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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

Answer is (a)

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) $\frac {x - \sqrt {x^2 -4}}{2}$

(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
$x= \frac {y + \sqrt {y^2 -4}}{2}$

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

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

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

Answer is (x)

Self explanatory

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) $(-\infty ,-1) \cup [0, \infty)$

Answer is (d)

$[x+1] \neq 0$

Hence domain is $(-\infty ,-1) \cup [0, \infty)$

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) $(-3, \infty) - {-1,-2}$

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 $(-3, \infty) - {-1,-2}$

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

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

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)

Answer is (b)

as |x| - x > 0 or |x| >x

so x< 0

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

(b)

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) $(-2,0) \cup (0,2)$

$-3(x-[x])^2+2(x−[x])+a^2=0$

let t=x-[x]. Now 0

$-3(t)^2+2(t)+a^2=0$

$3t^2 -2t-a^2=0$

$t=\frac { 2 \pm \sqrt {4 + 12a^2}}{6} = \frac {1 \pm \sqrt {1 + 3a^2}}{3}$

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)

(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 _______

(i) 39/2

Here

$f(x) = \frac {4^x}{4^x +2}$

$f(1-x) = \frac {2}{4^x +2}$

So $ f(x) + f(1-x)=1$

Now

$f(\frac {1}{40}) + f(\frac {2}{40}) + f(\frac {3}{40}) + .............. + f(\frac {39}{40})$

$= \frac {1}{2} ( \sum_{k=1}^{k=39} f(\frac {k}{40}) + \sum_{k=1}^{k=39} f(\frac {k}{40})= \frac {1}{2} ( \sum_{k=1}^{k=39} f(\frac {k}{40}) + \sum_{k=39}^{k=1} f(\frac {k}{40})$

$= \frac {1}{2} [\sum_{k=1}^{k=39} f(\frac {k}{40}) + f(\frac {40 -k}{40})] =\frac {39}{2}$

(ii) 1

f(x)= \frac {1}{2} (sin^2 x + sin^2(x + \pi/3) + cosx cos(x + \pi/3))$

Applying sin (A+B) and cos(A+B) identities, we get

f(x) =5/8

hence g(f(x))=1

(iii) $ (-\infty , -\frac {1}{e-1}] \cup [\frac {e}{e-1} , \infty)$

Let $f(x)=\frac {x^2 -6x +5}{x^2 -5x + 6}$

Then match the column A and B

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

(r) if 3 < x < 5, then f(x) < 0, then 2,4 is the answer

(s) if x> 5 then 0

Let $f(x)=\frac {x^2 -3x -10}{x^2 -2x -3}$

Then match the column A and B

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

(r) if 3 < x < 5, then f(x) < 0, then 2,4 is the answer

(s) if x> 5 then 0

(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.

(i) True

(ii) True

(iii) False

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