CBSE Sample Paper Mathematics Class XII (Term I) 121

SAMPLE PAPER
MATHEMATICS
A Highly Simulated Practice Questions Paper
for CBSE Class XII (Term I) Examination

Instructions
1. This question paper contains three sections - A, B and C. Each section is compulsory.
2. Section - A has 20 MCQs, attempt any 16 out of 20.
3. Section - B has 20 MCQs, attempt any 16 out of 20.
4. Section - C has 10 MCQs, attempt any 8 out of 10.
5. There is no negative marking.
6. All questions carry equal marks.

Maximum Marks : 40
Roll No. Time allowed : 90 min

Section A
In this section, attempt any 16 questions out of Questions 1-20. Each question is of 1 mark weightage.
1. If A is a 3 ´ 2 matrix, B is a 3 ´ 3 matrix and C is a 2 ´ 3 matrix, then the elements in A, B
and C are respectively
(a) 6, 9, 8 (b) 6, 9, 6 (c) 9, 6, 6 (d) 6, 6, 9
dy
2. If x = a sin q and y = a cos 2 q, then is equal to
dx
(a) -2 cos q (b) 2 cos q (c) 2 sinq (d) -2 sinq

3. The value of cosæç sin -1

1 1 ö
+ cos -1 ÷ is
è 2 2ø
(a) 0 (b) 1 (c) - 1 (d) None of these
æ 3ö
4. The principal value of sin -1 ç - ÷ is
SA

è 2 ø
2p p 4p 5p
(a) - (b) - (c) (d)
3 3 3 3
é1 -1ù 2
5. If A = ê ú , then A is
ë 0 4 û
é1 5ù é 1 -5 ù é0 16 ù é -5 1ù
(a) ê (b) ê (c) ê (d) ê
ë0 16úû ë0 16 û
ú ú
ë 1 -5 û
ú
ë 16 0û
 é a bù é a - bù
6. The product ê úê is equal to
ë - b aû ë b aúû
éa2 + b2 0 ù é( a + b) 2 0ù
(a) ê ú (b) ê ú
êë 0 a + b 2 úû
2
êë( a + b)
2
0úû
éa2 + b2 0ù é a 0ù
(c) ê 2 ú (d) ê ú
êë a + b2 0úû ë0 bû
é 2 3ù
7. If A = ê ú , then which of the following is true?
ë - 4 - 6û
(a) A( adj A) ¹ | A| I (b) A( adj A) ¹ ( adj A) A
(c) A( adj A) = ( adj A) A = | A| I (d) None of these
dy
8. If x + y = 1, then at (5, 5) is equal to
dx
(a) 1 (b) 0 (c) 2 (d) - 1
1 3 -2
9. If M 11 = - 40, M 12 = - 10 and M 13 = 35 of the determinant D = 4 - 5 6 , then the
3 5 2
value of D is
(a) - 80 (b) 60 (c) 70 (d) 100
1 a bc
10. If D = 1 b ca , then the minor M 31 is
1 c ab
(a) - c ( a - b 2 )
2
(b) c ( b 2 - a 2 ) (c) c ( a 2 + b 2 ) (d) c ( a 2 - b 2 )
é 0 0 4ù
11. The matrix P = ê 0 4 0ú is a
ê ú
êë 4 0 0úû
(a) square matrix (b) diagonal matrix
(c) unit matrix (d) None of these
dy 1
12. If y = (x 2 + 1) 2 , then at x = is equal to
dx 2
5 5 2
(a) (b) 5 (c) (d)
2 4 5

13. The equation of tangent to the curve y = x 2 + x - 2 at (1, 0) is given by

(a) 3x - y = 3 (b) 3x - y = -3 (c) x - 3y = 1 (d) None of these

14. Which one of the following statements is correct?

(a) e x is an increasing function
(b) e x is a decreasing function
(c) e x is neither an increasing nor a decreasing function
(d) e x is a constant function

15. If sin - 1 x = y, then

-p p -p p
(a) 0 £ y £ x (b) £y£ (c) 0 < y < p (d) <y<
2 2 2 2
 é 2( x + 1) 2x ù x
16. If A = ê ú is a singular matrix, then 2 is equal to
ë x x - 2û
(a) - 2 (b) - 3 (c) - 1 (d) 0
é1 2ù é x + 1 4ù
17. If A = ê ú and B = ê ú , then the value of y - x, if A 2 = B is
ë 3 1û ë 6 yû
(a) 0 (b) 1 (c) 2 (d) 3

18. If A is an invertible matrix of order 2, then det (A -1 ) is equal to

1
(a) det ( A) (b) (c) 1 (d) zero
det( A)
é 2 1ù é 4 7ù
19. If 2X + ê ú =ê ú , then X is equal to
ë 3 5û ë 1 1û
é1 3ù é1 3 ù é 1 -1 ù é 1 3ù
(a) ê ú (b) ê ú (c) ê ú (d) ê ú
ë -1 -2 û ë1 2 û ë3 -2 û ë -1 2 û

20. The greatest integer function f : R ® R, given by f (x) = [x] is

(a) one-one (b) onto
(c) both one-one and onto (d) neither one-one nor onto

Section B
In this section, attempt any 16 questions out of Questions 21-40. Each question is of 1 mark weightage.

21. If the curve ay + x 2 = 7 and x 3 = y, cut orthogonally at (1, 1), then the value of a is
(a) 1 (b) 0 (c) - 6 (d) 6
æ1 - x2 ö dy
22. If y = cos -1 ç 2
÷ , then at x = 1 is equal to
è1 + x ø dx
(a) 0 (b) 1 (c) 3 (d) - 1

23. If f (x) =|sin x| , then

(a) f is everywhere differentiable
(b) f is everywhere continuous but not differentiable at x = np , n Î Z
p
(c) f is everywhere continuous but not differentiable at x = (2 n + 1) , n Î Z
2
(d) None of the above
ì k cos x , if x ¹ p
ï 2 is continuous at x = p , when k equals
24. The function f (x) = í p - 2x
ï 3, p 2
if x =
ïî 2
(a) -6 (b) 6 (c) 5 (d) -5
dy
25. If y = x x , then is equal to
dx
x
(a) x (1 + log x) (b) x x (1 - log x)
(c) x(1 + log x) (d) None of these
 (log x)
26. The maximum value of is
x
2 1
(a) 1 (b) (c) e (d)
e e

27. If y = x (x - 3) 2 decreases for the values of x given by

3
(a) 1 < x < 3 (b) x < 0 (c) x > 0 (d) 0 < x <
2

28. The relation R in the set of natural numbers N defined as R = {(x, y) : y = x + 5 and x < 4}
is
(a) reflexive (b) symmetric (c) transitive (d) None of these

29. The maximum value of the function f (x) = x 3 + 2x 2 - 4 x + 6 exists at

(a) x = -2 (b) x = 1 (c) x = 2 (d) x = -1

30. Area of the triangle whose vertices are (a , b + c), (b , c + a) and (c, a + b), is
(a) 2 sq units (b) 3 sq units (c) 0 (d) None of these
é1 1ù
31. The set of all 2 ´ 2 matrices which is commutative with the matrix ê ú with respect to
ë1 0û
matrix multiplication is
é p qù é p qù é p - q pù ép q ù
(a) ê ú (b) ê ú (c) ê (d) ê
ë r r û ë q rû ë q r úû ë q p - qúû
- 1ö æ p öö
32. The value of tan - 1 æç -1æ 1 ö
÷ + cot ç
-1æ
÷ + tan ç sin ç - ÷ø ÷ is
è 3ø è 3ø è è 2 ø
p p
(a) (b)
6 12
p p
(c) - (d)
12 3
é1 1 ù 2
33. If A = ê ú and f ( x) = 1 - x , then f ( A) is
ë 0 - 1û
é1 1ù é0 0ù é1 2 ù
(a) ê (b) ê (c) ê (d) None of these
ë0 0úû ë0 0 ú
û ë3 4û
ú

34. For the set A = {1, 2, 3}, define a relation R in the set A as follows
R = {(1, 1), (2, 2), (3, 3), (1, 3)}
Then, the ordered pair to be added to R to make it the smallest equivalence relation is
(a) (1, 3) (b) (3, 1) (c) (2, 1) (d) (1, 2)
a h g
35. If D = h b f , then the cofactor A31 is
g f c
(a) - ( hc + fg) (b) hf - bg (c) fg + hc (d) hc - fg
dy
36. If y = log(sin e x ), then is equal to
dx
(a) e x cot( e x ) (b) -e x cot( e x )
(c) e x tan( e x ) (d) None of these
37. The number of all one-one functions from set A = {1, 2, 3} to itself is
(a) 2 (b) 6 (c) 3 (d) 1
dy
38. If y = log a x + log x a + log x x + log a a, then is equal to
dx
1 log a x
(a) + x log a (b) +
x x log a
1
(c) + x log a (d) None of these
x log a

d2y -b
39. If y = ax 3 + bx 2 + cx + d, then 2
at x = is equal to
dx 3a
(a) 1 (b) 2 (c) 3 (d) 0

40. The differential coefficient of sin (cos(x 2 )) w.r.t. x is.

(a) -2 x sin x 2 cos (cos x 2 ) (b) 2 x sin( x 2 ) cos ( x 2 )
(c) 2 x sin( x 2 ) cos ( x 2 ) cos x (d) None of these

Section C
In this section, attempt any 8 questions. Each question is of 1 mark weightage. Questions 46-50 are based
on Case-Study.
4 - x2
41. The function f (x) = is
4x - x 3
(a) discontinuous at only one point (b) discontinuous at exactly two points
(c) discontinuous at exactly three points (d) None of these

42. The equation of the tangent to the curve x 2 - 2yx + y 2 + 2x + y - 6 = 0 at (2, 2) is

(a) 2 y + x = 6 (b) 2 x + y = 6
(c) x + y = 4 (d) x = y
dy
43. If y x = e y - x , then is equal to
dx
1 + log y (1 + log y) 2
(a) (b)
y log y y log y
1 + log y (1 + log y) 2
(c) (d)
(log y) 2 log y

d2y dy
44. If y = Ae mx + Be nx and 2
- ( m + n) + mny = k, then k is equal to
dx dx
(a) 1 (b) 0
(c) -1 (d) None of these
dy
45. If cos y = x cos(a + y) with cos a ¹ 1, then is equal to
dx
sin 2 ( a + y) cos 2 ( a + y)
(a) (b)
sin a sin a
2
(c) sin ( a + y) sin a (d) None of these
CASE STUDY
If feasible solution of a LPP is given as follows:
Y

60
2x + y = 80
C
50

40

30

20 B(30, 20)

10

X′ X
O 10 20 30 A 50 60
Y′
x + y = 50

And the objective function is Z = 10500x + 9000y.

Based on above information, answer the following question.

46. The feasible solution consists

(a) (10, 10) (b) (50, 10)
(c) (0, 55) (d) (30, 21)

47. Objective function is maximum at the point

(a) (0, 0) (b) (30, 20)
(c) A (d) C

48. Objective function has the value 420000 at

(a) point A (b) point B
(c) point C (d) point O

49. Z| ( 20 , 20 ) - Z| ( 10 , 10 ) is
(a) 200000 (b) 195000
(c) 205000 (d) 190000

50. Sum of values of Z at all corner points is

(a) 1365000 (b) 1360000
(c) 1355000 (d) 1350000
 Answers
1. (b) 2. (d) 3. (a) 4. (b) 5. (b) 6. (a) 7. (c) 8. (d) 9. (a) 10. (d)
11. (a) 12. (a) 13. (a) 14. (a) 15. (b) 16. (c) 17. (b) 18. (b) 19. (a) 20. (d)
21. (d) 22. (b) 23. (b) 24. (b) 25. (a) 26. (d) 27. (a) 28. (c) 29. (a) 30. (c)
31. (d) 32. (c) 33. (b) 34. (b) 35. (b) 36. (a) 37. (b) 38. (d) 39. (d) 40. (a)
41. (c) 42. (b) 43. (d) 44. (b) 45. (b) 46. (a) 47. (b) 48. (a) 49. (b) 50. (a)

SOLUTIONS
1. The number of elements in m ´ n matrix is 8. Given, x + y = 1
equal to mn. Now, Differentiating both sides w.r.t. x, we get
2. Given, x = a sinq 1 1 dy dy y
dx + =0 Þ =-
\ = a cos q 2 x 2 y dx dx x
dq
y = a cos 2 q æ dy ö
and Now, ç ÷ = -1
è dx ø( 5, 5)
dy
\ = 2 a cos q ( - sin q )
dq 9. D = a11 A11 + a12 A12 + a13 A13
dy ( dy / dq ) -2 a cos q sin q
Now, = = = a11 M11 - a12 M12 + a13 M13
dx ( dx / dq ) a cos q
= -2 sinq = 1 × ( - 40 ) - 3( -10 ) + ( -2 ) (35 )
æ 1 1 ö = - 40 + 30 - 70
3. cos ç sin -1 + cos -1 ÷
è 2 2ø = - 80
æp pö æpö
= cos ç + ÷ = cos ç ÷ = 0 a bc
è4 4ø è2 ø 10. M31 = = ca2 - b2c = c( a2 - b2 )
b ca
æ 3ö
4. Let sin -1 ç - ÷ =q 11. A square matrix in which all non-diagonal
è 2 ø
elements are zero, is called a diagonal matrix
3 p p
Þ sinq = -
and - £ sin q £ and square matrix in which all diagonal
2 2 2 elements are 1 and rest are 0, is called unit
p matrix.
\ q=-
3
12. Given, y = ( x 2 + 1)2
æ 3 ö p
\ sin -1 ç - ÷=- dy d 2 d
è 2 ø 3 \ = ( x + 1)2 = 2 ( x 2 + 1) ( x 2 + 1)
dx dx dx
é 1 -1ù
5. Given, A = ê ú
ë0 4 û = 2 ( x 2 + 1)(2 x )
é 1 -1ù é 1 -1ù dy ö æ1 öæ 1ö æ5ö 5
= 2 ç + 1÷ ç2 ÷ = 2 ç ÷ =
\ A2 = A × A = ê úê ú
Now, ÷
ø
dx x = 1 è 4 ø è 2 ø è4ø 2
ë0 4 û ë0 4 û 2
é 1 + 0 -1 - 4 ù é 1 -5 ù
=ê ú=ê ú 13. Given curve is y = x 2 + x - 2
ë0 + 0 0 + 16 û ë0 16 û dy
\ = 2x +1
é a bù é a - bù é a2 + b2 - ab + baù dx
6. ê úê ú=ê 2 ú
ë - b aû ë b a û ë - ba + ab b + a û
2
æ dy ö
\ Slope of tangent at (1, 0) = ç ÷
è dx ø( 1, 0)
é a 2 + b2 0 ù
=ê ú
a + b2 û
2 = 2 (1) + 1 = 3
ë 0
Now, equation of tangent is given by
7. We know, if A is any square matrix of order n, y - 0 = 3 ( x - 1) = 3 x - 3
then A( adj A) = ( adj A) A = |A|× I
Þ 3x - y = 3
14. Let f ( x ) = ex Now, consider 0.6 Î R (codomain).
Now, on differentiating w.r.t. x. It is know that, f ( x ) = [ x ] is always an integer.
f ¢ ( x ) = ex > 0, "x Î R Thus, there does not exist any element x Î R
x
(domain) such that f ( x ) = 0.6.
So, f ( x ) = e is an increasing function.
Therefore f is not onto.
é-p pù
15. Range of sin - 1 x is ê , 21. We have, ay + x 2 = 7 and x 3 = y
ë 2 2 úû
-p p On differentiating w.r.t. x in both equations,
\ £y£ we get
2 2 dy dy
a× + 2 x = 0 and 3 x 2 =
16. Given, A is singular matrix. dx dx
\ |A| = 0 dy 2x dy
Þ =- and = 3x2
2 ( x + 1) 2x dx a dx
Þ =0 æ dy ö -2
x x -2 Þ ç ÷ = = m1
è dx ø( 1, 1) a
Þ 2 ( x + 1)( x - 2 ) - 2 x 2 = 0
æ dy ö
Þ 2 x2 - 2 x - 4 - 2 x2 = 0 and ç ÷ = 3 × 1 = 3 = m2
è dx ø( 1, 1)
Þ 2x + 4 = 0
Since, the curves cut orthogonally at (1, 1).
Þ x = -2
x \ m1 × m2 = - 1
Now, = -1 æ -2 ö
2 Þ ç ÷ ×3 = - 1
è a ø
é1 2 ù é x + 1 4ù
17. Given, A = ê ú and B = ê 6 \ a=6
ë 3 1 û ë y úû
æ 1 - x2 ö
Also, A2 = B 22. Given, y = cos -1 ç ÷
é 1 2 ù é 1 2 ù é7 4 ù è 1 + x2 ø
Now, A2 = ê úê ú=ê ú
ë3 1 û ë3 1 û ë6 7 û Let, x = tanq
Now, A = B2 Þ q = tan -1 x
é7 4 ù é x + 1 4 ù æ 1 - tan 2 q ö
Þ ê6 7 ú = ê 6 \ y = cos -1 ç ÷
ë û ë y úû è 1 + tan 2 q ø
On comparing both the matrices, we get = cos -1(cos 2 q ) = 2 q
x +1 = 7
\ y = 2 tan -1 x
Þ x = 6 and y = 7
\ y - x = 7 -6 = 1 dy 2
Now, =
18. We know, AA- 1 = I dx 1 + x 2
æ dy ö 2 2
\ |AA- 1| = |I| \ ç ÷ = = =1
è dx ø x =1 1 + 1 2
Þ |A||A- 1| = 1
1 23. Let u( x ) = sin x
Þ |A- 1| =
|A| v( x ) = |x|
\ f ( x ) = vou( x ) = v( u( x ))
é2 1ù é 4 7ù
19. Given, 2 X + ê = = v(sin x ) = |sin x|
ë3 5 úû êë 1 1úû
Q u( x ) = sin x is a continuous function and
é4 7 ù é2 1ù é 2 6ù
v( x ) = |x|is a continuous function
\ 2X = ê - =ê
ë1 1úû êë 3 5 û ë -2 -4 úû
ú
\ f ( x ) = vou( x ) is also continuous everywhere
1é 2 6ù é1 3 ù but v( x ) is not differentiable at x = 0
\ X= ê =ê
2 ë -2 -4 úû ú
ë -1 -2 û Þ f ( x ) is not differentiable where sin x = 0
Þ x = np, n Î Z
20. Range ( f ) = Integers ¹ R and [2 × 3 ] = [2 × 4 ] = 2
Hence, f ( x ) is continuous everywhere but not
Þ f is not one-one. differentiable at x = np, n Î Z.
 ì k cos x p 28. R = {(1, 6 ), (2 , 7 ), (3 , 8 )}
ï , if x ¹
24. We have, f ( x ) = í p - 2 x 2
R is not reflexive as (1, 1) Ï R.
ï 3, p
if x = R is not symmetric as (2, 7) Î R but (7, 2) Ï R.
ïî 2
p Now, since there is no pair in R such that ( x , y )
f ( x ) is continuous at x = and ( y , z) Î R, then ( x , z) cannot belong to it.
2
k cos x \ R is transitive.
\ lim =3
p- p - 2x Hence, R is neither reflexive, nor symmetric
x®
2 but transitive.
æp ö 29. Given functions,
k cos ç - h ÷
è2 ø f (x) = x3 + 2 x2 - 4x + 6
Þ lim =3
h ®0 æ p ö
p - 2 ç - h÷ and f ¢ (x) = 3x2 + 4x - 4
è2 ø
Now, for maximum or minimum of f ( x ), put
k sin h
Þ lim =3 f ¢ ( x ) = 0.
h ® 0 2h
Þ 3x2 + 4x - 4 = 0
k
Þ =3Þk =6 Þ 3x2 + 6x - 2 x - 4 = 0
2
Þ ( x + 2 )(3 x - 2 ) = 0
25. Given, y = x x 2
Þ x = -2 ,
Now, Taking log on both sides, we get 3
log y = x log x Now, f ¢ ¢ (x) = 6x + 4
Now, Differentiating both sides w.r.t. x, we get At x = -2 ,
1 dy x f ¢ ¢ ( -2 ) = 6( -2 ) + 4
= + log x
y dx x = -12 + 4 = - 8 < 0
dy [maximum]
Þ = y(1 + log x )
dx So, maximum value of the function f ( x ) exists
Þ
dy
= x x (1 + log x ) at x = -2.
dx a b+c 1
log x 1
26. Let y = 30. Area of triangle, D = b c+ a 1
x 2
c a+b 1
dy 1 1 1
Þ = - 2 log x + × 1
dx x x x = [ a {( c + a) - ( a + b)} - ( b + c)( b - c)
1 2
= 2 (1 - log x ) = 0 +1{b( a + b) - c( c + a)}]
x 1
= [ a( c - b) - b + c + ba + b - c2 - ac]
2 2 2
Þ x=e 2
d 2y 1
At x = e, <0 [maxima] = [ ac - ab - b2 + c2 + ba + b2 - c2 - ac]
dx 2 2
1 1
\ y= = ´ (0 ) = 0 sq units
e 2
27. We have, y = x( x - 3 )2 é p qù
31. Let A = ê ú be a matrix which commute
dy ë r sû
\ = x × 2 ( x - 3 ) × 1 + ( x - 3 )2 × 1
dx é1 1ù
with matrix B = ê ú.
= 2 x2 - 6x + x2 + 9 - 6x ë1 0 û
= 3 x 2 - 12 x + 9
Then, AB = BA
= 3 (x2 - 3x - x + 3)
é p qù é1 1ù é1 1ù é p qù
= 3 ( x - 3 )( x - 1) Þ ê r s ú ê1 =
ë ûë 0 úû êë1 0 úû ê r sú
ë û
+ – +
1 3 é +q
p pù é p + r q + sù
Þ êr+ s =
So, y = x( x - 3 )2 decreases for (1, 3). ë r úû êë p q úû
[since, y ¢ < 0 for all x Î(1, 3 ), hence y is Here, both matrices are equal, so we equate the
decreasing on (1, 3)] corresponding elements.
 \ p + q = p + r, p = q + s, r + s = p and r = q log x log a
38. Given, y = + +1+1
Þ r = q and s = p - q log a log x

ép q ù dy 1 log a
\ A= ê Þ = -
ë q p - qúû dx x log a x(log x )2

ép q ù 39. Given, y = ax 3 + bx 2 + cx + d
Hence, the required set is ê ú.
ë q p - qû \
dy
= 3 ax 2 + 2 bx + c
æ - 1ö æ p öö dx
- 1æ 1 ö - 1æ
32. tan - 1 ç ÷ + cot ç ÷ + tan ç sin ç - ÷ ÷ d 2y
è 3ø è 3ø è è 2 øø Þ = 6 ax + 2 b
-p p dx 2
= + + tan - 1( - 1) æ d 2y ö
6 3 æ -bö
Now, ç 2 ÷ = 6 a ç ÷ + 2 b = -2 b + 2 b = 0
é -1 æ - 1 ö - p - 1æ 1 ö p ù è dx ø b è 3a ø
êQ tan çè 3 ÷ø = 6 , cot çè 3 ÷ø = 3 ú x =-
3a
ê ú
ê æ- pö ú 40. Given, y = sin(cos x 2 )
êë and sin ç ÷ = -1 ú
è 2 ø û dy d
Therefore, = sin(cos x 2 )
-p p p -p dx dx
= + - = d
6 3 4 12 = cos(cos x 2 ) (cos x 2 )
dx
33. Given, f ( x ) = 1 - x 2 d
= cos(cos x )( - sin x 2 ) ( x 2 )
2
\ f ( A) = I - A2 dx
é 1 0 ù ì é 1 1 ù é 1 1 ùü = - sin x 2 cos(cos x 2 )(2 x )
\ f ( A) = ê ú - íê úê úý
ë0 1û îï ë0 -1û ë0 -1ûþ = - 2 x sin x 2 cos(cos x 2 )
é 1 0 ù é 1 0 ù é0 0 ù 4 - x2 (4 - x2 )
=ê ú-ê ú=ê ú 41. We have, f ( x ) = =
ë0 1û ë0 1û ë0 0 û 4x - x3 x (4 - x2 )
34. The given relation is (4 - x2 ) 4 - x2
= =
R = {(1, 1), (2, 2), (3, 3), (1, 3)} on the set 2
x (2 - x ) 2 x (2 + x ) (2 - x )
A = {1, 2, 3}.
Clearly, f ( x ) is discontinuous at exactly three
Clearly, R is reflexive and transitive. points x = 0 , x = - 2 and x = 2
To make R symmetric, we need (3, 1) as
42. Given, x 2 - 2 yx + y 2 + 2 x + y - 6 = 0
(1, 3) Î R.
On differentiating w.r.t. x, we get
\If (3, 1) Î R, then R will be an equivalence
relation. Hence, (3, 1) is the single ordered pair æ dy ö dy dy
2 x - 2 çy + x ÷ + 2 y + 2 + =0
which needs to be added to R to make it the è dx ø dx dx
smallest equivalence relation.
At (2, 2),
h g æ dy ö dy dy
35. A31 = ( - 1)3 + 1 M31 = = hf - bg 4 - 2 ç2 + 2 ÷ + 4 +2 + =0
b f è dx ø dx dx
36. Given, y = log(sin ex ) dy
Þ = -2
dy 1 dx
Þ = (cos ex )( ex ) Equation of tangent at (2, 2) is
dx sin( ex )
dy ( y - 2 ) = -2 ( x - 2 )
Þ = ex cot( ex )
dx Þ 2x + y = 6
37. If n( A) = x and n( B) = y, then number of 43. We have, y x = ey - x
one-one functions from A to B is given by y Px , Taking log both sides, we get
where x £ y. x log y = y - x
3 3! 3!
P3 = = Þ x (log y + 1) = y
(3 - 3 )! 0 ! y
Þ =x
= 3 ´2 ´1 = 6 [Q0 ! = 1] 1 + log y
 On differentiating w.r.t. x, we get dy 1 cos 2 ( a + y )
Þ = =
æ æ1ö ö dx dx sin a
ç (1 + log y ) - ç ÷ y ÷
ç è y ø ÷ dy dy
ç (1 + log y )2 ÷ dx = 1
46. From the given graph, we can clearly see that
çç ÷÷
è ø point (10, 10) is in the feasible region or
feasible solution.
log y dy
Þ =1 Hence, option (a) is correct.
(1 + log y )2 dx
47. To know the maximum value of Z, we need
dy (1 + log y )2
Þ = coordinates of all the corner points.
dx log y
We have, equation of lines x + y = 50 …(i)
44. Given, y = A emx + B enx …(i) and 2 x + y = 80 …(ii)
On differentiating twice w.r.t. x, we get For point A, put y = 0 into Eq. (ii),
dy d d 2 x + 0 = 80 Þ x = 40 Þ A( 40 , 0 )
= Aemx ( mx ) + Benx ( nx )
dx dx dx For point C, put x = 0 into Eq. (i),
= A emx m + B enx n …(ii) 0 + y = 50 Þ y = 50 Þ C(0 , 50 )
2
d y
Þ = m2 Aemx + n2 Benx …(iii) Now,
dx 2
Using Eqs. (i), (ii) and (iii), we get Corner Z = 10500 x + 9000 y
d 2y dy Points
2
- ( m + n) + mny
dx dx O(0, 0) 10500 ´ 0 + 9000 ´ 0 = 0
= m2 Aemx + n2 Benx - ( m + n) { Amemx + nBenx } A(40, 0) 10500 ´ 40 + 9000 ´ 0 = 42000
+ mn { Aemx + Benx }
2
d y dy 10500 ´ 30 + 9000 ´ 20 = 495000
B (30, 20)
i.e. 2
- ( m + n) + mny = 0 (Maximum)
dx dx
Þ k =0 C (0, 50) 10500 ´ 0 + 9000 ´ 50 = 450000

45. Given, cos y = x cos( a + y ) Hence, Z is maximum at B(30 , 20 ).

cos y 48. From the above table, we can see that objective
Þ x=
cos ( a + y ) function has value 420000 at point A.
dx d ì cos y ü 49. Z|( 20, 20)- Z|( 10, 10)
Þ = í ý
dy dy ïî cos ( a + y )þ = (10500 ´ 20 + 9000 ´ 20 )
cos( a + y ) ( - sin y ) - cos y( - sin( a + y ) 1) -(10500 ´ 10 + 9000 ´ 10 )
=
cos 2 ( a + y ) = 1000 [(105 ´ 2 + 90 ´ 2 ) - (105 + 90 )]
sin ( a + y - y ) sin a = 1000 [(210 + 180 ) - (195 )]
= 2
= 2
cos ( a + y ) cos ( a + y ) = 1000 [390 - 195 ] = 195000
[Qsin( A - B) = sin A cos B- cos A sin B] 50. Required sum = 0 + 420000 + 495000 + 450000
= 1365000