Revised marking scheme of class 12 maths.pdf

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1. Marking Scheme
Class XII Mathematics
Section: A (Multiple Choice Questions- 1 Mark each)
Q. No: Answer Hints/Solution
1 (b) In a symmetric matrix, the (i, j)
th
element is same as the (j, i)
th
element.
Hence,
2 (c)
3 (c)
Projection of on is zero

4 (a) .
5 (b) Given integral is of the form

6 (c)


7 (a)
8 (a)

sq. units
9 (d)

10 (c)

11 (d) .
12 (c)
13 (c) .
14 (b)
Since and are independent
15 (d)

16 (b)

17 (d) ,
.
18 (b) .
19 (b) Both A and R are true but R is not the correct explanation of A.

Hence, Domain of is .

20 (c) A is true but R is false.

2. Hence A is true.
Skew lines are lines in space which are neither parallel nor
intersecting. Hence R is false.
SECTION B (VSA questions of 2 marks each)
Q. No: Value point Marks
21



.
OR

To prove injection
To prove surjection


1





1
1
22 At any time, t let x cm be the length of a side of the equilateral triangle,
and let A be its area.




1


1
23 Let ,
Diagonal =
Required unit vector

OR
Given line is
Direction Ratios
Direction cosines,




1





1
24 ,







1
25

3. SECTION C
(Short Answer Questions of 3 Marks each)
26







1


1




27 P(not obtaining an odd person in a single round) = P(All three of
them throw tails or All three of them throw heads)

P(obtaining an odd person in a single round)

The required probability


.
OR
Let X denote the number of selected scientists who never commit error
in the work and reporting. X takes values 0,1,2



X 0 1 2
P(X)


Mean=








1








2

4. 28


Using


Adding (1) and (2)




OR





















1

1+1
29





Applying the condition and getting
Solution
OR

5. Substituting




























30 We have subject to the constraints
, , ,
.

The corner points of the feasible region are










1

6. Corner Points
A(0,200) 400
B(50,100) 250
C(20,40) 100
D(0,50) 100

Max (= 400) at x = 0, y = 200

1






31

By Partial fractions,












SECTION D
(Long answer type questions (LA) of 5 marks each)
32 Point of intersection of circle and line.



Required Area


sq. units



(Correct
Fig: 1
Mark)












1

7. Redefining
Area

square units
Correct
Fig: 1
Mark)














1


33 Let (, ) × . Then we have
a+b = b+a (by commutative property of addition of natural
numbers)
(, )(, )
Hence, R is reflexive.
Let (, ), (, ) × such that (a, b) R (c, d). Then
a+d = b+c
= (by commutative property of addition of
natural numbers
(, )(, )
Hence, R is symmetric.
Let (, ), (, ), (, ) × such that
(a, b) R (c, d) and (c, d) R (e, f).
Then a+d = b+c, c+f = d+e
=
=
(, )(, )
Hence, R is transitive.
Since, R is reflexive, symmetric and transitive, R is an
equivalence relation on × .


34 Vector equations of the given lines,

8. and
,
,
,

units.
OR
Eliminating t between the equations, we obtain the equation of the
path , which are the equations of the line passing
through the origin having direction ratios .
This line is the path of the rocket.
When t = 10 seconds, the rocket will be at the point (20, -40, 40).
Hence, the required distance from the origin at 10 seconds

The distance of the point (20, -40, 40) from the given line

km
1
½

½
½

1

1+½



1

½

1


2

½
35
,


Given system can be written as




½


2


1



½




1


SECTION E
(Case Studies/Passage based questions of 4 Marks each)
36

9. (i)
,
being a polynomial
function, is differentiable everywhere, hence, differentiable
in (1,5)
(ii)
Since, 2 is the critical point,
(iii)


In the Interval
Conclusion

+ve
is increasing

-ve
C(x) is decreasing

+ve
is increasing
OR
(iii)

are the critical values


Hence, by second derivative test 2 is a point of local
maximum. The local maximum value is
.

Hence, by second derivative test 3 is a point of local
minimum. The local minimum value is

We have
Also
5 is the point of absolute maximum and the absolute
maximum value of the function=56
1 is the point of absolute maximum and the absolute
maximum value of the function=24

1



1



1+1



1/2



½




1/2



1/2

37

(i)
(ii)











1

10. (iii) For the values of x less than 5, and close to 5 and for
the values of x greater than 5 and close to 5 .
Since there is only one critical point is possible (x=18 is not
possible) the volume is maximum at x=5. The maximum
volume is 2450 cubic inches.
OR
(iii)

, hence volume is
maximum when x=5. The maximum volume is 2450 cubic
inches.
1

½


1 ½


1 ½


½


38
Let the events be defined as:
E
1
:the examinee guesses , E
2
: copies the answer, E
3
: knows
the answer
A: the examinee answers correctively.
.
.
(i)




(ii) Required probability
.





1



1






2