Class 11: Functions – Multiple Choice Questions and Answers

$\displaystyle \textbf{Question 1. }\text{Let }A=\{1,2,3\},\ B=\{2,3,4\},\text{ then which of the following is a} \\ \text{function from }A\text{ to }B\ ?$
$\displaystyle \text{(a) }\{(1,2),(1,3),(2,3),(3,3)\} \qquad \text{(b) }\{(1,3),(2,4)\}$
$\displaystyle \text{(c) }\{(1,3),(2,2),(3,3)\} \qquad \text{(d) }\{(1,2),(2,3),(3,2),(3,4)\}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{In a function, every element of }A\text{ must have exactly one image in }B$
$\displaystyle \text{Option (c) satisfies this condition}$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

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

$\displaystyle \textbf{Question 3. }\text{Which one of the following is not a function?}$
$\displaystyle \text{(a) }\{(x,y):x,y\in R,\ x^2=y\}\qquad \text{(b) }\{(x,y):x,y\in R,\ y^2=x\}$
$\displaystyle \text{(c) }\{(x,y):x,y\in R,\ x=y^3\}\qquad \text{(d) }\{(x,y):x,y\in R,\ y=x^3\}$
$\displaystyle \text{Answer:}$
$\displaystyle y^2=x\Rightarrow y=\pm\sqrt{x}$
$\displaystyle \text{For one value of }x,\text{ there can be two values of }y$
$\displaystyle \therefore \text{It is not a function.}$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 4. }\text{If }f(x)=\cos(\log x),\text{ then }f(x^2)f(y^2)-\frac{1}{2}\left\{f\left(\frac{x^2}{y^2}\right)+f(x^2y^2)\right\}\text{ has the value}$
$\displaystyle \text{(a) }-2\qquad \text{(b) }-1\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x^2)=\cos(2\log x),\quad f(y^2)=\cos(2\log y)$
$\displaystyle f\left(\frac{x^2}{y^2}\right)=\cos(2\log x-2\log y)$
$\displaystyle f(x^2y^2)=\cos(2\log x+2\log y)$
$\displaystyle \frac{1}{2}\{\cos(A-B)+\cos(A+B)\}=\cos A\cos B$
$\displaystyle \therefore f(x^2)f(y^2)-\frac{1}{2}\left\{f\left(\frac{x^2}{y^2}\right)+f(x^2y^2)\right\}=0$
$\displaystyle \therefore \text{Correct option is (d) none of these.}$
$\\$

$\displaystyle \textbf{Question 5. }\text{If }f(x)=\cos(\log x),\text{ then }f(x)f(y)-\frac{1}{2}\left\{f\left(\frac{x}{y}\right)+f(xy)\right\}\text{ has the value}$
$\displaystyle \text{(a) }-1\qquad \text{(b) }\frac{1}{2}\qquad \text{(c) }-2\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\cos(\log x),\quad f(y)=\cos(\log y)$
$\displaystyle f\left(\frac{x}{y}\right)=\cos(\log x-\log y)$
$\displaystyle f(xy)=\cos(\log x+\log y)$
$\displaystyle \frac{1}{2}\{\cos(A-B)+\cos(A+B)\}=\cos A\cos B$
$\displaystyle \therefore f(x)f(y)-\frac{1}{2}\left\{f\left(\frac{x}{y}\right)+f(xy)\right\}=0$
$\displaystyle \therefore \text{Correct option is (d) none of these.}$
$\\$

$\displaystyle \textbf{Question 6. }\text{Let }f(x)=|x-1|.\text{ Then,}$
$\displaystyle \text{(a) }f(x^2)=[f(x)]^2\qquad \text{(b) }f(x+y)=f(x)f(y)$
$\displaystyle \text{(c) }f(|x|)=|f(x)|\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x^2)=|x^2-1|=|(x-1)(x+1)|$
$\displaystyle [f(x)]^2=|x-1|^2$
$\displaystyle \text{These are not always equal.}$
$\displaystyle f(x+y)\neq f(x)f(y)\text{ in general}$
$\displaystyle f(|x|)=||x|-1|\text{ and }|f(x)|=||x-1||=|x-1|$
$\displaystyle \text{These are not always equal.}$
$\displaystyle \therefore \text{Correct option is (d) none of these.}$
$\\$

$\displaystyle \textbf{Question 7. }\text{The range of }f(x)=\cos[x],\text{ for }-\pi/2<x<\pi/2\text{ is}$
$\displaystyle \text{(a) }\{-1,1,0\}\qquad \text{(b) }\{\cos1,\cos2,1\}\qquad \text{(c) }\{\cos1,-\cos1,1\}\qquad \text{(d) }[-1,1]$
$\displaystyle \text{Answer:}$
$\displaystyle -\frac{\pi}{2}<x<\frac{\pi}{2}$
$\displaystyle \therefore [x]\in\{-2,-1,0,1\}$
$\displaystyle f(x)=\cos[x]\in\{\cos2,\cos1,1\}$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 8. }\text{Which of the following are functions?}$
$\displaystyle \text{(a) }\{(x,y):y^2=x,\ x,y\in R\}\qquad \text{(b) }\{(x,y):y=|x|,\ x,y\in R\}$
$\displaystyle \text{(c) }\{(x,y):x^2+y^2=1,\ x,y\in R\}\qquad \text{(d) }\{(x,y):x^2-y^2=1,\ x,y\in R\}$
$\displaystyle \text{Answer:}$
$\displaystyle y=|x|$
$\displaystyle \text{For every value of }x,\text{ there is exactly one value of }y$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 9. }\text{If }f(x)=\log\left(\frac{1+x}{1-x}\right)\text{ and }g(x)=\frac{3x+x^3}{1+3x^2},\text{ then }f(g(x))\text{ is equal to}$
$\displaystyle \text{(a) }f(3x)\qquad \text{(b) }(f(x))^3\qquad \text{(c) }3f(x)\qquad \text{(d) }-f(x)$
$\displaystyle \text{Answer:}$
$\displaystyle f(g(x))=\log\left(\frac{1+g(x)}{1-g(x)}\right)$
$\displaystyle =\log\left(\frac{(1+x)^3}{(1-x)^3}\right)$
$\displaystyle =3\log\left(\frac{1+x}{1-x}\right)$
$\displaystyle =3f(x)$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 10. }\text{If }A=\{1,2,3\},\ B=\{x,y\},\text{ then the number of functions} \\ \text{that can be defined from }A\text{ into }B\text{ is}$
$\displaystyle \text{(a) }12\qquad \text{(b) }8\qquad \text{(c) }6\qquad \text{(d) }3$
$\displaystyle \text{Answer:}$
$\displaystyle n(A)=3,\quad n(B)=2$
$\displaystyle \text{Number of functions from }A\text{ into }B=2^3=8$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 11. }\text{If }f(x)=\log\left(\frac{1+x}{1-x}\right),\text{ then }f\left(\frac{2x}{1+x^2}\right)\text{ is equal to}$
$\displaystyle \text{(a) }(f(x))^2\qquad \text{(b) }(f(x))^3\qquad \text{(c) }2f(x)\qquad \text{(d) }3f(x)$
$\displaystyle \text{Answer:}$
$\displaystyle f\left(\frac{2x}{1+x^2}\right)=\log\left(\frac{1+\frac{2x}{1+x^2}}{1-\frac{2x}{1+x^2}}\right)$
$\displaystyle =\log\left(\frac{(1+x)^2}{(1-x)^2}\right)$
$\displaystyle =2\log\left(\frac{1+x}{1-x}\right)=2f(x)$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 12. }\text{If }f(x)=\cos(\log x),\text{ then value of }f(x)f(4)-\frac{1}{2}\left\{f\left(\frac{x}{4}\right)+f(4x)\right\}\text{ is}$
$\displaystyle \text{(a) }1\qquad \text{(b) }-1\qquad \text{(c) }0\qquad \text{(d) }\pm1$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\cos(\log x),\quad f(4)=\cos(\log4)$
$\displaystyle f\left(\frac{x}{4}\right)=\cos(\log x-\log4)$
$\displaystyle f(4x)=\cos(\log x+\log4)$
$\displaystyle \frac{1}{2}\{\cos(A-B)+\cos(A+B)\}=\cos A\cos B$
$\displaystyle \therefore \text{Given expression}=0$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 13. }\text{If }f(x)=\frac{2^x+2^{-x}}{2},\text{ then }f(x+y)f(x-y)\text{ is equals to}$
$\displaystyle \text{(a) }\frac{1}{2}\{f(2x)+f(2y)\}\qquad \text{(b) }\frac{1}{2}\{f(2x)-f(2y)\}$
$\displaystyle \text{(c) }\frac{1}{4}\{f(2x)+f(2y)\}\qquad \text{(d) }\frac{1}{4}\{f(2x)-f(2y)\}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x+y)f(x-y)=\frac{2^{x+y}+2^{-x-y}}{2}\cdot\frac{2^{x-y}+2^{-x+y}}{2}$
$\displaystyle =\frac{2^{2x}+2^{2y}+2^{-2y}+2^{-2x}}{4}$
$\displaystyle =\frac{1}{2}\left\{\frac{2^{2x}+2^{-2x}}{2}+\frac{2^{2y}+2^{-2y}}{2}\right\}$
$\displaystyle =\frac{1}{2}\{f(2x)+f(2y)\}$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 14. }\text{If }2f(x)-3f\left(\frac{1}{x}\right)=x^2\ (x\neq0),\text{ then }f(2)\text{ is equal to}$
$\displaystyle \text{(a) }-\frac{7}{4}\qquad \text{(b) }\frac{5}{2}\qquad \text{(c) }-1\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle 2f(2)-3f\left(\frac{1}{2}\right)=4$
$\displaystyle 2f\left(\frac{1}{2}\right)-3f(2)=\frac{1}{4}$
$\displaystyle \text{Solving these equations, we get }f(2)=-\frac{7}{4}$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 15. }\text{Let }f:R\to R\text{ be defined by }f(x)=2x+|x|.\text{ Then } \\ f(2x)+f(-x)-f(x)=$
$\displaystyle \text{(a) }2x\qquad \text{(b) }2|x|\qquad \text{(c) }-2x\qquad \text{(d) }-2|x|$
$\displaystyle \text{Answer:}$
$\displaystyle f(2x)=4x+2|x|$
$\displaystyle f(-x)=-2x+|x|$
$\displaystyle f(x)=2x+|x|$
$\displaystyle \therefore f(2x)+f(-x)-f(x)=2|x|$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 16. }\text{The range of the function }f(x)=\frac{x^2-x}{x^2+2x}\text{ is}$
$\displaystyle \text{(a) }R\qquad \text{(b) }R-\{1\}\qquad \text{(c) }R-\left\{-\frac{1}{2},1\right\}\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\frac{x(x-1)}{x(x+2)}=\frac{x-1}{x+2},\quad x\neq0,-2$
$\displaystyle \text{Let }y=\frac{x-1}{x+2}$
$\displaystyle yx+2y=x-1$
$\displaystyle x(y-1)=-(1+2y)$
$\displaystyle x=\frac{-(1+2y)}{y-1}$
$\displaystyle \therefore y\neq1$
$\displaystyle \text{Also, }x=0\text{ is not allowed, which gives }y=-\frac{1}{2}$
$\displaystyle \therefore \text{Range}=R-\left\{-\frac{1}{2},1\right\}$
$\displaystyle \therefore \text{Correct option is (c).}$

$\displaystyle \textbf{Question 17. }\text{If }x\neq1\text{ and }f(x)=\frac{x+1}{x-1}\text{ is a real function, then }f(f(f(2)))\text{ is}$
$\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }3\qquad \text{(d) }4$
$\displaystyle \text{Answer:}$
$\displaystyle f(2)=\frac{2+1}{2-1}=3$
$\displaystyle f(f(2))=f(3)=\frac{3+1}{3-1}=2$
$\displaystyle f(f(f(2)))=f(2)=3$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 18. }\text{If }f(x)=\cos(\log_e x),\text{ then }f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f(xy)+f\left(\frac{x}{y}\right)\right\}\text{ is equal to}$
$\displaystyle \text{(a) }\cos(x-y)\qquad \text{(b) }\log(\cos(x-y))\qquad \text{(c) }1\qquad \text{(d) }0$
$\displaystyle \text{Answer:}$
$\displaystyle f\left(\frac{1}{x}\right)=\cos\left(\log_e\frac{1}{x}\right)=\cos(-\log_e x)=\cos(\log_e x)$
$\displaystyle f\left(\frac{1}{y}\right)=\cos\left(\log_e\frac{1}{y}\right)=\cos(-\log_e y)=\cos(\log_e y)$
$\displaystyle \text{Let }\log_e x=A\text{ and }\log_e y=B.$
$\displaystyle \therefore f\left(\frac{1}{x}\right)=\cos A,\qquad f\left(\frac{1}{y}\right)=\cos B$
$\displaystyle f(xy)=\cos(\log_e xy)=\cos(\log_e x+\log_e y)=\cos(A+B)$
$\displaystyle f\left(\frac{x}{y}\right)=\cos\left(\log_e\frac{x}{y}\right)=\cos(\log_e x-\log_e y)=\cos(A-B)$
$\displaystyle \therefore f\left(\frac{1}{x}\right)f\left(\frac{1}{y}\right)-\frac{1}{2}\left\{f(xy)+f\left(\frac{x}{y}\right)\right\}$
$\displaystyle =\cos A\cos B-\frac{1}{2}\{\cos(A+B)+\cos(A-B)\}$
$\displaystyle =\cos A\cos B-\frac{1}{2}\{2\cos A\cos B\}$
$\displaystyle =\cos A\cos B-\cos A\cos B$
$\displaystyle =0$
$\displaystyle \therefore \text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 19. }\text{Let }f(x)=x,\ g(x)=\frac{1}{x}\text{ and }h(x)=f(x)g(x). \\ \text{ Then, }h(x)=1\text{ for}$
$\displaystyle \text{(a) }x\in R\qquad \text{(b) }x\in Q\qquad \text{(c) }x\in R-Q\qquad \text{(d) }x\in R,\ x\neq0$
$\displaystyle \text{Answer:}$
$\displaystyle h(x)=f(x)g(x)=x\cdot\frac{1}{x}=1$
$\displaystyle \text{This is defined only when }x\neq0$
$\displaystyle \therefore \text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 20. }\text{If }f(x)=\frac{\sin^4x+\cos^2x}{\sin^2x+\cos^4x}\text{ for }x\in R,\text{ then }f(2002)=$
$\displaystyle \text{(a) }1\qquad \text{(b) }2\qquad \text{(c) }3\qquad \text{(d) }4$
$\displaystyle \text{Answer:}$
$\displaystyle \sin^4x+\cos^2x=\sin^4x+1-\sin^2x$
$\displaystyle =1-\sin^2x+\sin^4x$
$\displaystyle \sin^2x+\cos^4x=\sin^2x+(1-\sin^2x)^2$
$\displaystyle =1-\sin^2x+\sin^4x$
$\displaystyle \therefore f(x)=1$
$\displaystyle \therefore f(2002)=1$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 21. }\text{The function }f:R\to R\text{ is defined by }f(x)=\cos^2x+\sin^4x.\text{ Then, }f(R)=$
$\displaystyle \text{(a) }[3/4,1)\qquad \text{(b) }(3/4,1]\qquad \text{(c) }[3/4,1]\qquad \text{(d) }(3/4,1)$
$\displaystyle \text{Answer:}$
$\displaystyle f(x)=\cos^2x+\sin^4x=1-\sin^2x+\sin^4x$
$\displaystyle \text{Let }\sin^2x=t,\text{ where }0\leq t\leq1$
$\displaystyle f(x)=t^2-t+1=\left(t-\frac{1}{2}\right)^2+\frac{3}{4}$
$\displaystyle \therefore f(R)=[3/4,1]$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 22. }\text{Let }A=\{x\in R:x\neq0,\ -4\leq x\leq4\}\text{ and } \\ f:A\to R\text{ be defined by }f(x)=\frac{|x|}{x}\text{ for }x\in A.\text{ Then }f(A)\text{ is}$
$\displaystyle \text{(a) }\{1,-1\}\qquad \text{(b) }\{x:0\leq x\leq4\}\qquad \text{(c) }\{1\}\qquad \text{(d) }\{x:-4\leq x\leq0\}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{|x|}{x}=1\text{ when }x>0$
$\displaystyle \frac{|x|}{x}=-1\text{ when }x<0$
$\displaystyle \therefore f(A)=\{1,-1\}$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 23. }\text{If }f:R\to R\text{ and }g:R\to R\text{ are defined by } \\ f(x)=2x+3\text{ and }g(x)=x^2+7,\text{ then the values of }x\text{ such that }g(f(x))=8\text{ are}$
$\displaystyle \text{(a) }1,2\qquad \text{(b) }-1,2\qquad \text{(c) }-1,-2\qquad \text{(d) }1,-2$
$\displaystyle \text{Answer:}$
$\displaystyle g(f(x))=(2x+3)^2+7$
$\displaystyle (2x+3)^2+7=8$
$\displaystyle (2x+3)^2=1$
$\displaystyle 2x+3=\pm1$
$\displaystyle \therefore x=-1,-2$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 24. }\text{If }f:[-2,2]\to R\text{ is defined by }f(x)=\begin{cases}-1,\text{ for }-2\leq x\leq0\\x-1,\text{ for }0\leq x\leq2\end{cases},\text{ then}$
$\displaystyle \{x\in[-2,2]:x\leq0\text{ and }f(|x|)=x\}=$
$\displaystyle \text{(a) }\{-1\}\qquad \text{(b) }\{0\}\qquad \text{(c) }\left\{-\frac{1}{2}\right\}\qquad \text{(d) }\phi$
$\displaystyle \text{Answer:}$
$\displaystyle x\leq0\Rightarrow |x|=-x$
$\displaystyle f(|x|)=f(-x)=-x-1$
$\displaystyle f(|x|)=x\Rightarrow -x-1=x$
$\displaystyle 2x=-1$
$\displaystyle x=-\frac{1}{2}$
$\displaystyle \therefore \{x\in[-2,2]:x\leq0\text{ and }f(|x|)=x\}=\left\{-\frac{1}{2}\right\}$
$\displaystyle \therefore \text{Correct option is (c).}$

$\displaystyle \textbf{Question 25. }\text{If }f(x)=\frac{10+x}{10-x},\ x\in(-10,10)\text{ and }f\left(\frac{200x}{100+x^2}\right)=2k\{f(x)\}^2,\text{ then }k=$
$\displaystyle \text{(a) }0.5\qquad \text{(b) }0.6\qquad \text{(c) }0.7\qquad \text{(d) }0.8$
$\displaystyle \text{Answer:}$
$\displaystyle f\left(\frac{200x}{100+x^2}\right)=\frac{10+\frac{200x}{100+x^2}}{10-\frac{200x}{100+x^2}}$
$\displaystyle =\frac{10(100+x^2)+200x}{10(100+x^2)-200x}$
$\displaystyle =\frac{1000+10x^2+200x}{1000+10x^2-200x}$
$\displaystyle =\frac{x^2+20x+100}{x^2-20x+100}$
$\displaystyle =\frac{(x+10)^2}{(x-10)^2}$
$\displaystyle =\left(\frac{x+10}{10-x}\right)^2$
$\displaystyle =\{f(x)\}^2$
$\displaystyle \text{Now, }f\left(\frac{200x}{100+x^2}\right)=2k\{f(x)\}^2$
$\displaystyle \Rightarrow \{f(x)\}^2=2k\{f(x)\}^2$
$\displaystyle \Rightarrow 2k=1$
$\displaystyle \Rightarrow k=\frac{1}{2}=0.5$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 26. }\text{If }f\text{ is a real valued function given by }f(x)=27x^3+\frac{1}{x^3}\text{ and } \\ \alpha,\beta\text{ are roots of }3x+\frac{1}{x}=12.\text{ Then,}$
$\displaystyle \text{(a) }f(\alpha)\neq f(\beta)\qquad \text{(b) }f(\alpha)=10\qquad \text{(c) }f(\beta)=-10\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle 3x+\frac{1}{x}=12$
$\displaystyle \left(3x+\frac{1}{x}\right)^3=27x^3+\frac{1}{x^3}+9\left(3x+\frac{1}{x}\right)$
$\displaystyle 12^3=f(x)+9(12)$
$\displaystyle f(x)=1728-108=1620$
$\displaystyle \therefore f(\alpha)=f(\beta)=1620$
$\displaystyle \therefore \text{Correct option is (d) none of these.}$
$\\$

$\displaystyle \textbf{Question 27. }\text{If }f(x)=64x^3+\frac{1}{x^3}\text{ and }\alpha,\beta\text{ are roots of }4x+\frac{1}{x}=3.\text{ Then,}$
$\displaystyle \text{(a) }f(\alpha)=f(\beta)=-9\qquad \text{(b) }f(\alpha)=f(\beta)=63\qquad \text{(c) }f(\alpha)\neq f(\beta)\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle 4x+\frac{1}{x}=3$
$\displaystyle \left(4x+\frac{1}{x}\right)^3=64x^3+\frac{1}{x^3}+12\left(4x+\frac{1}{x}\right)$
$\displaystyle 3^3=f(x)+12(3)$
$\displaystyle f(x)=27-36=-9$
$\displaystyle \therefore f(\alpha)=f(\beta)=-9$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 28. }\text{If }3f(x)+5f\left(\frac{1}{x}\right)=\frac{1}{x}-3\text{ for all non-zero }x,\text{ then }f(x)=$
$\displaystyle \text{(a) }\frac{1}{14}\left(\frac{3}{x}+5x-6\right)\qquad \text{(b) }\frac{1}{14}\left(-\frac{3}{x}+5x-6\right)$
$\displaystyle \text{(c) }\frac{1}{14}\left(-\frac{3}{x}+5x+6\right)\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle 3f(x)+5f\left(\frac{1}{x}\right)=\frac{1}{x}-3$
$\displaystyle 3f\left(\frac{1}{x}\right)+5f(x)=x-3$
$\displaystyle \text{Solving these two equations, we get}$
$\displaystyle 14f(x)=5x-\frac{3}{x}-6$
$\displaystyle \therefore f(x)=\frac{1}{14}\left(-\frac{3}{x}+5x-6\right)$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 29. }\text{If }f:R\to R\text{ be given by }f(x)=\frac{4^x}{4^x+2}\text{ for all }x\in R.\text{ Then,}$
$\displaystyle \text{(a) }f(x)=f(1-x)\qquad \text{(b) }f(x)+f(1-x)=0 \\ \text{(c) }f(x)+f(1-x)=1\qquad \text{(d) }f(x)+f(x-1)=1$
$\displaystyle \text{Answer:}$
$\displaystyle f(1-x)=\frac{4^{1-x}}{4^{1-x}+2}$
$\displaystyle =\frac{4}{4+2\cdot4^x}=\frac{2}{2+4^x}$
$\displaystyle f(x)+f(1-x)=\frac{4^x}{4^x+2}+\frac{2}{4^x+2}=1$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 30. }\text{If }f(x)=\sin[\pi^2]x+\sin[-\pi^2]x,\text{ where }[x]\text{ denotes the} \\ \text{greatest integer less than or equal to }x,\text{ then}$
$\displaystyle \text{(a) }f(\pi/2)=1\qquad \text{(b) }f(\pi)=2\qquad \text{(c) }f(\pi/4)=-1\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle [\pi^2]=9,\quad [-\pi^2]=-10$
$\displaystyle \therefore f(x)=\sin9x+\sin(-10x)=\sin9x-\sin10x$
$\displaystyle f\left(\frac{\pi}{2}\right)=\sin\frac{9\pi}{2}-\sin5\pi=1$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 31. }\text{The domain of the function }f(x)=\sqrt{2-2x-x^2}\text{ is}$
$\displaystyle \text{(a) }[-\sqrt3,\sqrt3]\qquad \text{(b) }[-1-\sqrt3,-1+\sqrt3]\qquad \\ \text{(c) }[-2,2]\qquad \text{(d) }[-2-\sqrt3,-2+\sqrt3]$
$\displaystyle \text{Answer:}$
$\displaystyle 2-2x-x^2\geq0$
$\displaystyle x^2+2x-2\leq0$
$\displaystyle (x+1)^2\leq3$
$\displaystyle -\sqrt3\leq x+1\leq\sqrt3$
$\displaystyle -1-\sqrt3\leq x\leq-1+\sqrt3$
$\displaystyle \therefore \text{Domain}=[-1-\sqrt3,-1+\sqrt3]$
$\displaystyle \therefore \text{Correct option is (b).}$

$\displaystyle \textbf{Question 32. }\text{The domain of definition of }f(x)=\sqrt{\frac{x+3}{(2-x)(x-5)}}\text{ is}$
$\displaystyle \text{(a) }(-\infty,-3]\cup(2,5)\qquad \text{(b) }(-\infty,-3)\cup(2,5)$
$\displaystyle \text{(c) }(-\infty,-3]\cup[2,5]\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+3}{(2-x)(x-5)}\geq0$
$\displaystyle \text{Critical points are }x=-3,2,5$
$\displaystyle x=2,5\text{ are not included as denominator becomes zero}$
$\displaystyle \therefore \text{Domain}=(-\infty,-3]\cup(2,5)$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 33. }\text{The domain of the function }f(x)=\sqrt{\frac{(x+1)(x-3)}{x-2}}\text{ is}$
$\displaystyle \text{(a) }[-1,2)\cup[3,\infty)\qquad \text{(b) }(-1,2)\cup[3,\infty)$
$\displaystyle \text{(c) }[-1,2]\cup[3,\infty)\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{(x+1)(x-3)}{x-2}\geq0$
$\displaystyle \text{Critical points are }x=-1,2,3$
$\displaystyle x=2\text{ is not included as denominator becomes zero}$
$\displaystyle \therefore \text{Domain}=[-1,2)\cup[3,\infty)$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 34. }\text{The domain of the function }f(x)=\sqrt{x-1}+\sqrt{3-x}\text{ is}$
$\displaystyle \text{(a) }[1,\infty)\qquad \text{(b) }(-\infty,3)\qquad \text{(c) }(1,3)\qquad \text{(d) }[1,3]$
$\displaystyle \text{Answer:}$
$\displaystyle x-1\geq0\Rightarrow x\geq1$
$\displaystyle 3-x\geq0\Rightarrow x\leq3$
$\displaystyle \therefore 1\leq x\leq3$
$\displaystyle \therefore \text{Domain}=[1,3]$
$\displaystyle \therefore \text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 35. }\text{The domain of definition of the function } \\ f(x)=\sqrt{\frac{x-2}{x+2}}+\sqrt{\frac{1-x}{1+x}}\text{ is}$
$\displaystyle \text{(a) }(-\infty,-2]\cup[2,\infty)\qquad \text{(b) }[-1,1]\qquad \text{(c) }\phi\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x-2}{x+2}\geq0\Rightarrow x\in(-\infty,-2)\cup[2,\infty)$
$\displaystyle \frac{1-x}{1+x}\geq0\Rightarrow x\in(-1,1]$
$\displaystyle \text{There is no common value of }x$
$\displaystyle \therefore \text{Domain}=\phi$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 36. }\text{The domain of definition of the function }f(x)=\log|x|\text{ is}$
$\displaystyle \text{(a) }R\qquad \text{(b) }(-\infty,0)\qquad \text{(c) }(0,\infty)\qquad \text{(d) }R-\{0\}$
$\displaystyle \text{Answer:}$
$\displaystyle |x|>0$
$\displaystyle \therefore x\neq0$
$\displaystyle \therefore \text{Domain}=R-\{0\}$
$\displaystyle \therefore \text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 37. }\text{The domain of definition of }f(x)=\sqrt{4x-x^2}\text{ is}$
$\displaystyle \text{(a) }R-\{0,4\}\qquad \text{(b) }R-(0,4)\qquad \text{(c) }(0,4)\qquad \text{(d) }[0,4]$
$\displaystyle \text{Answer:}$
$\displaystyle 4x-x^2\geq0$
$\displaystyle x(4-x)\geq0$
$\displaystyle \therefore 0\leq x\leq4$
$\displaystyle \therefore \text{Domain}=[0,4]$
$\displaystyle \therefore \text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 38. }\text{The domain of definition of }f(x)=\sqrt{x-3-2\sqrt{x-4}}-\sqrt{x-3+2\sqrt{x-4}}\text{ is}$
$\displaystyle \text{(a) }[4,\infty)\qquad \text{(b) }(-\infty,4]\qquad \text{(c) }(4,\infty)\qquad \text{(d) }(-\infty,4)$
$\displaystyle \text{Answer:}$
$\displaystyle x-4\geq0\Rightarrow x\geq4$
$\displaystyle x-3-2\sqrt{x-4}=(\sqrt{x-4}-1)^2\geq0$
$\displaystyle x-3+2\sqrt{x-4}=(\sqrt{x-4}+1)^2\geq0$
$\displaystyle \therefore \text{Domain}=[4,\infty)$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 39. }\text{The domain of the function }f(x)=\sqrt{5|x|-x^2-6}\text{ is}$
$\displaystyle \text{(a) }(-3,-2)\cup(2,3)\qquad \text{(b) }[-3,-2)\cup[2,3)$
$\displaystyle \text{(c) }[-3,-2]\cup[2,3]\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle 5|x|-x^2-6\geq0$
$\displaystyle \text{For }x\geq0,\ 5x-x^2-6\geq0$
$\displaystyle x^2-5x+6\leq0\Rightarrow 2\leq x\leq3$
$\displaystyle \text{For }x<0,\ -5x-x^2-6\geq0$
$\displaystyle x^2+5x+6\leq0\Rightarrow -3\leq x\leq-2$
$\displaystyle \therefore \text{Domain}=[-3,-2]\cup[2,3]$
$\displaystyle \therefore \text{Correct option is (c).}$

$\displaystyle \textbf{Question 40. }\text{The range of the function }f(x)=\frac{x}{|x|}\text{ is}$
$\displaystyle \text{(a) }R-\{0\}\qquad \text{(b) }R-\{-1,1\}\qquad \text{(c) }\{-1,1\}\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x}{|x|}=1\text{ when }x>0$
$\displaystyle \frac{x}{|x|}=-1\text{ when }x<0$
$\displaystyle \therefore \text{Range}=\{-1,1\}$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 41. }\text{The range of the function }f(x)=\frac{x+2}{|x+2|},\ x\neq-2\text{ is}$
$\displaystyle \text{(a) }\{-1,1\}\qquad \text{(b) }\{-1,0,1\}\qquad \text{(c) }\{1\}\qquad \text{(d) }(0,\infty)$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x+2}{|x+2|}=1\text{ when }x>-2$
$\displaystyle \frac{x+2}{|x+2|}=-1\text{ when }x<-2$
$\displaystyle \therefore \text{Range}=\{-1,1\}$
$\displaystyle \therefore \text{Correct option is (a).}$
$\\$

$\displaystyle \textbf{Question 42. }\text{The range of the function }f(x)=|x-1|\text{ is}$
$\displaystyle \text{(a) }(-\infty,0)\qquad \text{(b) }[0,\infty)\qquad \text{(c) }(0,\infty)\qquad \text{(d) }R$
$\displaystyle \text{Answer:}$
$\displaystyle |x-1|\geq0$
$\displaystyle \therefore \text{Range}=[0,\infty)$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$

$\displaystyle \textbf{Question 43. }\text{Let }f(x)=\sqrt{x^2+1}.\text{ Then, which of the following is correct?}$
$\displaystyle \text{(a) }f(xy)=f(x)f(y)\qquad \text{(b) }f(xy)\geq f(x)f(y)$
$\displaystyle \text{(c) }f(xy)\leq f(x)f(y)\qquad \text{(d) none of these}$
$\displaystyle \text{Answer:}$
$\displaystyle f(xy)=\sqrt{x^2y^2+1}$
$\displaystyle f(x)f(y)=\sqrt{x^2+1}\sqrt{y^2+1}$
$\displaystyle =\sqrt{x^2y^2+x^2+y^2+1}$
$\displaystyle \therefore f(xy)\leq f(x)f(y)$
$\displaystyle \therefore \text{Correct option is (c).}$
$\\$

$\displaystyle \textbf{Question 44. }\text{If }[x]^2-5[x]+6=0,\text{ where }[.]\text{ denotes the greatest integer function, then}$
$\displaystyle \text{(a) }x\in[3,4]\qquad \text{(b) }x\in(2,3]\qquad \text{(c) }x\in[2,3]\qquad \text{(d) }x\in[2,4)$
$\displaystyle \text{Answer:}$
$\displaystyle [x]^2-5[x]+6=0$
$\displaystyle ([x]-2)([x]-3)=0$
$\displaystyle \therefore [x]=2\text{ or }[x]=3$
$\displaystyle [x]=2\Rightarrow x\in[2,3)$
$\displaystyle [x]=3\Rightarrow x\in[3,4)$
$\displaystyle \therefore x\in[2,4)$
$\displaystyle \therefore \text{Correct option is (d).}$
$\\$

$\displaystyle \textbf{Question 45. }\text{The range of }f(x)=\frac{1}{1+2\cos x}\text{ is}$
$\displaystyle \text{(a) }[1/3,1]\qquad \text{(b) }(-\infty,-1]\cup[1/3,\infty)\qquad \text{(c) }[-1,1/3]\qquad \text{(d) }[-1/3,1]$
$\displaystyle \text{Answer:}$
$\displaystyle -1\leq\cos x\leq1$
$\displaystyle -2\leq2\cos x\leq2$
$\displaystyle -1\leq1+2\cos x\leq3$
$\displaystyle 1+2\cos x\neq0$
$\displaystyle \therefore 1+2\cos x\in[-1,0)\cup(0,3]$
$\displaystyle \therefore \frac{1}{1+2\cos x}\in(-\infty,-1]\cup[1/3,\infty)$
$\displaystyle \therefore \text{Correct option is (b).}$
$\\$