CBSE 12TH MATHEMATICS - Online Test

Answer : Option C
Explaination / Solution:

Particular solution of a given differential equation does not contain arbitrary constants i.e. a,b ,c etc since the value of those arbitrary can be found out by subsituting the given values.

Answer : Option B
Explaination / Solution:

Shortest distance between the lines x−x1a1=y−y1b1=z−z1c1and x−x1a2=y−y1b2=z−z1c2is given by: 

Answer : Option C
Explaination / Solution:

Let matrix A is of order m x n , and matrix B is of order p x q . then , the product AB is defined only when n = p. that’s why, If A and B are any two matrices, thenAB may or may not be defined.

Answer : Option C
Explaination / Solution:

Required area :
 = ∫12y.dx =∫12logxdx =[xlogx−x]21= 2log2−2−log1+1= 2log2−1 = 2log2−loge= log4−loge =  log(4/e)

Answer : Option A
Explaination / Solution:

The optimal value of the objective function Z = ax + by may or may not exist, if the feasible region for a LPP is unbounded. This is because the maximum or minimum value of the objective function may not exist.Even if it exists it must occur in a corner pointof the feasible region.

Answer : Option B
Explaination / Solution:

A relation R on a non empty set A is said to be symmetric if fx Ry⇔yRx, for all x , y ∈R Clearly ,x2+ y2= 1 is same as y2+x2= 1 for all x , y ∈R. Therefore ,R is symmetric.

Answer : Option C
Explaination / Solution:

Answer : Option C
Explaination / Solution:

f′(x)=3x2+3x+3
=3(x2+x+1)
=3{(x+12)2+34}⩾94.

Hence an increasing function.

Answer : Option C
Explaination / Solution:

Answer : Option A
Explaination / Solution:

If P1(x1, y1, z1) and P2(x2, y2, z2) are any two points, then the vector joining P1 and P2 is the vector P1P2. then:  P1P2−→−−  =