MP Board Class 12 Sample Paper 2025 Maths

माध्यमिक शिक्षा मंडल,
मध्य प्रदेश

sample Paper
2025

dsoy vH;kl gsrq uewuk ç'u i=
Sample Question Paper for Practice only
gk;j lsds.Mjh ijh{kk −2025
Higher Secondary Examination −2025
fo"k; − mPp xf.kr
Subject Name −Higher Mathematics
(Hindi & English Versions)
Total Questions Total Printed Pages Time Maximum Marks
23 13 3 Hour 80

funsZ'k :
¼i½ lHkh ç'u vfuok;Z gSaA

¼ii½ ç'u la[;k 1 ls 5 rd ds miç'u çR;sd 1 vad ds gSaA

¼iii½ ç'u la[;k 6 ls 15 rd çR;sd 2 vad ds gSaA

¼iv½ ç'u la[;k 16 ls 19 rd çR;sd 3 vad ds gSaA

¼v½ ç'u la[;k 20 ls 23 rd çR;sd 4 vad ds gSaA

Instructions %
(i) All questions are compulsory.
(ii) Sub-questions of Question numbers 1 to 5 carry 1 mark each.
(iii) Question numbers 6 to 15 carry 2 marks each.
(iv) Question numbers 16 to 19 carry 3 marks each.
(v) Question numbers 20 to 23 carry 4 marks each.

1

(1) lgh fodYi pqudj fyf[k, : 1x6=6
𝑖) ;fn 𝐴 = {1,2,3} gks rks vo;o (1,2) okys rqY;rk laca/kksa dh la[;k gS &
𝑎) 0 𝑏) 1 𝑐) 2 𝑑) 3

𝑖𝑖) 𝑠𝑖𝑛−1 dh eq[; 'kk[kk dk ifjlj gS&
𝜋 𝜋
𝑎) (0, 𝜋) 𝑏) [− , ]
2 2

𝑐) 𝑅 𝑑) (0,2𝜋)

𝑖𝑖𝑖) ;fn 𝐴 = [ 3 1
] gS] rks 𝐴2 dk eku gS &
−1 2
3 −1 10 5
𝑎) [ ] 𝑏) [ ]
1 2 −5 3
6 2 −3 1
𝑐) [ ] 𝑑) [ ]
−2 4 −1 −2

𝑖𝑣) ;fn | 𝑥 2
|=|
6 2
| gks rks 𝑥 cjkcj gS%
18 𝑥 18 6

𝑎) 6 𝑏) ± 6 𝑐) − 6 𝑑) 0
𝑣) ;fn nks lfn'kksa 𝑎⃗ rFkk 𝑏⃗⃗ ds chp dk dks.k 𝜃 gS ,oa 𝑎⃗. 𝑏⃗⃗ = 0, rc 𝜃 cjkcj gS &
𝜋 𝜋
𝑎) 1 𝑏) 𝑐) 𝑑) 𝜋
4 2

𝑣𝑖) og fcUnq ftlls js[kk 𝑟⃗ = −𝑖̂ + 2𝑘̂ + 𝜇(4𝑖̂ + 4𝑗̂ + 4𝑘̂) xqtjrh gS &
𝑎) (−1,2,0) 𝑏) (4,4,4) 𝑐) (0,0,0) 𝑑) (−1,0,2)

Choose and write correct option -

𝑖) If 𝐴 = {1,2,3} then number of equivalence relation with element (1,2) is&
𝑎) 0 𝑏) 1 𝑐) 2 𝑑) 3

2

𝑖𝑖) Range of principal value of 𝑠𝑖𝑛−1 is
𝜋 𝜋
𝑎) (0, 𝜋) 𝑏)[− , ]
2 2

𝑐)𝑅 𝑑)(0,2𝜋)
3 1
𝑖𝑖𝑖) If 𝐴 = [ ] then value of s 𝐴2 is
−1 2
3 −1 10 5
𝑎) [ ] 𝑏) [ ]
1 2 −5 3
6 2 −3 1
𝑐) [ ] 𝑑) [ ]
−2 4 −1 −2

𝑥 2 6 2
𝑖𝑣) If | |=| | then value of 𝑥 is
18 𝑥 18 6

𝑎) 6 𝑏) ±6 𝑐) -6 𝑑) 0

𝑣) If angle between ⃗⃗⃗⃗and
𝑎 𝑏⃗⃗ is 𝜃 and 𝑎⃗. 𝑏⃗⃗ = 0, then 𝜃 is equal to&
𝜋 𝜋
𝑎) 1 𝑏) 𝑐) 𝑑) 𝜋
4 2

𝑣𝑖) The point through which the line⃗⃗⃗𝑟 = −𝑖̂ + 2𝑘̂ + 𝜇(4𝑖̂ + 4𝑗̂ + 4𝑘̂) passes is −
𝑎) (−1,2,0) 𝑏) (4,4,4) 𝑐) (0,0,0) 𝑑) (−1,0,2)

(2) fjDr LFkkuksa dh iwfrZ dhft, & 1x6=6
𝑖) leqPp; 𝐴 ij ifjHkkf"kr lac/a k 𝑅 … … … dgykrk gS ;fn izR;sd
𝑎 ∈ 𝐴 ds fy, (𝑎, 𝑎) ∈ R
𝑖𝑖) ;fn 𝐸 ,oa 𝐹 Lora= ?kVuk, gS 𝑃(𝐹) ≠ 0 rks 𝑃(𝐸 ∣ 𝐹) = … … ..

𝑖𝑖𝑖) ;fn 𝐴 ,d fo"ke lefer vkO;wg gS rks 𝐴 = … … ..
𝑖𝑣) ;fn dksà oxZ vkO;wg vO;qRØe.kh; gS rc |𝐴| =……gSA

𝑣) 𝑙𝑜𝑔|𝑠𝑒𝑐𝑥| dk vodyu xq.kkad………gSA

𝑣𝑖) Qyu 𝑓 ds çkar esa ,d fcUnq 𝑐 ftl ij 𝑓 ′ (𝑐) = 0 Qyu 𝑓 dk… . . … ..fcUnq
dgykrk gSSA
3

Fill in the blanks -
i) Relation 𝑅 defined on set A for every 𝑎 ∈ 𝐴 (𝑎, 𝑎) ∈ 𝑅 then relation 𝑅 is called……
ii) If 𝐸 and 𝐹 are independent events, 𝑃(𝐹) ≠ 0 then 𝑃(𝐸 ∣ 𝐹) = … … ..
iii) If A is a skew symmetric matrix then A = … ……
iv) If A is a square non invertable matrix then |A| is . . . . . . . ..
v) Differential coefficient of 𝑙𝑜𝑔|𝑠𝑒𝑐𝑥| is … …
vi) A point 𝒄 in the domain of a function 𝑓 at which 𝑓 ′ (𝑐) = 0 is called
a … … . point of the function 𝑓.

(3) lR;@vlR; fyf[k, & 1x6=6
i) 𝑓 (𝑥) = 2𝑥 }kjk iznRr Qyu 𝑓: 𝑁 → 𝑁 ,dSdh gSA
ii) fdlh rRled vkO;wg ds fod.kZ ds lHkh vo;o leku gksrs gSA
iii) 𝑓 (𝑥) = |𝑥| }kjk iznRr Qyu larr gksrk gS A
iv) 𝑐𝑜𝑡 −1 (√3) dk eq[; eku 3𝜋
6
gSA

v) vody lehdj.k dy
dx
= 𝑒 𝑥+𝑦 dk O;kid gy 𝑒 𝑥 + 𝑒 −𝑦 = 𝐶 gSSA

vi) pkj dksfV okys fdlh vody lehdj.k ds O;kid gy esa mifLFkr LosPN vpjksa dh
la[;k 4 gksrh gSA
𝐖𝐫𝐢𝐭𝐞 𝐭𝐫𝐮𝐞 𝐚𝐧𝐝 𝐟𝐚𝐥𝐬𝐞 -

i) The function 𝑓: 𝑁 → 𝑁 given by 𝑓 (𝑥) = 2𝑥 is one − one.
ii) All the elements of the diagonal of an identity matrix are equal.
iii) The function given by𝑓 (𝑥) = |𝑥| is continuous.
iv) Principal value of 𝑐𝑜𝑡 −1 (√3) is 3𝜋
6

v)The differential equation dy
dx
= 𝑒 𝑥+𝑦 has a general solution 𝑒 𝑥 + 𝑒 −𝑦 = 𝐶

vi)The number of arbitrary constants present in the general solution of a differential
equation of order four is 4.

4

(4) lgh tksM+h cukb;s & 1x7=7
LrEHk v LrEHk c
i) ∫ 𝑡𝑎𝑛𝑥 𝑑𝑥 𝑥 𝑎2
a)
2
√𝑥 2 + 𝑎2 + 2 𝑙𝑜𝑔|𝑥 + √𝑥 2 + 𝑎2 | + 𝑐
1
ii) ∫ √𝑥 2−𝑎2 𝑑𝑥 b) 𝑙𝑜𝑔|𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥| + 𝑐
1
iii) ∫ √𝑎2−𝑥 2 𝑑𝑥 c) 𝑙𝑜𝑔|𝑠𝑒𝑐𝑥| + 𝑐

1 𝑎+𝑥
iv) ∫ √𝑥 2 + 𝑎2 𝑑𝑥 d) 𝑙𝑜𝑔 | | +c
2𝑎 𝑎−𝑥
𝑥
v) ∫ 𝑠𝑒𝑐𝑥 𝑑𝑥 e) 𝑠𝑖𝑛−1 +c
𝑎
1 𝑥 𝑎2
vi) ∫ 𝑎2−𝑥 2 𝑑𝑥 f) √𝑥 2 − 𝑎2 − 𝑙𝑜𝑔|𝑥 + √𝑥 2 − 𝑎2 | + 𝑐
2 2

vii) ∫ √𝑥 2 − 𝑎2 𝑑𝑥 g) 𝑙𝑜𝑔|𝑥 + √𝑥 2 − 𝑎2 |+c

Match the correct column -
𝐶𝑜𝑙𝑢𝑚𝑛 𝐴 𝐶𝑜𝑙𝑢𝑚𝑛 𝐵
i. ∫ 𝑡𝑎𝑛𝑥 𝑑𝑥 𝑥 𝑎2
a)
2
√𝑥 2 + 𝑎2 + 2 𝑙𝑜𝑔|𝑥 + √𝑥 2 + 𝑎2 | + 𝑐
1
ii. ∫ √𝑥 2−𝑎2 𝑑𝑥 b) 𝑙𝑜𝑔|𝑠𝑒𝑐𝑥 + 𝑡𝑎𝑛𝑥| + 𝑐
1
iii. ∫ √𝑎2−𝑥 2 𝑑𝑥 c) 𝑙𝑜𝑔|𝑠𝑒𝑐𝑥| + 𝑐

1 𝑎+𝑥
iv. ∫ √𝑥 2 + 𝑎2 𝑑𝑥 d) 𝑙𝑜𝑔 | | +c
2𝑎 𝑎−𝑥
𝑥
v. ∫ 𝑠𝑒𝑐𝑥 𝑑𝑥 e) 𝑠𝑖𝑛−1 +c
𝑎
1 𝑥 𝑎2
vi. ∫ 𝑎2−𝑥 2 𝑑𝑥 f) √𝑥 2 − 𝑎2 − 𝑙𝑜𝑔|𝑥 + √𝑥 2 − 𝑎2 | + 𝑐
2 2

vii. ∫ √𝑥 2 − 𝑎2 𝑑𝑥 g) 𝑙𝑜𝑔|𝑥 + √𝑥 2 − 𝑎2 |+c

5

(5) ,d okD;@'kCn esa mRrj fyf[k, & 1x7=7
i)∫ 𝑠𝑖𝑛2 𝑥 𝑑𝑥 dk eku fyf[k,A
𝑑𝑦
ii) vody lehdj.k + (𝑠𝑖𝑛𝑥)𝑦 = 𝑐𝑜𝑠𝑥 dk lekdyu xq.kd fyf[k,A
𝑑𝑥

𝑑2𝑦 3 𝑑𝑦 2 𝑑𝑦
iii) vody lehdj.k ( ) + ( ) + 𝑠𝑖𝑛 ( ) + 1 = 0 dh ?kkr fyf[k,A
𝑑𝑥 2 𝑑𝑥 𝑑𝑥

iv) 𝑓(𝑥) = 𝑥 2 , 𝑥 ∈ 𝑅 ls iznRr Qyu 𝑓 dk fuEure eku fyf[k,A
v) lfn'k 𝑎⃗= 𝑖̂ − 2𝑗̂ dk ekikad fyf[k,A
vi) ;fn 𝑎⃗ = 2𝑖̂ − 𝑗̂ , 𝑏⃗⃗ = 𝑖̂ + 2𝑗̂ rks 𝑎⃗ × 𝑏⃗⃗ dk eku fyf[k,A
vii) ;fn 𝐸 rFkk 𝐹 nks Lora= ?kVuk,a gks rc P(E ∩ F) dk eku D;k gksxk?

𝐖𝐫𝐢𝐭𝐞 𝐚𝐧𝐬𝐰𝐞𝐫 𝐢𝐧 𝐨𝐧𝐞 𝐰𝐨𝐫𝐝/𝐬𝐞𝐧𝐭𝐞𝐧𝐜𝐞 -
i. Write the value of ∫ 𝑠𝑖𝑛2 𝑥 𝑑𝑥.
𝑑𝑦
ii. Write the integrating factor of the differential equation + (𝑠𝑖𝑛𝑥)𝑦 =
𝑑𝑥

𝑐𝑜𝑠𝑥.
𝑑2 𝑦 3 𝑑𝑦 2
iii. Write the degree of the differential equation ( ) +( ) +
𝑑𝑥 2 𝑑𝑥
𝑑𝑦
𝑠𝑖𝑛 ( ) + 1 = 0.
𝑑𝑥

iv. Write the minimum value of function 𝑓 given by 𝑓(𝑥) = 𝑥 2 , 𝑥 ∈ 𝑅
v. Write the modulus of the vector 𝑎⃗= 𝑖̂ − 2𝑗̂.

vi. If 𝑎⃗ = 2𝑖̂ − 𝑗̂ , 𝑏⃗⃗ = 𝑖̂ + 2𝑗̂ then write the value of 𝑎⃗ × 𝑏⃗⃗
vii. If 𝐸 and 𝐹 are two independent events, then what will be value of 𝑃(𝐸 ∩ 𝐹)

(6) fl) dhft, fd çk—r la[;kvk dss leqPp; 𝑁 esa 𝑅 = {(𝑥, 𝑦): 𝑦 = 𝑥 + 5 rFkk 𝑥 < 4} }kjk
iznRr laca/k 𝑅 u rks lefer gS] u rks LorqY; gS vkSj u ladzked gSA 2

6

Prove that the relation given by a𝑅 = {(𝑥, 𝑦): 𝑦 = 𝑥 + 5 aand 𝑥 < 4} in the set of
natural numbers 𝑁 is neither symmetric nor reflexive nor transitive.

vFkok@OR
fl) dhft, fd 𝑓(𝑥) = 𝑥1 }kjk ifjHkkf"kr Qyu 𝑓: 𝑅∗ → 𝑅∗,dSdh rFkk vkPNknd gS
tgka 𝑅∗ lHkh v'kwU; okLrfod la[;kvksa dk leqPp; gSA

1
Prove that the function defined by 𝑓(𝑥) = , 𝑓: 𝑅∗ → 𝑅∗ is one-one and onto
𝑥
where 𝑅∗ is the set of all nonzero real numbers.

(7) fn, x;s Qyu dks ljyre :i esa fyf[k,A 2
𝑐𝑜𝑠𝑥 −3𝜋 𝜋
𝑡𝑎𝑛−1 ( ), <𝑥<
1−𝑠𝑖𝑛𝑥 2 2

Write given function in the simplest form
𝑐𝑜𝑠𝑥 −3𝜋 𝜋
𝑡𝑎𝑛−1 ( ), <𝑥<
1−𝑠𝑖𝑛𝑥 2 2

vFkok@OR
1−𝑐𝑜𝑠𝑥
𝑡𝑎𝑛−1 √ , 0 < 𝑥 < 𝜋 dk eku Kkr dhft, A
1+𝑐𝑜𝑠𝑥

1−𝑐𝑜𝑠𝑥
Find the value of 𝑡𝑎𝑛−1 √ ,0 <𝑥 <𝜋
1+𝑐𝑜𝑠𝑥

(8) fl) dhft, & 2
63 3 5
𝑡𝑎𝑛−1 ( ) = 𝑐𝑜𝑠 −1 ( ) + 𝑠𝑖𝑛−1 ( )+
16 5 13

Prove that −
63 3 5
𝑡𝑎𝑛−1 ( ) = 𝑐𝑜𝑠 −1 ( ) + 𝑠𝑖𝑛−1 ( )+
16 5 13

vFkok@OR

7

fl) dhft, &
√1 + 𝑠𝑖𝑛𝑥 + √1 − 𝑠𝑖𝑛𝑥 𝑥 𝜋
𝑡𝑎𝑛−1 ( )= , 𝑥 ∈ (0, )
√1 + 𝑠𝑖𝑛𝑥 − √1 − 𝑠𝑖𝑛𝑥 2 2
Prove that −
√1 + 𝑠𝑖𝑛𝑥 + √1 − 𝑠𝑖𝑛𝑥 𝑥 𝜋
𝑡𝑎𝑛−1 ( )= , 𝑥 ∈ (0, )
√1 + 𝑠𝑖𝑛𝑥 − √1 − 𝑠𝑖𝑛𝑥 2 2
𝑑2 𝑦
] (9) ;fn 𝑦 = 𝑡𝑎𝑛 𝑥 rks 𝑑𝑥 2 dk eku Kkr dhft,A
−1
2

−1 𝑑2 𝑦
If 𝑦 = 𝑡𝑎𝑛 𝑥 then find the value of
𝑑𝑥 2

vFkok@OR
;fn 𝑦 + 𝑠𝑖𝑛𝑦 = 𝑐𝑜𝑠𝑥 rks 𝑑𝑦
𝑑𝑥
Kkr dhft,A
𝑑𝑦
If 𝑦 + 𝑠𝑖𝑛𝑦 = 𝑐𝑜𝑠𝑥 then find
𝑑𝑥

(10)
𝜋
𝑓(𝑥) = −𝑠𝑖𝑛𝑥, 𝑥 ∈ (0, ) }kjk iznRr Qyu ds LFkkuh; mPpre vkSj LFkkuh; fuEure
2

eku Kkr dhft,A 2
Find the local maximum and local minimum values of the function given by
𝜋
𝑓(𝑥) = −𝑠𝑖𝑛𝑥, 𝑥 ∈ (0, )
2

vFkok@OR
fn[kkb, fd iznRr Qyu 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 4𝑥 tgk a𝑥 ∈ 𝑅, 𝑅 ij o?kZeku Qyu gSA

Show that the given function 𝑓(𝑥) = 𝑥 3 − 3𝑥 2 + 4𝑥 where 𝑥 ∈ 𝑅 is an increasing
function on 𝑅.
(11) ,d ifjorZu'khy ?ku dk fdukjk 3𝑐𝑚/𝑠 dh nj ls c< jgk gSA ?ku dk vk;ru fdl
nj ls c< jgk gS tcfd fdukjk 10 lseh yack gS A 2

The edge of a variable cube is increasing at the rate of 3𝑐𝑚/𝑠. At what rate is the
volume of the cube increasing when the edge is 10 cm long?

8

vFkok@OR
𝑓(𝑥) = 3𝑥 + 4𝑥 − 12𝑥 2 + 12 }kjk iznRr Qyu ds LFkkuh; mPpre vkSj LFkkuh;
4 3

fuEure eku Kkr dhft,A
Find the local maximum and local minimum values of the function given
by 𝑓(𝑥) = 3𝑥 4 + 4𝑥 3 − 12𝑥 2 + 12.

2
(12) vody lehdj.k 𝑑𝑦
𝑑𝑥
=
1+𝑦
1+𝑥 2
dk O;kid gy Kkr dhft,A 2
𝑑𝑦 1+𝑦 2
Find the general solution of the differential equation =
𝑑𝑥 1+𝑥 2

vFkok @OR
vody lehdj.k (𝑒 𝑥 + 𝑒 −𝑥 )𝑑𝑦 − (𝑒 𝑥 − 𝑒 −𝑥 )𝑑𝑥 = 0 dk O;kid gy Kkr dhft,A
Find the general solution of the differential equation
(𝑒 𝑥 + 𝑒 −𝑥 )𝑑𝑦 − (𝑒 𝑥 − 𝑒 −𝑥 )𝑑𝑥 = 0

(13) lfn'k 5𝑖̂ − 𝑗̂ + 2𝑘̂ ds vuqfn'k ,d lfn'k Kkr dhft, ftldk ifjek.k 8 bdkbZ gSA 2
Find a vector in the direction of the vector 5𝑖̂ − 𝑗̂ + 2𝑘̂ which has magnitude
8 units.

vFkok@OR
lfn'k 𝑎⃗ vkSj 𝑏⃗⃗ bl izdkj gS] fd |𝑎⃗| = 3 vkSj |𝑏⃗⃗| = √32 rFkk 𝑎⃗ × 𝑏⃗⃗ ,d ek=d
lfn'k gS] 𝑎⃗ vkSj 𝑏⃗⃗ ds chp dk dks.k Kkr dhft,A
√2
The vector 𝑎⃗ and 𝑏⃗⃗ are such that,|𝑎⃗| = 3 and |𝑏⃗⃗| = and 𝑎⃗ × 𝑏⃗⃗ is a unit vector,
3
find the angle between 𝑎⃗ and 𝑏⃗⃗.

(14) ;fn ,d js[kk tks 𝑥, 𝑦 vkSj 𝑧 v{kksa ds lkFk Øe'k% 90𝑜 , 135𝑜 vkSj 45𝑜 dks.k cukrh gS
rks ml js[kk ds fnd~&dkslkbu Kkr dhft, 2
If a line makes angles 90𝑜 , 135𝑜 and 45𝑜 with axes 𝑥, 𝑦 and 𝑧 respectively then
find the direction cosines of the line
9

vFkok @OR
js[kk ;qXe 𝑟⃗ = 3𝑖̂ + 2𝑗̂ − 4𝑘̂ + 𝜆(𝑖̂ + 2𝑗̂ + 2𝑘̂) vkSj 𝑟⃗ = 5𝑖̂ − 2𝑗̂ + 𝜇(3𝑖̂ + 2𝑗̂ + 6𝑘̂)
ds chp dk dks.k Kkr dhft,A
Find the angle between the pair of lines 𝑟⃗= 3𝑖̂ + 2𝑗̂ − 4𝑘̂ + 𝜆(𝑖̂ + 2𝑗̂ + 2𝑘̂) and
𝑟⃗= 5𝑖̂ − 2𝑗̂ + 𝜇(3𝑖̂ + 2𝑗̂ + 6𝑘̂) .

(15) ;fn 𝑎⃗ = 2𝑖̂ + 𝑗̂ + 3𝑘̂ vkSj 𝑏⃗⃗ = 3𝑖̂ + 5𝑗̂ − 2𝑘̂ gks rks |𝑎⃗ × 𝑏⃗⃗| Kkr dhft,A 2
If 𝑎⃗ = 2𝑖̂ + 𝑗̂ + 3𝑘̂ and 𝑏⃗⃗ = 3𝑖̂ + 5𝑗̂ − 2𝑘̂ then find |𝑎⃗ × 𝑏⃗⃗|.

vFkok@OR
lfn’k fof/k ls ,d f=Hkqt dk {ks=Qy Kkr dhft, ftlds 'kh"kZ fcanq
𝐴(1,1,1) , 𝐵(1,2,3) vkSj 𝐶(2,3,1) gSaA
Find the area of a triangle by vector method whose vertices are 𝐴(1,1,1) , 𝐵(1,2,3)
and 𝐶(2,3,1).
𝑐𝑜𝑠 ∝ −𝑠𝑖𝑛 ∝
(16) ;fn 𝐴 = [ ] rFkk 𝐴 + 𝐴′ = 𝐼 rks ∝ dk eku Kkr dhft,A 3
𝑠𝑖𝑛 ∝ 𝑐𝑜𝑠 ∝
𝑐𝑜𝑠 ∝ −𝑠𝑖𝑛 ∝
If A= [ ] and 𝐴 + 𝐴′ = 𝐼 then find the value of ∝
𝑠𝑖𝑛 ∝ 𝑐𝑜𝑠 ∝

vFkok@OR
;fn 2𝑋 − 𝑌 = [4 6] rFkk 𝑋 − 2𝑌 = [ 2 3
] rks 𝑋 rFkk 𝑌 Kkr dhft,A
8 0 −2 9
4 6 2 3
If 2𝑋 − 𝑌 = [ ] and 𝑋 − 2𝑌 = [ ] then find 𝑋 and 𝑌
8 0 −2 9
(17) lekdyu dk iz;ksx djrs gq, oØ 𝑦 2 = 9𝑥 js[kkvksa 𝑥 = 2, 𝑥 = 4 ,oa 𝑥&v{k ls f?kjs
{ks= dk izFke prqFkk±'k esa {ks=Qy Kkr dhft,A 3
By using integration find the area of the region bounded by the curve 𝑦 2 = 9𝑥
lines 𝑥 = 2, 𝑥 = 4 and 𝑥 axis in the first quadrant.

vFkok@OR

lekdyu dk iz;ksx djrs gq, nh?kZo`Rr 9𝑥 2 + 16𝑦 2 = 144 ls f?kjs {ks= dk {ks=Qy Kkr
dhft,A
By using integration find the area enclosed by the ellipse 9𝑥 2 + 16𝑦 2 = 144.
10

(18) vkys[kh; fof/k }kjk fuEu jSf[kd çksxzkeu leL;k dks gy dhft,∶
fuEu O;ojks/kksa ds varxZr 𝑥 + 3𝑦 ≤ 60, 𝑥 + 𝑦 ≥ 10, 𝑥 ≤ 𝑦, 𝑥 ≥ 0, 𝑦 ≥ 0
𝑧 = 3𝑥 + 9𝑦 dk U;wure eku Kkr dhft,A 3
S
Solve the following linear programming problem by graphical method :
Minimise 𝑧 = 3𝑥 + 9𝑦 subject to the constraints :
𝑥 + 3𝑦 ≤ 60, 𝑥 + 𝑦 ≥ 10, 𝑥 ≤ 𝑦, 𝑥 ≥ 0, 𝑦 ≥ 0
vFkok@OR
vkys[kh; fof/k }kjk fuEu jSf[kd çksxzkeu leL;k dks gy dhft,A
fuEu O;ojks/kksa ds varxZr 𝑥 + 𝑦 ≤ 50, 3𝑥 + 𝑦 ≤ 90, 𝑥 ≥ 0, 𝑦 ≥ 0
𝑧 = 4𝑥 + 𝑦 dk vf/kdrehdj.k dhft,:
Solve the following linear programming problem by graphical method :
Maximise 𝑧 = 4𝑥 + 𝑦 subject to the constraints :
𝑥 + 𝑦 ≤ 50, 3𝑥 + 𝑦 ≤ 90, 𝑥 ≥ 0, 𝑦 ≥ 0

(19) ,d FkSys A esa 3 lQsn ,oa 4 yky xsna s gS vkSj FkSys B eas 5 lQsn ,oa 6 yky xsna s gSA
bu FkSykas esa ls ,d xsan fudkyh tkrh gS vkSj ;g yky ik;h tkrh gS rks FkSys B ls bl
xsan ds fudkyus dh çkf;drk Kkr dhft, A 3
A bag A contains 3 white and 4 red balls and bag B contains 5 white and 6 red
balls. If a ball is taken out from these bags and it is found to be red, then find the
probability of taking out this ball from the bag B.
vFkok @OR
,d ik¡lk rhu ckj mNkyk tkrk gSA çkf;drk Kkr dhft, fd de ls de ,d ckj
fo"ke vad vk,
A dice is thrown three times. Find the probability that an odd number comes up
at least once.
11

(20) 𝐾 ds ml eku dks Kkr dhft, ftlls iznRr Qyu 4
1−𝑐𝑜𝑠𝑘𝑥
, यदि 𝑥 ≠ 0
𝑓(𝑥) = { 𝑥𝑠𝑖𝑛𝑥
1 𝑥 = 0 ij larr gksA
, यदि 𝑥 = 0
2

Find the value of 𝐾 for which the given function
1−𝑐𝑜𝑠𝑘𝑥
, if 𝑥 ≠ 0
𝑓(𝑥) = {1 𝑥𝑠𝑖𝑛𝑥 is continuous at 𝑥 = 0
, if 𝑥 = 0
2

vFkok @OR

;fn 𝑦 = (𝑡𝑎𝑛−1 𝑥)2 rks n'kkZb, fd (𝑥 2 + 1)2 𝑦2 + 2𝑥(𝑥 2 + 1)𝑦1 = 2
If 𝑦 = (𝑡𝑎𝑛−1 𝑥)2 then show that (𝑥 2 + 1)2 𝑦2 + 2𝑥(𝑥 2 + 1)𝑦1 = 2
(21) fn, x;s lehdj.k fudk; dks vkO;wg fof/k ls gy dhft, & 4
𝑥 + 2𝑦 − 3𝑧 = −4
2𝑥 + 3𝑦 + 2𝑧 = 2
3𝑥 − 3𝑦 − 4𝑧 = 11
S Solve given system of equations by matrix method .
𝑥 + 2𝑦 − 3𝑧 = −4
2𝑥 + 3𝑦 + 2𝑧 = 2
3𝑥 − 3𝑦 − 4𝑧 = 11

vFkok @OR
1 2 −3
;fn 𝐴 = [2 3 2 ] rks 𝐴−1 Kkr dhft,A
3 −3 −4
1 2 −3
If 𝐴 = [2 3 2 ] then find 𝐴−1
3 −3 −4
𝜋
(22) fl) dhft, ∫0 𝑙𝑜𝑔𝑠𝑖𝑛𝑥𝑑𝑥 = − 𝜋2 𝑙𝑜𝑔2
2 4
𝜋
𝜋
Prove that ∫02 𝑙𝑜𝑔𝑠𝑖𝑛𝑥𝑑𝑥 = − 𝑙𝑜𝑔2
2

12

vFkok@OR
(𝑥 2 +1)𝑒 𝑥
∫ (𝑥+1)2 𝑑𝑥 dk eku Kkr dhft,A

(𝑥 2 +1)𝑒 𝑥
Find the value of ∫ 𝑑𝑥
(𝑥+1)2

(23) fuEufyf[kr js[kkvksa ds chp dh U;wure~ nwjh Kkr dhft,& 4
𝑟⃗ = (𝑖̂ + 2𝑗̂ + 3𝑘̂ ) +++λ(2𝑖̂ + 3𝑗̂ + 4𝑘̂);
𝑟⃗ = (2𝑖̂ + 4𝑗̂ + 5𝑘̂) +++μ(3𝑖̂ + 4𝑗̂ + 5𝑘̂ )
Find the shortest distance between following lines.
𝑟⃗ = (𝑖̂ + 2𝑗̂ + 3𝑘̂ ) +++λ(2𝑖̂ + 3𝑗̂ + 4𝑘̂);
𝑟⃗ = (2𝑖̂ + 4𝑗̂ + 5𝑘̂) +++μ(3𝑖̂ + 4𝑗̂ + 5𝑘̂ )

vFkok@OR
ljy js[kk dk dkrÊ; ,oa lfn'k :i Kkr dhft, tks fcUnq (−2,4, −5) ls tkrh gS
𝑥+3 𝑦−4 𝑧+8
rFkk js[kk 3
=
5
=
8
ds lekUrj gSA
Find the Cartesian and vector form of the straight line which passes through the
𝑥+3 𝑦−4 𝑧+8
point (−2,4, −5) and parallel to the line = =
3 5 8

13

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