The inverse of y = 5^logx is:

Total number of possible relation = $2^{n^2}=2^4=16$  

Favourable relations = $?,\left\{\left(x,x\right)\right\},\left\{\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right)\right\}$

$\left\{\left(x,x\right),\left(y,y\right),\left(x,y\right),\left(y,x\right)\right\}$

Probability =  $\frac{5}{16}$  

$l={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dx+}{\displaystyle \underset{-1/2}{\overset{1}{\int }}\left|2x\right|dx}$

let  $l_1={\displaystyle \underset{-1/2}{\overset{1}{\int }}\left[2x\right]dxput2x=t\Rightarrow dx=\frac{dt}{2}}$

$\therefore l=l_1+l_2=0+\frac{5}{8}=\frac{5}{8}\Rightarrow 8l=5$

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}.

35. Given, f={(ab, a+b): a, b $\in$ z}

Let a=1 and b=1;                   a, b $\in$ z.

So, ab=1 × 1=1

a+b=1+1=2.

So, we have the order pair (1,2).

Now, let a= –1 and b= –1; a, b $\in$ z

So, ab=(–1) × (–1)=1

a+b=(–1)+(–1)= –2

So, the ordered pair is (1, –2).

?The element 1 has two image i.e., 2 and –2.

Hence, f is not a function.