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WorksheetsFORMULA QUIZ : CLASS 12 : 28 dec 2022
Total questions: 48
Worksheet time: 24mins
If, a ⊥ b , then
a = b
a ∥ b
a . b = 0
a , b = 5
If a line makes α, β , and , γ ,
angle with x, y and z axes then -
sin2α+ sin2β+ sin2γ = 1
sin2α+ sin2β+ sin2γ = 2
cos2α+ cos2β+ cos2γ = 2
cos2α+ cos2β+ cos2γ = 1
∫cosec x cot x dx =
cosec x + C
- cosec x+ C
cot x + c
- cot x + C
For - any - matrix A, A(adj A)=
∣A∣ I
∣A∣2
∣adj A∣
∣adj A∣2
If for any matrix A , A' = - A , then
A is symmetric
A is skew symmetric
A is invertible
A is singular
∫ xx2−1 dx =
sin−1x + c
cos −1 x + c
sec −1 x + c
tan−1x + C
Projection − of a , on b
∣∣∣b ∣∣∣a . b
∣∣∣a ∣∣∣a . b
∣∣∣b ∣∣∣a
∣∣∣a ∣∣∣ b
Unit − vector ⊥ to, both
a and b is
∣∣∣a ∣∣∣a
∣∣∣a × b ∣∣∣a × b
∣∣∣a × b ∣∣∣
∣∣∣a × b ∣∣∣1
(a,b)∈R ⟹ ( b, a) ∈ R, then
R is symmetric
R is transitive
R is equivqalence
None of these
∫tanx dx =
cot x
− log cos x
coeec2x
sec x tanx
A' - is - transpose - of matrix, then
( AB)' = A'B'
(AB)' = AB
(AB)' = - AB
(AB)' = B'A'
dxdcotx =
cosec2x
− cosec2x
cosecx . cotx
sec x. tanx
Matrix − A = A n×n , then
∣adj A∣=
∣A∣
∣A∣n
∣A∣n−1
I
bijective function is -
One - one function
Onto function
One one and onto both
None of these
dxd(ax)=
axloga
logaax
a
x
A and B, are - independent, then
P(A∩B)= P(B)P(A)
P(A∩B)= P(A).P(B)
P(A∩B)= P(A)+P(B)
P(A+B)= P(A) + P(B)
sin−1(1+x22x)=
2sin−1 x
2tan−1 x
2cot−1 x
2sec−1 x
logab=
1
logeblogea
logealogeb
0
(x−a)(x−b)2p(x) =
(x−a)A+(x−b)B
(x−a)A+(x−b)2B
(x−a)A+(x−b)B+(x−b)2c
(x−a)A+(x−b)2Bx+ c
dxdtanx
secx .tan x
sec2x
cosecx. cot x
− cosec2x
By, Bayes′−theorem, P(AE1)=
P(A)P(E1A)+P(E2)P(E2A)P(A)P(E1A)
P(A)P(AE1)+P(E2)P(E2A)P(A)P(AE1)
P(E1)P(E1A)+P(E2)P(E2A)P(E1)P(E1A)
P(E1)P(E1A)−P(E2)P(E2A)P(E1)P(E1A)
Matrix, A = An×n ,
then ∣kA∣
k∣A∣
kn∣A∣
kn−1∣A∣
I
∫sinx dx =
- cos x+ C
cos x + C
cot x + C
sec x .tan x + C
∫ x2−a2dx=
2a1log∣∣∣∣x+ax−a∣∣∣∣+C
2a1log∣∣∣∣x−ax+a∣∣∣∣+C
a1tan−1(ax)+C
sin−1(ax)+C
If (a, a)∈ R, then − R − is
Reflexive
Symmetric
Transitive
None of these
( Am×n) × (B p×q ) is
possible if
m = q
n = q
n = m
n = p
If , f′′(x) > 0 , then
f is maximum
f is increasing
f is minimum
f is decreasing
∫ex[ f(x) + f′(x)] dx =
ex+ C
f(x)+C
exf′(x) +C
exf(x)+C
∣∣∣a ∣∣∣a is
Projection
of a , on bVector , ⊥ to a
unit − vector , along , a
angle − between, a and b
If , dxdy+Py =Q , then −
Integrating factor is -
∫P dx
∫Q dx
e∫Pdx
e∫Q dx
∫baf(x) dx∫abf(x) dx =
0
1
-1
2
If for every element y of codomain there exists an element
x in domain such that f(x) = y, then
f is one one
f is onto
f is transpose
f is continuous
Unit - matrix , may also be called
Null matrix
Row matrix
Column matrix
Scalar matrix
∫sec x dx =
log tanx + c
cosec2x+C
log(tanx)+C
log(sec x+ tanx) +C
We solve ∫ ax2+bx+ c(ax+b)dx , by
Applying - partial - fractions
applying − product − rule
Nr = A dx.d (ax2+ bx+c) + B
putting : ax2+ bx+c= t
direction ratios of - P and Q are (a, b, c ) and
(p, q, r), then , the direction ratios, of PQ are -
(p+a, q+b, r+c)
(p/a, q/b, r/c)
(p - a, q - b, r - c)
None of these
Row matrix has how many rows -
1
2
3
4
Matrix − A
is − invertible − if
∣A∣ =0
∣A∣= I
∣A∣=0
None of these
dxd(log x)=
1
x
x1
0
Plane - passing - through - point −(x1, y1, z1) is
A(x− x1)+ B (y − y1)
+C (z−z1)= 0
ax+by+cz=1
ax− x1= by− y1= cz−z1
none of these
tan−1(1−xyx+y)=
tan−1x− tan−1y
2tan−1x+ 2tan−1y
tan−1x+ tan−1y
tan−1(x+y )
while - dealing with - homogeneous - differential
- equation we -
evalute - integrating - factor
put , y = vx
arrange - in the form , f(x) dx = g(y) dy
None - of - these
If A ,B , C, D are vertices of a parallelogram, then its area is
∣∣∣AB×CD∣∣∣
21∣∣∣AB×CD∣∣∣
∣∣∣AB×AC∣∣∣
21∣∣∣AB×AC∣∣∣
OB− OA =
BA
AB
∣∣∣AA∣∣∣
0
dxd(logeaax)=
ax
logeaax
axlogea
0
Vector - equation - of - plane, passing through
03 - non - collinear - points
a , b and c is −
( r − a). [ (b − a )×(c − a )]
= 0r [ (b − a )×(c − a )]= 0
( r − a)= 0
( r − a). [ (b + a )×(c + a )]= 0
Area, between
x = f(y), and, y =a, andy = b, is −
∫ab dy
∫abxdy
∫aby dx
∫abdx
y − y1 = − (dxdy)(x1, y1)1 ( x − x1)
represents
Tangent − at − (x1,y1)
minimummaximum − at − (x1,y1)
decreasingincreasing − at − (x1,y1)
Normal − at − (x1,y1)
