MATHEMATICS Complete Syllabus of Class XII

Model Test Paper-2(for School / Board Exams.)

MATHEMATICS

(Complete Syllabus of Class XII)

MM : 100 Time : 3 Hrs.

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472

CODE

A

GENERAL INSTRUCTIONS :

(i) All questions are compulsory.

(ii) Questions number 1 to 6 are very short answer type questions and carry 1 mark each.

(iii) Questions number 7 to 19 are short answer type questions and carry 4 marks each.

(iv) Questions number 20 to 26 are long answer type questions and carry 6 marks each.

(v) There is no overall choice. However, an internal choice has been provided in four questions of 4 marks,

and two questions of 6 marks each. You have to attempt only one of the given choices in such questions.

(vi) Use log tables if necessary, use of calculator is not allowed.

SECTION-A

1. If f(x) is a real valued function such that f(x) = x3 + x2 + 4x + 4. Show that f(x) is injective function. [1]

2. Find the values of –1

7cos cos

6

⎛ ⎞⎜ ⎟⎝ ⎠

[1]

3. Let

2 4 3

– – 1 0 –1

2 –5 6

A B

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

and

3 4 1

2 –1 0 2

5 5 6

A B

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

. Find the value of matrices A and B. [1]

4. Evaluate: 7–

dx

x x

⌠⎮⌡ [1]

5. Let ˆ ˆ ˆ ˆ ˆ ˆ– , 3 4 5a i j k b i j k �

�

. Find perpendicular component vector of b�

on a�

. [1]

6. Find the integrating factor of the differential equation: [1]

cos sin cos 1dy

x x y x x xdx

(2)

Model Test Paper-2 Subjective Test for Class-XII (Mathematics)

SECTION-B

7. Show that the lines – 3 1 2

2 –3 1

x y z and – 7 7

–3 1 2

x y z are coplanar. Also find the point of

intersection. [4]

OR

Find the equation of plane passing through the intersection of the planes x + y – 2z = 1 and x + 3y – z = 4

and also passing through (1, 0, 1).

8. The probability of solving a question by three students are 1 1 1, ,

2 4 6respectively. Find the probability of question

being solved. [4]

9. Find the equation of tangent to the curve – 7

– 3 – 4

xy

x x at the point where it cuts the x-axis. [4]

10. Evaluate:

2

9

2

sec

sec tan

xdx

x x∫ . [4]

OR

Evaluate :

3

3 2

tan tan

tan 3 tan 2tan 6

x xdx

x x x

∫ .

11. Prove that –1 –1

1 – 1– 1tan – cos

4 21 1–

x xx

x x

. [4]

12. If (x) = 8x – 3 and g(x) = 3x – 8, then find (fog)–1(x). [4]

OR

Discuss the commutativity and associativity of the binary operation on R defined by

*4

aba b for all a, b R.

13. Prove that 0a b c b c a c a b � � �

� � � � � �

. [4]

14. Evaluate:

3

6

1 tan

dx

x

∫ . [4]

15. Find the values of a and b if the function 1

3

2

3 ; 0

1 ; 0x

x

f xax bx

xx

⎧⎪⎪ ⎨ ⎛ ⎞ ⎪ ⎜ ⎟⎜ ⎟⎪ ⎝ ⎠⎩

is continuous at x = 0. [4]

OR

Verify Lagrange’s mean value theorem for the function f(x) = x(x–1)(x–2) in 1

0,2

⎡ ⎤⎢ ⎥⎣ ⎦

.

(3)

Subjective Test for Class-XII (Mathematics) Model Test Paper-2

16. Solve: sin logd

y xy x x xdx

[4]

17. Express the matrix

–1 2 3

4 –1 5

3 –2 4

A

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

as the sum of a symmetric and a skew symmetric matrix. [4]

18. If 1 3

2 1A

⎡ ⎤ ⎢ ⎥⎣ ⎦

satisfy the equation A2 – kA – 5I = 0, then find k and also A–1. [4]

19. Find dy

dx, when

2

–1 1sin

2

xy

[4]

SECTION-C

20. Find the area bounded by curves 22 –y x and y x and the x-axis. [6]

OR

Find the area bounded by x2 = 4y and the line x = 4y – 2.

21. By using elementary row transformation find A–1 where

2 –3 3

2 2 3

3 –2 2

A

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

. [6]

22. The postmaster of a local post office wishes to hire extra helpers during the Deepawali season, because of

a large increase in the volume of mail handling and delivery. Because of the limited office space and the

budgetary conditions, the number of temporary helpers must not exceed 10. According to past experience,

a man can handle 300 letters and 80 packages per day, on the average, and a woman can handle 400 letters

and 50 packets per day. The postmaster believes that the daily volume of extra mail and packages will be

no less than 3400 and 680 respectively. A man receives Rs. 225 a day and a woman receives Rs. 200 a

day. How many men and women helpers should be hired to keep the pay-roll at a minimum? Formulate an

LPP and solve it graphically. [6]

23. Find the image of point (1, 3, 4) in the plane 2x – y + z + 3 = 0. [6]

OR

Show that the lines ˆ ˆ ˆ ˆ ˆ ˆ–3 5 –3 5r i j k i j k �

and ˆ ˆ ˆ ˆ ˆ ˆ– 2 5 – 2 5r i j k i j k �

are

coplanar. Also find the equation of plane containing the lines.

24. 40% students of a college reside in hostel and the remaining reside outside. At the end of the year, 50% of

the hosteliers got A grade while from outside students, only 30% got A grade in the examination. At the end

of the year, a student of the college was chosen at random and was found to get A grade. What is the

probability that the selected student was a hostelier? [6]

25. Obtain the differential equation of all circles of radius r. [6]

26. A jet of enemy is flying along the curve y = x2 + 2 and a soldier is placed at the pont (3, 2). Find the minimum

distance between the soldier and the jet. [6]

� � �

(4)


MATHEMATICS

SECTION-A

1. f(x) = x3 + x2 + 4x + 4

Let x1, x

2 A

f(x1) = f(x

2)

3 2 3 2

1 1 1 2 2 24 4 4 4x x x x x x

3 3 2 2

1 2 1 2 1 2– – 4 – 0x x x x x x

2 2

1 2 1 1 2 2 1 2– 4 0x x x x x x x x⎡ ⎤ ⎣ ⎦

x1 = x

2

So, f(x) is injective function.

2.–1

7cos cos

6

⎛ ⎞⎜ ⎟⎝ ⎠

–1

cos cos6

⎛ ⎞⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠

–1

cos –cos6

⎛ ⎞ ⎜ ⎟⎝ ⎠

–1

cos cos –6

⎡ ⎤⎛ ⎞ ⎜ ⎟⎢ ⎥⎝ ⎠⎣ ⎦

–1

5cos cos , 0,

6

5

6

Model Test Paper-2(for School / Board Exams.)

MM : 100 Time : 3 Hrs.

Regd. Office : Aakash Tower, Plot No.-4, Sec-11, MLU, Dwarka, New Delhi-110075

Ph.: 011-47623456 Fax : 011-47623472

CODE

A

SOLUTIONS

(5)


3.

2 4 3

– – 1 0 –1

2 –5 6

A B

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

...(i)

3 4 1

2 –1 0 2

5 5 6

A B

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

...(ii)

On adding (i) and (ii)

5 8 4

0 0 1

7 0 12

A

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

10 16 8

2 0 0 2

14 0 24

A

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

From (ii),

3 4 1 10 16 8

–1 0 2 – 0 0 2

5 5 6 14 0 24

B

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

–7 –12 –7

–1 0 0

–9 5 –18

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

4.7–

dxI

x x ∫ 7 –6

1–

dx

x x

∫

Put 1 – x–6 = t

6x– 7 dx = dt

1

6

dtI

t ∫

6

1 1 1ln ln 1–

6 6t C C

x

⎛ ⎞ ⎜ ⎟⎝ ⎠

5. Perpendicular component vector b�

on a�

b�

– (Component vector of b�

along a�

)

ˆ ˆ– ·b b a a� �

ˆ ˆ ˆ ˆ ˆ ˆ– –ˆ ˆ ˆ ˆ ˆ ˆ3 4 5 – 3 4 5 · ·

3 3

i j k i J ki j k i j k

⎡ ⎤ ⎢ ⎥⎣ ⎦

3 4 – 5ˆ ˆ ˆ ˆ ˆ ˆ3 4 5 – –

3i j k i j k

⎛ ⎞ ⎜ ⎟⎝ ⎠

2 2 2ˆ ˆ ˆ ˆ ˆ ˆ3 4 5 – –

3 3 3i j k i j k

⎛ ⎞ ⎜ ⎟⎝ ⎠

7 10 17ˆ ˆ ˆ

3 3 3i j k

(6)


6. cos sin cos 1dy

x x y x x xdx

sin cos 1

cos cos

dy x x xy

dx x x x x

sin cos

cos.

x x xdxPdx

x xI F e e

∫∫

1tan x dx

xe

∫

log sec logx x

e

log secsec

x x

e x x

SECTION-B

7. Given lines are

– 3 1 2

2 –3 1

x y z and

– 7 7

–3 1 2

x y z

Any points on above lines are

2 3, – 3 –1, – 2 and –3 7, , 2 – 7

Since given lines are coplanar so they must be intersecting.

2 3 –3 7, – 3 –1 , – 2 2 – 7

2 3 – 4 0 ...(i)

3 1 0 ...(ii)

– 2 5 0 ...(iii)

Solving (i) and (ii)

1

3 4 –12 – 2 2 – 9

1

7 –14 –7

= –1, = 2

Values of and satisfy the 3rd equation so, they are Coplanar.

Point of intersection is (1, 2, –3).

OR

Equation of plane passing through the intersection of x + y – 2z = 1 and x + 3y – z = 4 is

(x + y – 2z – 1) + (x + 3y – z – 4) = 0

1 1 3 –2 – –1– 4 0x y z

(7)


It passes through (1, 0, 1)

1 1 –2 – –1– 4 0

1 – 2 – – 1– 4 0

–4 –2 = 0

2 1

– –4 2

So, equation of plane is

3– – 1 0

2 2 2

x y z

x – y – 3z + 2 = 0

8. Let E1, E

2, E

3 be the events of solving a question by 1st, 2nd and 3rd student.

1 2 3

1 1 1, ,

2 4 6P E P E P E

1 1 11–

2 2P E

2 1 31–

4 4P E

3

1 51–

6 6P E

P(E) = Probability of question being solved

= 1 – Probability of question not being solved

1 2 31– P E E E

1 3 51–

2 4 6

5 111–

16 16

9. – 7

– 3 – 4

xy

x x

It cuts the x-axis at (7, 0)

2 2

– 3 – 4 – – 7 – 3 – 4

– 3 – 4

x x x x xdy

dx x x

2 2

– 3 – 4 – – 7 2 – 7

– 3 – 4

x x x x

x x

dy

dx at (7, 0) is

2 2

7 – 3 7 – 4 – 0

7 – 3 7 – 4

4 3 1

16 9 12

(8)


Equation of tangent at (7, 0) is

– 0 1

– 7 12

y

x

12 y = x – 7

x – 12y – 7 = 0

10.

2

9

2

sec

sec tan

xI dx

x x

∫

Let secx + tanx = t

2sec tan secx x x dx dt

or, sec tan

sec

dtx x dx

x

sec

sec

dt dtt dx x dx

x t ⇒

Now, sec2x – tan2x = 1

(secx + tanx) (secx – tanx) = 1

1

sec – tanx xt

secx + tanx = t

1 1

sec2

x t

t

⎛ ⎞ ⎜ ⎟⎝ ⎠

1

1 1

2

dx dt

t tt

⎛ ⎞ ⎜ ⎟⎝ ⎠

2

2

1dt

t

9

2

1 1 1

2t dt

t tI

t

⎛ ⎞ ⎜ ⎟⎝ ⎠ ∫

2

9

2

11

1

2

tdt

t

∫

9 13– –2 2

1

2t t dt ∫

7 11– –2 2

1 2 2– –

2 7 11t t C

⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦

7 11

– –2 2

1 1– sec tan – sec tan7 11

x x x x C

(9)


OR

3

3 2

tan tan

tan 3 tan 2tan 6

x xI dx

x x x

∫

3 23 2 6

t dt

t t t

∫2

Put tan

sec

x t

x dx dt

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

23 2 3

t dt

t t t

∫

23 2

t dt

t t

∫

Using partial fraction,

22 3 23 2

t A Bt C

t tt t

t = A(t2 + 2) + (Bt + C) (t + 3)

Put t = –3, –3 = 11A 3

–11

A

Equating coefficient of t2 and t

30 –

11A B B A ⇒

9 21 3 1– 3 1–

11 11B C C B ⇒

2

3 2

3 11 11–11 3 2

t

I dtt t

∫

2

3 1 3 2– ln 311 11 2

tt dt

t

∫

2

42

3 1 3 3– ln 311 11 2 2

t

t dtt

∫

2 2

3 3 2 2– ln 311 22 112 2

t dt dtI t

t t

∫ ∫

2 –13 3 2 1– ln 3 ln 2 tan11 22 11 2 2

tt t C

2 –13 3 2 tan– ln tan 3 ln tan 2 tan11 22 11 2

xx x C

(10)


11.–1 –1

1 – 1– 1tan – cos

4 21 1–

x xx

x x

x = cos2

–1

1 cos2 – 1– cos2tan

1 cos2 1– cos2

–1

2 cos – 2 sintan

2 cos 2 sin

–1

cos – sintan

cos sin

–1

1– tantan

1 tan

–1

tan – tan4

tan

1 tan ·tan4

–1

tan tan –4

⎛ ⎞ ⎜ ⎟⎝ ⎠

–4

–1

1– cos

4 2x

12. f(x) = 8x – 3 g(x) = 3x – 8

fog(x) = f(g(x)) = f(3x – 8) = 8 (3x – 8) – 3

= 24x – 64 – 3

= 24x – 67

Let y = 24x – 67

24x = y + 67

167

24x y

–11

6724

fog x x

OR

* ,4

aba b a b R

Commutative law

* * . ,4 4

ba abb a a b a b R

(11)


* is commutative for every a and b

Associative law

Let a, b, c R.

4* * *

4 4 16

abc

ab abca b c c

⎛ ⎞ ⎜ ⎟⎝ ⎠

4* * *

4 4 16

abc

bc abca b c a

⎛ ⎞ ⎜ ⎟⎝ ⎠

a*(b*c) = (a*b)*c

Hence, * is associative over R.

13. a b c b c a c a b � � �

� � � � � �

a b a c b c b a c a c b � � � �

� � � � � � � �

– – –a b a c b c a b a c b c � � � �

� � � � � � � �

= 0

14.

3

6

1 tan

dxI

x

∫

3

6

1 tan –6 3

dxI

x

⎛ ⎞ ⎜ ⎟

⎝ ⎠

∫...(i) –

b b

a a

f x dx f a b x dx

⎡ ⎤⎢ ⎥ ⎢ ⎥⎣ ⎦

∫ ∫∵

3 3

6 6

tan

1 cot tan 1

dx xI dx

x x

∫ ∫ ...(ii)

Adding (i) and (ii)

3

6

1 tan2

1 tan

xI dx

x

∫

3

3

6

6

–3 6 6

dx x

∫

12

I

(12)


15. 1

3

2

3 ; 0

1 ; 0x

x

f xax bx

xx

⎧⎪⎪ ⎨ ⎛ ⎞ ⎪ ⎜ ⎟⎜ ⎟⎪ ⎝ ⎠⎩

f(0) = 3

1

3

20 0

lim 0 lim 1h

h h

ah bhf h

h

⎛ ⎞ ⎜ ⎟⎜ ⎟⎝ ⎠

3

2

1ln 1

0

l im

a h b h

h h

h

e

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

For limit to exist, we have

3

20

lim 0h

ah bh

h

2

0

lim 0h

a bh

h

, which is possible only when a = 0

3

2

1ln 1

0 0

lim 0 lim

ah bh

h h

h h

f h e

⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠

0

ln 1lim ·

b

h

bhb e

bhe

For function to be continuous at x = 0 we have,

0 0f f

eb = 3

b = ln3 and a 0.

OR

f(x) = x(x – 1) (x – 2)

This function is a polynomial in x, it is continuous as well as differentiable for all x R.

f(x) is continuous in 1

0,2

⎡ ⎤⎢ ⎥⎣ ⎦

and differentiable in 1

0,2

⎛ ⎞⎜ ⎟⎝ ⎠

.

Thus f(x) satisfies conditions of Lagrange’s mean value theorem on 1

0,2

⎡ ⎤⎢ ⎥⎣ ⎦

.

Hence there exists at least one real number c in 1

0,2

⎛ ⎞⎜ ⎟⎝ ⎠

such that

1

– 02

1– 0

2

f f

f c

⎛ ⎞⎜ ⎟⎝ ⎠

(13)


–1 –1 – 2 – 2f x x x x x x x

= 3x2 – 6x + 2

2

1 1 1–1 – 2 – 0

2 2 23 – 6 2

1

2

c c

⎛ ⎞⎛ ⎞⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠ 1 –3 3

–2 2 4

12c2 – 24c + 5 = 0

24 576 – 240 24 336 24 18.3

24 24 24c

42.3 5.7or

24 24c

= 1.7 or 0.23

So, c = 0.23 1

0,2

⎡ ⎤ ⎢ ⎥⎣ ⎦

Hence, Lagrange’s mean value theorem is verified.

16. sin logd

y xy x x xdx

sin logdy

y y x x x xdx

2 sin logdy

y x x x xdx

sin log – 2dy

x x x x ydx

2sin log –

dy yx x

dx x

2

sin logdy

y x xdx x

This is linear differential equation

2, sin logP Q x x

x

2

.

dxPdxxI F e e∫∫

2cos 2xe x

(14)


2 2· · sin logy x x x x dx ∫

2 2

1 2

sin logx xdx x xdx

I I

∫ ∫

2 2

1sin –cos 2 cosI x xdx x x x xdx ∫ ∫

2– cos 2 sin – sinx x x x x dx

⎡ ⎤ ⎢ ⎥⎣ ⎦∫2

– cos 2 sin 2cosx x x x x

2

2cosI x x dx ∫

3 31log · –

3 3

x xx dx

x ∫

3 31log –

3 3 3

x xx c

3 3

log –3 9

x xx c

So, solution is

32 2

· – cos 2 sin 2cos log3

xy x x x x x x x

3

–9

xc

17.

–1 2 3

4 –1 5

3 –2 4

A

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

–1 4 3

2 –1 –2

3 5 4

A

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

Let

–1 2 3 –1 4 31

4 –1 5 2 –1 –22 2

3 –2 4 3 5 4

A AP

⎛ ⎞⎡ ⎤ ⎡ ⎤ ⎜ ⎟⎢ ⎥ ⎢ ⎥ ⎜ ⎟⎢ ⎥ ⎢ ⎥

⎜ ⎟⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎝ ⎠

–2 6 61

6 –2 32

6 3 8

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

–1 3 3

3 –1 3/2

3 3/2 4

⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥⎣ ⎦

(15)


–1 2 3 –1 4 3

1 1– 4 –1 5 – 2 –1 –2

2 23 –2 4 3 5 4

Q A A

⎧ ⎫⎡ ⎤ ⎡ ⎤⎪ ⎪⎢ ⎥ ⎢ ⎥ ⎨ ⎬⎢ ⎥ ⎢ ⎥⎪ ⎪⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦⎩ ⎭

0 –2 0 0 –1 01

2 0 7 1 0 7/22

0 –7 0 0 –7/2 0

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

Now, A = P + Q

–1 3 3 0 –1 0

3 –1 –3/2 1 0 7/2

3 3/2 4 0 –7/2 0

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

18.1 3

2 1A

⎡ ⎤ ⎢ ⎥⎣ ⎦

21 3 1 3 7 6

2 1 2 1 4 7A

⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦ ⎣ ⎦

A2 – kA – 5I = 0

7 6 3 5 0

– – 04 7 2 0 5

k k

k k

⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦ ⎣ ⎦

2 – 6 – 3 0 0

4 – 2 2 – 0 0

k k

k k

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥

⎣ ⎦ ⎣ ⎦

2 – k = 0 k = 2

–1

1.A adj A

A

1 31– 6 –5

2 1A

1 –2 1 –3adj.

–3 1 –2 1

T

A⎡ ⎤ ⎡ ⎤

⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

–1

1 –31–

–2 15A

⎡ ⎤ ⎢ ⎥

⎣ ⎦

1 3–5 5

2 1–

5 5

⎡ ⎤⎢ ⎥

⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(16)


19.

2

–1 1sin

2

xy

Put x2 = cos2

Then,

2

–1 –1 –11 cos2 2cossin sin sin cos

2 2y

–1 2 2

sin sin 1– 1–y

2 21–y

2 –2dy d

ydx dx

–

dy dydx dx

Now, x2 = cos2

2 –2sin2dx

xd

4– sin2 – 1–dx x

d x x

–1 2

4

1 –– cos2 1–

dy xy xdx x

–1 2

24 –1

cos

12 1– ·sin

2

dy x x

dx xx

SECTION-C

20. 22 – ,y x y x

2 2

, 02,

– , 0

x xx y y

x x

⎧ ⎨ ⎩

Point of intersection of both curvesO

x y2 2

+ = 2

y x = | |

III

x2 + x2 = 2

x2 = 1

x = ±1

(17)


Area of region (I) is

1 2

2

0 1

2 –xdx x dx ∫ ∫

0

1 22

2 –1

1

2 – sin2 2 2

x x xx

1 10 – –

2 2 42

⎡ ⎤ ⎢ ⎥⎣ ⎦

1 1– sq. units

2 42

⎛ ⎞ ⎜ ⎟⎝ ⎠

Total area = 1 1

2 – 1– 2 sq. units2 4 22

⎡ ⎤ ⎡ ⎤ ⎢ ⎥ ⎢ ⎥⎣ ⎦⎣ ⎦

OR

x2 = 4y is the equation of parabola

–2

x = y4 – 2

x y2

= 4x = 4y – 2 is the equation of line.

Solving both the equation,

x2 = x + 2

x2 – x – 2 = 0

x2 – 2x + x – 2 = 0

(x – 2) (x + 1) = 0

x = 2 or –1

Area

2 2 2

–1 –1

2–

4 4

x xdx dx

∫ ∫

2 2

–1

2 –

4

x xdx

∫

22 3

–1

12 –

4 2 3

x xx

⎡ ⎤ ⎢ ⎥

⎢ ⎥⎣ ⎦

1 8 1 12 4 – – 2 –

4 3 2 3

⎡ ⎤ ⎢ ⎥⎣ ⎦

1 18 – – 3

4 2

⎡ ⎤ ⎢ ⎥⎣ ⎦

1 1 1 9 95 – sq. units

4 2 4 2 8

⎛ ⎞ ⎜ ⎟⎝ ⎠

(18)


21. A = IA

2 –3 3 1 0 0

2 2 3 0 1 0

3 –2 2 0 0 1

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥ ⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

or,

2 – 3 3 1 0 0

0 5 0 – 1 1 0

3 – 2 2 0 0 1

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

{R2 R

2 – R

1}

2 – 3 3 1 0 0

o r , 0 5 0 – 1 1 0

1 1 – 1 – 1 0 1

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

{R3 R

3 – R

1}

1 – 4 4 2 0 – 1

o r , 0 5 0 – 1 1 0

1 1 – 1 – 1 0 1

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

{R1 R

1 – R

3}

1 –4 4 2 0 –1

or, 0 1 0 –1/5 1/5 0

1 1 –1 –1 0 1

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

2 2

1

5R R⎧ ⎫⎨ ⎬⎩ ⎭

1 – 4 4 2 0 – 1

o r 0 1 0 – 1/ 5 1/ 5 0

0 5 – 5 – 3 0 2

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

{R3 R

3 – R

1}

1 – 4 4 2 0 – 1

o r 0 1 0 – 1/5 1/5 0

0 1 – 1 – 3 /5 0 2 /5

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

3 3

1

5R R⎧ ⎫⎨ ⎬⎩ ⎭

1 0 0 – 2 / 5 0 3 / 5

o r 0 1 0 – 1/ 5 1/ 5 0

0 1 – 1 – 3 / 5 0 2 / 5

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

1 1 3

1

4R R R⎧ ⎫ ⎨ ⎬⎩ ⎭

1 0 0 – 2 / 5 0 3 / 5

o r 0 1 0 – 1/ 5 1/ 5 0

0 0 – 1 – 2 / 5 – 1/ 5 2 / 5

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

{R3 R

3 – R

2}

1 0 0 – 2 / 5 0 3 / 5

o r 0 1 0 – 1/ 5 1 / 5 0

0 0 1 2 / 5 1 / 5 – 2 / 5

A

⎡ ⎤ ⎡ ⎤⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎢ ⎥ ⎢ ⎥⎣ ⎦ ⎣ ⎦

{R3 (–1)R

3 }

–1

2 3– 05 5

1 1or A – 0

5 5

2 1 2–

5 5 5

⎡ ⎤⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎢ ⎥⎣ ⎦

(19)


22.

(0, 172

(

(0, 10)

B(6, 4)

A(0, 685

(

3 + 4 = 34

xy

x y + = 10

8 + 5 = 68x y

X

Y

Let the number of men and women hired be x and y respectively.

To minimize, Z = 225 x + 200 y

10x y

300 400 3400 3 4 34x y x y ⇒

80 50 680 8 5 68x y x y ⇒

, 0x y

68 68At 0, , 200 2720

5 5Z

⎛ ⎞ ⎜ ⎟⎝ ⎠

At 6, 4 , 1350 800 2150Z

34 34At , 0 , 225 45 34 2550

5 5Z

⎛ ⎞ ⎜ ⎟⎝ ⎠

Minimum Z = Rs. 2150 at (6, 4)

23. Let P be a given point. If Q be a point such that the lines joining these points is bisected by the plane.

2 – 3 0x y z

Equation of line through P and perpendicular to the plane is

–1 – 3 – 4

2 –1 1

x y z

M

Q

P(1, 3, 4)

Coordinate of 2 1, – 3, 4Q r r r

Point M is the midpoint of PQ.

So, 1, – 3, 42 2

r rM r⎛ ⎞ ⎜ ⎟⎝ ⎠

(20)


2 1 – – 3 4 3 02 2

r rr

⎛ ⎞ ⎜ ⎟⎝ ⎠

r = –2

Q(–3, 5, 2)

OR

1 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ–3 5 , –3 5a i j k b i j k

�

2 2ˆ ˆ ˆ ˆ ˆ ˆ ˆ– 2 5 , – 2 5a i j k b i j k

�

2 1 1 2

ˆ ˆ ˆ

ˆ ˆ– · 2 · –3 1 5

–1 2 5

i j k

a a b b i j � �

� �

ˆ ˆ ˆ ˆ ˆ2 · –5 10 – 5i j i j k

= –10 + 10 = 0

So, lines are coplanar

Perpendicular vector 1 2

ˆ ˆ ˆˆ –5 10 – 5n b b i j k � �

ˆ ˆ ˆ–5 – 2i j k

Equation of plane is

ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ· – 2 – 2 · –3 5 0r i j k i j k i j k �

or x – 2y + z = 0

24. Let E1 : Student resides in the hostel

E2 : Student resides outside the hostel

1 2

40 60,

100 100P E P E

A: Getting a grade in the examination

1 2

50 30,

100 100

A AP P

E E

⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟

⎝ ⎠ ⎝ ⎠

1

11

1 2

1 2

AP P E

EEP

A A AP P E P P E

E E

⎛ ⎞⎜ ⎟

⎛ ⎞ ⎝ ⎠⎜ ⎟⎛ ⎞ ⎛ ⎞⎝ ⎠ ⎜ ⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

50 40

10100 100

50 40 30 60 19

100 100 100 100

(21)


25. Let the centre of circle be (a, b)

Equation of circle is

2 2 2– –x a y b r ...(i)

2 – 2 – 0dy

x a y bdx

...(ii)

22

21 – 0

d y dyy b

dxdx

⎛ ⎞ ⎜ ⎟⎝ ⎠

...(iii)

2

1

2

1– –

yy b

y

From (ii), 21 1

2

1

–

y y

x ay

Putting these value in (i)

2 22 2 2

1 1 12

2 2

2 2

1 1y y y

ry y

or,

3 22 22

21

dy d yr

dx dx

⎡ ⎤ ⎛ ⎞⎛ ⎞ ⎢ ⎥ ⎜ ⎟⎜ ⎟ ⎜ ⎟⎝ ⎠⎢ ⎥ ⎝ ⎠⎣ ⎦

26. Let P(x, y) be the position of jet and the soldier is placed at A(3, 2)

2 2– 3 – 2AP x y ...(i)

As y = x2 + 2 y – 2 = x2 ...(ii)

Let AP2 = (x – 3)2 + x4 = z

32 – 3 4

dzx x

dx

2

2

212 2

d zx

dx

0 1dz

xdx

⇒

2

2

d z

dx at x = 1 = 14 > 0

z is minimum when x = 1

When x = 1, y = 3

Distance = 2 23 –1 1

5

� � �

MATHEMATICS Complete Syllabus of Class XII

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Transcript of MATHEMATICS Complete Syllabus of Class XII