MATHS MCQ

$f(x) = \cos (9x) + \cos ( - 10x)$$ = \cos (9x) + \cos (10x)$

$ = 2\cos \left( {\frac{{19x}}{2}} \right)\cos \left( {\frac{x}{2}} \right)$

$f\left( {\frac{\pi }{2}} \right) = 2\cos \left( {\frac{{19\pi }}{4}} \right)\cos \left( {\frac{\pi }{4}} \right)$;

$f\left( {\frac{\pi }{2}} \right) = 2 \times \frac{{ - 1}}{{\sqrt 2 }} \times \frac{1}{{\sqrt 2 }} = - 1$.

From (i), clearly $f(x)$ is defined for those values of $x$ for which ${\log _{10}}\left[ {\frac{{5x - {x^2}}}{4}} \right] \ge 0$

==> $\left( {\frac{{5x - {x^2}}}{4}} \right) \ge {10^0} \Rightarrow \left( {\frac{{5x - {x^2}}}{4}} \right) \ge 1$

==> ${x^2} - 5x + 4 \le 0$ ==> $(x - 1)(x - 4) \le 0$

Hence domain of the function is $[1, 4].$

==> ${x^2} > 1,$ ==> $x < - 1{\rm{ \,or\, }}x > 1$ and $3 + x > 0$

$\therefore$ $x > - 3$ and $x \ne - 2$

$\therefore$ ${D_f} = ( - 3,\, - 2) \cup ( - 2,\, - 1) \cup (1,\,\infty )$.

If we want the range of $f(x)$ to be real then, $0 \leq 2-2 x-x^{2}$

$\Longrightarrow x^{2}+2 x \leq 2$

$\Longrightarrow x^{2}+2 x+1 \leq 3$

$\Longrightarrow(x+1)^{2} \leq 3$

$\Longrightarrow-\sqrt{3} \leq x+1 \leq \sqrt{3}$

$\Longrightarrow-\sqrt{3}-1 \leq x \leq \sqrt{3}-1$

So the domain of $x$ is $[-1-\sqrt{3},-1+\sqrt{3}]$

==> ${x^2} - 6x + 6 \ge 1$ ==> $(x - 5)(x - 1) \ge 0$

This inequality holds if $x \le 1$ or $x \ge 5$. 

Hence, the domain of the function will be $( - \infty ,\,1] \cup [5,\,\infty )$.

We know that, $0 \le {\cos ^2}x \le 1$ at $\cos x = 0,\,$  $f(x) = 1$ and

at $\cos x = 1$, $f(x) = \sqrt 2 $

$\therefore$ $1 \le x \le \sqrt 2 $==>$x \in [1,\,\,\sqrt 2 ]$.

$ = [x:f(x) = 0$ and $g(x) = 0]$ $ = {[f(x)]^2} + {[g(x)]^2} = 0$.

We have, $\left[x^2\right]=x+1$

Clearly, Eq.$(i)$ $\Rightarrow x$ is an integer $\quad \ldots (ii)$

$x^2-\left\{x^2\right\} =x+1$

$\Rightarrow \quad x^2-x-1 =\left\{x^2\right\}$

$Eq.(iii) \Rightarrow x^2-x-1 < 1$

$\Rightarrow \quad-1 < x < 2 \quad \ldots$ (iv)

From Eqs.$(ii)$ and $(iv)$ $\Rightarrow$ possible values of $x$ are $0$ and $1 .$

But $0$ and $1$ do not satisfy Eq.$(i)$.

$\therefore$ Equation $\left[x^2\right]=x+1$ has no soltuion.

Let $f(x)=a x+b$

$f(1)=a+b=5 \quad$...(i)

$f(2)=2 a+b=7 \quad \ldots$ (ii)

On solving Eqs. $(i)$ and $(ii)$, we get

$a =2, b=3$

$\therefore \quad f(x) =2 x+3$

$f(12) =2(12)+3=27$

We have,

$\left[a^k\right]>[a]^k$

It is true only $k \leq \frac{1}{a-[a]}+1$

$\frac{1}{a-[a]}+1 \rightarrow \infty$

$\therefore k$ is smaller than $\infty$.

We have, $\log _{\{x\}}[x]+\log _{[x]}\{x\}$

$\log _{\{x\}}[x]$ is defined if $[x] > 0, x \notin$ integer

$\begin{array}{lc}\therefore & x > 1 \\ \log _{[x]}\{x\} \text { is defined if }[x] > 1, x \notin \text { integer } \\ \because & x > 2\end{array} .$

From Eqs.$(i)$ and $(ii)$, $x > 2$

$\therefore$ Option (c) $x \in(2,3)$ satisfied and in

option (d) $x \in(3,5)$ not satisfied because $4$ is an integer.

$ -\sin 5 x \in[-1,1] $

$ 7-\sin 5 x \in[6,8] $

$ \frac{1}{7-\sin 5 x} \in\left[\frac{1}{8}, \frac{1}{6}\right]$

$ a \in f(a)$

That means 'a' will connect with subset which contain element ' $a$ '.

Total options for 1 will be $2^6$. (Because $2^6$ subsets contains $1$)

Similarly, for every other element

Hence, total is $2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6 \times 2^6=2^{42}$

Ans. $2+42=44$

$ \text { Domain : } x^2-25 \geq 0 \Rightarrow x \in(-\infty,-5] \cup[5, \infty) $

$ 4-x^2 \neq 0 \Rightarrow x \neq\{-2,2\} $

$ x^2+2 x-15>0 \Rightarrow(x+5)(x-3)>0 $

$ \Rightarrow x \in(-\infty,-5) \cup(3, \infty) $

$ \therefore x \in(-\infty,-5) \cup[5, \infty) $

$ \alpha=-5 ; \beta=5 $

$ \therefore \alpha^2+\beta^3=150$

$x \in\left(5, \frac{41}{5}\right] $

$3 \alpha+10 \beta=97 $

Option ($1$)

$x+1 > 0 \Rightarrow x > -1$

$x+1 \neq 1 \Rightarrow x \neq 0 \text { and } x > 0$

Denominator

$x^2-2 x-3 \neq 0$

$(x-3)(x+1) \neq 0$

$x \neq-1,3$

So Ans $(2, \infty)-\{3\}$

$=5+2 \sqrt{6+x-x^2}$

$y^2=5+2 \sqrt{\frac{25}{4}-\left(x-\frac{1}{2}\right)^2}$

$y_{\max }=\sqrt{5+5}=\sqrt{10}$

$y_{\min }=\sqrt{5}$

$\left(\frac{5}{37}, \frac{2}{5}\right]$

By cross multiplying

$y x^2-8 x y+12 y-x^2-2 x-1=0$

$x^2(y-1)-x(8 y+2)+(12 y-1)=0$

Case $1, y \neq 1$

$D \geq 0$

$\Rightarrow(8 y+2)^2-4(y-1)(12 y-1) \geq 0$

$\Rightarrow y(4 y+21) \geq 0$

$y \in\left(-\infty, \frac{-21}{4}\right] \cup[0, \infty)-\{1\}$

Case $2, y =1$

$x^2+2 x+1=x^2-8 x+12$

$10 x=11$

$x =\frac{11}{10} \quad$ So, $y$ can be 1

Hence $y \in\left(-\infty, \frac{-21}{4}\right] \cup[0, \infty)$

Let $g(x)=x^2-x+1$

$=\left(x-\frac{1}{2}\right)^2+\frac{3}{4}$

$\because\left| x ^2- x +1\right| \text { and }\left[ x ^2- x +2\right]$

Both have minimum value at $x =1 / 2$

$\Rightarrow \text { Minimum } f ( x )=\frac{3}{4}+0$

$=\frac{3}{4}$

$f(x)-g(x)=\sqrt{x}-\sqrt{1-x},$ domain $[0,1]$

$g(x)-f(x)=\sqrt{1-x}-\sqrt{x},$ domain $[0,1]$

$\frac{f(x)}{g(x)}=\frac{\sqrt{x}}{\sqrt{1-x}},$ domain $[0,1)$

$\frac{g(x)}{f(x)}=\frac{\sqrt{1-x}}{\sqrt{x}},$ domain $(0,1]$

So, common domain is $(0,1)$

${2^{(x\, + 2)({x^2} - 5x + 6)}}\, = \,{2^0}\, \Rightarrow \,x\, = \, - \,2,2,3$

$A\, = \,\{  - 2\,,\,2\,,3\} $

$B\, = \,\{ x\, \in \,Z\,:\, - \,3\,\, < \,\,2x\, - \,1\, < \,9\} $

$B\, = \,\{ 0,1,2,3,4\} $

Hence, $A\times B$ has is $15$ elements. So number of subsets of $A\times B$ is $2^{15}$

given $|a-b| \geq 2$ so if

$a=1  b=3,4,5,6  \rightarrow  4 \times 2=8$

$a=1  b=4,5,6 \rightarrow  3 \times 2=6$

$a=1  b=5,6  \rightarrow 2 \times 2=4$

$a=1  b=6  \rightarrow  2 \times 1=2$

$i.e. $Total elements in $X$ is ${ }^{20} C _6$

Now for $n ( Y )$,

range of $R$ has exactly one element $i.e$. second elements must be constant in $R$ and since $R$ must have 6 element so it is not possible to satisfy both condition so $n ( Y )=0$.

$n ( z ) \quad  1 \rightarrow 3,4,5,6$

$2 \rightarrow 4,5,6$

$3 \rightarrow 1,5,6$

$4 \rightarrow 1,2,6$

$5 \rightarrow 1,2,3$

$6 \rightarrow 1,2,3,4$

no. of relation that are function will be

$={ }^4 C _1 \times{ }^3 C _1 \times{ }^3 C _1 \times{ }^3 C _1 \times{ }^3 C _1 \times{ }^4 C _1$

$=(4 \times 3 \times 3)^2= k ^2$

i.e. $k =36$

$f^{\prime}(x)=\frac{\left(x^2+2 x+4\right)(2 x-3)-\left(x^2-3 x-6\right)(2 x+2)}{\left(x^2+2 x+4\right)^2}$

$f^{\prime}(x)=\frac{5 x(x+4)}{\left(x^2+2 x+4\right)^2}$

$f^{\prime}(x): \frac{+, \quad+}{-4}, \quad f(0)=-\frac{3}{2}, \quad \lim _{x \rightarrow \pm \infty} f(x)=1$

Range : $\left[-\frac{3}{2}, \frac{11}{6}\right]$, clearly $f(x)$ is into

$=\frac{4^{ x }}{4^{ x }+2}+\frac{4 / 4^{ x }}{\frac{4}{4^{ x }}+2}$

$=\frac{4^{ x }}{4^{ x }+2}+\frac{4}{4+2.4^{ x }}$

$=\frac{4^{ x }}{4^{ x }+2}+\frac{2}{2+4^{ x }}$

$=1$

$\text { so, } f\left(\frac{1}{40}\right)+f\left(\frac{2}{40}\right)+\ldots+f\left(\frac{39}{40}\right)-f\left(\frac{1}{2}\right)$

$=19+f\left(\frac{1}{2}\right)-f\left(\frac{1}{2}\right)=19$

==> Range $ = (1,\,7/3]$.

$\therefore$ $x > - 3$ and $(x + 1)(x + 2) \ne 0,$  $i.e.$  $x \ne - 1,\, - 2$

$\therefore$ Domain$ = ( - 3,\,\infty ) - \{ - 1,\, - 2\} $.

$n(A × B) = n(A) . n(B) = 3 × 3 = 9.$

$\therefore A × (B  \cup C) = {a, b} × {c, d, e}$

$= {(a, c), (a, d), (a, e), (b, c), (b, d), (b, e)}.$

= $R \times (P \cap Q) = (R \times P) \cap (R \times Q)$ = $(R \times Q) \cap (R \times P)$.

$A \times B = B \times A$ is true, if $A = B$.

$(A \cup B) \times (A \cap B) = \{ (1,\,3),\,(2,3),(3,3),(8,3)\} $.

$(A - B) \times (A \cap B) = \{ (3,\,2);\,(3,\,5)\} $.

$B \cap C = \{4\}$

$\therefore$ $A × (B \cap C) = \{(2, 4); (3, 4)\}.$

Hence, $n\,[(A \times B) \cap (B \times A)] = 4$.

So, the total number of subsets of $A \times A$ is ${2^9}$ and a subset of $A \times A$ is a relation over the set $A$.

$x=1 \rightarrow y=1+2=3$

$x=2 \rightarrow y=2+2=4$

$x=3 \rightarrow y=3+2=5$

$x=4 \rightarrow y=4+2=6$

$x=5 \rightarrow y=5+2=7$

$R=\{(1,3),(2,4),(3,5),(4,6),(5,7)\}$

Here in $(4,6)$ ,$6$ does not belong to either $X$or$Y$

So $R_1$ is not a relation between $X$ and $Y$

In $R _4$, since it contains an element $(7,9)$ which relates set $Y$ to set $Y$, while rest elements relate set $X$ to $Y$.

$\therefore R _4$ is not a relation between $X$ and $Y$

Since every subset of $A × B$ defines a relation from $A$ to $B$, number of relation from $A$ to $B$ is equal to number of subsets of $A \times B = {2^6} = 64$.

$ = \{ (4,\,1),\,(5,\,2),\,(6,\,3),\,.....\} $.

But $\log {x^2}$ defined for all real values of $x$, also $\log |x|$ is also defined $\forall $ real $x$. 

Hence $\log {x^2}$and $2\log |x|$ are identical functions.

$ = 4 + 3x + 3{x^2} + 4{x^3} = f(x)$.

$f(-x) = -2x + |-x|$ = $\,-2x + \,|x|$

$f(x) = 2x + |x|$ ==> $f(2x) + f( - x) - f(x)$

$ = 4x + 2|x| + |x| - 2x - 2x - |x|$$ = 2\,\,|x|$.

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