Relations and Functions

 $[x{]}^2+2[x+2]-7=0$

$\begin{array}{rr}& \Rightarrow [x{]}^2+2[x]+4-7=0\\ & \Rightarrow \left[x\right]=1,-3\\ & \Rightarrow x\in \left[\mathrm{1,2}\right)\cup [-3,-2)\end{array}$

 $\frac{7}{2}\left(1+cos2\theta \right)-\frac{3}{2}\left(1-cos2\theta \right)-2co{s}^{2}2\theta =2$

Put cos 2 $\theta$= t

Equation 2t²– 5t = 0, t (2t – 5) = 0

$t=0,\frac{5}{2}$

cos 20 = 0, 0 < 2 < 4

$x_1+x_2=2\left(ta{n}^2\theta +co{t}^2\theta \right)\Rightarrow ?$

$=2\left(1+1\right)+2\left(1+1\right)+2\left(1+1\right)+2\left(1+1\right)=16$

Kindly go through the solution

This is a Short Answer Type Question as classified in NCERT Exemplar

Sol:

$\begin{array}{l}Givenfunction,f(x)=cosx\forall x\in R\\ Let\left[-\frac{\pi }{2},\frac{\pi }{2}\right]\in f(x)\\ f\left(-\frac{\pi }{2}\right)=cos\left(-\frac{\pi }{2}\right)=cos\frac{\pi }{2}=0\\ cos\left(\frac{\pi }{2}\right)=cos\frac{\pi }{2}=0\\ But-\frac{\pi }{2}\ne \frac{\pi }{2}\\ \therefore ,f(x)isnotone-one.\\ Now,f(x)=cosx,\forall x\in Risnotontoasthereisnopre-imageforanyrealnumber.\\ Whichdoesnotbelongtotheintervals[-1,1],therangeofcosx.\end{array}$

36. Given, A={9,10,11,12,13}.

f(x)=the highest prime factor of n.

and f: A → N.

Then, f(9)=3 [? prime factor of 9=3]

f (10)=5 [? prime factor of 10=2,5]

f(11)=11 [? prime factor of 11 = 11]

f(12)=3 [? prime factor of 12 = 2, 3]

f(13)=13 [? prime factor of 13 = 13]

?Range of f=set of all image of f(x) = {3,5,11,13}.

A binary operation * on {a, b} is a function from {a, b} * {a, b} → {a, b}

i.e., * is a function from { (a, a), (a, b), (b, a), (b, b)} → {a, b}.

Hence, the total number of binary operations on the set {a, b} is 2⁴ i.e., 16.

The correct answer is B.

It is given that,

$f:R\to R$ is defined as $f\left(x\right)=\{\begin{array}{ccc}\hfill 1\hfill & \hfill x>0\hfill & \hfill \hfill \\ \hfill 0\hfill & \hfill x=0\hfill & \hfill \hfill \\ \hfill -1\hfill & \hfill x<0\hfill & \hfill \hfill \end{array}$

Also, $g:R\to R$ is defined as $g\left(x\right)=\left[x\right]$ , where [x] is the greatest integer less than or equal to x.

Now, let $x\in \left(0,1\right)$

Then, we have:

$\left[x\right]=1$ if $x=1$ and $\left[x\right]=0$ if $0<x<1$

$\begin{array}{l}\therefore fog\left(x\right)=f\left(g\left(x\right)\right)=f\left(\left[x\right]\right)=\{\begin{array}{cc}\hfill f\left(1\right)\hfill & \hfill if,x=1\hfill \\ \hfill f\left(0\right)\hfill & \hfill if,x\in \left(0,1\right)\hfill \end{array}=\left\{\left(1,"if,x=1"\right),\left(0,:if,x\in \left(0,1\right)"\right):\right\}\\ gof\left(x\right)=g\left(f\left(x\right)\right)\\ =g\left(1\right)\left[x>0\right]\\ =\left[1\right]=1\end{array}$

Thus, when $x\in \left(0,1\right)$ , we have $fog\left(x\right)=0and,gof\left(x\right)=1.$

Hence, fog and gof do not coincide in (0, 1).

Therefore, option (B) is correct.

2, Therefore, option (B) is correct.

It is clear that 1 is reflexive and symmetric but not transitive.

Therefore, option (A) is correct.

It is given that A = {–1, 0, 1, 2}, B = {–4, –2, 0, 2}

Also, it is given that $f,g:A\to B$ are defined by $f\left(x\right)=x^2-x,x\in A$ and $g\left(x\right)=2x-\frac{1}{2}-1,x\in A$ .

It is observed that:

$\begin{array}{l}f\left(-1\right)=\left(1^2\right)-\left(-1\right)=1+1=2\\ g\left(-1\right)=2\left(-1\right)-\frac{1}{2}-1=2\left(\frac{3}{2}\right)-1=3-1=2\\ \Rightarrow f\left(-1\right)=g\left(-1\right)\\ f\left(0\right)={\left(0\right)}^{\wedge }2-0=0\\ g\left(0\right)=2\left(0\right)-\frac{1}{2}-1=2\left(\frac{1}{2}\right)-1=1-1=0\\ \Rightarrow f\left(0\right)=g\left(0\right)\\ f\left(1\right)={\left(1\right)}^{\wedge }2-1=1-1=0\\ g\left(1\right)=2a-\frac{1}{2}-1=2\left(\frac{1}{2}\right)-1=1-1=0\\ \Rightarrow f\left(1\right)=g\left(1\right)\\ f\left(2\right)={\left(2\right)}^{\wedge }2-2=4-2=2\\ g\left(2\right)=2\left(2\right)-\frac{1}{2}-1=2\left(\frac{3}{2}\right)-1=3-1=2\\ \Rightarrow f\left(2\right)=g\left(2\right)\\ \therefore f\left(a\right)=g\left(a\right)\forall a\in A\end{array}$

Hence, the functions f and g are equal.