(4)

Graph of the function f(x)

The sum of local minimum values at A and B

$=\frac{7}{3}–\frac{11}{72}=\frac{157}{72}$.

(1)

We have, $f\left(x\right)=\frac{2^x}{2^x+\sqrt{2}}, x\in R$

$\Rightarrow f\left(x\right)+f(1–x)=\frac{2^x}{2^x+\sqrt{2}}+\frac{2^{1–x}}{2^{1–x}+\sqrt{2}}$

        $=\frac{2^x}{2^x+\sqrt{2}}+\frac{2}{2+\sqrt{2}·2^x}=1$

Now, $S=\sum _{k=1}^{81}f\left(\frac{k}{82}\right)=f\left(\frac{1}{82}\right)+f\left(\frac{2}{82}\right)+...+f\left(\frac{81}{82}\right)$

$\Rightarrow S=f\left(\frac{1}{82}\right)+...+f(1–\frac{2}{82})+f(1–\frac{1}{82})$

$=f\left(\frac{1}{82}\right)+f(1–\frac{1}{82})+f\left(\frac{2}{82}\right)+f(1–\frac{2}{82})+...+ \mathrm{upto} 40 +f\left(\frac{41}{82}\right)$

$=\underset{40 \mathrm{times}}{\underbrace{(1+1+1+...+1)}}+\frac{2^{1/2}}{2^{1/2}+\sqrt{2}}=40+\frac{1}{2}=\frac{81}{2}$.

(3)

As f(x) is onto, hence A is range of f(x).

Now, $f\text{'}\left(x\right)=6x^2–30x+36=6(x–2)(x–3)$

for extremum, f(2) = 16 – 60 + 72 + 7 = 35; f(3) = 54 – 135 + 108 + 7 = 34; f(0) = 7; F(1) = 30.

Hence, range $\in$ [7, 35] = A

Also, for range of g(x), g(x) = $1-\frac{1}{x^{2025}+1}\in [0,1)=B$

 $\therefore$  S = {0, 7, 8, ..., 35}

Hence, n(S) = 30

(3)

Let $f_1\left(x\right)={\log }_5(18x–x^2–77)$

$\therefore 18x–x^2–77>0$

$\Rightarrow x^2–18x+77<0$

$\Rightarrow (x–11)(x–7)<0$

$\Rightarrow x\in (7,11) \therefore \alpha =7, \beta =11$

Also, let $f_2\left(x\right)={\log }_{(x–1)}\left(\frac{2x^2+3x–2}{x^2–3x–4}\right)$

$\therefore x–1>0 \Rightarrow x>1, x–1\ne 1 \Rightarrow x\ne 2$,

$\frac{2x^2+3x–2}{x^2–3x–4}>0 \Rightarrow \frac{(2x–1)(x+2)}{(x–4)(x+1)}>0$

$\Rightarrow x\in (4,\infty ) \therefore \gamma =4$

Now, ${\alpha }^2+{\beta }^2+{\gamma }^2$ = 49 + 121 + 16 = 186.

(1)

$18x^2-11x+1>0$

$(2x-1)(9x-1)>0$

$x<\frac{1}{9}\text{ or }\frac{1}{2}<x$

$\text{Also }10x^2-17x+7>0$

$(x-1)(10x-7)>0$

$x<\frac{7}{10}\text{ or }1<x$

$\text{and }10x^2-17x+7\ne 1$

$x\in (-\infty ,\frac{1}{9})\cup (\frac{1}{2},\frac{7}{10})\cup (1,\infty )-\left\{\frac{6}{5}\right\}$

$90(a+b+c+d+e)=90(\frac{1}{9}+\frac{1}{2}+\frac{7}{10}+1+\frac{6}{5})$

$=10+45+63+90+108=316$

(4)

$-1\le \frac{5-x}{2x+3}\le 1\&10-x>0,10-x\ne 1$

$\left|\frac{{\displaystyle 5-x}}{{\displaystyle 2x+3}}\right|$$\ge 1\&x<10\&x\ne 9$

${(5-x)}^2-{(2x+3)}^2\le 0\&x<10\&4x\ne 9$

$(x+8)(3x-2)\ge 0\&x<10\&x\ne 9$

$\Rightarrow (-\infty ,-8]\cup [\frac{{\displaystyle 2}}{{\displaystyle 3}},10)-\left\{9\right\}$

$\Rightarrow (\alpha +\beta +\gamma +\delta )=6(-8+\frac{{\displaystyle 2}}{{\displaystyle 3}}+10+9)$

$=70$

(3)

$x\in (0,\frac{\pi }{2})\Rightarrow y=1+1+1+1=4$

$x\in (\frac{\pi }{2},\pi )\Rightarrow y=1-1-1-1=-2$

$x\in (\pi ,\frac{3\pi }{2})\Rightarrow y=-1-1+1+1=0$

$x\in (\frac{3\pi }{2},2\pi )\Rightarrow y=-1+1-1-1=-2$

$\therefore \text{ Range of }y\text{ is }\{-2,0,4\}$

$\text{Required sum}=-2+0+4=2$

(4)

$\mathbf{\text{Statement 1}}\text{: }$$f\left(x\right)$$=\frac{x}{1+\left|x\right|}$

$f\left(x\right)$$=$$\{\begin{array}{cc}\frac{x}{1+x}& x\ge 0\\ \frac{x}{1-x}& x<0\end{array}$

$\text{f(x) is one-one}$

$\mathbf{\text{Statement 2}}\text{: }$$f\left(x\right)$$=\frac{x^2+4x-30}{x^2-8x+18},f\left(0\right)=\frac{-30}{18}=\frac{-5}{3}$

$\frac{-5}{3}=\frac{x^2+4x-30}{x^2-8x+18}$

$\text{On solving }x=0,-1$

$\Rightarrow f\left(0\right)=f(-1)=\frac{-5}{3}$

$\therefore$ $f\left(x\right)\text{ is many-one}$

(1)

$m=\frac{9\times 10\times 19}{6}=15\times 19$

$3f\left(x\right)+2f\left(\frac{15}{x}\right)=5x$

$\text{Replace }x\text{ by }\frac{15}{x}$

$3f\left(\frac{15}{x}\right)+2f\left(x\right)=\frac{75}{x}$

$9f\left(x\right)-4f\left(x\right)=15x-\frac{150}{x}$

$5f\left(x\right)=15x-\frac{150}{x}$

$f\left(x\right)=3x-\frac{30}{x}$

$f\left(5\right)=15-\frac{30}{5}=9$

$f\left(2\right)=6-15=-9$

$f\left(5\right)-f\left(2\right)=18$

(1)

$A:\left|\right|x-3|-3|\le 1$

$\Rightarrow -1\le |x-3|-3\le 1$

$\Rightarrow 2\le$$|x-3|$$\le 4$

$\Rightarrow 2\le (x-3)\le 4\text{ or }-4\le (x-3)\le -2$

$\Rightarrow 5\le x\le 7\text{ or }-1\le x\le 1$

$A=\{-1,0,1,5,6,7\}$

$B\Rightarrow x=4,|x-2|=1\Rightarrow x=3\text{ or }1\left(\text{reject}\right)$

$\Rightarrow B=\{3,4\}$

$\text{Number of onto functions from }A\text{ to }B=2^6-2=62$

(3)

$\text{Let }{x}^2-10x+85=\lambda$

$\therefore$ $\text{Domain for first term}$

$\lambda >0...\left(1\right)$

$\&7-{\log }_2\lambda >0\Rightarrow \lambda <2^7...\left(2\right)$

$\&{\log }_5(7-{\log }_2\lambda )>0\Rightarrow \lambda <2^6...\left(3\right)$

$\therefore$ $\text{from (1), (2) \& (3)}$

$0<\lambda <2^6$

$0<x^2-10x+85<64$

$\Rightarrow x\in (3,7)...\left(A\right)$

$\&\text{ domain for second term }-1\le \frac{3x-7}{x-17}\le 1$

$\Rightarrow x\in [-5,6]...\left(B\right)$

$\text{From (A) \& (B), domain of function will be }(3,6]$

$\Rightarrow \alpha =3,\beta =6$

$\Rightarrow \alpha +\beta =9$

(2)

$f(x+y)=f\left(x\right) f\left(y\right)\Rightarrow f\left(x\right)=a^x$

$(\because f(1)=7\Rightarrow a^1=7)$

$\text{So }f\left(x\right)=7^x$

$\text{Now}$

$g(x+y)=g\left(xy\right)(\text{put }y=1)$

$\Rightarrow g(x+1)=g\left(x\right)$

$\text{so }g\left(1\right)=g\left(2\right)=g\left(3\right)=\cdots =g\left(n\right)=1$

$\text{Given }\sum _{x=1}^n\frac{f\left(x\right)}{g\left(x\right)}=19607$

$\sum _{x=1}^n\frac{7^x}{1}=19607$

$\Rightarrow 7\left(\frac{{\displaystyle 7^n-1}}{{\displaystyle 7-1}}\right)=19607$

$7^n-1=\frac{6}{7}\times 19607$

$7^n=16807$

$\Rightarrow n=5$

(1)

$f\left(x\right)={\left[x\right]}^2-\left[x\right]-6=\left(\right[x]+2)\left(\right[x]-3)$

$\left(1\right)f\left(x\right)>0\Rightarrow \left[x\right]\in (-\infty ,-2)\cup (3,\infty )$

$\Rightarrow x\in (-\infty ,-2)\cup [4,\infty )$

$\left(2\right)f\left(x\right)<0\Rightarrow \left[x\right]\in (-2,3)$

$\Rightarrow x\in [-1,3)$

$\text{option (2) is correct}$

$\left(3\right){\int }_0^{2}f\left(x\right)dx={\int }_0^1(0-0-6)dx+{\int }_1^2(1-1-6)dx$

$=-6-6$

$=-12$

$\left(4\right)f\left(x\right)=0\Rightarrow \left[x\right]=3\text{ or }\left[x\right]=-2$

$\text{infinitely many solutions}$

(2)

$g\left(2\right)=13$

$f\left(g\right(2\left)\right)=f\left(13\right)$

$\text{Now }4g\left(f\right(x\left)\right)=3x^2-32x+72$

$4\left[3f^2\right(x)+2f(x)-3]=3x^2-32x+72$

$\text{Let }f\left(x\right)=t$

$12t^2+8t-(3x^2-32x+84)=0$

$f\left(x\right)=\frac{-8\pm \sqrt{64+48(3x^2-32x+84)}}{24}$

$f\left(x\right)=\frac{-8\pm 4(3x-16)}{24}$

$\because$ $f\left(0\right)=-3$ $\therefore$ $\text{we take +ve sign}$

$\therefore$ $f\left(x\right)=\frac{-8+4(3x-16)}{24}$

$\therefore$ $f\left(13\right)=\frac{-8+4·23}{24}=\frac{84}{24}=\frac{7}{2}$

(2)

Ans.    ${}^{10}C_7\times 7!$

Explanation:

Number of elements in set $A=7$

Number of elements in set $B=10$

Selection of 7 elements from set B is ${}^{10}C_7$

and these elements are related one-one to set A in $7!$ ways.

$\therefore$  Total one-one functions from set A to set B $={}^{10}C_7\times 7!$

(3)

Ans.   24

Explanation:

The total number of injective mappings from the set containing 3 elements into the set containing 4 elements is ${}^{4}P_3=4!=24$

(3)

Ans.    $\{4,-4\},\varphi$

Explanation:

Since for $f^{-1}\left(17\right)=x$

$\Rightarrow$                 $f\left(x\right)=17\text{ or }{x}^2+1=17$

$\Rightarrow$                      $x=\pm 4$

or             $f^{-1}\left(17\right)=\{4,-4\}$

and for   $f^{-1}(-3)=x$

$\Rightarrow$                  $f\left(x\right)=-3$

$\Rightarrow$               $x^2+1=-3$

$\Rightarrow$                      $x^2=-4$

Hence,  $f^{-1}(-3)=\varphi$

(2)

Ans.   6

Explanation:

Number of one-one functions = 3 × 2 × 1 = 6

(3)

Ans.    0

Explanation:

Total number of elements in set $A=5$

Total number of elements in set $B=6$

As, the number of bijections from A to B is possible only when $n\left(A\right)\ge n\left(B\right)$

But here, $n\left(A\right)<n\left(B\right)$

Hence, the total number of bijections from A to B is 0.

(3)

Ans.   $\mathit{x}$

Explanation:

               $f\left(x\right)={(3-x^3)}^{\frac{1}{3}}$

         $f\circ f\left(x\right)=f\left[f\right(x\left)\right]$

                       $={\{3-{\left[{(3-x^3)}^{\frac{{\displaystyle 1}}{{\displaystyle 3}}}\right]}^3\}}^{\frac{1}{3}}$

                       $={(3-3+x^3)}^{\frac{1}{3}}$

                       $={\left(x^3\right)}^{\frac{1}{3}}$

                       $=x$

(1)

Ans.     ${\mathbf{3}}^{\mathbf{9}}$

Explanation:

The number of binary operations that can be defined on a set of $n$ elements is $n^{n^2}$

Given $n=3$,

$\therefore$  Number of binary operations $=3^{3^2}=3^9$

(1)

Ans.      $\frac{\mathbf{4}}{\mathbf{3}}$

Explanation:

Let $e$ be the identity element in $Q^{+}$ with respect to binary operation $\circ$ such that,

       $a\circ e=a=e\circ a,$  $\forall a\in Q^{+}$

       $\frac{ae}{2}=a$ and  $\frac{ea}{2}=a$

        $e=2,$  $\forall a\in Q^{+}$

Thus, 2 is the identity element in $Q^{+}$

Let $b\in Q^{+}$ be the inverse of 3, then

        $3\circ b=e=b\circ 3$

        $\frac{3b}{2}=2$ and  $\frac{b\left(3\right)}{2}=2$

         $b=\frac{4}{3}$

Thus, $\frac{4}{3}$ is the inverse.

(1)

Ans.   $\frac{\mathbf{x}\mathbf{+}\mathbf{4}}{\mathbf{3}}$

Explanation:

                $f\left(x\right)=3x-4$

Let                $y=f\left(x\right)$

                     $y=3x-4$

               $y+4=3x$

             $\frac{y+4}{3}=x$

Replacing $y$ by $x$,

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

(2)

Ans.    injective

(2)

Ans.      (B) and (E) only

Explanation:

$Ex-f\left(x\right)=\frac{1}{\left|x\right|}$ is not continuous at every point.

(2)

Ans.  ${}^{\mathbf{8}}\mathit{C}_{\mathbf{5}}\mathbf{\times }\mathbf{5}\mathbf{!}$

Explanation:

No. of elements in set A = 5

No. of elements in set B = 8

For one-one mapping,

5 elements can be selected out of 8 elements of set B in ${}^{8}C_5$ ways.

Hence, the number of one-one mappings from A to B is ${}^{8}C_5\times 5!$

(2)

Ans.       injective

Explanation:

For one-one,      $f\left(n_1\right)=f\left(n_2\right)$

$\Rightarrow$                 $2n_1+3=2n_2+3$

$\Rightarrow$                       $2n_1=2n_2$

$\Rightarrow$                         $n_1=n_2$

$\Rightarrow f$ is one-one.

Let                            $y=f\left(x\right)=2n+3$

$\Rightarrow$                     $y-3=2n$

$\Rightarrow$                           $n=\frac{y-3}{2}\notin N$

$\Rightarrow$  So, $f\left(n\right)$ is not onto.

(2)

Ans.    27

Explanation:

Here,  $A=\{a,b,c\}$  and  $B=\{4,5,6\}$

$\therefore$  $n\left(A\right)=3$  and  $n\left(B\right)=3$

So, number of functions from A to B $=3^3=3\times 3\times 3=27$

(1)

Ans.    9

Explanation:

                               $f(-1)=3(-1)=-3$

                                   $f\left(2\right)={\left(2\right)}^2=4$

                                   $f\left(4\right)=2\times 4=8$

$\Rightarrow$  $f(-1)+f\left(2\right)+f\left(4\right)=-3+4+8=9$

(2)

Ans.    ${\mathbf{2}}^{\mathbf{n}}\mathbf{-}\mathbf{2}$

Explanation:

Each element in A can map onto any one of the two elements of B.

Total possible functions $=2^n$

Total number of surjections $=2^n-2$