4

Sample Paper
Sample Paper
Solved
xf.kr
MATHEMATICS
Time Allowed : 3 hours Maximum Marks : 80

General Instructions: Same instruction as given in the Sample Paper 1.

SECTION - A 20 × 1 = 20
This section comprises multiple choice questions (MCQs) of 1 mark each.

1. How many relations are possible in set A  0 2b −2 

 
such that n(A) = 2? 4. The matrix A =  3 1 3  is a
(a) 4 (b) 8 3a 3 −1
(c) 16 (d) 32 1  
2. The graph shown below depicts: symmetric matrix. Then which of the
Y following is not correct?
p
(a) Product of a and b is 1.
2 (b) Product of a and b is –1.
5
(c) Sum of a and b is .
6
−13
(d) Difference of a and b is . 1
6
7 6 x
X'
–1 0 1
X
5. If one root of the equation 2 x 2 = 0
x 3 7
is x = −9, then the other two roots are:
(a) 6, 3 (b) 6, –3
(c) 2, 7 (d) 2, 6 1
p 6. Which of the following is not a cofactor of
2 1 0 4 
Y'  
(a) y = sin–1x (b) y = cos–1x A = 3 5 −1 ?
(c) y = cosec–1x (d) y = tan–1x 1 0 1 2 
 
 1
3. The value of tan–1 (1) + cos–1  − 2 
(a) 11 (b) 2
 1 (c) –1 (d) 0 1
+ sin  −  is:
–1
 2 7. x
∫x (1 + log x ) dx is:

(a) 2π (b) 3π
3 4 (a) xx + c (b) 1 + c
3π 2 x
(c) (d) π 1
5 5 (c) log x + c (d) x + c 1

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 8. If f(x) = |x + 1| + |x – 1| then which of the 14. Th equation of a line passing through
following is true regarding the continuity (–1, 1, 0) and parallel to the line
and differentiability of f(x) at x = 1?
x − 3 2y − 1 3 − z
(a) f(x) is continuous at x = 1 but not = is:
3 2 1
differentiable at x = 1
(b) f(x) is not continuous at x = 1 but x −1 y −1 z −0
(a) = =
differentiable at x = 1 3 1 −1
(c) f(x) is not continuous at x = 1 and not
differentiable at x = 1 x −1 y +1 z −0
(b) = =
(d) f(x) is continuous as well as 3 1 −1
differentiable at x = 1 1
x +1 y −1 z −0
9. The area (in sq. units) of the shaded region (c)
3
=
1
=
−1
as shown in the figure is:
Y x= 4 x +1 y +1 z −0
(d) = = 1
3 1 −1

15. The equation that best describes the

X′ X
0 shaded area in given graph is:

Y
x = y2 x=y
Y'

X' X
16
(a) 32
0
(b)
3 3
(c) 4 (d) 16 1 Y'
10. The integrating factor of the differential
(a) x > y (b) x < y
dx
equation (y log y) + x = 2 log y, is given
dy (c) x ≤ y (d) x ≥ y 1
by:
16. A class consists of 80 students. 25 of them
(a) log y (b) log (log y) are girls and 55 are boys. 10 of them are
(c) y (d) ey 1 athletic and the remaining are scholarly
11. The integrating factor of the differential while 20 of them are introverts. The
dy probability of selecting an introvert athletic
equation x – y = 2x2 is:
dx
girl is:
1
(a) logx (b) 8 3
x2 (a) (b)
653 256
1
(c) e–x (d) 1 5 1
x (c) (d) 1
→ 512 802
12. For any vector a, the value of
→

→

→

17. The angle between the straight lines
| a . i |2 + | a . j |2 + | a . k |2 is:
1− x y+2 3− z
(a) a (b) a2 = =
−1 2 3
(c) 1 (d) 0 1
  x +1 2− y z +3
13. The projection of the vector i − j on the and = = is:
2 −5 4
 
vector i + j is:
(a) 30° (b) 45°
(a) 1 (b) –1
(c) 0 (d) 2 1 (c) 60° (d) 90° 1

52 Mathematics Class XII

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 → → → → → → → → → → → → −45
18. If | a × b |2 + | a . b |2 = 144 and | a | = 4 , then and | c | = 4, then a . b + b . c + c . a =
2
→ → → → → → → −45
| b | is equal to: a.b + b. c + c .a = .
2
(a) 16 (b) 8 → → →
Reason (R): For any three vectors a , b , c ,
(c) 3 (d) 12 1 → → → → → →
| a + b + c |2 = | a |2 + | b |2 + | c |2
Assertion-Reason Based Questions → → → → → → → → →
| a |2 + | b |2 + | c |2+ 2( a . b + b . c + c . a ).
Question numbers 19 and 20 are Assertion-Reason
based questions carrying 1 mark each. Two 1
statements are given, one labelled Assertion (A)
and the other labelled Reason (R). Select the 20. Assertion (A): The coordinates of the foot
correct answer from the options (a), (b), (c) and of perpendicular from the
(d) as given below. point (0, 2, 2) on line

(a) Both (A) and (R) are true and (R) is the correct x −3 y+1 z+4
= = are
explanation of (A). 5 2 3

(b) Both (A) and (R) are true but (R) is not the  159 20 125 
correct explanation of (A).  38 , 38 , 38  .

(c) (A) is true but (R) is false. →

Reason (R): If P( α ) be a point and
(d) (A) is false but (R) is true. → → →
r = a+ λ b be the vector
→ → →
19. Assertion (A): Three vectors a , b and c equation of a line, then
→ → → →
satisfy the condition a + b + c = 0 equation of a perpendicular
→ → →
→ → → →
a + b + c = 0 and if | a | = 5, | b | = 2 from the point on a line r
→ → →
is a − α+ λ b . 1

SECTION - B 5 × 2 = 10
This section comprises very short answer (VSA) type questions of 2 marks each.

21. Prove that: 23. Find two numbers whose sum is 48 and
 2π   2π  −1  3π 
sin−1  sin  + cos −1  cos  + tan  tan  whose product is as large as possible.  2
 3   3   4 
24. Anuj is working on adjusting a drone's
= 3π
4 flight path to make sure it follows the right

OR direction. He knows that the dot product of

Prove that: sec2 ( tan–12) + cosec2 (cot–13) = 15 the vector ˆ

i + ˆj + kˆ with a unit vector along
2 the sum of two other vectors 2ˆ
i + 4 ˆj – 5kˆ

22. Check, if (1, 2), (1, 5) and (2, 4) are collinear and λˆ
i + 2ˆj + 3kˆ , is equal to one. He
using determinant. If not, what is the area needs to find the value of l. Help him to
of triangle formed by them?
evaluate l. 2
OR
x x −a x −b x −5 y+2 z
25. Show that the lines = = and
If f(x) = x + a 0 x − c , then find the 7 −5 1
x +b x +c 0
x y z
= = are perpendicular. 2
value of f(0). 2 1 2 3

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 SECTION - C 6 × 3 = 18
This section comprises short answer (SA) type questions of 3 marks.

  30. Prashant is designing two parallel

26. Consider a function f :  0, π  ® R given pipelines in 3D space and needs to ensure
 2
they are correctly spaced apart to avoid
 π interference. He wants to calculate the
by f(x) = sin x and g :  0,  ® R given by shortest distance between the path of the
 2 two pipelines. The paths are represented
g(x) = cos x. Show that, both f and g are by the vectors equation:
→
one – one, but f + g is not one – one. 3     
r = i + j + λ( 2i − j + k )

→
27. If x 1 + y + y 1 + x = 0 where,
.
x y, then and r = 2i + j − k + µ( 4i − 2 j + 2k )

1
prove that dy = − 3 Find the shortest distance between these
dx (1 + x )2 two lines. 3
π /4
28. Evaluate: ∫0 log (1 + tan x ) dx 31. In a hostel, 60% students read Hindi
newspaper, 40% read English newspaper
OR and 20% read both Hindi and English
x newspaper. A student is selected at
Find: ∫ ( x 2 + 1)( x − 1) dx 3 random.
(A) Find the probability that a student
reads neither Hindi nor English
29. Amit is designing a park in his new house newspaper.
and wants to find the area of a specific (B) If a student reads Hindi newspaper,
section bounded by certain curves to find the probability that the student
reads English newspaper also.
ensure proper landscaping. He needs to
OR
calculate the area enclosed by the lines

Suppose a girl throws a die. If she gets 1 or
y = |x + 3|, x = –6, x = 0 and y = 0. Find the 2, she tosses a coin three times and notes
area of this region. the number of tails. If she gets 3, 4, 5 or 6,
she tosses a coin once and notes whether
OR
a ‘head’ or ‘tail’ is obtained. If she obtained

Find the area of region bounded by y = – 1, exactly one ‘tail’, what is the probability
y = 2, x = y3 and x = 0. 3 that she threw 3, 4, 5 or 6 with the die? 3

SECTION - D 4 × 5 = 20
This section comprises long answer (LA) type questions of 5 marks each.
2 4 bottom is drawn away from the wall along
32. For the matrix A =   , verify that the ground. How quickly does the height
1 −3
on the wall drop when the ladder's foot is
(A–1)' = (A')–1.
4 cm from the wall? Also, find the speed of
OR falling ladder when bottom is drawn away
at a rate of 4 cm/s. 5
Using matrix method, solve the following
system of equations: 34. Make a rough sketch of the region
x – y + z = 4, 2x + y –3z = 0, x + y + z = 2 5
{(x, y) : 0 ≤ y ≤ x2, 0 ≤ y ≤ x, 0 ≤ x ≤ 2} and find
the area of the region using integration. 5
33. Prove that the volume of the largest cone
that can be inscribed in a sphere of radius 35. (A) The sides of an equilateral triangle are
8 increasing at the rate of 2 cm/sec.
R is of the volume of the sphere. Find the rate at which the area increases
27
when the side is 10 cm.
OR
(B) Show that, the function
A 5 m long ladder is leaned up against f(x) = x3– 3x2 + 6x – 100 increasing on R.
a wall. At a rate of 2 cm/s, the ladder's  5

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 SECTION - E 3 × 4 = 12
This section comprises 3 case study based questions of 4 marks each.

36. Case Study – 1 Meena wanted to use that concept to help

students figure out how the area of the 3D

Jyoti CNC is the largest CNC (Computer
model is composed of 3 straight lines.
Numerical Control) machine manufacturing
On the below graph (cross section of the 3D
company of India. Their unit in Bhubaneswar,
Odisha has three machine operators A, B, model), O is the origin. The shaded region R

and C. The operators supervise the machines is defined by three inequalities. One of the

while they execute the tasks and make any inequality is x + y £ 6.

necessary adjustments to produce a better Y

result. Their main focus is to minimise defects
7
as it increases the cost of operations.
6
The first operator A produces 1% defective 5
items, whereas the other two operators B 4

x+
and C produce 5% and 7% defective items,

y
=
3

6
respectively. 2 R

Machine % of the time on the 1

Operators job X' X
O 1 2 3 4 5 6 7 8
A 50%
Y'
B 30%

C 20% Based on the given information, answer

the following questions:
Based on the given information, answer
the following questions: (A) Given that the point (x, y) is in the

(A) What is the probability that item region R. What is the maximum value of
produced by the operator C is x + 2y? 1
defective? 1 (B) Given that the point (x, y) is in the
(B) What is the total probability of region R, then find the minimum value of
producing a defective item in the 6x − y. 1
factory? 1
(C) Find the other two inequalities plotted
(C) What is the conditional probability
on the graph.
that the defective item is produced by
the operator A? OR
OR What is the area of the region R? 2
The factory in charge wants to do a 38. Case Study – 3
quality check. During inspection he

A veterinary doctor was examining a sick cat
picks an item from the stockpile at
brought by a pet lover. When it was brought
random. If the chosen item is defective,
then what is the probability that it is to the hospital, it was already dead. The pet
not produced by operator C? 2 lover wanted to find its time of death. He

37. Case Study – 2 took the temperature of the cat at 11.30 pm

One very useful application of Linear which was 94.6°F. He took the temperature
programming is using a graphical method again after one hour; the temperature was
for solving problems in two variables. Mrs. lower than the first observation. It was 93.4°F.

Sample Paper 4 55

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 The room in which the cat was put is always
at 70°F. The normal temperature of the cat is
taken as 98.6°F when it was alive. The doctor
estimated the time of death using Newton's
law of cooling which is governed by the
dT
differential equation: ∝ ( T − 70 ), where
dt
70°F is the room temperature and T is the Based on the given information, answer the
temperature of an object at time t. following questions:
Substituting the two different observations (A) State the degree and order of the given
of T and t made, in the solution of the differential equation. 2
(B) Write the solution of the differential
dT
differential equation = k ( T − 70 ) where dT
dt equation = k ( T − 70 ) given that
dt
k is a constant of proportion, time of death
t = 0 when T = 72. 2
is calculated.

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 SOLUTIONS
SAMPLE PAPER - 4

SECTION - A
1. (c) 16 Since A is symmetric matrix, i.e., AT = A
Explanation: Number of relations with 2
È 0 3 3a ˘ È 0 2b -2˘
Í ˙ Í ˙
elements = 2 22 Í2b 1 3 ˙ = Í3 1 3˙
= 24 Í-2 3 -1˙ Í3a 3 -1˙
Î ˚ Î ˚
= 16
On comparing corresponding elements, we get
ttConcept Applied 2b = 3 and 3a = −2
¬¬
Number of relations from A to A
= 2 Number of elements in A × A = 2 × 2
2 3
Þ a=– ,b=
= 2n
2 3 2
−2 3
2. (a) y = sin–1x i.e., ab = × = −1
3 2
Explanation: The graph represents
−2 3
y = sin–1x also, a+b= +
3 2
whose domain is [–1, 1]
È p p˘ −4 + 9 5
and range is Í– , ˙ = =
Î 2 2˚ 6 6
3π
3. (b) and a–b=
−2 3
−
4 3 2
Explanation:

−4 − 9 −13
  1  1   = =
tan (1) + cos −1  −  + sin−1  −  
−1 6 6

  2   2  
Caution
 π  π  π Students should remember that if two matrices are
= tan  tan  + cos −1  − cos  + sin−1  − sin 
−1

 4  3  6 equal then each element of one matrix is equal to
π π π corresponding element of the other matrix.
= +π− −
4 3 6
[∵ sin–1(–q) = – sin–1 q and cos–1(–q) = p – cos–1 q]
5. (c) 2, 7
Explanation:
3π 7 6 x
=
4
2 x 2 =0
Caution x 3 7
 In such type of questions, first simplify each term
7(7x – 6) – 6(14 – 2x) + x (6 – x2) = 0
independently and always try to convert all given
– x3 + 67x – 126 = 0
inverse trigonometric functions.
x3 – 67x + 126 = 0
4. (a) Product of a and b is 1. (x + 9)(x2 – 9x + 14) = 0
Explanation: [ x = –9 is one root]
(x + 9)(x –2) (x – 7) = 0
È 0 2b -2˘
Í ˙ Hence, the other two roots are 2 and 7.
Given, A= Í3 1 3˙
Í3a 3 -1˙
Î ˚ 6. (d) 0
Explanation: Given,
È 0 3 3a ˘ 1 0 4 
Í ˙
AT = Í2b 1 3 ˙

A = 3 5 −1

Í-2 3 -1˙ 0 1 2 
Î ˚  

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 C11 = (–1)2 (10 + 1) = 11, 32
C12 = (–1)3 (6) = –6, C13 = (–1)4 = 3,
9. (a)
3
C21 = (–1)3 (–4) = 4, C22 = (–1)4 (2) = 2,
Explanation: Given curves are x = y2 and x = 4.
C23 = (–1)5 (1) = –1, C31 = (–1)4 (–20) = –20,
So, their points of intersection are (4, 2) and
C32 = (–1)5 (–1 – 12) = 13, C33 = (–1)6(5) = 5. (4, –2).
i.e., 11, 2 and –1 are all cofactors of A. So, required area,
Hence, 0 is not a cofactor of A. 4
A = 2∫ y.dx
7. (a) x + cx
0
Explanation: Let xx = z. 4
Then, x log x = log z = 2∫ x dx
1
Þ (1 + log x) dx = z dz 0

1  3
4
\ I = ∫ z . dz
z 2  x 2 0
=
x 3
= ∫ 1 dz = z + c = x + c 2
8. (a) f(x) is continuous at x = 1 but not differentiable  3  2
at x = 1 =  4 2 − 0  × × 2
Explanation: Continuity at x = 1   3
f(1) = |1 + 1| + |1 – 1|
= |2| + |0| 32
= sq. units
= 2 + 0 = 2 3
LHL = lim f ( x )
x Æ 1-
10. (a) log y
dx
= lim | x + 1| + | x - 1| Explanation: Given, (y log y)
+ x = 2 log y
x Æ 1- dy
= lim ( x + 1) - ( x - 1)
x Æ 1- dx x 2

⇒ + =
= lim ( x + 1 - x + 1) = 2 dy y log y y
x Æ 1-
This is a linear differential equation of the form
RHL = lim f ( x )
x Æ 1+ dx
+ Px = Q,
= lim | x + 1| + | x - 1| dy
x Æ 1+ 1 2
where, P= and Q =
= lim ( x + 1) + ( x - 1) y log y y
x Æ 1+
1
= lim ( x + 1 + x - 1) ∫ dy
I.F. = e ∫
P dy
x Æ1 + ∴ = e y log y
= lim (2 x ) = 2 × 1 = 2 Put, log y = t
x Æ 1+ 1
dy = dt

Q f(1) = lim f ( x ) = lim f ( x ) y
- +
x Æ1 xÆ 1
1
Thus, f(x) is continuous at x = 1 ∫ dt
⇒ I.F. = e t = elog t = t = log y
Differentiability at x = 1
f (1 + h ) - f (1) 1
LHD = lim 11. (d)
hÆ0 - h x
Explanation: Given, x
dy – y = 2x2
2-2
= lim =0 dx
h Æ 0- h
dy y
f (1 + h ) - f (1) Equation can be rewritten as − = 2x
RHD = lim dx x
hÆ0 + h Comparing this with standard linear equation
2(1 + h ) - 2
= lim dy
h Æ 0+ h + Py = Q(x), we get
dx
2h 1
= lim =2 P(x) = – , Q (x) = 2x
h Æ 0+ h x
1
Q LHD ≠ RHD
Integrating Factor = I.F. = e ∫ = e ∫x
P dx − dx

Thus, f(x) is not differentiable at x = 1. 1
log 1
\ The function f(x) = |x + 1| + |x – 1| is continuous = e–logx = e x
=
at x = 1 but not differentiable at x = 1. x

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 ttConcept Applied dy 16. (c)
5
¬¬
For linear differential equation of type + Py = Q(x), 512
dx
I.F. = e ∫
P( x )dx
Explanation: A : Event that selected student is
an introvert athletic girl.
12. (b) a2 P(E) =
25
, for the event E ‘‘student is a girl’’
" 80
Explanation: Let
a = xit + yjt + zkt. Then,
10
→ P(F) = , for the event F ‘‘student is athletic’’
 →
 →

a .i = x ; a . j = y and a .k = z 80
20
→ ^ → ^ → ^ P(G) = , for the event G ‘‘student is introvert.’’
⇒ | a . i |2 + | a . j |2 + | a .k |2 = x2 + y2 + z2 80
→ \ P(A) = P(E) × P(F) × P(G)
= | a |2 = a2
25 10 20 5
13. (c) 0 = × × = ,
80 80 80 512
→ ^ ^ ^ ^ ^
a = i − j = 1 i −1 j+ 0 k as E, F, G are independent events.
Explanation: Let

→ ^ ^ ^ ^ ^ ttConcept Applied
b = i + j = 1 i + 1 j+ 0 k ¬¬
For mutually independent events, P(A ∩ B) = P(A) P(B).
→ → → →
Projector of vector a on the b = 1 ( a . b ) 17. (d) 90°
→
|b| Explanation: Given lines are:
→ → ^ ^ ^ ^ ^ ^ 1− x y +2 3− z x +1 2 − y z + 3
l1 : = = and l2 : = =
Now, a . b = ( i − j + 0 k ).( i + j + 0 k ) −1 2 3 2 −5 4
= 1 + (–1) + 0 These equations can be re-written as:
= 0 x −1 y + 2 z − 3 x +1 y − 2 z + 3
→ l1 : = = and l2 : = =
1 2 −3 2 5 4
Also, magnitude of b = 12 + 12 + 02
→ On comparing with standard equation of line
|b| = 1 +1 + 0 we get, a1 = 1, b1 = 2, c1 = –3 and a1 = 2; b2 = 5,
c2 = 4
= 2
∴ Angle between the given lines is given by,
→ → → →
1 1×2 + 2 ×5 + −3 × 4
Projection of a on b = (a . b) cos =
→
|b| 12 + 22 + ( −3)2 × 22 + 52 + 42

1 2 + 10 − 12
= (0) = =0
2 14 45
= 0
Þ cos = 0
x +1 y −1 z −0
Þ = cos–1(0) = cos–1(cos 90°)
14. (c) = =

Þ = 90°
3 1 −1

Explanation: Given,
x − 3 2y − 1 3 − z
= = Concept Applied tt
3 2 1 x – x1 y – y1 z – z1 x – x2 y – y2 z – z2
These equations can be re-written as: ¬¬ If l1 : = = and l2 : = =
a1 b1 c1 a2 b2 c2
1
y− x – x2 y – y2 z – z2
x −3 2 = z −3 = =
= a2 b2 c2
, then the angle between l1 and l2 is given by,
3 1 −1
Clearly, direction ratios of this line are a1a2 + b1b2 + c1c2
proportional to 3, 1, –1. So, the direction ratios cos q =
a1 + b1 + c1 a2 + b2 + c2
2 2 2 2 2 2
of the parallel line are also proportional to
3, 1, –1. 18. (c) 3
∴ The required line passes through (–1, 1, 0) and Explanation: Given,
its direction ratios are proportional to 3, 1, –1. → → → →
| a × b |2 + | a . b |2 = 144
So, its equation is: → → → →
x +1 y −1 z − 0 | a |2| b |2 sin2 θ + | a |2| b |2 cos2 θ = 144
= =
3 1 −1
 → → 
15. (c) x £ y
→ →
∵| a × b | = | a || b | sin θ 
Explanation: As point (0, 1) lies in the given  → → → → 
shaded area. The inequality x £ y represents the  | a . b | = | a || b | cos θ
 
given graph.

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 ( ) 20. (d) (A) is false but (R) is true.
→ →

Þ | a |2 . | b |2 sin2 θ + cos2 θ = 144
→

Þ 16 | b |2 = 144 Explanation: Let L be the foot of the

→
|b| = 3 perpendicular drawn from the point P(0, 2, 2) to

Þ
the given line.
ttConcept Applied
→ → → → x − 3 y +1 z + 4
|a×b|=
¬¬ | a | | b | sin θ Let = = =λ
→ →
5 2 3
→ →
a . b = | a | . | b | cos θ ⇒ x = 5λ + 3, y = 2λ – 1, z = 3λ – 4
\ coordinates of L be (5λ + 3, 2λ – 1, 3λ – 4)
19. (a) Both (A) and (R) are true and (R) is the correct
Therefore, direction ratios of PL are proportional
explanation of (A).
Æ Æ Æ to 5λ + 3 – 0, 2λ – 1 – 2, 3λ – 4 – 2.
Explanation: Given, | a | = 5, | b | = 2 and| c | = 4.

Æ Æ Æ Æ i.e., 5λ + 3, 2λ – 3, 3λ – 6
It is also given, a + b + c = 0 Direction ratios of the given line are proportional
Æ Æ Æ
2
⇒ |a+ b+ c | = 0 to 5, 2, 3.

Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Given, PL is perpendicular to the given line.
⇒ a .( a + b + c ) + b .( a + b + c ) + c .( a + b + c ) = 0
\ 5(5λ + 3) + 2(2λ – 3) + 3(3λ – 6) = 0
Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ Æ
⇒ a . a + b . b + c . c + 2( a . b ) + 2( b . c ) + 2( c . a ) = 0
⇒ 25λ + 15 + 4λ – 6 + 9λ – 18 = 0
2
Æ
2
Æ
2
Æ Æ Æ Æ Æ Æ Æ ⇒ 38λ = 9
⇒ | a | + | b | + | c | + 2( a . b + b . c + c . a ) = 0

Æ Æ Æ Æ
⇒ 52 + 2 2+ 4 2 + 2( a . b + b . c + c . a ) = 0

Æ Æ

⇒ λ= 9
38
Putting λ = 9 in (5λ + 3, 2λ – 1, 3λ – 4) the
Æ Æ Æ Æ Æ Æ
⇒ 2( a . b + b . c + c . a ) = –(25 + 4 + 16)

Æ Æ Æ Æ Æ Æ -45 38

⇒ a.b+ b.c + c .a =  159 −20 −125 
2 coordinates of L are  , ,
 38 38 38 
SECTION - B
OR
21. MARKS BREAKDOWN Ch 2
MARKS BREAKDOWN Ch 2
Topic: Principal Value of Trigonometric Function
Key Steps: Topic: Principal Value of Trigonometric Function

✔ Simplify LHS within principal value range (1m) Key Steps:

✔ Prove required result (1m)

✔ Use inverse trigonometric properties to simplify
LHS (1m)
✔ Reach LHS = RHS stage (1m)
LHS
Let tan–1 2 = α
      ⇒ tan α = 2
= sin−1 sin 2π + cos −1 cos 2π + tan−1 tan 3π and cot–1 3 = b
 3   3   4 
⇒ cot b = 3
  π   2π   π   sec2 (tan–1 2) + cosec2 (cot–13)
= sin−1 sin  π −   + + tan−1 tan  π −  
= sec2 α + cosec2 b
  3   3   4  
= 1 + tan2 α + 1 + cot2 b
 = 2 + (2)2 + (3)2
 π  2π  − π 
= sin−1  sin  + + tan−1 tan    = 15
 3 3   4 
Hence, proved.
π 2π π
= + −
3 3 4 22. MARKS BREAKDOWN Ch 4
π 3π Topic: Condition of Collinearity
= π − = = RHS
4 4
Key Steps:
Hence, proved. ✔ Explain condition of collinearity (1/2m)

ttConcept Applied ✔ Rewrite given points in matrix form (1/2m)

✔ Find determinant (1/2m)
¬¬
2π
3
π π
does not lie between − and .
2 2
✔ Calculate area (1/2m)

¬¬
sin (p – q) = sin q For points to be collinear, area of triangle with
¬¬
tan (p – q) = – tan q these points as vertices should be zero.

Sample Paper 4 93

https://amzn.to/4fBClzA
 Now, Let P be the product of these numbers. Then,
1 2 1 P = xy = x (48 – x) = 48x – x2
1
∆ = 1 5 1 dP d2 P
2 Þ = 48 – 2x and = –2
2 4 1 dx dx 2
1 dP
= [1 (5 − 4) − 2 (1 − 2) + 1 (4 − 10)] The critical points of P are given by = 0.
2 dx
1
= |1+2−6| dP
2 ∴ =0 ⇒ 48 – 2x = 0 ⇒ x = 24
1 3 dx
= |− 3 | = ≠0
2 2  d2 P 
Also,  2 = –2 < 0.
i.e., points are not collinear and area of triangle  dx  x =24
3
formed by them is sq. units. So, P is maximum where x = 24.
2
Putting x = 24 in (i), we get y = 48 – 24 = 24.
Caution Hence, the required numbers are both equal
 Since, area of triangle is a positive quantity, we always to 24.
take the absolute value of the determinant.
24. MARKS BREAKDOWN Ch 10
OR
Topic: Scalar Product
MARKS BREAKDOWN Ch 4 Key Steps:
Topic: Determinant of a Matrix ✔ Assume three vectors as a, b & c
→ → →
(1/2m)
✔ Apply magnitude property of vector (1/2m)
Key Steps:
✔ Calculate λ (1m)
✔ Put x = 0 in given matrix (1/2m) →

✔ Expand given matrix along R (1m)

Let a = it + tj + kt ,
1
→
✔ Calculate f(0) (1/2m) b = 2it + 4tj – 5kt ,
→
0 x -a x -b c = λit + 2tj + 3kt
Æ Æ
f (x) = x + a 0 x -c
Here, b + c = (2 + λ) it + 6tj - 2kt
x +b x +c 0 Æ Æ
and, |b +c | = (2 + λ) 2 + 36 + 4
0 0-a 0-b

Þ f (0) = 0 + a 0 0-c = l2 + 4l + 44
0+b 0+c 0 It is given that
Æ Æ Æ
a .( b + c )
0 -a - b Æ Æ =1
|b +c |
= a 0 -c
(i j k) . [(λ 2) i 6tj − 2kt]
t + t + t + t +
b c 0 Þ =1
λ 2 + 4λ + 44
Expanding along R1 λ+2+6−2
f(0) = 0 (0 + c2) + a (0 + bc) Þ =1
λ 2 + 4λ + 44
– b (ac – 0)
= abc – acb = 0 Þ (λ + 6)2 = λ2 + 4λ + 44
Þ 8λ = 8, or λ = 1
23. MARKS BREAKDOWN Ch 6
25. MARKS BREAKDOWN Ch 11
Topic: Maxima and Minima
Key Steps: Topic: Angle Between Two Lines
✔ Define a function according to question (1/2m) Key Steps:
✔ Differentiate given function (1/2m) ✔ Compare both lines with standard form & get
✔ Find critical points (1/2m)
required values
✔ Apply condition & prove that lines are
(1m)

✔ Calculate point of maxima at which number is perpendicular (1m)

larger (½m) Tip: If two lines with direction ratio a1, b1, c1 and
Let the number be x and y. Then, a2, b2, c2 are perpendicular then, a1a2 + b1b1 +
c1c2 = 0.
x + y = 48 (given) ....(i)

94 Mathematics Class XII

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 x - x1 y − y1 z − z1 x −5 y+2 z x y z
Two lines = = and = = and = =
a1 b1 c1 7 −5 1 1 2 3

x − x2 z − z2
Here, a1 = 7, b1 = –5, c1 = 1 and a2 = 1, b2 = 2, c2 = 3
y − y2
= = are perpendicular if
a2 b2 c2 So, a1a2 + b1b2 + c1c2 = (7)(1) + (–5)(2) + (1)(3)

a1a2 + b1b2 + c1c2 = 0. = 7 – 10 + 3 = 0
We have two lines as, Hence, the given two lines are perpendicular.

SECTION - C
26. Þ = y 2 (1 + x)
x2 (1 + y)
MARKS BREAKDOWN Ch 1
Þ x2 + x2y
= y2 + xy2
Topic: Types of Functions Þ 2 2
x – y + xy (x – y)
=0
Key Steps: Þ (x – y) (x + y) + xy (x – y)
=0
✔ Prove f is one-one (1m) Þ (x – y) (x + y + xy)
=0
✔ Prove g is one-one (1m) Þ =0x + y + xy [ x ≠ y]
✔ Prove f + g is not one-one (1m)
Þ y = −
x
1+x
To Prove: f is one-one dy (1 + x) .1 − x (1)
π Þ =− f p
Let x, y be any two elements in <0, F . Then, dx (1 + x) 2
2
1
f(x) = f(y) = −
(1 + x) 2
⇒ sin x = sin y Hence, proved.
⇒ x=y
28. MARKS BREAKDOWN Ch 7
Thus, f is one-one.
To Prove: g is one-one Topic: Definite Integrals
g(x) = g(y) Key Steps:
⇒ cos x = cos y ✔ Rewrite integrating function (1/2m)
⇒ x=y ✔ Simplify integration (1m)
Thus, g is one-one. ✔ Integrate function (1/2m)
To Prove: f + g is not one-one ✔ Apply limits & simplify result (1m)
(f + g)(0) = f(0) + g(0) Caution: Before applying the definite integral
= sin 0 + cos 0 property, check whether f(x) is continuous on
= 0 + 1 = 1 given limit.
π Ê ˆ Ê ˆ π
(f + g) d n = f p + g p 4
2 ÁË 2 ˜¯ ÁË 2 ˜¯ Let I= y0 log (1 + tan x) dx ...(i)
π π π
= sin + cos π
2 2 log =1 + tan d − x nG dx
4
= 1 + 0 = 1
= y0 4
π
π 1 − tan x
\ 0 and have same images. = y0 4
log d1 + n dx
2 1 + tan x
So, f + g is not one-one. π
2
= y04 log d 1 + tan x n dx
27. MARKS BREAKDOWN Ch 5 π π
4 4
Topic: Differentiation = y0 log 2 dx − y0 log (1 + tan x) dx
Key Steps: ...(ii)
Adding equations (i) and (ii), we get
✔ Simplify given eq. (1m) π

✔ Differentiate functions (1m) Þ 2I = y0 4

log 2 dx
✔ Prove required relation (1m) π

= log 2 5x?04 = log 2 $

π
x 1+y+y 1+x = 0 4
π
Þ x 1+y = −y 1+x Þ I= $ log 2
8

Sample Paper 4 95

https://amzn.to/4fBClzA
 OR First we sketch the graph of y = |x + 3|
We know that
MARKS BREAKDOWN Ch 7
ÏÔ x + 3, if x + 3 ≥ 0
Topic: Indefinite Integrals y = |x + 3| = Ì
ÓÔ-( x + 3), if x + 3 < 0
Key Steps:
✔ Simplify given integral function (1m)
ÔÏ x + 3, if x + 3 ≥ 0
✔ Apply method of partial fraction in given integral
Þ y = |x + 3| = Ì
function (1m) ÓÔ- x - 3, if x + 3 < 0
✔ Integrate given function & simplify it (1m)
Caution: Check the type of function & then So, we have y = x + 3 for x ³ – 3 and y = − x – 3
carefully apply partial fraction formula for x < −3
x Also, y = x + 3 is a straight line which cuts x-axis
Let I = ∫ ( x2 + 1)( x − 1) dx and y-axis at (−3, 0) and (0, 3), respectively.
Shaded region is the required area.
x A Bx + C
Let 2
= +
( x + 1)( x − 1) x − 1 x2 + 1 Y

Þ x = A (x2 + 1) + (Bx + C) (x – 1) B C 3
Equating coefficient of x2, x and constant, we

y
3

=
x= – 6 x+ 2

–x
get =

–
y 1

3
A + B = 0 ...(i) A P O(0, 0)
X' X
−B + C = 1 ...(ii) –6 –5 –4 –3 –2 –1 1 2 3
A – C = 0 ...(iii)
Solving equations (i), (ii) and (iii), we get Y'

1 1 Required area = Area of ABPA + Area of PCOP

A=C= and B = –
2 2 −3 0
= ∫ ( − x − 3) dx + ∫ ( x + 3) dx
1 −6 −3
( − x + 1)
x 1

\ = +2 −3 0
( x 2 + 1)( x − 1) 2( x − 1) x2 + 1  x2   x2 
=  − − 3x  +  + 3x 
x  2  − 6  2  −3
Þ ∫
dx
( x 2 + 1)( x − 1)
= 1 1 ( − x + 1)  9    9 
∫ 2( x − 1) dx + 2 ∫ 2
x +1
dx =  − + 9 − ( −18 + 18) + 0 −  − 9 
 2 
    2  
1
= ∫
1 1 x 1 1
dx − ∫ dx + ∫ dx  9 9
2 ( x − 1) 2 x +1 2 x +1
2 2
=  − −  + ( 9 + 9)
 2 2
1
= ∫
1 1 2x 1 1
dx − ∫ dx + ∫ dx
2 ( x − 1) 4 x +1 2 x +1 = 18 – 9 = 9 sq. units
2 2

1 1 1 OR
= log| x − 1 | − log| x 2 + 1 | + tan−1 x + C
2 4 2

29. MARKS BREAKDOWN Ch 8

MARKS BREAKDOWN Ch 8
Topic: Area Bounded by Curves and Lines
Topic: Area Bounded by Curves and Lines
Key Steps:
Key Steps:
✔ Make sketch of region bounded by given curves ✔ Make sketch of region bounded by given curves
(1/2m) (1/2m)
✔ Write required points of intersection (1/2m) ✔ Write required points of intersection (1/2m)
✔ Write formula for required area & apply ✔ Write formula for required area & apply
integration with suitable limit (1m) integration with suitable limit (1m)
✔ Apply limit & calculate required area (1m) ✔ Apply limit & calculate required area (1m)
 x , if x ≥ 0
Tip: | x |=  The intersection of curve x = y3 and lines
− x , if x < 0
y = – 1 and y = 2 are P(– 1, – 1) and Q(8, 2).

96 Mathematics Class XII

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 Y
→ → →
x = y3 | b×( a 2 − a 1 )|
d = → ...(i)
|b|
y= 2 A(0, 2)
→ → ^ ^ ^ ^ ^
Q(8, 2) a2 − a1 = (2 i + j − k ) − ( i + j )
^ ^
= i − k
We have,
it tj kt
→ → →
X¢
O(0, 0)
X b × ( a2 − a1 ) = 2 – 1 1
1 0 –1
(– 1, – 1) = it + 3tj + kt
→ → →
y= – 1 P Þ | b × ( a2 − a1 )| = 1 + 9 + 1 = 11
B(0, – 1)
and |b | = 4+1+1 = 6
So, from (i), we have
Y¢
 11  11
d=   , or units.
 6  6
∴ Required Area = Area of shaded region OAQO  
+ Area of shaded region OBPO
2 0
= ∫ y3 dy + ∫ y3 dy 31.
0 −1 MARKS BREAKDOWN Ch 13

2 0 Topic: Conditional Probability

 y4   y4 
=   +   Key Steps:
 4 0  4  −1
✔ (A) Write given events & calculate required
16 1 probability (11/2m)
=
4
+ −
4 ✔ (B) Calculate required probability (11/2m)

Let H: Event that the students read Hindi

17
= sq. units newspaper
4 E: Event that the students read English
newspaper
30. MARKS BREAKDOWN Ch 11 60 3
Here, P(H) = =
100 5
Topic: Shortest Distance Between two Lines
40 2
Key Steps: P(E) = =
✔ Rewrite given eq. of lines & find DR's of given line 100 5
(1/2m) 20 1
and P(H ∩ E ) = =
✔ Write formula for shortest distance between 2 100 5
parallel lines & simplify it (1/2m)
✔ Find required distance (2m) P(H ∩ E) = 1 – P(H ∪ E)
(A)

The given lines are 1 – [P(H) + P(E) – P(H ∩ E)]

=
"
3 2 1
r = it + tj + λ (2it − tj + kt) = 1 − < + − F
" 5 5 5
andr = 2it + tj − kt + µ (4it − 2tj + 2kt) 4 1
= 1 − =
= 2it + tj − kt + v (2it − tj + kt), 5 5
where v = 2µ P(E ∩ H)
(B) P(E|H) =
P(H)
These two lines pass through the points having
" " 1
position vectors a = it + tj ; a = 2 it + tj − kt
1 2 5 1
respectively. = =
"
3 3
Both lines are parallel to the vector b = 2 it − tj + kt 5
Hence, the distance (d) between the two parallel ttConcept Applied
lines is given by ¬¬
H ∩ E ≠ H ∩ E. This is why P(H ∩ E) ≠ 1 – P(H ∩ E).

Sample Paper 4 97

https://amzn.to/4fBClzA
 OR
Now, P(E1|A) =
( ) ( )
P E1 .P A | E1
( ) ( ) (
P(E1 ).P A | E1 + P E2 .P A | E2 )
MARKS BREAKDOWN Ch 13
2 1
Topic: Bayes' Theorem ×

P(E1|A) = 3 2
Key Steps: 2 1 1 3
× + ×
✔ List all event & calculate probabilities for them 3 2 3 8
(1m)
1 1
✔ Apply Bayes' theorem & substitute values (1m)
✔ Calculate required probability (1m)
P(E1|A) = = 3
3
1 1 8+3
+

Let E1 = Event that the girl threw 3, 4, 5 or 6 3 8 24
E2 = Event that the girl threw 1, 2. P(E1|A) = 1 × 24
and A = Event that the girl obtained exactly one 3 11
tail. 8
=
4 2 2 1 11
P(E1) = = , P(E2) = =
6 3 6 3
ttConcept Applied
1 3 ¬¬
When a coin is tossed three times, then the sample

P(A|E1) = , P(A|E2) =
2 8 space is {HHH, HHT, HTH, HTT, THH, THT, TTH, TTT}.

SECTION - D
1
32. MARKS BREAKDOWN Ch 4 Þ (A')– 1 = adj (Al )
| Al |
Topic: Inverse of Matrix
1 −3 −1
= > H
Key Steps: − 10 − 4 2
✔ Find value of |A| (1m)
1 3 1
✔ Calculate adjoint of matrix (3m) = > H ...(ii)
10 4 − 2
✔ Prove required result (1m)
From (i) and (ii), we get
Here, |A| = – 6 – 4 = – 10 ≠ 0 (A– 1)' = (A')– 1
–1
So, A exists. 1 3 1
= > H
Now, A11 = – 3, A12 = – 1, A21 = – 4, A22 = 2 10 4 − 2
Hence, proved.
−3 −4
So, adj A = > H OR
−1 2
1 MARKS BREAKDOWN Ch 4
Þ A– 1 = (adj A)
|A | Topic: Solutions of Simultaneous Linear Equations
1 −3 −4
= > H Key Steps:
− 10 − 1 2 ✔ Write system of linear eqs. in matrix form as
1 3 4 AX = B (1/2m)
= > H ✔ Find value of |A| (1/2m)
10 1 − 2
✔ Calculate adjoint & inverse of matrix
1 3 1
Hence, (A– 1)' = > H ...(i) (11/2m+ 11/2m)
10 4 − 2 ✔ Find value of x, y & z (1m)
2 1 Tip: Unique solution of equation AX = B is given by
Further, A' = > H
4 −3 X = A–1B, where |A| ≠ 0.

Þ |A'| = – 6 – 4 = – 10 ≠ 0 The matrix equation for the given system of

So (A') –1
exists. equations is
AX = B ...(i)
Here, A11 = – 3, A12 = – 4; A21 = – 1, A22 = 2 SRS1 − 1 V SRSxWWV SRS4WVW
1WW
SS WW S W S W
−3 −1 1 − 3WW, X = SSSyWWW, B = SSS0WWW
adj (A') = > H where A = SS2
So, SS W SS WW SS WW
−4 2 S1 1 1WW SzW S2W
T X T X T X

98 Mathematics Class XII

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 Here |A| = 1(1 + 3) + 1(2 + 3) + 1(2 – 1) DE = AD – AE = h – R
= 4 + 5 + 1 = 10 ≠ 0 In right DCDE, we have
\ A– 1 exists. R2 = (h – R)2 + r2
A11 = 4; A12 = – 5 A13 = 1 Þ r2 = 2hR – h2
A21 = 2; A22 = 0 A23 = – 2 Volume (V) of the cone
A31 = 2; A32 = 5 A33 = 3 1
= πr 2 h
SRS 4 2 2WW
V 3
SS W 1
So, adj A = SS− 5 0 5WWW = πh [2hR – h 2]
SS W 3
S 1 − 2 3WW 2 1
T
1
X = πRh 2 − πh 3
Þ A– 1 = adj A 3 3
|A | We shall determine for what value of h, the
RS V volume of the inscribed cone is maximum.
S 4 2 2WW
1 SS W Now,
= S− 5 0 5WWW
10 SSS W dV 4
S 1 − 2 3WW = πRh − πh 2
T X dh 3
dV
From (i), we have Equating, to zero, we have
dh
X = A– 1B 4
πRh = ph2
SRSxWVW SRS 4 2 2WWSS4WW
VR V 3
SS WW 4
1 SS WS W Þ h = R
Þ SSyWW = SS− 5 0 5WWWSSS0WWW 3
SS WW 10 SS WS W 2
Sz W S 1 − 2 3WWSS2WW d V 4
T X TR XT X Now = πR − 2πh
RS VW SS 16 + 0 + 4VWW dh 2 3
SSxWW d2 V 4 8
SSyWW 1 SS WW
Þ f 2p = πR − πR < 0
Þ SS WW = 10 SSS− 20 + 0 + 10WWW dh h = 4 R 3 3
SSz WW SS 4 + 0 + 6WW 3 4
T X TR X Thus, R is maximum, when h = R .
SS 20VWW RSS 2VWW 3
1 SS W S W Since we have,
= S− 10WWW = SSS− 1WWW
10 SSS W S W r2 = 2hR – h2
S 10WW SS 1WW
T X T X 2
Þ x = 2, y = – 1, z = 1 is the required solution of 4 4 
= 2 × R×R–
the given system of equations. 3  3 R 

33. 8 2 16 2 8 2
MARKS BREAKDOWN Ch 6 = R – R = R
3 9 9
Topic: Maxima and Minima 1 2
Now, volume of cone = pr h
Key Steps: 3
✔ Assume radius & height of inscribed cone (1/2m)
= 1 π× 8 R 2 × 4 R
✔ Calculate volume of cone (1m)
3 9 3
✔ Find 2 derivative of given function & calculate
nd

maximum value (2m) =

8 4 3
× πR
✔ Prove required result (1½m) 27 3

Let R be the radius of the sphere. 8

Hence, volume of cone = volume of sphere.
Let r and h be the radius of the base and height 27
of the inscribed cone, respectively. OR
A
MARKS BREAKDOWN Ch 6
Topic: Rate of Change of Quantities

h Key Steps:

E
✔ Apply Pythagoras theorem & calculate value of y
R (1m)
✔ Differentiate y w.r.t. time (1m)
B D
r
C ✔ Calculate required rate (1m)
✔ Calculate rate with new information (1m)
✔ Compare it to first value of rate (1m)

Sample Paper 4 99

https://amzn.to/4fBClzA
 Let y m be the height of the wall at which the
\ Area of shaded region
ladder touches. Also, let the foot of ladder be 1 2
2
x m away from the wall. = ∫ x dx + ∫ x dx
0 1
Then, by Pythagoras theorem, we have:
1 2
2 2
x + y = 25 [Length of the ladder = 5m] x  x2  3
=   +  
 3  0  2 1
Þ y = 25 − x 2
1   1
Then, the rate of change of height (y) with =  − 0 + 2 − 
respect to time (t) is given by, 3   2

dy = 1 + 3
=
−x . dx 3 2
dt 25 − x 2 dt

= 2 + 9 = 11 sq. units
6 6
dx
It is given that = 2 cm/s
dt
35. MARKS BREAKDOWN Ch 6
dy −2 x
\ = Topic: Rate of Change of Quantities and
dt 25 − x 2
Increasing Decreasing Function
Now, when x = 4 cm, we have:
Key Steps:
dy
=
−2 ×4
= −
8 ✔ (A) Apply property of equilateral triangle &
dt 2 3 calculate area (½m)
25 − 4 Differentiate area w.r.t. time (1m)
Hence, the height of the ladder on the wall is Calculate rate at which area is increasing
(1m)
8
decreasing at the rate of
3
cm/s. ✔(B) Rewrite given function (½m)
dx dy Differentiate function (1m)
−4 x
If = 4 cm/s, then = i.e., the Prove that given function is increasing on R
dt dt 25 − x 2 (1m)
ladder will fall twice as quickly. (A) Let 'a' be the side of an equilateral triangle
34. MARKS BREAKDOWN Ch 8 and A be the area of an equilateral triangle.

Topic: Area Bounded by Curves and Lines

da
Then, = 2 cm/sec
dt
Key Steps:
✔ Sketch the region bounded by given curves (1m)
We know that area of an equilateral triangle,
✔ Write required points of intersection (1m)
A=
3 2
a
✔ Write the formula for required area (1m) 4
✔ Calculate required area (2m) On differentiating both sides w.r.t 't', we get
Given curves are: dA 3 da
0 ≤ y ≤ x2; 0 ≤ y ≤ x, 0 ≤ x ≤ 2. = ×2a ×
dt 4 dt
On solving first two curves, the point of
intersection is (1, 1) and (0, 0). dA 3
= ×2 ×10 ×2 [Given a = 10]
Y
dt 4

= 10 3 cm2/sec
2
y=x 4 y=x Hence, the area is increasing at the rate of

10 3 cm2/sec.
2 (2, 2) (B) Given, f(x) = x3 – 3x2 + 6x – 100
On differentiating both sides w.r.t. x, we get
X' X f '(x) = 3x2 – 6x + 6
= 3x2 – 6x + 3 + 3
–2 0 1 2

= 3(x2 – 2x + 1) + 3
–2
= 3(x – 1)2 + 3 > 0
x=2
\ f '(x) > 0
Y' This shows that function f(x) is increasing on R.

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 SECTION - E
36. Ch 13 Caution
MARKS BREAKDOWN
 Read the case carefully, to get an idea, about the
Topic: Bayes' Theorem probabilities given and what is needed to be calculated
Key Steps: to get derived result.
✔ (A) Write all given information & calculate
required probability (1m) 37. Ch 12
✔ (B) Calculate total probability (1m) MARKS BREAKDOWN
✔ (C) Write formula for Bayes' theorem (1m) Topic: Graphical Method of solving an LPP
Substitute values (1/2m)
Determine required probability (1/2m) Key Steps:
OR ✔ (A) Find corner points (1/2m)
Write formula for Bayes' theorem (1m) Calculate maximum value (1/2m)
Substitute values (1/2m) ✔ (B) Find corner points (1/2m)
Determine required probability (1/2m) Calculate minimum value (1/2m)
✔ (C) Find inequality for line 1 (1m)
Let A be the event that A is on the job; Find inequality for line 2 (1m)
B be the event that B is on the job; OR
C be the event that C is on the job and Write corner points (1/2m)
D be the event that item is defective Calculate area of region (11/2m)
P(A) = 50% = 0.5; P(B) = 30% = 0.3; P(C) = 20% = 0.2
(A) Corner points of R are (4, 2), (2, 1), and (2, 4).
P(D|A) = 1% = 0.01; P(D|B) = 5% = 0.05;
P(D|C) = 7% = 0.07 Corner Point z = x + 2y
7
(A) P(D|C) = 7% = z(4, 2) 4+2×2=4+4=8
100
z(2, 1) 2+2×1=2+2=4
(B) P(D) = P(A)P(D|A)+ P(B)P(D|B)+ P(C)P(D|C)
    = 0.5 × 0.01 + 0.3 × 0.05 + 0.2 × 0.07 z(2, 4) 2 + 2 × 4 = 2 + 8 = 10
   
= 0.005 + 0.015 + 0.014 = 0.034 Hence, maximum value of (x + 2y) is 10 at
= 17 the point (2, 4).
500 (B) Corner points of R are (4, 2), (2, 1) and (2, 4).

(C) P(A|D) =
( )
P(A)P D|A
z = 6x – y
  P(A)P (D|A ) + P(B)P (D|B) + P(C)P (D|C ) Corner Point

z(4, 2) 6 × 4 – 2 = 22
0.5 × 0.01
= z(2, 1) 6 × 2 – 1 = 11
0.5 × 0.01 + 0.3 × 0.05 + 0.2 × 0.07
5 z(2, 4) 6×2–4=8
=
5 + 15 + 14
Hence, minimum value of (6x – y) is 8 at the
5 point (2, 4)
=
34
(C) Y Line 1
OR
7

P (Defective item is not produced by C) = P(C | D)
= 1 − P(Defective item is produced by C) 6
P(C)P(D|C) 5
P(C|D) =
P(A)P(D|A) + P(B)P(D|B) + P(C)P(D|C) 4 Line 2
x+
y

0.2 × 0.07 3


P(C|D) =
6

0.5 × 0.01 + 0.3 × 0.05 + 0.2× 0.07 2 R

14 1
=
5 + 15 + 14 X' X
O 1 2 3 4 5 6 7 8
14 7
= =
34 17 Y'

7 10 Line 1: This line is parallel to y-axis and

P(C|D) = 1 – =
17 17 touches x-axis at point x = 2.

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 So, equation of this line 1 is x = 2 or x – 2 =0. 38. MARKS BREAKDOWN Ch 9
Since, the feasible region R lies to the right
Topic: Solutions of Differential Equations
side of x = 1, this inequality is x ≥ 2.
Key Steps:
Line 2: For every point lying on the line 2, ✔ (A) Write degree of differential eq. (1m)
Write order of differential eq. (1m)
x-coordinates is double the y-coordinate.
✔ (B) Simplify given differential eq. (1/2m)
\ Equation of this line 2 is x = 2y or x – 2y = 0. Integrate both sides & find general solution
(1m)
Since, (3, 2) lies in region R, substituting Apply given initial condition (1/2m)
(3, 2) in x – 2y = 0, we get (A) Given differential equation is
3 – 2(2) = 3 – 4 = – 1 ≤ 0 dT
= k ( T − 70 )
dt
\Inequality of x = 2y is x – 2y ≤ 0 or x ≤ 2y. dT
Here, highest order derivative is , =whose
k ( T − 70 )
Hence, x ≥ 2 and x ≤ 2y are the other two power is 1. dt
inequalities. So, the degree of given differential equation
is 1.
OR
(B) Given, differential equation is,
Corner points of R are (4, 2), (2, 1) and (2, 4). dT
= k(T – 70)
Area of triangle with coordinates (4, 2), (2, 1) dt

and (2, 4) is given as, dT
⇒ = k dt
1 T − 70
A = | 4(1 – 4) + 2(4 – 2)
2 On integrating both sides, we get
+ 2(2 – 1) |
dT
∫ T − 70 = ∫ k dt

1
  = | – 12 + 4 + 2 | ⇒ log |T – 70| = kt + c
2 Given that, t = 0, T = 72
1 \ log |72 – 70| = 0 + c
  = ×6 ⇒ c = log 2
2
So, the solution of given differential equation
= 3 sq. units is: log |T – 70| = kt + log 2

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