CBSE 11TH MATHEMATICS - Online Test

Given A={1,2,3,4} and B={4,5,6,7}

A-B set will contain those elements which are present in A but not in B

we can see A∩B={4}

Now A−B=A−(A∩B)

so A−B={1,2,3,4}−{4}={1,2,3}

The slope of the line joining the points (a,0) and (0,b) is [b-0]/[0-a] = -(b/a)

Hence the equation of the line is y = (-b/a)x +b

i.e; ay = -bx +ab

Substituting the x coordinate 3a in the place of x in the above equation we get y = -2b

Hence (3a,-2b) is another point on the line.

[(p∧∼q)∧(∼p)]∨[(p∧∼q)∧q)]     Since p∧(q∨r)≡(p∧q)∨(p∧r)

F V F = F     Since       p∧∼p=F

Hence contracdiction

s = (ae,0) and T = (-ae,0) and B = (0,b)

Since it is an equilateral triangle, ST2 = TB2

This implies 4a2e2 = a2e2 + b2

3a2e2 = b2 

3a2e2 = a2(1 - e2)

3e2  = 1 - e2

Therefore e = 1/2