Maths M.C.Q (1 Marks)

Therefore $\log \,|{x^2} - 9|\,$ does not exist at $x = - \,3,\,\,3.$

Hence domain of function is $R - \left\{ { - \,3,\,\,3} \right\}.$

$i.e.,$ if $x > 0$ and $x \ne 1$

==> $x \in (0,\,1) \cup (1,\,\infty ).$

Obviously, minimum value is $-2$ and maximum $\infty $.

Hence range of function is $[-2, \infty].$

==> $3 + x > 0$==> $x \ne n\pi + {( - 1)^n}0$

==> $x \ne n\pi $. Domain of $f(x)  = R - \{ n\pi ,\,\,n \in I\} $.

==> $y = {x^x}\, \Rightarrow \,\,\,\log y = x\log x$ and $6 - x \ge 0$==>$x \ge 4$ and $x \le 6$

$\therefore $ Domain of $f(x)$ = $[4,\,\,6]$.

So the option $(b)$ is correct

Domain of function $\sqrt {\sin 2x} $ is $[n\pi ,\,n\pi + \pi /2]$.

$i.e.,$  $x \in ( - \,\infty ,\, - \,2)\, \cup \,(2,\,\infty )$.

or ${x^2} - 5x + 6 \le 0$ or $(x - 2)\,(x - 3) \le 0$.

Hence $2 \le x \le 3.$

$\Rightarrow \,\,|\,\,x\,\,|\, < 3$ or $ - 3 < x < 3.$

Hence domain is $( - \,3,\,\,2].$

$1 - x \ge 0\,\, \Rightarrow \,\,x \le 1,\,\,x \ne 0$

Hence domain is $[ - 1,\,1] - \{ 0\} $.

Clearly $f(x)$ is defined, if $4 + x \ge 0$ ==> $x \ge - 4$

$4 - x \ge 0$ ==> $x \le 4$

$x(1 - x) \ge 0$ ==> $x \ge 0$ and $x \le 1$

$\therefore$ Domain of $f = ( - \infty ,\,4] \cap [ - 4,\,\infty ) \cap [0,\,1]$$ = [0,\,1]$.

But $2 < {e^x} < 3$ ==> $3 < ({e^x} + 1) < 4$

==> $\frac{1}{4} < \frac{1}{{1 + {e^x}}} < \frac{1}{3}$

$\therefore$ Range of $f(x) = \left( {\frac{1}{4},\,\frac{1}{3}} \right)$.

$\because$ Required interval $ = ( - \infty ,\,0) \cup (1,\,\infty )$.

therefore $x \in ( - \,\infty ,\, - 1)\, \cup \,(1,\,\,\infty ).$

$|x|\,\, > x$ but $|x|\,\, = x$ for  $x $ positive and $|x|\,\, > x$ for  $ x $ negative. 

So, domain will be $( - \,\infty ,\,\,0)$.

Domain of ${y_1} = \sqrt {{x^2} - 1} \, \Rightarrow \,\,{x^2} - 1 \ge 0\,\, \Rightarrow \,\,{x^2} \ge 1$

$x \in ( - \,\infty ,\,\,\infty ) - ( - 1,\,\,1)$ and Domain of ${y_2}$ is real number, 

$\therefore $ Domain of $f(x) = ( - \infty ,\,\,\infty ) - ( - 1,\,\,1)$.

==>$5x - 3 - 2{x^2} \ge 0$ or $(x - 1)\left( {x - \frac{3}{2}} \right) \ge 0$

$\therefore$ $D \in \,[1,\,3/2]$.

For minimum $\cos (bx + c) = - 1$

from $(i)$, $f(x) = - a + d = (d - a)$

For maximum $\cos (bx + c) = 1$

from $(i)$, $f(x) = a + d = (d + a)$

$\therefore$ Range of $f(x) = [d - a,\,\,d + a]$

therefore range of $f(x)$ is $[ - \sqrt 2 ,\,\,\sqrt 2 ].$

$\therefore \,\, - \sqrt 2 \le f(x) \le \sqrt 2 $ and $[ - 1,\,\,1]\, \subset \,[ - \sqrt 2 ,\sqrt 2 ]$.

Hence $f(x)$ lies in $\left[ {\frac{1}{3},\,\,1} \right]$.

and ${\sec ^2}\theta \ge 1$   for  $\theta > \frac{\pi }{3}$ , $\sec \theta \ge 2$

==> ${\sec ^2}\theta \ge 4$.  Required interval $ = [2,\,\,\infty )$.

because $[x]$ is a greatest integer whose value is equal to or less than zero.

$\because {\log _{10}}(1 - x) \ne 0$==> $1 - x \ne 1$==> $x \ne 0$

Again $1 - x > 0$ ==> $1 > x$ ==> $x < 1$

All these can be combined as $ - 2 \le x < 0$ and $0 < x < 1$.

$ = n((A \cap B) \times (B \cap A)) = n(A \cap B).n(B \cap A)$

$ = n(A \cap B).n(A \cap B) = (99)(99) = {99^2}$.

==> $x = \frac{1}{8}[14P + 6]$, $(x \in Z)$

==> $x$ = $\frac{1}{4}(7P + 3)$

==> $x = 6, 13, 20, 27, 34, 41, 48,.….$

$\therefore $ Solution set $= {6, 20, 34, 48,...}$ $\cup$   ${13, 27, 41, ....} $= $[6] \cup   [13]$,

where $[6], [13]$ are equivalence classes of $6$ and $13$ respectively.

$ \Rightarrow \,\,{x^2}(1 - y) + 2\,(17 - y)\,x + (7y - 71) = 0$

For real value of $x,\,\,{B^2} - 4AC \ge 0$

$ \Rightarrow \,\,{y^2} - 14y + 45 \ge 0\,\, \Rightarrow y \ge 9,\,\,y \le 5$ .

==> ${x^2} + 14x + 9 = {x^2}y + 2xy + 3y$

==> ${x^2}(y - 1) + 2x(y - 7) + (3y - 9) = 0$

Since $x$ is real, $\therefore$ $4{(y - 7)^2} - 4(3y - 9)(y - 1) > 0$

==> $4({y^2} + 49 - 14y) - 4(3{y^2} + 9 - 12y) > 0$

==> $4{y^2} + 196 - 56y - 12{y^2} - 36 + 48y > 0$

==> $8{y^2} + 8y - 160 < 0$

==> ${y^2} + y - 20 < 0$

==> $(y + 5)(y - 4) < 0$;  $y$ lies between $-5$ and $4.$

Domain of $f:$

We know that, $0 \leq \mathrm{x}-[\mathrm{x}]<1$ for all $\mathrm{x} \in \mathrm{R}$

and $x-[x]=0$ for $x \in Z$

$\mathrm{So}, 0<\mathrm{x}-[\mathrm{x}]<1$ for all $\mathrm{x} \in \mathrm{R}-\mathrm{Z}$

Hence, domain of $\mathrm{f}=\mathrm{R}-\mathrm{Z}$

Range of $f:$

We have,

$0<\mathrm{x}-[\mathrm{x}]<1 \text { for all } \mathrm{x} \in \mathrm{R}-\mathrm{Z}$

$\Rightarrow 0<\sqrt{\mathrm{x}-[\mathrm{x}]}<1$ for all $\mathrm{x} \in \mathrm{R}-\mathrm{Z}$

$\Rightarrow 1<\frac{1}{\sqrt{x-|x|}}<\infty$ for all $x \in R-Z$

$\Rightarrow 1<\mathrm{f}(\mathrm{x})<\infty$ for all $\mathrm{x} \in \mathrm{R}-\mathrm{Z}$

Hence, range of $\mathrm{f}=(1, \infty)$

==> $x \ne \pm \sqrt 4 $ and ${x^3} - x > 0 \Rightarrow x({x^2} - 1) > 0$

==> $x > 0,\,x > 1$ 

$\therefore$ $D = ( - 1,\,0) \cup (1,\,\infty ) - \{ \sqrt 4 \} $

$i.e.,$ $D = ( - 1,\,0) \cup (1,\,2) \cup (2,\,\infty )$.

so total number of common elements in $(\mathrm{A} \times \mathrm{B})$

and $\mathrm{B} \times \mathrm{A} \Rightarrow \mathrm{n}((\mathrm{A} \times \mathrm{B}) \cap(\mathrm{B} \times \mathrm{A}))=2500$