JEE Main and Advanced questions on relations and functions
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Multiple Choice Questions
Question 1
If the function f: R-> R defined as $f(x) =|x|(x -sin x)$, then which of the following statement is true
(a) f is bijective
(b) f is one-one but not Onto
(c) f is onto but not one-one
(d) f is neither one-one nor onto Answers
Answer is (a)
Question 2
If the function f:$[1,\infty)$ -> [2,\infy)$ defined as $f(x) =x + \frac {1}{x}$, then $f^{-1}$ x is
(a) $\frac {x}{1+x^2}$
(b) $x + \frac {1}{x}$
(c) $\frac {x + \sqrt {x^2 -4}}{2}$
(d) $\frac {x - \sqrt {x^2 -4}}{2}$ Answers
(c)
$ y=x + \frac {1}{x}$
$ xy=x^2 +1$
$x^2 -xy + 1=0$
$x = \frac {y \pm \sqrt {y^2 -4}}{2}$
Here checking the domain and range
$x= \frac {y + \sqrt {y^2 -4}}{2}$
Question 3
The function $f:[0,\infty)$ -> R given by $f(x)= \frac {x}{x+1}$ is
(a) one -one and onto
(b) one-one but not onto
(c) onto but not one-one
(d) Neither one-one nor onto Answers
Answer is (b)
$f(x)= \frac {x}{x+1}$
$\frac {df(x)}{dx}= \frac {1}{(x+1)^2}$
This is positive, So f(x) is a increasing function and it is one-one function
for onto
$y=\frac {x}{x+1}$
$x = \frac {y}{1-y}$
So y cannot be 1,So Range is subset of codomain
Hence not onto function
Question 4
if $g(f(x))=sin^2 x$ and $f(g(x)) =(sin \sqrt x)^2$ then
(a) $f(x) =x^2$ and $g(x) = sin \sqrt x$
(b) $f(x) =sinx $ and $g(x) = |x|$
(c) $f(x) =sin^2 x $ and $g(x) = \sqrt x$
(d) f and g cannot be determined Answers
Answer is (x)
Self explanatory
Question 5
The function $f(x)=[x] sin (\frac {\pi}{[x+1]})$ where [.] denotes the greatest integer function. The domain of the function f is
(a) $(-\infty ,\infty) - {-1}$
(b) $[0 ,\infty)$
(c) $(-\infty ,-1)$
(d) $(-\infty ,-1) \cup [0, \infty)$ Answers
Answer is (d)
$[x+1] \neq 0$
Hence domain is $(-\infty ,-1) \cup [0, \infty)$
Question 6
The domain of the function $f(x)=\frac {log_2 (x+3)}{x^2 + 3x +2}$?
(a) $R - {-1,-2,-3}$
(b) $R - {-1,-2}$
(c) $(-2,\infty)$
(d) $(-3, \infty) - {-1,-2}$ Answers
Answer: (d)
Here x+ 3 > 0 or x > -3
also $x^2 + 3x +2= (x+1)(x+2)$,so x cannot take the value -1 and -2
Hence the domain is $(-3, \infty) - {-1,-2}$
Question 7
Consider the below two relations
A= {(x,y)|x,y are real numbers and x=wy for some rational number}
B= {(m/n,p/q)|m,n, p and q are integers such that n and q are not zero and qm=pn$
(a) A is a equivalence relation but B is not
(b) A and B both are equivalence relation
(c) A is not a equivalence relation but B is a equivalence relation
(d) A and B both are not equivalence relations Answers
for A
xAy => x=wy
yAx => y=wx
and they are not equal, so it is not symmetric
For B
$\frac {m}{n} A \frac {m}{n}$ = mn=mn which is true for all m,n in integers
Hence Reflexive
if $\frac {m}{n} A \frac {p}{q}$ => mq=pn
This can be written s
np=mq i.e $\frac {p}{q} A \frac {m}{n}$
Hence symmetric
Now if $\frac {m}{n} A \frac {p}{q}$ => mq=pn and $\frac {p}{q} A \frac {r}{s}$ => ps=rq or rq=ps
then
(qm)(ps)= (pn)(rq)
ms=nr
ie.$\frac {m}{n} A \frac {r}{s}$
Hence transitive
So B is a equivalence relations
hence (c) is the correct answer
Question 8
let $f(x) = \frac {1}{\sqrt {|x| - x}}$, the domain of f(x) is
(a) $(-\infty ,\infty)$
(b) $(-\infty ,0)
(c) $(-\infty ,\infty) - {0}$
(d) $(0 ,\infty) Answers
Answer is (b)
as |x| - x > 0 or |x| >x
so x< 0
Question 9
Let f: R -> [-1/2,1/2] and $f(x) =\frac {x}{x^2 +1}$ then f(x) is
(a) one-one but not onto
(b) onto but not one -one
(c) one-one and onto
(d) neither one-one nor onto Answers
(b)
Question 10
If $a \in R$ and the equation $-3(x-[x])^2+2(x−[x])+a^2=0$ (where [x] denotes the greatest integer ) has a no integral solution , then all possible values of a lie in the interval
(a)(1,2)
(b)(-2,-1)
(c)$(-1,0) \cup (0,1)$
(d) $(-2,0) \cup (0,2)$ Answers
$-3(x-[x])^2+2(x−[x])+a^2=0$
let t=x-[x]. Now 0
Now equation becomes
$-3(t)^2+2(t)+a^2=0$
$3t^2 -2t-a^2=0$
$t=\frac { 2 \pm \sqrt {4 + 12a^2}}{6} = \frac {1 \pm \sqrt {1 + 3a^2}}{3}$
it can be positive only as otherwise it will be lessor than t<0
$t= \frac {1 + \sqrt {1 + 3a^2}}{3}$
Now
$\frac {1 + \sqrt {1 + 3a^2}}{3} < 1$
$\sqrt {1 + 3a^2} < 2$
$1 + 3a^2 < 4$
$a^2 < 1$
Also a cannot be zero as then it will have integral solutions
So answer is (c)
Fill in the blanks
Question 11
(i) Let f :[0,1] -> R be defined by
$f(x) = \frac {4^x}{4^x +2}$
Then the value of
$f(\frac {1}{40}) + f(\frac {2}{40}) + f(\frac {3}{40}) + .............. + f(\frac {39}{40})$ is ______
(ii) if $f(x)= \frac {1}{2} (sin^2 x + sin^2(x + \pi/3) + cosx cos(x + \pi/3))$ and g(5/8) =1, then g(f(x)) is _____
(iii) The domain of the function $cos^{-1} (log_e (\frac {x}{x-1}))$ is _______ Answers
Question 12
Let $f(x)=\frac {x^2 -6x +5}{x^2 -5x + 6}$
Then match the column A and B Answers
The graph of the function is given below
$f(x)=\frac {x^2 -6x +5}{x^2 -5x + 6}=\frac {(x-1)(x-5)}{(x-2)(x-3)}$
(p) if 0< x <1 then 0
(q) if 1 < x < 2, then f(x) < 0, then 2,4 is the answer
(r) if 3 < x < 5, then f(x) < 0, then 2,4 is the answer
(s) if x> 5 then 0
Question 13
Let $f(x)=\frac {x^2 -3x -10}{x^2 -2x -3}$
Then match the column A and B Answers
The graph of the function is given below
$f(x)=\frac {x^2 -3x -10}{x^2 -2x -3}=\frac {(x+2)(x-5)}{(x+1)(x-3)}$
(p) if x < -2 then 0
(q) if -2 < x < -1, then f(x) < 0, then 2,4 is the answer
(r) if 3 < x < 5, then f(x) < 0, then 2,4 is the answer
(s) if x> 5 then 0
True and False
Question 14
(i) The function f(x)=sin(log(x + \sqrt {x^2 +1})) is odd function
(ii)If the function $f:[1, \infty) -> [1, \infty)$ is defined by $f(x)=2^{x(x−1)}, then $f^{−1}(x) is given as $\frac {1}{2}[1 + \sqrt {1+4log_2 x}]$
(iii) The relation R on the set A = {1, 2, 3} defined as R = {{1, 1), (1, 2), (2, 1), (3, 3)} is reflexive, symmetric and transitive. Answers
