TBSE Class 12 Model Question Paper 2021 Maths

Model Question
Class - XII : Mathematics : 80 Marks : 2020-2021
!Ó˲yàÈüÈܲ : ≤Ã!ï˛!ê˛ ≤Èϟ¿Ó˚ ÙylÈüÈ1 : 1 x 20 = 20
I) Tick the correct option : 1x5
1. Î!ò A ~ܲ!ê˛ Óà≈ Ùƒy!ê˛∆: •Î˚ñ ï˛ˆÏÓ A–1 ~Ó˚ x!hflÏc ÌyܲˆÏÓ Î!ò ~ÓÇ ~ܲӰÙye Î!ò
a) A !§Aà%°yÓ˚ Ùƒy!ê˛∆: •Î˚
b) A ll‰ !§Aà%°yÓ˚ Ùƒy!ê˛∆: •Î˚
c) A, 2 x 2 ܲˆÏÙÓ˚ Ùƒy!ê˛∆: •Î˚
d) A, 3 x 3 ܲˆÏÙÓ˚ Ùƒy!ê˛∆: •Î˚–
2. y = 4x ÓˆÏܲÓ˚ (4,4) !Ó®%ˆÏï˛ x!˲°ˆÏ¡∫Ó˚ ≤ÃÓîï˛y •ˆÏ°y
2

1 1
a) –2 (b) 2 (c) (d) −
2 2

3. Î!ò a . b = a × b •Î˚ñ ï˛ˆÏÓ a ~ÓÇ b ~Ó˚ ÙôƒÓï˛#≈ ˆÜ˛yî •ˆÏ°y û

π π π
a) b) c) d) ~à%ˆÏ°yÓ˚ ˆÜ˛yl!ê˛•z lÎ˚–
6 3 2
dx
4. ∫ 4 x − x = 2

x−2 x−2
(a ) sin −1 +c (b) − sin −1 +c
2 2

x−2
(c) cos −1 +c d) ~à%ˆÏ°yÓ˚ ˆÜ˛yl!ê˛•z lÎ˚–
2

5. Î!ò Ùƒy!ê˛∆: B ≤Ã!ï˛§Ù ~ÓÇ !Ó≤Ã!ï˛§Ù í˛z˲Î˚•z •Î˚ñ ï˛ˆÏÓ
a) B •ˆÏ°y ~ܲ!ê˛ Ü˛î≈ Ùƒy!ê˛∆:
b) B •ˆÏ°y ~ܲ!ê˛ ~ܲܲ (Identity) Ùƒy!ê˛∆:
c) B •ˆÏ°y ¢)lƒ Ùƒy!ê˛∆:
d) ~à%ˆÏ°yÓ˚ ˆÜ˛yl!ê˛•z lÎ˚–
II) Fill in the blanks :
x − 4 y − 4 z −1
6. = = §Ó˚° ˆÓ˚áy yz §Ùï˛°ˆÏܲ ˆÎ !Ó®%ˆÏï˛ ˆSÈò ܲˆÏÓ˚ ï˛yÓ˚ fiÌylyAܲ •ˆÏ°y ûûûûû–
2 −3 5

−1 2π
7. sin sin
3
~Ó˚ Ù%რÙyl •ˆÏ°y ûûûû–

−1 2
8. Î!ò sin sin + cos −1 x = 1 •Î˚ñ ï˛ˆÏÓ x = ûûûû–
5
x y z x−5 y−2 z−3
9. = = ~ÓÇ = = §Ó˚°ˆÏÓ˚áy Î%àˆÏ°Ó˚ ÙôƒÓï˛#≈ ˆÜ˛yî •ˆÏ°y ûûûû–
2 2 1 4 1 8

10. Î!ò A ~ܲ!ê˛ 3x3 ܲˆÏÙÓ˚ llÈüÈ!§Aà%°yÓ˚ Ùƒy!ê˛∆: ~ÓÇ |A|=5 •Î˚ñ ï˛ˆÏÓ |adj A|=_____.
Answer the short questions
sin 2 x
11. Ùyl !lî≈Î˚ ܲˆÏÓ˚y / ∫ 1 − cos xdx
12. Ùyl !lî≈Î˚ ܲˆÏÓ˚y / ∫ e (log sin x + cot x ) dx
x

π
2
3
13. !l!ò≈T˛ §ÙyܲˆÏ°Ó˚ ôÙ≈ ÓƒÓ•yÓ˚ ܲˆÏÓ˚ Ùyl !lî≈Î˚ ܲˆÏÓ˚y / ∫ x cos xdx
π
−
2

14. Î!ò Ùƒy!ê˛∆: A, 3 x 4 ܲˆÏÙÓ˚ ~ÓÇ Ùƒy!ê˛∆: Bñ m x n ܲˆÏÙÓ˚ ~Ó˚)˛õ Ùƒy!ê˛∆: •Î˚ ÎyˆÏï˛ ATB ~ÓÇ BAT í˛z˲Î˚ §ÇK˛yï˛
•Î˚ñ ï˛ˆÏÓ m ~ÓÇ n ~Ó˚ Ùyl !lî≈Î˚ ܲˆÏÓ˚y–
5 7 8
15. !Óhfl,Ïï˛ ly ܲˆÏÓ˚ 15 17 18 !lî≈yÎ˚ˆÏܲÓ˚ Ùyl !lî≈Î˚ ܲˆÏÓ˚y–
20 24 26
16. xˆÏ˛õ«˛Ü˛ fÈüÈ~Ó˚ x=0 !Ó®%ˆÏï˛ §hs˝ï˛y ˛õÓ˚#«˛y ܲˆÏÓ˚yñ ˆÎáyˆÏl f !l¡¨Ó˚)ˆÏ˛õ §ÇK˛yï˛
sin
, x≠0
f ( x) = x
0 , x=0

2 3
d2y dy
17. !l¡¨!°!áï˛ xÓܲ° §Ù#ܲÓ˚î!ê˛Ó˚ Ü˛Ù ~ÓÇ âyï˛ !lî≈Î˚ ܲˆÏÓ˚y / + =0
dx 2 dx

18. ‘p’ ~Ó˚ ˆÎ ÙyˆÏlÓ˚ çlƒ a = 3ɵi + 2 ɵj + 9kɵ ~ÓÇ b = ɵi − 2 p ɵj + 3kɵ ˆË˛QÓ˚ ò%!ê˛ §ÙˆÏÓ˚á ï˛y !lî≈Î˚ ܲˆÏÓ˚y–
1 4R 3
19. ˆÏòGÎ˚y xyˆÏSÈ ˆÎ A ~ÓÇ B ò%!ê˛ âê˛ly ˆÎáyˆÏl P(A) = , P(B)= ~ÓÇ P(AUB)= – A ~ÓÇ B âê˛ly ò%!ê˛ fl∫yô#l
2 3 5
!ܲly ˛õÓ˚#«˛y ܲˆÏÓ˚y–
20. ~ܲ!ê˛ ˆVÑ˛yܲ¢)lƒ Ù%oy ò%•zÓyÓ˚ ê˛§‰ ܲÓ˚y •ˆÏ°y– P(F/E) !lî≈Î˚ ܲˆÏÓ˚yñ ˆÎáyˆÏl
E : Ü˛Ù˛õˆÏ«˛ ~ܲ!ê˛ ˆ•í˛ í˛zë˛yÓ˚ âê˛ly
F : í˛z˲Î˚ ê˛ˆÏ§•z ˆ•í˛ í˛zë˛yÓ˚ âê˛ly–
!Ó˲yàÈüÈá / ≤Ã!ï˛!ê˛ ≤Èϟ¿Ó˚ Ùyl 2 / 2x6=12
−1 cos x
21. tan
1 = sin x
ˆÜ˛ §Ó˚°ï˛Ù xyܲyˆÏÓ˚ ≤Ãܲy¢ ܲˆÏÓ˚y–
22. logx cosxˆÜ˛ x ~Ó˚ §yˆÏ˛õˆÏ«˛ xÓܲ°l ܲˆÏÓ˚y–
x2 y 2
23. + = 1 í˛z˛õÓ,ˆÏ_Ó˚ í˛z˛õ!Ó˚!fiÌï˛ ~Ùl !Ó®%§Ù)• !lî≈Î˚ ܲˆÏÓ˚y ˆÎ !Ó®%à%ˆÏ°yˆÏï˛ flõ¢≈ܲ x xˆÏ«˛Ó˚ §Ùyhs˝Ó˚y°–
9 16
24. a = 3iɵ + ɵj − 4kɵ ~ÓÇ b = 6ɵi + 5 ɵj − 2kɵ í˛z˲Î˚ ˆË˛QˆÏÓ˚Ó˚ í˛z˛õÓ˚ °¡∫ ~Ùl ~ܲ!ê˛ ˆË˛QÓ˚ !lî≈Î˚ܲˆÏÓ˚y ÎyÓ˚ Ùyl 3
~ܲܲ–
25. 3x–y+2z–4=0 ~ÓÇ x+y+z–2=0 §Ùï˛°mˆÏÎ˚Ó˚ ˆSÈò§Ó˚°ˆÏÓ˚áyàyÙ# ~ÓÇ (2, 2, 1) !Ó®%àyÙ# §Ùï˛ˆÏ°Ó˚ §Ù#ܲÓ˚î !lî≈Î˚
ܲˆÏÓ˚y–

1 1
26. ~ܲ!ê˛ !l!ò≈T˛ §Ù§ƒy A ~ÓÇ B ~Ó˚ ˛õˆÏ«˛ fl∫yô#l˲yˆÏÓ §Ùyôyl ܲÓ˚yÓ˚ §Ω˛yÓly ÎÌyܲˆÏÙ ~ÓÇ – Î!ò ï˛yˆÏòÓ˚ í˛z˲Î˚•z
2 3
§Ù§ƒy!ê˛ fl∫yô#l˲yˆÏÓ §Ùyôyl ܲÓ˚ˆÏï˛ ˆã˛T˛y ܲˆÏÓ˚ñ ï˛ˆÏÓ §Ù§ƒy!ê˛ §Ùyôyl •GÎ˚yÓ˚ §Ω˛yÓly !lî≈Î˚ ܲˆÏÓ˚y–
!Ó˲yàÈüÈà : Each Question Carries 4 Marks : 4x6=24
27. ≤ÃÙyî ܲˆÏÓ˚y ˆÎ ñ
1 2 1 4
tan − 1 + tan −1 = sin −1
4 9 2 5

3 3 −1
28. A = −2 − 2 1 Ùƒy!ê˛∆:ˆÏܲ ~ܲ!ê˛ ≤Ã!ï˛§Ù ~ÓÇ ~ܲ!ê˛ !Ó≤Ã!ï˛§Ù Ùƒy!ê˛∆ˆÏ:Ó˚ §Ù!T˛Ó˚)ˆÏ˛õ ≤Ãܲy¢ ܲˆÏÓ˚y–
−4 − 5 2

2 dy
29. Î!ò x 1 + y + y 1 + x = 0 •Î˚ñ ï˛ˆÏÓ ≤ÃÙyî ܲˆÏÓ˚y ˆÎñ (1 + x ) + 1= 0
dx
D xÌÓy
2
x d2y dy
Î!ò y = x log a + bx •Î˚ñ ï˛ˆÏÓ ≤ÃÙyî ܲˆÏÓ˚yñ x3 2 = x − y
dx dx

π
2
x sin x cos x
30. Ùyl !lî≈Î˚ ܲˆÏÓ˚y / ∫ dx
0
sin 4 x + cos 4 x

D xÌÓy
≤ÃÙyî ܲˆÏÓ˚y ˆÎñ
1
log(1 + x ) π
∫ 1+ x 2
dx = log 2
0
8

31. !l¡¨!°!áï˛ xÓܲ° §Ù#ܲÓ˚î!ê˛Ó˚ !ӈϢ£Ï §Ùyôyl !lî≈Î˚ ܲˆÏÓ˚y /
dy
− 3 y cot x = sin 2 x
dx

π
ˆòGÎ˚y xyˆÏSÈ ˆÎñ y = 2, Îál x =
2
32. ÙˆÏl ܲˆÏÓ˚y ˆÎñ COVID-19 xƒy!rê˛ˆÏçl ˆê˛ˆÏfiê˛Ó˚ å˛õÓ˚#«˛yÓ˚ä !lË≈˛Ó˚ˆÏÎyàƒï˛y !l¡¨!°!áï˛Ë˛yˆÏÓ í˛zˆÏÕ‘á ܲÓ˚y •°ÈüÈ
~ܲ!ê˛ ˛õÓ˚#«˛yÎ˚ ˆÏÎ §Ü˛° ˆ°yܲ COVID-19 xyܲyhs˝ñ ï˛yˆÏòÓ˚ 95% ~Ó˚ ˆ«˛ˆÏe ˆÓ˚yà!ê˛Ó˚ ¢ly=˛ •Î˚ñ !ܲv 5% x¢ly=˛˛
ˆÌˆÏܲ ÎyÎ˚– ˆÎ §Ü˛° ˆ°yܲ COVID-19 Ù%=˛ñ ï˛yˆÏòÓ˚ 99% ~Ó˚ ˛õÓ˚#«˛yÎ˚ COVID-19 ˆlˆÏà!ê˛Ë˛ (–ve) !lô≈y!Ó˚ï˛ •Î˚ !ܲv
1% COVID-19 ˛õ!ç!ê˛Ë˛ (+ve) ôÓ˚y ˛õˆÏí˛¸– ~ܲ!ê˛ !Ó¢y° çl§ÇáƒyÓ˚ ÙˆÏôƒ Ùye 0.1% ˆ°yˆÏܲÓ˚ COVID-19 xyˆÏSÈñ ï˛y
•ˆÏï˛ ~ܲçl ˆ°yܲ ΈÏÌFSÈ˲yˆÏÓ !lÓ≈yã˛l ܲˆÏÓ˚ COVID-19 ˛õÓ˚#«˛y ܲÓ˚ˆÏï˛ ˆòGÎ˚y •ˆÏ°y ~ÓÇ ˆÓ˚yà !Óòƒy!ÓÍ (Pathologise)
!Ó˚ˆÏ˛õyê≈˛ !òˆÏ°l ˆÎ ï˛yÓ˚ COVID-19 +ve– ˙ ˆ°yܲ!ê˛Ó˚ ≤ÃÜ,˛ï˛˛õˆÏ«˛ COVID-19 xyܲyhs˝ •GÎ˚yÓ˚ §Ω˛yÓly !lî≈Î˚ ܲˆÏÓ˚y–

!Ó˲yàÈüÈâ / Each Carries Quesiton 6 Marks / 6x4=24
33. Ùƒy!ê˛∆: ˛õÂô!ï˛ ≤ÈÏÎyˆÏà !l¡¨!°!áï˛ ˜Ó˚!áܲ §Ù#ܲÓ˚îà%ˆÏ°y §Ùyôyl ܲˆÏÓ˚y /
x − y + 2z = 7
3 x + 4 y − 5 z = −5
2 x − y + 3 z = 12
xÌÓy
!lî≈yÎ˚ˆÏܲÓ˚ ôÙ≈ ≤ÈÏÎ˚yà ܲˆÏÓ˚ñ ˆòáyG ˆÎ
(b + c) 2 ba ca
ab (c + a ) 2 cb = 2abc(a + b + c)3
ac bc (a + b)2

34.x2=y x!ôÓ,_ ~ÓÇ y=x+2 myÓ˚y §#ÙyÓÂô xM˛°!ê˛Ó˚ ~ܲ!ê˛ á§í˛¸y !ã˛e xAܲl ܲˆÏÓ˚y– §Ùyܲ°l ˛õÂô!ï˛ ≤ÈÏÎ˚yà ܲˆÏÓ˚ ~•z
§#ÙyÓÂô xM˛°!ê˛Ó˚ ˆ«˛eÊ˛° !lî≈Î˚ ܲˆÏÓ˚y–
x −1 y − 2 z − 3 x−2 y−4 z −5
35. = = ~ÓÇ = = §Ó˚° ˆÓ˚áy ò%!ê˛Ó˚ ˆË˛QÓ˚ xyܲyˆÏÓ˚ §Ù#ܲÓ˚î !°á– xï˛˛õÓ˚
2 3 4 3 4 5
§Ó˚°ˆÏÓ˚áy ò%!ê˛Ó˚ ÙôƒÓï˛#≈ «%˛oï˛Ù ò)Ó˚c !lî≈Î˚ ܲˆÏÓ˚y–

xÌÓy
P(3, 2, 1) !Ó®% ˆÌˆÏܲ 2x–y+z=1 §Ùï˛ˆÏ°Ó˚ í˛z˛õÓ˚ x!AÜ˛ï˛ x!˲°ˆÏ¡∫Ó˚ ˜òâ≈ƒ ~ÓÇ ˛õyò!Ó®%Ó˚ fiÌylyAܲ !lî≈Î˚ ܲˆÏÓ˚y– §Ùï˛ˆÏ°
P !Ó®%Ó˚ ≤Ã!ï˛!Ó¡∫ !Ó®%Ó˚ fiÌylyAܲ G !lî≈Î˚ ܲˆÏÓ˚y–

4R
36. ˆòáyG ˆÎñ R Óƒy§yô≈ !Ó!¢T˛ ~ܲ!ê˛ ˆày°ˆÏܲ xhs˝!°≈!áï˛ §Ó≈Ó,•Í xyÎ˚ï˛ˆÏlÓ˚ °¡∫ Ó,_yܲyÓ˚ ¢AÜ%˛Ó˚ í˛zFã˛ï˛y •ˆÏ°y – xyˆÏÓ˚y
3
8
ˆòáyG ˆÎñ ¢AÜ%˛!ê˛Ó˚ §Ó≈Ó,•Í xyÎ˚ï˛l •ˆÏ°y ˆày°Ü˛!ê˛Ó˚ xyÎ˚ï˛ˆÏlÓ xÇ¢–
27

D x!ï˛!Ó˚=˛ ≤ß¿yÓ°#

Model Question
Class - XII : Mathematics : 80 Marks : 2020-2021
Section-A : Each Question Carries È1 Mark : 1 x 20 = 20
I) Answer the correct option : 1x5
–1
1. If A is a squrare matrix, then A exists if and only if
a) A is singular
b) A is non-singular
c) A, is of order 2 x 2
d) A, is of order 3 x 3
2. y2= 4x at the point (4,4) is
1 1
a) –2 (b) 2 (c) (d) −
2 2

3. If a . b = a × b then, the angle between a and b is

π π π
a) b) c) d) none of these
6 3 2
dx
4. ∫ 4 x − x 2 = equals

x−2 x−2
(a ) sin −1 +c (b) − sin −1 +c
2 2

x−2
(c) cos −1 +c d) none of these
2

5. If the matrix B is both symmetric and skew-symmetric, then
a) B is a diagonal matrix
b) B is an identity matrix
c) B is zero matirx
d) none of these
II) Fill in the blanks : 1x5
x − 4 y − 4 z −1
6. The coordiantes of the point where the line = = cuts the yz– plane is _______ .
2 −3 5
−1 2π
7. The principal value of sin sin is _______.
3

−1 2
8. If sin sin + cos −1 x = 1 , then x =
5
x y z x−5 y−2 z−3
9. The angle between the pair of lines = = and = = is _______ .
2 2 1 4 1 8

10. Let A be a non singular matrix of order 3x3 and |A|=5. Then |adj A|=_____.
III) Answer the short type questions : 1x10
sin 2 x
11. Evaluate / ∫ 1 − cos xdx
12. Evaluate / ∫ e (log sin x + cot x ) dx
x

π
2
3
13. Evalute, by using the property of definite integral / ∫ x cos xdx
π
−
2

14. If matrix A, is of order 3 x 4 and matrix B is of order m x n such that ˛ ATB and BAT are both
defined. Then determine the values of m and n.

5 7 8
15. Without expanding, find the value of the determinant 15 17 18
20 24 26
16. Examine the continuity of the function fÈ at x=0, where f is defined by
sin
, x≠0
f ( x) = x
0 ,
x=0

2 3
d2y dy
17. Find the order and degree of the following differential equation / + =0
dx 2 dx

18. Find the value of ‘p’ for which the vectors a = 3ɵi + 2 ɵj + 9kɵ and b = ɵi − 2 p ɵj + 3kɵ are collinear.
1 3 3
19. Given that the events A and B are such that P(A) = , P(B)= , P(AUB)= – Check whether
2 5 5
the events A and B are independent or not.
20. A fair coin is tossed two times. Find P(F/E) , where
E : At least one head appears
F : Heads on both tosses.
Section : B : Each Question Carries 2 Marks : 2x6=12
−1 cos x
21. Express than tan in the simplest form.
1 = sin x
22. Differentiate logx cosx with respect to x .
x2 y 2
23. Find points on the ellipse + = 1 at which the tangents are parallel to ˛ x- axis.
9 16
24. Find a vector of magnitude 3, which is perpendicular to both the vectors a = 3iɵ + ɵj − 4kɵ and

b = 6ɵi + 5 ɵj − 2kɵ .
25. Find the equation of the plane passing through the intersection of the planes 3x–y+2z–4=0 and
x+y+z–2=0 and the point (2, 2, 1) .

1 1
26. Probability of solving a specific problem independently by A and B are and respectively. If
2 3
both of them try to solve the problem independently, find the probability that the problem is solved.
Section C : Each Question Carries 4 Marks : 4x6=24
27. Prove that,
1 2 1 4
tan − 1 + tan −1 = sin −1
4 9 2 5

3 3 −1
28. Express the matrix A = −2 − 2 1 as the sum of a symmetric and a skew-symmetric matrix.
−4 − 5 2

2 dy
29. If x 1 + y + y 1 + x = 0 , prove that (1 + x ) + 1= 0
dx
* OR
2 2
x 3 d y dy
If y = x log , prove that x 2
= x −y
a + bx dx dx

π
2
x sin x cos x
30. Evaluate / ∫ dx
0
sin 4 x + cos 4 x

* OR
Prove that
1
log(1 + x ) π
∫ 1+ x 2
dx = log 2
0
8

31. Find the particular solution of the diffrential equation
dy
− 3 y cot x = sin 2 x
dx
π
given that y = 2, when x =
2
32. Suppose that the reliability of COVID-19 Antigen test is specified as follows :
Of people having COVID-19, 95% of the test detect disease but 5% got undetected. Of
people free of COVID-19, 99% of the test are judged COVID-19 negative but 1% diagnosed as
showing COVID-19 ˛positive . From a large population of which only 0.1% have COVID-19 , one
person is selected at random, given the COVID-19 ˛test, and the pathologist reports him/her as COVID-
19 positive. Determine the probility that the selected person actually has COVID-19.

Section D : Each Quesiton Carries 6 Marks : 6x4=24
33. Solve the following system of linear equations by matrix method :
x − y + 2z = 7
3 x + 4 y − 5 z = −5
2 x − y + 3 z = 12
OR
Using properites of determinants, show that

(b + c) 2 ba ca
ab (c + a ) 2 cb = 2abc(a + b + c)3
ac bc (a + b)2

34. Draw a rough sketch of the region enclosed by the parabola x2 = y and the line y = x+2 . Also find
the area of the region, using method of integration.
x −1 y − 2 z − 3 x−2 y−4 z −5
35. Write the vector equations of the following lines = = and = =
2 3 4 3 4 5
and hence determine the shortest distance between them,

OR
Find the length and the foot of the perpendicular from the poing P (3, 2, 1) to the plane 2x–y+z =1 .
Also, find the image of the point P in the plane.

36. Show that the height of the right circular cone of maximum volume that can be inscribed in a sphere
4R 8
of radius R is . Also, show that the maximum volume of the cone is of the volume of the
3 27
sphere.

* Additional Question