CBSE 12TH MATHEMATICS - Online Test

Answer : Option C
Explaination / Solution:

=[sint]π/2−π/2=sinπ2−sin(−π2)=1+1=2

Answer : Option B
Explaination / Solution:

Position vectors of any point in space are usually calculated from the origin, so we  write OA−→−,OB−→−,OC−→− three points A ,B,C  in space to represent the origin as the initial point.

Answer : Option C
Explaination / Solution:

Answer : Option D
Explaination / Solution:

(12√)+(12√)=2–√

Answer : Option B
Explaination / Solution:

limx→02sinx−sin2xx3 =limx→02sinx(1−cosx)x3 =2limx→0sinxx.1−cosxx2 =2.1.12=1

Answer : Option B
Explaination / Solution:

The power or index of the highest ordered derivative in the polynomial is the degree of the differential equation provided equation is in the standard form.

Answer : Option A
Explaination / Solution:

since we know Direction cosines of a line are coefficient of i,j,k of a unit vector along that line,first find a vector  PQ−→−=(x2−x1)iˆ+(y2−y1)jˆ+(z2−z1)kˆ then to convert  it unit vector divide by its magnitute |PQ−→−| the coefficient of this unit vector will be x2−x1d,y2−y1d,z2−z1d

Answer : Option C
Explaination / Solution:

Required area :

Answer : Option B
Explaination / Solution:

It is an identity matrix with the order 2*2, I2=[1001]

Answer : Option D
Explaination / Solution:

f ‘ (x) =0 
$\Rightarrow$ f (x) is constant in ( 0 , 1 ) and also in ( 2, 4 ). But this does not mean that f ( x) has the same value in both the intervals . However , if f ( c ) =f ( d ) , where c ∈( 0 , 1 ) and d ∈ ( 2, 4) then f ( x ) assumes the same value at all x ∈ ( 0 ,1 ) U (2, 4 ) and hence f is a constant function.