CBSE 11TH MATHEMATICS - Online Test

Answer : Option D
Explaination / Solution:

For symmetric distribution, median=Q1+Q32=20+402=30

Answer : Option A
Explaination / Solution:

when a point M which is in the 2nd quadrant is reflected in the origin, its image is formed in the 4th quadrant whose coordinates are (x,-y)

Hence the image of the point (1,-2) is (1,-2)

Answer : Option B
Explaination / Solution:

∼(p∨q)≡∼p∧∼q De Morgan's law

Answer : Option C
Explaination / Solution:

putting the value y=c into parabola,we get

x2=c-1

or x2-(c-1)=0

here discriminat=4c−4−−−−−√

line y=c is tangent when discriminant is equal to 0.

putting disriminant =0 we get c=1.

(0, c) will be a point on the parabola. 

Answer : Option B
Explaination / Solution:

Answer : Option B
Explaination / Solution:

Replace n = 1 we have 1/2. When n = 2 we have 1/2+1/6=2/3,( by the principle of mathematical induction when n = 1 we need to take one term from LHS. and for n = 2 we need to take the sum of two terms from the LHS. .....)

Answer : Option C
Explaination / Solution:
No Explaination.

Answer : Option D
Explaination / Solution:

General equation of the sphere is x2+y2+z2+2gx+2fy+2hz+d=0---------------------1)

Since 1) passes through the point (0,0,0) using this in 1) we get d=0

Similarly 1) passes through ( a , 0 , 0 ) , ( 0 , b , 0 ) ( 0 , 0 , c ) using these values in 1)

a2+2ag=0=>a(a+2g)=0b2+2bf=0=>b(b+2f)=0c2+2ch=0=>c(c+2h)=0

But as abc≠0  So , a ≠0 ,b ≠0 ,c ≠0

So from above equations , we have a =  - 2g , b= - 2f , c = -2h

 centre is (-f ,-g , -h) = ( a/2 , b/2 , c/2 )

 

Answer : Option A
Explaination / Solution:

Given xyz=30

We have the possible values of x ,y,z are the following triads 

1,1,30 
1,2,15 
1,3,10 
1,5,6 
2,3,5 
First one can have 3!/2! = 3 ways and the remaining four triads can have 3! combinations 
Hence total combinations = 3 + 4*3! = 27 

Answer : Option A
Explaination / Solution: