) and (c) is true for all permissible values.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{The range of }f(x)=\cos[x],\text{ for }-\frac{\pi}{2}<x<\frac{\pi}{2}\text{ is}$
$\displaystyle \text{(a) }\{-1,1,0\}\qquad\text{(b) }\{\cos1,\cos2,1\}$
$\displaystyle \text{(c) }\{\cos1,-\cos1,1\}\qquad\text{(d) }[-1,1]$
$\displaystyle \text{Answer:}$
$\displaystyle -\frac{\pi}{2}<x<\frac{\pi}{2}\Rightarrow -1.57\ldots<x<1.57\ldots$
$\displaystyle \therefore [x]\in\{-2,-1,0,1\}$
$\displaystyle \therefore f(x)\in\{\cos(-2),\cos(-1),\cos0,\cos1\}$
$\displaystyle =\{\cos2,\cos1,1\}$
$\displaystyle \therefore \text{Range}(f)=\{\cos1,\cos2,1\}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Which of the following are functions?}$
$\displaystyle \text{(a) }\{(x,y):y^2=x,\ x,y\in R\}$
$\displaystyle \text{(b) }\{(x,y):y=|x|,\ x,y\in R\}$
$\displaystyle \text{(c) }\{(x,y):x^2+y^2=1,\ x,y\in R\}$
$\displaystyle \text{(d) }\{(x,y):x^2-y^2=1,\ x,y\in R\}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In (b), }y=|x|\text{ gives exactly one value of }y\text{ for every }x\in R.$
$\displaystyle \text{In (a), (c) and (d), suitable values of }x\text{ give two values of }y.$
$\displaystyle \therefore \text{Only (b) represents a function.}$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{If }f(x)=\log\left(\frac{1+x}{1-x}\right)\text{ and }g(x)=\frac{3x+x^3}{1+3x^2},$
$\displaystyle \text{then }f(g(x))\text{ is equal to}$
$\displaystyle \text{(a) }f(3x)\qquad\text{(b) }\{f(x)\}^3\qquad\text{(c) }3f(x)\qquad\text{(d) }-f(x)$
$\displaystyle \text{Answer:}$
$\displaystyle f(g(x))=\log\left(\frac{1+g(x)}{1-g(x)}\right)$
$\displaystyle =\log\left(\frac{1+3x^2+3x+x^3}{1+3x^2-3x-x^3}\right)$
$\displaystyle =\log\left(\frac{(1+x)^3}{(1-x)^3}\right)$
$\displaystyle =3\log\left(\frac{1+x}{1-x}\right)=3f(x)$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{If }A=\{1,2,3\},\ B=\{x,y\},\text{ then the number of functions}$
$\displaystyle \text{that can be defined from }A\text{ into }B\text{ is}$
$\displaystyle \text{(a) }12\qquad\text{(b) }8\qquad\text{(c) }6\qquad\text{(d) }3$
$\displaystyle \text{Answer:}$
$\displaystyle n(A)=3\quad\text{and}\quad n(B)=2.$
$\displaystyle \text{Each element of }A\text{ can be mapped to either of the }2\text{ elements of }B.$
$\displaystyle \therefore \text{Number of functions from }A\text{ into }B=2^3=8.$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{If }f(x)=\log\left(\frac{1+x}{1-x}\right),\text{ then }f\left(\frac{2x}{1+x^2}\right)\text{ is equal to}$
$\displaystyle \text{(a) }\{f(x)\}^2\qquad\text{(b) }\{f(x)\}^3\qquad\text{(c) }2f(x)\qquad\text{(d) }3f(x)$
$\displaystyle \text{Answer:}$
$\displaystyle f\left(\frac{2x}{1+x^2}\right)=\log\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right)$
$\displaystyle =\log\left(\frac{1+x^2+2x}{1+x^2-2x}\right)$
$\displaystyle =\log\left(\frac{(1+x)^2}{(1-x)^2}\right)$
$\displaystyle =2\log\left(\frac{1+x}{1-x}\right)$
$\displaystyle =2f(x)$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{If }f(x)=\cos(\log x),\text{ then value of }f(x)f(4)-\frac{1}{2}\left\{f\left(\frac{x}{4}\right)\right.$
$\displaystyle \left.+f(4x)\right\}\text{ is}$
$\displaystyle \text{(a) }1\qquad\text{(b) }-1\qquad\text{(c) }0\qquad\text{(d) }\pm1$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a=\log x\text{ and }b=\log4.$
$\displaystyle f(x)=\cos a,\qquad f(4)=\cos b$
$\displaystyle f\left(\frac{x}{4}\right)=\cos(a-b),\qquad f(4x)=\cos(a+b)$
$\displaystyle \text{Required value}=\cos a\cos b-\frac{1}{2}\{\cos(a-b)+\cos(a+b)\}$
$\displaystyle \text{Using }\cos(a-b)+\cos(a+b)=2\cos a\cos b,$
$\displaystyle \text{Required value}=\cos a\cos b-\cos a\cos b=0.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{If }f(x)=\frac{2^x+2^{-x}}{2},\text{ then }f(x+y)f(x-y)\text{ is equal to}$
$\displaystyle \text{(a) }\frac{1}{2}\{f(2x)+f(2y)\}\qquad\text{(b) }\frac{1}{2}\{f(2x)-f(2y)\}$
$\displaystyle \text{(c) }\frac{1}{4}\{f(2x)+f(2y)\}\qquad\text{(d) }\frac{1}{4}\{f(2x)-f(2y)\}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x+y)f(x-y)=\frac{2^{x+y}+2^{-x-y}}{2}\cdot\frac{2^{x-y}+2^{-x+y}}{2}$
$\displaystyle =\frac{1}{4}\left(2^{2x}+2^{2y}+2^{-2y}+2^{-2x}\right)$
$\displaystyle =\frac{1}{2}\left\{\frac{2^{2x}+2^{-2x}}{2}+\frac{2^{2y}+2^{-2y}}{2}\right\}$
$\displaystyle =\frac{1}{2}\{f(2x)+f(2y)\}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{If }2f(x)-3f\left(\frac{1}{x}\right)=x^2\ (x\ne0),\text{ then }f(2)\text{ is equal to}$
$\displaystyle \text{(a) }-\frac{7}{4}\qquad\text{(b) }\frac{5}{2}\qquad\text{(c) }-1\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Putting }x=2,\quad 2f(2)-3f\left(\frac{1}{2}\right)=4\qquad ...(1)$
$\displaystyle \text{Putting }x=\frac{1}{2},\quad 2f\left(\frac{1}{2}\right)-3f(2)=\frac{1}{4}\qquad ...(2)$
$\displaystyle \text{Multiplying (1) by }2,\quad4f(2)-6f\left(\frac{1}{2}\right)=8$
$\displaystyle \text{Multiplying (2) by }3,\quad6f\left(\frac{1}{2}\right)-9f(2)=\frac{3}{4}$
$\displaystyle \text{Adding,}\quad-5f(2)=8+\frac{3}{4}=\frac{35}{4}$
$\displaystyle \therefore f(2)=-\frac{7}{4}$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Let }f:R\to R\text{ be defined by }f(x)=2x+|x|.\text{ Then}$
$\displaystyle f(2x)+f(-x)-f(x)=$
$\displaystyle \text{(a) }2x\qquad\text{(b) }2|x|\qquad\text{(c) }-2x\qquad\text{(d) }-2|x|$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Given, }f(x)=2x+|x|.$
$\displaystyle f(2x)=4x+|2x|=4x+2|x|$
$\displaystyle f(-x)=-2x+|-x|=-2x+|x|$
$\displaystyle \therefore f(2x)+f(-x)-f(x)$
$\displaystyle =4x+2|x|-2x+|x|-(2x+|x|)$
$\displaystyle =2|x|$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{The range of the function }f(x)=\frac{x^2-x}{x^2+2x}\text{ is}$
$\displaystyle \text{(a) }R\qquad\text{(b) }R-\{1\}\qquad\text{(c) }R-\left\{-\frac12,1\right\}$
$\displaystyle \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\frac{x(x-1)}{x(x+2)}=\frac{x-1}{x+2},\qquad x\ne0,-2.$
$\displaystyle \text{Let }y=\frac{x-1}{x+2}.$
$\displaystyle yx+2y=x-1$
$\displaystyle \Rightarrow x(y-1)=-(1+2y)$
$\displaystyle \Rightarrow x=\frac{1+2y}{1-y},\qquad y\ne1.$
$\displaystyle \text{Also, }x\ne0\Rightarrow 1+2y\ne0\Rightarrow y\ne-\frac12.$
$\displaystyle \therefore \text{Range of }f=R-\left\{-\frac12,1\right\}.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{If }x\ne1\text{ and }f(x)=\frac{x+1}{x-1}\text{ is a real function, then}$
$\displaystyle f(f(f(2)))\text{ is}$
$\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }3\qquad\text{(d) }4$
$\displaystyle \text{Answer:}$
$\displaystyle f(2)=\frac{2+1}{2-1}=3$
$\displaystyle f(f(2))=f(3)=\frac{3+1}{3-1}=2$
$\displaystyle f(f(f(2)))=f(2)=3$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 31: }\text{If }f(x)=\cos(\log_e x),\text{ then }f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f(xy)\right.$
$\displaystyle \left.+f\left(\frac{x}{y}\right)\right\}\text{ is equal to}$
$\displaystyle \text{(a) }\cos(x-y)\qquad\text{(b) }\log(\cos(x-y))$
$\displaystyle \text{(c) }1\qquad\text{(d) }\cos(x+y)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }a=\log_e x\text{ and }b=\log_e y.$
$\displaystyle f\left(\frac{1}{x}\right)=\cos(-a)=\cos a,\qquad f\left(\frac{1}{y}\right)=\cos(-b)=\cos b$
$\displaystyle f(xy)=\cos(a+b),\qquad f\left(\frac{x}{y}\right)=\cos(a-b)$
$\displaystyle \text{Required value}=\cos a\cos b-\frac{1}{2}\{\cos(a+b)+\cos(a-b)\}$
$\displaystyle \text{Using }\cos(a+b)+\cos(a-b)=2\cos a\cos b,$
$\displaystyle \text{Required value}=\cos a\cos b-\cos a\cos b=0.$
$\displaystyle \therefore \text{None of the given options is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 32: }\text{Let }f(x)=x,\ g(x)=\frac{1}{x}\text{ and }h(x)=f(x)\,g(x).\text{ Then, }h(x)=1\text{ for}$
$\displaystyle \text{(a) }x\in R\qquad\text{(b) }x\in Q\qquad\text{(c) }x\in R-Q\qquad\text{(d) }x\in R,\ x\ne0$
$\displaystyle \text{Answer:}$
$\displaystyle h(x)=f(x)\,g(x)=x\left(\frac{1}{x}\right)=1,\qquad x\ne0$
$\displaystyle \text{Since }g(x)=\frac{1}{x}\text{ is not defined at }x=0,$
$\displaystyle h(x)=1\text{ for all }x\in R,\ x\ne0.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 33: }\text{If }f(x)=\frac{\sin^4x+\cos^2x}{\sin^2x+\cos^4x}\text{ for }x\in R,\text{ then }f(2002)=$
$\displaystyle \text{(a) }1\qquad\text{(b) }2\qquad\text{(c) }3\qquad\text{(d) }4$
$\displaystyle \text{Answer:}$
$\displaystyle \sin^4x+\cos^2x=\sin^4x+\cos^2x(\sin^2x+\cos^2x)$
$\displaystyle =\sin^4x+\sin^2x\cos^2x+\cos^4x$
$\displaystyle =\sin^2x(\sin^2x+\cos^2x)+\cos^4x$
$\displaystyle =\sin^2x+\cos^4x$
$\displaystyle \therefore f(x)=1$
$\displaystyle \therefore f(2002)=1.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 34: }\text{The function }f:R\to R\text{ is defined by }$
$\displaystyle f(x)=\cos^2x+\sin^4x.\text{ Then, }f(R)=$
$\displaystyle \text{(a) }\left[\frac34,1\right)\qquad\text{(b) }\left(\frac34,1\right]$
$\displaystyle \text{(c) }\left[\frac34,1\right]\qquad\text{(d) }\left(\frac34,1\right)$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=1-\sin^2x+\sin^4x$
$\displaystyle \text{Let }t=\sin^2x.\text{ Then }0\leq t\leq1.$
$\displaystyle f(x)=1-t+t^2=\left(t-\frac12\right)^2+\frac34$
$\displaystyle \therefore \frac34\leq f(x)\leq1$
$\displaystyle \therefore f(R)=\left[\frac34,1\right].$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 35: }\text{Let }A=\{x\in R:x\ne0,\,-4\leq x\leq4\}\text{ and}$
$\displaystyle f:A\to R\text{ be defined by }f(x)=\frac{|x|}{x}\text{ for }x\in A.\text{ Then }f(A)\text{ is}$
$\displaystyle \text{(a) }\{1,-1\}\qquad\text{(b) }\{x:0\leq x\leq4\}$
$\displaystyle \text{(c) }\{1\}\qquad\text{(d) }\{x:-4\leq x\leq0\}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\frac{|x|}{x}=\begin{cases}1,&x>0\\-1,&x<0\end{cases}$
$\displaystyle \text{Since }A=[-4,0)\cup(0,4],\text{ both cases occur.}$
$\displaystyle \therefore f(A)=\{-1,1\}.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 36: }\text{If }f:R\to R\text{ and }g:R\to R\text{ are defined by}$
$\displaystyle f(x)=2x+3\text{ and }g(x)=x^2+7,\text{ then the values of }x\text{ such that}$
$\displaystyle g(f(x))=8\text{ are}$
$\displaystyle \text{(a) }1,2\qquad\text{(b) }-1,2\qquad\text{(c) }-1,-2\qquad\text{(d) }1,-2$
$\displaystyle \text{Answer:}$
$\displaystyle g(f(x))=(2x+3)^2+7$
$\displaystyle \text{Given, }g(f(x))=8.$
$\displaystyle (2x+3)^2+7=8$
$\displaystyle \Rightarrow (2x+3)^2=1$
$\displaystyle \Rightarrow 2x+3=\pm1$
$\displaystyle \Rightarrow x=-1\text{ or }x=-2$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 37: }\text{If }f:[-2,2]\to R\text{ is defined by}$
$\displaystyle f(x)=\begin{cases}-1,&-2\leq x\leq0\\x-1,&0\leq x\leq2,\end{cases}\text{ then}$
$\displaystyle \{x\in[-2,2]:x\leq0\text{ and }f(|x|)=x\}=$
$\displaystyle \text{(a) }\{-1\}\qquad\text{(b) }\{0\}\qquad\text{(c) }\left\{-\frac12\right\}\qquad\text{(d) }\phi$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }x\leq0,\quad |x|=-x\geq0.$
$\displaystyle \therefore f(|x|)=|x|-1=-x-1$
$\displaystyle \text{Given, }f(|x|)=x.$
$\displaystyle -x-1=x$
$\displaystyle \Rightarrow 2x=-1$
$\displaystyle \Rightarrow x=-\frac12$
$\displaystyle \text{Also, }-\frac12\in[-2,2]\text{ and }-\frac12\leq0.$
$\displaystyle \therefore \{x\in[-2,2]:x\leq0\text{ and }f(|x|)=x\}=\left\{-\frac12\right\}.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 38: }\text{If }e^{f(x)}=\frac{10+x}{10-x},\ x\in(-10,10)\text{ and}$
$\displaystyle f(x)=k\,f\left(\frac{200x}{100+x^2}\right),\text{ then }k=$
$\displaystyle \text{(a) }0.5\qquad\text{(b) }0.6\qquad\text{(c) }0.7\qquad\text{(d) }0.8$
$\displaystyle \text{Answer:}$
$\displaystyle e^{f(x)}=\frac{10+x}{10-x}$
$\displaystyle \Rightarrow f(x)=\log\left(\frac{10+x}{10-x}\right)$
$\displaystyle \text{Let }t=\frac{200x}{100+x^2}.$
$\displaystyle f(t)=\log\left(\frac{10+t}{10-t}\right)$
$\displaystyle =\log\left(\frac{10(100+x^2)+200x}{10(100+x^2)-200x}\right)$
$\displaystyle =\log\left(\frac{x^2+20x+100}{x^2-20x+100}\right)$
$\displaystyle =\log\left(\frac{(x+10)^2}{(x-10)^2}\right)$
$\displaystyle =2\log\left(\frac{10+x}{10-x}\right)=2f(x)$
$\displaystyle \therefore f(x)=\frac12f\left(\frac{200x}{100+x^2}\right)$
$\displaystyle \therefore k=\frac12=0.5.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 39: }\text{If }f\text{ is a real valued function given by}$
$\displaystyle f(x)=27x^3+\frac{1}{x^3}\text{ and }\alpha,\beta\text{ are roots of }3x+\frac1x=12.\text{ Then,}$
$\displaystyle \text{(a) }f(\alpha)\ne f(\beta)\qquad\text{(b) }f(\alpha)=10$
$\displaystyle \text{(c) }f(\beta)=-10\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle 3x+\frac1x=12$
$\displaystyle \text{Cubing both sides,}$
$\displaystyle \left(3x+\frac1x\right)^3=12^3$
$\displaystyle 27x^3+\frac1{x^3}+3\left(3x\right)\left(\frac1x\right)\left(3x+\frac1x\right)=1728$
$\displaystyle 27x^3+\frac1{x^3}+9(12)=1728$
$\displaystyle \Rightarrow 27x^3+\frac1{x^3}=1620$
$\displaystyle \therefore f(\alpha)=f(\beta)=1620.$
$\displaystyle \therefore \text{Options (a), (b) and (c) are false.}$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 40: }\text{If }f(x)=64x^3+\frac{1}{x^3}\text{ and }\alpha,\beta\text{ are the roots of}$
$\displaystyle 4x+\frac{1}{x}=3.\text{ Then,}$
$\displaystyle \text{(a) }f(\alpha)=f(\beta)=-9\qquad\text{(b) }f(\alpha)=f(\beta)=63$
$\displaystyle \text{(c) }f(\alpha)\ne f(\beta)\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle 4x+\frac{1}{x}=3$
$\displaystyle \text{Cubing both sides,}$
$\displaystyle \left(4x+\frac{1}{x}\right)^3=27$
$\displaystyle 64x^3+\frac{1}{x^3}+3(4x)\left(\frac{1}{x}\right)\left(4x+\frac{1}{x}\right)=27$
$\displaystyle 64x^3+\frac{1}{x^3}+12(3)=27$
$\displaystyle \Rightarrow 64x^3+\frac{1}{x^3}=-9$
$\displaystyle \therefore f(\alpha)=f(\beta)=-9.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 41: }\text{If }3f(x)+5f\left(\frac{1}{x}\right)=\frac{1}{x}-3$
$\displaystyle \text{for all non-zero }x,\text{ then }f(x)=$
$\displaystyle \text{(a) }\frac{1}{14}\left(\frac{3}{x}+5x-6\right)$
$\displaystyle \text{(b) }\frac{1}{14}\left(-\frac{3}{x}+5x-6\right)$
$\displaystyle \text{(c) }\frac{1}{14}\left(-\frac{3}{x}+5x+6\right)\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle 3f(x)+5f\left(\frac{1}{x}\right)=\frac{1}{x}-3\qquad ...(1)$
$\displaystyle \text{Replacing }x\text{ by }\frac{1}{x},$
$\displaystyle 3f\left(\frac{1}{x}\right)+5f(x)=x-3\qquad ...(2)$
$\displaystyle \text{Multiplying (1) by }3\text{ and (2) by }5,$
$\displaystyle 9f(x)+15f\left(\frac{1}{x}\right)=\frac{3}{x}-9$
$\displaystyle 25f(x)+15f\left(\frac{1}{x}\right)=5x-15$
$\displaystyle \text{Subtracting the first equation from the second,}$
$\displaystyle 16f(x)=5x-\frac{3}{x}-6$
$\displaystyle \therefore f(x)=\frac{1}{16}\left(-\frac{3}{x}+5x-6\right).$
$\displaystyle \therefore \text{None of the given expressions equals }f(x).$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 42: }\text{If }f:R\to R\text{ be given by }f(x)=\frac{4^x}{4^x+2}$
$\displaystyle \text{for all }x\in R.\text{ Then,}$
$\displaystyle \text{(a) }f(x)=f(1-x)\qquad\text{(b) }f(x)+f(1-x)=0$
$\displaystyle \text{(c) }f(x)+f(1-x)=1\qquad\text{(d) }f(x)+f(x-1)=1$
$\displaystyle \text{Answer:}$
$\displaystyle f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}$
$\displaystyle =\frac{4}{4+2(4^x)}=\frac{2}{2+4^x}$
$\displaystyle \therefore f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{2}{4^x+2}$
$\displaystyle =\frac{4^x+2}{4^x+2}=1$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 43: }\text{If }f(x)=\sin[\pi^2]x+\sin[-\pi^2]x,\text{ where }[x]$
$\displaystyle \text{denotes the greatest integer less than or equal to }x,\text{ then}$
$\displaystyle \text{(a) }f\left(\frac{\pi}{2}\right)=1\qquad\text{(b) }f(\pi)=2$
$\displaystyle \text{(c) }f\left(\frac{\pi}{4}\right)=-1\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle [\pi^2]=9\text{ and }[-\pi^2]=-10$
$\displaystyle \therefore f(x)=\sin9x-\sin10x$
$\displaystyle f\left(\frac{\pi}{2}\right)=\sin\frac{9\pi}{2}-\sin5\pi$
$\displaystyle =1-0=1$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 44: }\text{The domain of the function }f(x)=\sqrt{2-2x-x^2}\text{ is}$
$\displaystyle \text{(a) }[-\sqrt3,\sqrt3]\qquad\text{(b) }[-1-\sqrt3,-1+\sqrt3]$
$\displaystyle \text{(c) }[-2,2]\qquad\text{(d) }[-2-\sqrt3,-2+\sqrt3]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }f(x)\text{ to be real, }2-2x-x^2\geq0.$
$\displaystyle \Rightarrow x^2+2x-2\leq0$
$\displaystyle \Rightarrow (x+1)^2\leq3$
$\displaystyle \Rightarrow -\sqrt3\leq x+1\leq\sqrt3$
$\displaystyle \Rightarrow -1-\sqrt3\leq x\leq-1+\sqrt3$
$\displaystyle \therefore \text{Domain of }f=[-1-\sqrt3,-1+\sqrt3].$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 45: }\text{The domain of definition of }f(x)=\sqrt{\frac{x+3}{(2-x)(x-5)}}\text{ is}$
$\displaystyle \text{(a) }(-\infty,-3]\cup(2,5)\qquad\text{(b) }(-\infty,-3)\cup(2,5)$
$\displaystyle \text{(c) }(-\infty,-3]\cup[2,5]\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }f(x)\text{ to be real, }\frac{x+3}{(2-x)(x-5)}\geq0.$
$\displaystyle \text{The critical points are }-3,\ 2,\ 5.$
$\displaystyle \begin{array}{c|ccccccc}x&(-\infty,-3)&-3&(-3,2)&2&(2,5)&5&(5,\infty)\\ \hline \text{Sign}&+&0&-&\text{undefined}&+&\text{undefined}&-\end{array}$
$\displaystyle \therefore \text{Domain of }f=(-\infty,-3]\cup(2,5).$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 46: }\text{The domain of the function }f(x)=\sqrt{\frac{(x+1)(x-3)}{x-2}}\text{ is}$
$\displaystyle \text{(a) }[-1,2)\cup[3,\infty)\qquad\text{(b) }(-1,2)\cup[3,\infty)$
$\displaystyle \text{(c) }[-1,2]\cup[3,\infty)\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }f(x)\text{ to be real, }\frac{(x+1)(x-3)}{x-2}\geq0.$
$\displaystyle \text{The critical points are }-1,\ 2,\ 3.$
$\displaystyle \begin{array}{c|ccccccc}x&(-\infty,-1)&-1&(-1,2)&2&(2,3)&3&(3,\infty)\\ \hline \text{Sign}&-&0&+&\text{undefined}&-&0&+\end{array}$
$\displaystyle \therefore \text{Domain of }f=[-1,2)\cup[3,\infty).$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 47: }\text{The domain of definition of the function }$
$\displaystyle f(x)=\sqrt{x-1}+\sqrt{3-x}\text{ is}$
$\displaystyle \text{(a) }[1,\infty)\qquad\text{(b) }(-\infty,3)$
$\displaystyle \text{(c) }(1,3)\qquad\text{(d) }[1,3]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }f(x)\text{ to be real, both radicands must be non-negative.}$
$\displaystyle x-1\geq0\quad\text{and}\quad3-x\geq0$
$\displaystyle \Rightarrow x\geq1\quad\text{and}\quad x\leq3$
$\displaystyle \Rightarrow 1\leq x\leq3$
$\displaystyle \therefore \text{Domain of }f=[1,3].$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 48: }\text{The domain of definition of the function }$
$\displaystyle f(x)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}}\text{ is}$
$\displaystyle \text{(a) }(-\infty,-2]\cup[2,\infty)\qquad\text{(b) }[-1,1]$
$\displaystyle \text{(c) }\phi\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the first radical, }\frac{x-2}{x+2}\geq0,\quad x\ne-2.$
$\displaystyle \Rightarrow x\in(-\infty,-2)\cup[2,\infty)$
$\displaystyle \text{For the second radical, }\frac{1-x}{1+x}\geq0,\quad x\ne-1.$
$\displaystyle \Rightarrow x\in(-1,1]$
$\displaystyle \text{These two sets have no common element.}$
$\displaystyle \therefore \text{Domain of }f=\phi.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 49: }\text{The domain of definition of the function }f(x)=\log|x|\text{ is}$
$\displaystyle \text{(a) }R\qquad\text{(b) }(-\infty,0)\qquad\text{(c) }(0,\infty)\qquad\text{(d) }R-\{0\}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }\log|x|\text{ to be defined, }|x|>0.$
$\displaystyle \Rightarrow x\ne0$
$\displaystyle \therefore \text{Domain of }f=R-\{0\}.$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 50: }\text{The domain of definition of }f(x)=\sqrt{4x-x^2}\text{ is}$
$\displaystyle \text{(a) }R-[0,4]\qquad\text{(b) }R-(0,4)\qquad\text{(c) }(0,4)\qquad\text{(d) }[0,4]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }f(x)\text{ to be real, }4x-x^2\geq0.$
$\displaystyle \Rightarrow x(4-x)\geq0$
$\displaystyle \Rightarrow 0\leq x\leq4$
$\displaystyle \therefore \text{Domain of }f=[0,4].$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 51: }\text{The domain of definition of }f(x)=$
$\displaystyle \sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-3+2\sqrt{x-4}}\text{ is}$
$\displaystyle \text{(a) }[4,\infty)\qquad\text{(b) }(-\infty,4]$
$\displaystyle \text{(c) }(4,\infty)\qquad\text{(d) }(-\infty,4)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }\sqrt{x-4}\text{ to be real, }x\geq4.$
$\displaystyle \text{Let }t=\sqrt{x-4}.\text{ Then }t\geq0\text{ and }x=t^2+4.$
$\displaystyle x-3-2\sqrt{x-4}=t^2+1-2t=(t-1)^2\geq0$
$\displaystyle x-3+2\sqrt{x-4}=t^2+1+2t=(t+1)^2\geq0$
$\displaystyle \text{Thus, both outer square roots are defined for every }x\geq4.$
$\displaystyle \therefore \text{Domain of }f=[4,\infty).$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 52: }\text{The domain of the function }f(x)=\sqrt{5|x|-x^2-6}\text{ is}$
$\displaystyle \text{(a) }(-3,-2)\cup(2,3)\qquad\text{(b) }[-3,-2)\cup[2,3)$
$\displaystyle \text{(c) }[-3,-2]\cup[2,3]\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }f(x)\text{ to be real, }5|x|-x^2-6\geq0.$
$\displaystyle \text{Let }t=|x|.\text{ Then }x^2=t^2\text{ and }t\geq0.$
$\displaystyle 5t-t^2-6\geq0$
$\displaystyle \Rightarrow t^2-5t+6\leq0$
$\displaystyle \Rightarrow (t-2)(t-3)\leq0$
$\displaystyle \Rightarrow 2\leq t\leq3$
$\displaystyle \Rightarrow 2\leq|x|\leq3$
$\displaystyle \therefore x\in[-3,-2]\cup[2,3].$
$\displaystyle \therefore \text{Domain of }f=[-3,-2]\cup[2,3].$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 53: }\text{The range of the function }f(x)=\frac{x}{|x|}\text{ is}$
$\displaystyle \text{(a) }R-\{0\}\qquad\text{(b) }R-\{-1,1\}\qquad\text{(c) }\{-1,1\}\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{The function is defined for }x\ne0.$
$\displaystyle f(x)=\frac{x}{|x|}=\begin{cases}1,&x>0\\-1,&x<0\end{cases}$
$\displaystyle \therefore f(x)\text{ can take only the values }-1\text{ and }1.$
$\displaystyle \therefore \text{Range of }f=\{-1,1\}.$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 54: }\text{The range of the function }f(x)=\frac{x+2}{|x+2|},\ x\ne-2\text{ is}$
$\displaystyle \text{(a) }\{-1,1\}\qquad\text{(b) }\{-1,0,1\}\qquad\text{(c) }\{1\}\qquad\text{(d) }(0,\infty)$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\frac{x+2}{|x+2|}$
$\displaystyle \text{If }x>-2,\text{ then }|x+2|=x+2\text{ and }f(x)=1.$
$\displaystyle \text{If }x<-2,\text{ then }|x+2|=-(x+2)\text{ and }f(x)=-1.$
$\displaystyle \therefore f(x)\text{ can take only the values }-1\text{ and }1.$
$\displaystyle \therefore \text{Range of }f=\{-1,1\}.$
$\displaystyle \therefore \text{Option (a) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 55: }\text{The range of the function }f(x)=|x-1|\text{ is}$
$\displaystyle \text{(a) }(-\infty,0)\qquad\text{(b) }[0,\infty)$
$\displaystyle \text{(c) }(0,\infty)\qquad\text{(d) }R$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }|x-1|\geq0\text{ for every }x\in R,\text{ we have }f(x)\geq0.$
$\displaystyle \text{Also, }f(1)=0\text{ and every positive real value is attained by }f.$
$\displaystyle \therefore \text{Range of }f=[0,\infty).$
$\displaystyle \therefore \text{Option (b) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 56: }\text{Let }f(x)=\sqrt{x^2+1}.\text{ Then, which of the following is correct?}$
$\displaystyle \text{(a) }f(xy)=f(x)f(y)\qquad\text{(b) }f(xy)\geq f(x)f(y)$
$\displaystyle \text{(c) }f(xy)\leq f(x)f(y)\qquad\text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle f(xy)=\sqrt{x^2y^2+1}$
$\displaystyle f(x)f(y)=\sqrt{x^2+1}\sqrt{y^2+1}$
$\displaystyle =\sqrt{x^2y^2+x^2+y^2+1}$
$\displaystyle \text{Since }x^2+y^2\geq0,$
$\displaystyle x^2y^2+1\leq x^2y^2+x^2+y^2+1.$
$\displaystyle \therefore \sqrt{x^2y^2+1}\leq\sqrt{x^2y^2+x^2+y^2+1}.$
$\displaystyle \therefore f(xy)\leq f(x)f(y).$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 57: }\text{If }[x]^2-5[x]+6=0,\text{ where }[\,\cdot\,]\text{ denotes}$
$\displaystyle \text{the greatest integer function, then}$
$\displaystyle \text{(a) }x\in[3,4]\qquad\text{(b) }x\in(2,3]$
$\displaystyle \text{(c) }x\in[2,3]\qquad\text{(d) }x\in[2,4)$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }[x]=n.\text{ Then }n^2-5n+6=0.$
$\displaystyle \Rightarrow (n-2)(n-3)=0$
$\displaystyle \Rightarrow n=2\text{ or }n=3$
$\displaystyle [x]=2\Rightarrow 2\leq x<3$
$\displaystyle [x]=3\Rightarrow 3\leq x<4$
$\displaystyle \therefore x\in[2,3)\cup[3,4)=[2,4).$
$\displaystyle \therefore \text{Option (d) is correct.}$
$\displaystyle \\$

$\displaystyle \textbf{Question 58: }\text{The range of }f(x)=\frac{1}{1-2\cos x}\text{ is}$
$\displaystyle \text{(a) }\left[\frac13,1\right]\qquad\text{(b) }\left[-1,\frac13\right]$
$\displaystyle \text{(c) }(-\infty,-1]\cup\left[\frac13,\infty\right)\qquad\text{(d) }\left[-\frac13,1\right]$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }y=\frac{1}{1-2\cos x}.$
$\displaystyle \Rightarrow \cos x=\frac{y-1}{2y},\qquad y\ne0$
$\displaystyle \text{Since }-1\leq\cos x\leq1,\text{ we require}$
$\displaystyle \left|\frac{y-1}{2y}\right|\leq1.$
$\displaystyle \Rightarrow (y-1)^2\leq4y^2$
$\displaystyle \Rightarrow 3y^2+2y-1\geq0$
$\displaystyle \Rightarrow (3y-1)(y+1)\geq0$
$\displaystyle \Rightarrow y\leq-1\text{ or }y\geq\frac13.$
$\displaystyle \therefore \text{Range of }f=(-\infty,-1]\cup\left[\frac13,\infty\right).$
$\displaystyle \therefore \text{Option (c) is correct.}$
$\displaystyle \\$

$\displaystyle \text{VERY SHORT ANSWER QUESTIONS}$

$\displaystyle \textbf{Question 1: }\text{If }A=\{1,2,4\},\ B=\{2,4,5\}\text{ and }C=\{2,5\},\text{ write}$
$\displaystyle (A-C)\times(B-C).$
$\displaystyle \text{Answer:}$
$\displaystyle A-C=\{1,4\}\text{ and }B-C=\{4\}$
$\displaystyle \therefore (A-C)\times(B-C)=\{(1,4),(4,4)\}$
$\displaystyle \\$

$\displaystyle \textbf{Question 2: }\text{If }n(A)=3,\ n(B)=4,\text{ then write }n(A\times A\times B).$
$\displaystyle \text{Answer:}$
$\displaystyle n(A\times A\times B)=n(A)\times n(A)\times n(B)$
$\displaystyle =3\times3\times4=36$
$\displaystyle \therefore n(A\times A\times B)=36$
$\displaystyle \\$

$\displaystyle \textbf{Question 3: }\text{If }R\text{ is a relation defined on the set }Z\text{ of integers by the rule}$
$\displaystyle (x,y)\in R\Leftrightarrow x^2+y^2=9,\text{ then write domain of }R.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+y^2=9,\qquad x,y\in Z$
$\displaystyle \text{The possible ordered pairs are }(\pm3,0)\text{ and }(0,\pm3).$
$\displaystyle \therefore \text{Domain}(R)=\{-3,0,3\}$
$\displaystyle \\$

$\displaystyle \textbf{Question 4: }\text{If }R=\{(x,y):x,y\in Z,\ x^2+y^2\leq4\}\text{ is a relation defined}$
$\displaystyle \text{on the set }Z\text{ of integers, then write domain of }R.$
$\displaystyle \text{Answer:}$
$\displaystyle x^2+y^2\leq4,\qquad x,y\in Z$
$\displaystyle \text{Since }y^2\geq0,\text{ we have }x^2\leq4.$
$\displaystyle \therefore x\in\{-2,-1,0,1,2\}$
$\displaystyle \therefore \text{Domain}(R)=\{-2,-1,0,1,2\}$
$\displaystyle \\$

$\displaystyle \textbf{Question 5: }\text{If }R\text{ is a relation from set }A=\{11,12,13\}\text{ to set}$
$\displaystyle B=\{8,10,12\}\text{ defined by }y=x-3,\text{ then write }R^{-1}.$
$\displaystyle \text{Answer:}$
$\displaystyle y=x-3$
$\displaystyle x=11\Rightarrow y=8,\qquad x=12\Rightarrow y=9,\qquad x=13\Rightarrow y=10$
$\displaystyle \text{Since }9\notin B,\quad R=\{(11,8),(13,10)\}.$
$\displaystyle \therefore R^{-1}=\{(8,11),(10,13)\}$
$\displaystyle \\$

$\displaystyle \textbf{Question 6: }\text{Let }A=\{1,2,3\}\text{ and }R=\{(a,b):|a^2-b^2|\leq5,\ a,b\in A\}.$
$\displaystyle \text{Then write }R\text{ as set of ordered pairs.}$
$\displaystyle \text{Answer:}$
$\displaystyle |1^2-3^2|=|3^2-1^2|=8>5$
$\displaystyle \text{All the other ordered pairs in }A\times A\text{ satisfy }|a^2-b^2|\leq5.$
$\displaystyle \therefore R=\{(1,1),(1,2),(2,1),(2,2),(2,3),(3,2),(3,3)\}$
$\displaystyle \\$

$\displaystyle \textbf{Question 7: }\text{Let }R=\{(x,y):x,y\in Z,\ y=2x-4\}.\text{ If }(a,-2)\text{ and}$
$\displaystyle (4,b^2)\in R,\text{ then write the values of }a\text{ and }b.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }(a,-2)\in R,\quad -2=2a-4.$
$\displaystyle \therefore 2a=2\Rightarrow a=1$
$\displaystyle \text{Since }(4,b^2)\in R,\quad b^2=2(4)-4=4.$
$\displaystyle \therefore b=\pm2$
$\displaystyle \therefore a=1\text{ and }b=\pm2$
$\displaystyle \\$

$\displaystyle \textbf{Question 8: }\text{If }R=\{(2,1),(4,7),(1,-2),\ldots\},\text{ then write the linear relation}$
$\displaystyle \text{between the components of the ordered pairs of the relation }R.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the linear relation be }y=mx+c.$
$\displaystyle \text{Using }(2,1)\text{ and }(4,7),\quad m=\frac{7-1}{4-2}=3.$
$\displaystyle \therefore y=3x+c$
$\displaystyle \text{Using }(2,1),\quad 1=3(2)+c\Rightarrow c=-5.$
$\displaystyle \therefore y=3x-5$
$\displaystyle \text{Also, }(1,-2)\text{ satisfies }y=3x-5.$
$\displaystyle \therefore \text{The required linear relation is }y=3x-5.$
$\displaystyle \\$

$\displaystyle \textbf{Question 9: }\text{If }A=\{1,3,5\}\text{ and }B=\{2,4\},\text{ list the elements of }R,\text{ if}$
$\displaystyle R=\{(x,y):x,y\in A\times B\text{ and }x>y\}.$
$\displaystyle \text{Answer:}$
$\displaystyle A\times B=\{(1,2),(1,4),(3,2),(3,4),(5,2),(5,4)\}$
$\displaystyle \text{The ordered pairs satisfying }x>y\text{ are }(3,2),(5,2)\text{ and }(5,4).$
$\displaystyle \therefore R=\{(3,2),(5,2),(5,4)\}$
$\displaystyle \\$

$\displaystyle \textbf{Question 10: }\text{If }R=\{(x,y):x,y\in W,\ 2x+y=8\},\text{ then write the domain}$
$\displaystyle \text{and range of }R.$
$\displaystyle \text{Answer:}$
$\displaystyle 2x+y=8\Rightarrow y=8-2x$
$\displaystyle \text{Since }x,y\in W,\quad x=0,1,2,3,4.$
$\displaystyle \therefore R=\{(0,8),(1,6),(2,4),(3,2),(4,0)\}$
$\displaystyle \therefore \text{Domain}(R)=\{0,1,2,3,4\}$
$\displaystyle \text{Range}(R)=\{0,2,4,6,8\}$
$\displaystyle \\$

$\displaystyle \textbf{Question 11: }\text{Let }A\text{ and }B\text{ be two sets such that }n(A)=3\text{ and }n(B)=2.$
$\displaystyle \text{If }(x,1),(y,2),(z,1)\text{ are in }A\times B,\text{ write }A\text{ and }B.$
$\displaystyle \text{Answer:}$
$\displaystyle (x,1),(y,2),(z,1)\in A\times B$
$\displaystyle \therefore x,y,z\in A\text{ and }1,2\in B.$
$\displaystyle \text{Since }n(A)=3\text{ and }n(B)=2,$
$\displaystyle \therefore A=\{x,y,z\}\text{ and }B=\{1,2\}$
$\displaystyle \\$

$\displaystyle \textbf{Question 12: }\text{Let }A=\{1,2,3,5\},\ B=\{4,6,9\}\text{ and }R\text{ be a relation from }A$
$\displaystyle \text{to }B\text{ defined by }R=\{(x,y):x-y\text{ is odd}\}.\text{ Write }R\text{ in roster form.}$
$\displaystyle \text{Answer:}$
$\displaystyle x-y\text{ is odd when one of }x,y\text{ is odd and the other is even.}$
$\displaystyle \therefore R=\{(1,4),(1,6),(2,9),(3,4),(3,6),(5,4),(5,6)\}$
$\displaystyle \\$

$\displaystyle \textbf{Question 13: }\text{Write the range of the real function }f(x)=|x|.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Since }|x|\geq0\text{ for every }x\in R,$
$\displaystyle \text{and }|x|\text{ can take every non-negative real value,}$
$\displaystyle \therefore \text{Range of }f=[0,\infty).$
$\displaystyle \\$

$\displaystyle \textbf{Question 14: }\text{If }f\text{ is a real function satisfying}$
$\displaystyle f\left(x+\frac1x\right)=x^2+\frac1{x^2}\text{ for all }x\in R-\{0\},$
$\displaystyle \text{then write the expression for }f(x).$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let }t=x+\frac1x.$
$\displaystyle t^2=\left(x+\frac1x\right)^2=x^2+\frac1{x^2}+2$
$\displaystyle \Rightarrow x^2+\frac1{x^2}=t^2-2.$
$\displaystyle \therefore f(t)=t^2-2.$
$\displaystyle \text{Replacing }t\text{ by }x,\text{ we get}$
$\displaystyle f(x)=x^2-2.$
$\displaystyle \\$

$\displaystyle \textbf{Question 15: }\text{Write the range of the function }f(x)=\sin[x],\text{ where}$
$\displaystyle -\frac{\pi}{4}\leq x\leq\frac{\pi}{4}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }-\frac{\pi}{4}\leq x<0,\quad [x]=-1.$
$\displaystyle \therefore f(x)=\sin(-1)=-\sin1.$
$\displaystyle \text{For }0\leq x\leq\frac{\pi}{4},\quad [x]=0.$
$\displaystyle \therefore f(x)=\sin0=0.$
$\displaystyle \therefore \text{Range of }f=\{-\sin1,0\}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 16: }\text{If }f(x)=\cos[\pi^2]x+\cos[-\pi^2]x,\text{ where }[x]\text{ denotes}$
$\displaystyle \text{the greatest integer less than or equal to }x,\text{ then write the value of }f(\pi).$
$\displaystyle \text{Answer:}$
$\displaystyle [\pi^2]=9\quad\text{and}\quad[-\pi^2]=-10.$
$\displaystyle \therefore f(x)=\cos9x+\cos(-10x)=\cos9x+\cos10x.$
$\displaystyle f(\pi)=\cos9\pi+\cos10\pi$
$\displaystyle =-1+1=0.$
$\displaystyle \therefore f(\pi)=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 17: }\text{Write the range of the function }f(x)=\cos[x],\text{ where}$
$\displaystyle -\frac{\pi}{2}<x<\frac{\pi}{2}.$
$\displaystyle \text{Answer:}$
$\displaystyle -\frac{\pi}{2}<x<\frac{\pi}{2}\Rightarrow -1.57<x<1.57.$
$\displaystyle \therefore [x]\in\{-2,-1,0,1\}.$
$\displaystyle \therefore f(x)\in\{\cos(-2),\cos(-1),\cos0,\cos1\}.$
$\displaystyle \text{Since }\cos(-\theta)=\cos\theta,$
$\displaystyle \therefore \text{Range of }f=\{\cos2,\cos1,1\}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 18: }\text{Write the range of the function }f(x)=e^{x-[x]},\ x\in R.$
$\displaystyle \text{Answer:}$
$\displaystyle x-[x]\text{ is the fractional part of }x.$
$\displaystyle \therefore 0\leq x-[x]<1.$
$\displaystyle \text{Since }e^t\text{ is an increasing function,}$
$\displaystyle 1\leq e^{x-[x]}<e.$
$\displaystyle \therefore \text{Range of }f=[1,e).$
$\displaystyle \\$

$\displaystyle \textbf{Question 19: }\text{Let }f(x)=\frac{\alpha x}{x+1},\ x\ne-1.\text{ Then write the value of }\alpha$
$\displaystyle \text{satisfying }f(f(x))=x\text{ for all }x\ne-1.$
$\displaystyle \text{Answer:}$
$\displaystyle f(f(x))=\frac{\alpha f(x)}{f(x)+1}$
$\displaystyle =\frac{\alpha\left(\frac{\alpha x}{x+1}\right)} {\frac{\alpha x}{x+1}+1}$
$\displaystyle =\frac{\alpha^2x}{(\alpha+1)x+1}.$
$\displaystyle \text{Given, }f(f(x))=x.$
$\displaystyle \therefore \frac{\alpha^2x}{(\alpha+1)x+1}=x$
$\displaystyle \Rightarrow \alpha^2=(\alpha+1)x+1.$
$\displaystyle \text{For this to hold for all }x,\quad \alpha+1=0.$
$\displaystyle \therefore \alpha=-1.$
$\displaystyle \\$

$\displaystyle \textbf{Question 20: }\text{If }f(x)=1-\frac1x,\text{ then write the value of }f\left(f\left(\frac1x\right)\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f\left(\frac1x\right)=1-\frac{1}{1/x}=1-x.$
$\displaystyle \therefore f\left(f\left(\frac1x\right)\right)=f(1-x)$
$\displaystyle =1-\frac1{1-x}$
$\displaystyle =\frac{1-x-1}{1-x}$
$\displaystyle =\frac{-x}{1-x}=\frac{x}{x-1}.$
$\displaystyle \therefore f\left(f\left(\frac1x\right)\right)=\frac{x}{x-1}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 21: }\text{Write the domain and range of the function }f(x)=\frac{x-2}{2-x}.$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\frac{x-2}{-(x-2)}=-1,\qquad x\ne2.$
$\displaystyle \therefore \text{Domain of }f=R-\{2\}.$
$\displaystyle \text{Since }f(x)=-1\text{ for every }x\ne2,$
$\displaystyle \therefore \text{Range of }f=\{-1\}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 22: }\text{If }f(x)=4x-x^2,\ x\in R,\text{ then write the value of}$
$\displaystyle f(a+1)-f(a-1).$
$\displaystyle \text{Answer:}$
$\displaystyle f(a+1)=4(a+1)-(a+1)^2$
$\displaystyle =4a+4-a^2-2a-1=-a^2+2a+3.$
$\displaystyle f(a-1)=4(a-1)-(a-1)^2$
$\displaystyle =4a-4-a^2+2a-1=-a^2+6a-5.$
$\displaystyle \therefore f(a+1)-f(a-1)$
$\displaystyle =(-a^2+2a+3)-(-a^2+6a-5)$
$\displaystyle =8-4a.$
$\displaystyle \therefore f(a+1)-f(a-1)=8-4a.$
$\displaystyle \\$

$\displaystyle \textbf{Question 23: }\text{If }f,g,h\text{ are real functions given by }f(x)=x^2,\ g(x)=\tan x$
$\displaystyle \text{and }h(x)=\log_e x,\text{ then write the value of }(h\circ g\circ f)\left(\sqrt{\frac{\pi}{4}}\right).$
$\displaystyle \text{Answer:}$
$\displaystyle f\left(\sqrt{\frac{\pi}{4}}\right)=\left(\sqrt{\frac{\pi}{4}}\right)^2=\frac{\pi}{4}.$
$\displaystyle g\left(\frac{\pi}{4}\right)=\tan\frac{\pi}{4}=1.$
$\displaystyle h(1)=\log_e1=0.$
$\displaystyle \therefore (h\circ g\circ f)\left(\sqrt{\frac{\pi}{4}}\right)=0.$
$\displaystyle \\$

$\displaystyle \textbf{Question 24: }\text{Write the domain and range of function }f(x)\text{ given by}$
$\displaystyle f(x)=\frac{1}{\sqrt{x-|x|}}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }f(x)\text{ to be defined, }x-|x|>0.$
$\displaystyle \text{If }x\geq0,\quad x-|x|=x-x=0.$
$\displaystyle \text{If }x<0,\quad x-|x|=x-(-x)=2x<0.$
$\displaystyle \therefore x-|x|\text{ is never positive for any }x\in R.$
$\displaystyle \therefore \text{Domain of }f=\phi.$
$\displaystyle \text{Since the domain is empty, the range is also empty.}$
$\displaystyle \therefore \text{Range of }f=\phi.$
$\displaystyle \\$

$\displaystyle \textbf{Question 25: }\text{Write the domain and range of }f(x)=\sqrt{x-[x]}.$
$\displaystyle \text{Answer:}$
$\displaystyle x-[x]\text{ is the fractional part of }x.$
$\displaystyle \therefore 0\leq x-[x]<1\text{ for every }x\in R.$
$\displaystyle \therefore \text{Domain of }f=R.$
$\displaystyle 0\leq\sqrt{x-[x]}<1$
$\displaystyle \therefore \text{Range of }f=[0,1).$
$\displaystyle \\$

$\displaystyle \textbf{Question 26: }\text{Write the domain and range of function }f(x)\text{ given by}$
$\displaystyle f(x)=\sqrt{[x]-x}.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For }f(x)\text{ to be real, }[x]-x\geq0.$
$\displaystyle \text{But, by definition of the greatest integer function, }[x]\leq x.$
$\displaystyle \therefore [x]-x\leq0.$
$\displaystyle \text{Hence }[x]-x=0\text{ is required.}$
$\displaystyle \Rightarrow [x]=x$
$\displaystyle \text{This holds whenever }x\in Z.$
$\displaystyle \therefore D(f)=Z.$
$\displaystyle \text{For every }x\in Z,\quad f(x)=\sqrt0=0.$
$\displaystyle \therefore R(f)=\{0\}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 27: }\text{Let }A\text{ and }B\text{ be two sets such that }n(A)=p\text{ and }n(B)=q,$
$\displaystyle \text{write the number of functions from }A\text{ to }B.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Each element of }A\text{ can be mapped to any one of the }q\text{ elements of }B.$
$\displaystyle \text{Since }A\text{ contains }p\text{ elements,}$
$\displaystyle \therefore \text{Number of functions from }A\text{ to }B=q^p.$
$\displaystyle \\$

$\displaystyle \textbf{Question 28: }\text{Let }f\text{ and }g\text{ be two functions given by}$
$\displaystyle f=\{(2,4),(5,6),(8,-1),(10,-3)\}\text{ and}$
$\displaystyle g=\{(2,5),(7,1),(8,4),(10,13),(11,-5)\}.$
$\displaystyle \text{Find the domain of }f+g.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Domain of }f=\{2,5,8,10\}.$
$\displaystyle \text{Domain of }g=\{2,7,8,10,11\}.$
$\displaystyle \text{Domain of }(f+g)=\text{Domain of }f\cap\text{Domain of }g.$
$\displaystyle \therefore \text{Domain of }(f+g)=\{2,8,10\}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 29: }\text{Find the set of values of }x\text{ for which the functions}$
$\displaystyle f(x)=3x^2-1\text{ and }g(x)=3+x\text{ are equal.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the functions to be equal, }f(x)=g(x).$
$\displaystyle \therefore 3x^2-1=3+x$
$\displaystyle \Rightarrow 3x^2-x-4=0$
$\displaystyle \Rightarrow (3x-4)(x+1)=0$
$\displaystyle \Rightarrow x=\frac43\text{ or }x=-1.$
$\displaystyle \therefore \text{Required set of values of }x=\left\{-1,\frac43\right\}.$
$\displaystyle \\$

$\displaystyle \textbf{Question 30: }\text{Let }f\text{ and }g\text{ be two real functions given by}$
$\displaystyle f=\{(0,1),(2,0),(3,-4),(4,2),(5,1)\}\text{ and}$
$\displaystyle g=\{(1,0),(2,2),(3,-1),(4,4),(5,3)\}.$
$\displaystyle \text{Find the domain of }fg.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Domain of }f=\{0,2,3,4,5\}.$
$\displaystyle \text{Domain of }g=\{1,2,3,4,5\}.$
$\displaystyle \text{Domain of }fg=\text{Domain of }f\cap\text{Domain of }g.$
$\displaystyle \therefore \text{Domain of }fg=\{2,3,4,5\}.$
$\displaystyle \\$