CBSE Board-XII Maths - 2015 Eng. - Career Point / 17 MATHEMATICS CBSE-XII-2015 EXAMINATION CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : , Email:

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MATHEMATICS CBSE-XII-2015 EXAMINATION

CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

CAREER POINT

MATHEMATICS Paper & Solution

Time : 3 Hrs. Max. Marks : 100 General Instruction : (i) All questions are compulsory. There are 26 questions in all. (ii) This question paper has three sections : Section A, Section B, Section C.

Section A

1. Find the sum of the order and the degree of the following differential equation :

y = x 3

dxdy

+ 2

2

dxyd

Sol. y = x 3

dxdy

+ 2

2

dxyd

order of equation = 2 degree of equation = 1 hence sum of order and degree = 3 2. Find the solution of the following differential equation :

x )y1( 2+ dx + y )x1( 2+ dy = 0 Sol. By data

x )y1( 2+ dx + y )x1( 2+ dy = 0

∴ 2x1

x

+dx +

2y1

y

+dy = 0

Integrating, we get ∫+

dxx1

x2

+ ∫+

dyy1

y2

= 0

2x1+ + 2y1+ = C

3. If A =

θθ−θθ

cossinsincos

, then for any natural number n, find the value of Det (An).

Sol. Q A =

θθ−θθ

cossinsincos

Now A2 = A.A =

θθ−θθ

⋅

θθ−θθ

cossinsincos

cossinsincos

A2 =

θθθθ−θθθ−θ

22

22

sincoscossin2cossin2sincos

A2 =

θθ−θθ

2cos2sin2sin2cos

Similarly we can say that

An =

θθ−θθ

ncosnsinnsinncos

Code : 65/3/A

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Now det (An) =

θθ−θθ

ncosnsinnsinncos

= cos2nθ + sin2nθ = 1

4. Find a vector of magnitude 171 which is perpendicular to both of the vectors →a = k3j2i −+ and

→b = k2ji3 +− .

Sol. Let →c be any vector perpendicular to

→a and

→b then

→c = (

→a ×

→b )

→c = {( k3j2i −+ ) × ( k2ji3 +− )}

→c =

213321

kji

−− = i – 11j – 7k

Now vector of magnitude 171 in the direction of →c is given by

= magnitude. c

= 171491211

)k7j11i(++

−−

= i – 11 j – 7k 5. Find the angle between the lines 2x = 3y = –z and 6x = –y = –4z. Sol. The equation of straight lines are

21

0x − =

31

0y − =10z

−− and

61

0x − =10y

−− =

410z

−

−

The given lines are parallel to the vectors

→a =

21 i +

31 j – k and

→b =

61 i – j –

41 k

→a =

61 (3i + 2j – 6k) and

→b =

121 (2i – 12 j – 3k)

Now angle between these two vectors is given by

cos θ = |b||a|

b.a→→

cos θ = 914443649

)k3j12i2).(k6j2i3(++++−−−+

cos θ = 15749

18246 +− = 0

Hence both lines are perpendicular.

6. In a triangle OAC, if B is the mid-point of side AC and →

OA =→a ,

→OB =

→b , then what is

→OC ?

Sol. If O is origin then position vector of point A, B and C are →a ,

→b and

→c

w.r.t. origin then

→

OA =→a ,

→OB =

→b and

→OC =

→c

and if →c divide AB in the ratio 1 : 1

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CAREER POINT

then →OC =

2OBOA→→

+

→OC =

2ba→→

+

Section B

7. Evaluate : ∫π

+

2/

02

2

xsin31dxxcos

Sol. I = ∫π

+

2/

02

2

xsin31dxxcos dx

I = ∫π

+−

2/

02

2

xsin31xsin1 dx

I = ∫π

++−

2/

02 dx

)xsin31(1.

34

31

I = – 31

∫π 2/

0

dx + 34

∫π

+

2/

02 dx

xsin311

I = –31 .

2π +

34

∫π

+

2/

02

2

xtan41dxxsec

Let 2 tan x = t

dt21dxxsec2 =

I = –6π +

34 .

21

∫∞

+02t1

dt

I = –6π +

32 . (tan–1 t ∞

0)

I = –6π +

32 ×

2π

I = –6π +

3π ⇒ I =

6π

8. Evaluate : ∫π

+

+4/

0x2sin3xcosxsin dx

Sol. I = dx)xcosx(sin13

xcosxsin4/

02∫

π

−−++

I = dx)xcosx(sin4

xcosxsin4/

02∫

π

−−+ …(1)

Let sin x – cos x = t (cos x + sin x) dx = dt at x = 0, t = 0 – 1 = –1

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CAREER POINT

at x = 4π , t = 0

I = ∫−

−

0

12t4

dt

+

−+

=−∫ c

xaxan

a21dx

xa1

thatknowwe

22 l

I = 22

1×

0

1t2t2n

−

−+

l

I = 41

+−

−1212n1n ll

I = 41 – ln

31 ⇒ I =

41 + ln3

9. Let →a = k2j4i ++ ,

→b = k7j2i3 +− and

→c = k4ji2 +− . Find a vector

→d which is perpendicular to both

→a and

→b and

→c .

→d = 27.

Sol. Given that →d is perpendicular to both

→a and

→b then

→d is given by

→d = λ(

→a ×

→b ) …(1)

→d = λ

723241kji

−

→d = λ{i(28 + 4) – j(7 – 6) + k(–2 – 12)}

→d = λ{32i – j – 14k} …(2)

again given that →c .

→d = 27

λ( k4ji2 +− ).( k14ji32 −− ) = 27 λ(64 + 1 – 56) = 27 9λ = 27 ⇒ λ = 3 Hence from equation (2) the resultant

vector →d is

→d = 3(32i – j – 14k)

10. Find the shortest distance between the following lines :

→r = ( k3j2i ++ ) + λ( k4j3i2 ++ )

→r = ( k5j4i2 ++ ) + µ( k8j6i4 ++ )

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

5x – 3y + 4z + 9 = 0 and is parallel to the line 5z5

43y

21x

−−

=−

=− .

Sol. From the given equations, we observe that

first line passes through →a = k3j2i ++

and parallel to →α = 2 k4j3i ++

and second line passes through →b = k5j4i2 ++

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CAREER POINT

and parallel to →β = 4 k8j6i ++

Because both lines are parallel then shortest distance between both lines are given by

SD =||

)ab(→

→→→

α

α×− …(1)

where →b –

→a = k5j4i2 ++ – k3j2i −−

= k2j2i ++ …(2)

→SD =

1694)k4j3i2()k2j2i(

++++×++

=29

ki2 −

=295

OR Equation of plane through the intersection of the given planes 2x + y – z = 3 and 5x – 3y + 4z + 9 = 0 is 2x + y – z – 3 + λ(5x – 3y + 4z + 9) = 0 ...(i) (2 + 5λ)x + (1 – 3λ)y + (–1 + 4λ)z + (–3 + 9λ) = 0

as per data this plane is parallel to the straight line 2

1x − = 4

3y − = 5z5

−−

Then normal of plane is perpendicular to straight line 2

1x − = 4

3y − = 5

5z −

⇒ a1a2 + b1b2 + c1c2 = 0 2(2 + 5λ) + 4(1– 3λ) + 5(– 1+ 4λ) = 0 4 + 10λ + 4 – 12λ – 5 + 20λ = 0

18λ + 3 = 0 ⇒ λ =183

−

λ = –61

Now from equation (1) the equation of plane is 2x + y – z – 3 –61 (5x – 3y + 4z + 9) = 0

Multiply by 6 we get 12x + 6y – 6z – 18 – 5x + 3y – 4z – 9 = 0 7x + 9y – 10z – 27 = 0 11. A man takes a step forward with probability 0.4 and backward with probability 0.6, Find the probability that

at the end of 5 steps, he is one step away from he starting point. OR

Suppose a girl throws a die. If she gets a 1 or 2, she tosses a coin three times and notes the number of 'tails'. If she gets 3, 4, 5 or 6, she tosses a coin once and notes whether a 'head' or 'tail' is obtained. If she obtained exactly one 'tail', what is the probability that she threw 3, 4, 5 or 6 with the die?

Sol. Prob. of step forward p = .4 Prob. of step backward q = .6 So by binomial distribution (p + q)5 = 5C0.p5 + 5C1.p4.q1 + 5C2.p3.q2 + 5C3.p2.q3 + 5C4.p1.q4 + 5C5.q5 Prob. for man so that he is one-step away from starting point will be = 5C2.p3.q2 + 5C3.p2.q3 = 10.(.4)3.(.6)2 + 10.(.4)2.(.6)3

= 10.(.4)2.(.6

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= 10.(.16).(.36) = (1.6) . (.36) = .576

OR

E1

E2

E1 is the event that on throwing a dice 1 or 2 comes

So 31

62)E(P 1 ==

E2 is the event that on throwing a dice 3, 4, 5, 6 comes

So 32

64)E(P 2 ==

If A is event that on throwing coin only one tail. Then for Prob. of exactly one tail if he got 3, 4, 5, or 6 is be

P(E2/A) =)E/A(P).E(P)E/A(P).E(P

)E/A(P).E(P

2211

22

+

P(A/E2) = 21 , P(A/E1) = 8

3

P(A/E2) =

83.

31

21.

32

21.

32

+=

81

31

31

+=

31 ×

1124 =

118

12. Using elementary row operations (transformations), find the inverse of the following matrix :

013321210

OR

If A =

−−

087806760

, B =

021201110

, C =

−32

2, then calculate AC, BC and (A + B) C. also verify that

(A + B) C = AC + BC.

Sol. Given that A =

013321210

we write A = I A

So A100010001

013321210

=

operate R1 ↔ R2 we get

A100001010

013210321

=

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CAREER POINT

operate R3 → R3 – 3R1, we get

A130001010

950210321

−=

−−

operate R1 →R1 – 2R2 and R3 →R3 + 5R2 we get

A135001012

100210101

−

−=

−

operate R1 → R1 + R3 and R2 → R2 – 2R3

A135269

123

100010001

−−−

−=

Hence A–1 =

−−−

−

135269

123

OR

A = ,087806760

−− B = ,

021201110

C =

−32

2

Now AC =

1333 32

2

087806760

××

−

−−

=

=

++++−

+−

30129

0161424012

21120 …(1)

BC =

1333 32

2

021201110

××

−

=

−=

+−+++−

281

042602320

…(2)

Now (A+B) C =

−

+

−−

32

2

021201110

087806760

=

−

−−

32

2

0681005870

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CAREER POINT

=

=

++++−+−

282010

012163001024140

…(3)

Now add (1) and (2)

AC + BC =

=

−+

282010

281

30129

…(4)

= (A + B) C Now from (3) & (4) (A + B) C = AC + BC Hence proved

13. Discuss the continuity an differentiability of the function f(x) = |x| + |x – 1| in the interval (–1, 2). Sol. f(x) = |x| + |x – 1| Defined this function we get

f(x) =

>−≤≤

<+−

1x1x21x01

0x1x2

Now test continuity at x = 0 and 1 at x = 0 f(0) = 1 LHL at x = 0 f(0 – 0) =

0hlim

→– 2(0 – h) + 1 = +1

RHL at x = 0 f(0 + 0) =

0hlim

→1 = 1

Q f(0) = f(0 – 0) = f(0 + 0) So f(x) continuous at x = 0 Now at x = 1 f(1) = 1 LHL at x = 1

0hlim

→f(1 – h) =

0hlim

→1 = 1

RHL at x =1 0h

lim→

f(1 + h) = 0h

lim→

2(1 + h) – 1 = 1

Q f(1) = f(1 – 0) = f(1 + 0) hence f(x) continuous at x = 1 So we can say that f(x) continuous in (– 1, 2)

Differentiability → Q f(x) =

>−≤≤

<+−

1x1x21x01

0x1x2

f(0) = 1 & f(1) = 1 at x = 0

L f ′(0) = 0h

lim→ h

)0(f)h0(f−

−−

= 0h

lim→ h

11)h0(2−

−+−− = –2

Rf ′(0) = 0h

lim→ h

)0(f)h0(f −+

= 0h

lim→ h

11− = 0

Lf ′(0) ≠ Rf ′ (0)

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Hence f(x) not differentiable at x = 0 So we can say that f(x) not differentiable in (–1, 2)

14. If x = a (cos 2t + 2t sin 2t) and y = a (sin 2t – 2t cos 2t), then find 2

2

dxyd .

Sol. Given that x = a(cos 2t + 2t sin 2t) Diff. w.r. to t

dtdx = a(–2sin 2t + 4t cos 2t + 2sin 2t)

dtdx = 4at cos2t ...(1)

Again y = a(sin 2t – 2tcos 2t) D.w.r. to t

dtdy = a[2cos2t – 2cos2t + 4t sin 2t]

dtdy = 4at sin 2t ...(2)

then dxdy =

dt/dxdt/dy Now from (1) & (2)

dxdy =

t2cosat4t2sinat4

dxdy = tan 2t

Again diff. w.r. to x

2

2

dxyd = 2sec22t .

dxdt

Put dxdt from (1)

2

2

dxyd =

t2cos22 ×

t2cosat41

2

2

dxyd =

at21 . sec32t

15. If (ax + b) ey/x = x, then show that

x3

2

2

dxyd =

2

ydxdyx

−

Sol. Given that (ax + b)ey/x = x

⇒ ey/x = bax

x+

…(1)

Diff. w.r. to x

⇒ ey/x =

−

2x

ydxdyx

= 2)bax(axbax

+−+

From (1)

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CAREER POINT

bax

x+

× 2x

ydxdyx −

= 2)bax(b+

xdxdy – y =

baxbx+

…(2)

Again differentiate w.r. to x

x 2

2

dxyd +

dxdy –

dxdy = 2)bax(

abxb).bax(+

−+

x 2

2

dxyd = 2

2

)bax(b+

Now from equ. (2)

x 2

2

dxyd =

2

x

ydxdyx

−

⇒ x32

2

dxyd =

2

ydxdyx

−

Hence proved

16. Evaluate : dx)xsinx(xxcosxxsin

∫ +−

OR

Evaluate : dx)1x)(1x(

x2

3

∫ +−

Sol. I = dx)xsinx(xxcosxxsin

∫ +−

I = dx

xxsin1x

xcosxxsin2

∫

+

−

Let 1 +x

xsin = t

2xxsinxcosx − dx = dt

I = – ∫ tdt

I = – ln |t| + c

I = – lnx

xsin1+ + c

I = ln xsinx

x+

+ c

OR

I = dx)1x)(1x(

x2

3

∫ +−

I = dx)1x)(1x(

11x2

3

∫ +−+−

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CAREER POINT

I = dx)1x)(1x(

)1xx)(1x(2

2

∫ +−++− + dx

)1x)(1x(1

2∫ +−

I = dx1x1x

2

2

∫ ++ + dx

1xx

2∫ ++ dx

)1x)(1x(1

2∫ +−

I = x + dx1x

x2∫ +

+ dx)1x)(1x(

12∫ +−

I = x + I1 + I2 …(1)

where I1 = dx1x

x2∫ +

Let x2 + 1 = t

I1 = 21

∫ dtt1 2x dx = dt

I1 = 21

ln |t| + C1 x dx =21 dt

I1 = 21

ln (x2 + 1) + C1 …(2)

Now I2 = ∫ +−dx

)1x)(1x(1

2 …(3)

Let )1x)(1x(

12 +−

=1x

A−

+1xCBx

2 ++ …(4)

=)1x)(1x(

)1x(Cx)1x(B)1x(A2

2

+−−+−++

Compare numerate A(x2 + 1) + Bx(x – 1) + C(x – 1) = 1

at x = 1 2A = 1 ⇒ 21A =

at x = 0 A – C = 1 ⇒ C = A – 1 ⇒ 21C −=

at x = –1 2A + 2B – 2C = 1

1 + 2B + 1 = 1 ⇒ 21B −=

From (4) )1x)(1x(

12 +−

=21 .

1x1−

–21

1x1x

2 ++

I2 = dx)1x)(1x(

12∫ +−

=21

∫ −1x1 dx –

21 dx

1xx

2∫ +–

21 dx

1x1

2∫ +

I2 = 21

ln(x – 1) – 41

ln(x2 + 1) –21 tan–1x +

21 …(5)

Now from equ (1), (4) and (5)

I = x +21

ln(x2 + 1) +21

ln|x – 1| – 41

ln(x2 + 1)

I = x +41

ln(x2 + 1) +21

ln|x – 1| – 21 tan–1 x + C

17. There are 2 families A and B. There are 4 men, 6 women and 2 children in family A, and 2 men, 2 women

and 4 children in family B. The recommended daily amount of calories is 2400 for men, 1900 for women, 1800 for children and 45 grams of proteins for men, 55 grams for women and 33 grams for children. Represent the above information using matrices. Using matrix multiplication, calculate the total requirement

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CAREER POINT

of calories and proteins for each of the 2 families. What awareness can you create among people about the balanced diet from this question?

Sol. The members of the two families can be represented by the 2 × 3 matrix

F =

422264cwm

A

B

and the recommended daily allowance of calories and proteins for each member can be represented by 3 × 2

R =

2400 45 1900 55 1800 33

m w c

calories proteins

The total requirement of calories and proteins for each of the two families is given by the matrix multiplication

FR =

422264

3355

18001900

452400

=

++++++++13211090720038004800

663301803600114009600A

B

=

3321580057624600A

B

Hence family A required 24600 calories and 576 gm proteins and family B requires 15800 calories and 332 gm proteins.

18. Evaluate : tan

π

+

−

451tan2 1

Sol. Q tan

π

+

−

451tan2 1

Q 2 tan–1 x = tan–12x1

x2−

So tan

π

+

−

451tan2 1 = tan

+−

×−− 1tan

2511

512

tan 11

= tan

+× −− 1tan

2425

52tan 11

= tan

+ −− 1tan

125tan 11

= tan

×−

+−

11251

1125

tan 1

= tan

×−

712

1217tan 1

= tan

−

717tan 1

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CAREER POINT

=7

17

19. Using properties of determinants, prove that

c2cb2ba2a

3

3

3

= 2(a – b)(b – c)(c – a)(a + b + c).

Sol. Let ∆ = 3

3

3

3

3

3

cc2bb2aa2

c2cb2ba2a

=

∆ = 2 3

3

3

cc1bb1aa1

operate R2 → R2 – R3 R3 → R3 – R1 we get

∆ = 2 33

33

3

acac0abab0

aa1

−−−−

∆ = 2 )acac)(ac(ac0)abab)(ab(ab0

aa1

22

22

3

++−−++−−

∆ = 2 (b – a) (c – a) acca10abba10

aa1

22

22

3

++++

operate R2 → R2 – R3

∆ = 2 (b – a) (c – a) acca10

acabcb00aa1

22

22

3

++−+−

∆ = 2 (b – a) (c – a) acca10

)cba)(cb(00aa1

22

3

++++−

∆ = 2 (b – a) (c – a) (b – c) (a + b + c) Hence proved

Section C

20. An urn contains 5 red and 2 black balls. Two balls are randomly drawn, without replacement. Let x represent the number of black balls drawn. What are the possible values of X? Is X a random variable? If yes, find the mean and variance of X.

Sol.

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CAREER POINT

5 Red 2 Black 1 2

urn.

(i) x = 0, 1, 2 (ii) yes, x is a random variable it varies from 0 to 2

422

4220

4220)x(P

210xballx

blackof.No=

(a) P(0) = P(R1) × P

1

2

RR =

75 ×

64 =

4220

(b) P(1) = P(R1) × P

1

2

RB + P(B1) × P

1

2

BR

=75 ×

62 +

72 ×

65 =

4220

(c) P(2) = P(B1) × P

1

2

BB =

72 ×

61 =

422

(iii) Mean ( x ) = x1P1 + x2P2 + x3P3

=

×

42020 +

×

42021 +

×

4222

=4224 =

2112

(iv) Variance =∑=

−2

0x

2)xx)(x(P = P(0).2

42120

− +

P(1).2

42121

− + P(2)

2

42122

−

=4220 × 2)42(

144 +4220 × 2)42(

900 +422 × 2

2

)42()72( = 3)42(

21. Solve the following linear programming problem graphically. Minimise z = 3x + 5y subject to the constraints x + 2y ≥ 10 x + y ≥ 6 3x + y ≥ 8 x, y ≥ 0 Sol. The objective function is z = 3x + 5y The constraints are x + 2y ≥ 10 …(1) x + y ≥ 6 …(2) 3x + y ≥ 8 …(3) x, y ≥ 0 …(4)

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CAREER POINT

A 4

5

B(0, 8)

C(1, 5)D(2, 4)

8

6

4

2

10

10 8 6 4 2 12

3

E(10, 0)

The shaded unbounded region is the feasible region of the given LPP. The vertices of the unbounded feasible. Region are B(0, 8) C(1, 5) and D(2, 4) and E(10, 0) at B(0, 8) z = 3 × 0 + 5 × 8 = 40 at C(1, 5) z = 3 × 1 + 5 × 5 = 28 at D(2, 4) z = 3 × 2 + 5 × 4 = 26 at E(10, 0) z = 3 × 10 + 5 × 0 = 30 The minimum of 40, 28, 26, 30 is 26. We draw the graph of the inequality 3x + 5y < 26 The corresponding equation is 3x + 5y < 26

and this passes through L

526,0 and M

0,

326

The open half plane of 3x + 5y < 26 is shown in the figure. This half plane has no point in common with the feasible region

∴ Minimum value of z = 26 and occurs when x = 2 and y = 4 22. Determine whether the relation R defined on the set R of all real numbers as R = {(a, b) : a, b ∈ R and

a – b + 3 ∈ S, where S is the set of all irrational numbers}, is reflexive, symmetric and transitive. OR

Let A = R × R and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Prove that * is commutative and associative. Find the identity element for * on A. Also write the inverse element of the element (3, – 5) in A.

Sol. Let R be the given relation and defined as R = {(a, b) : a, b, ∈ R and a – b + 3 ∈s} where S is set of irrational number then (i) Reflexive : Let a∈R then a – a = 0 ∈R a – a + 3 = 3 ∈S So aRa therefore R is reflexive (ii) symmetric Let a, b ∈R and Let a = 3 , b = 1 then aRb ⇒ 3 – 1 + 3 ⇒ = 2 3 – 1 ∈S

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CAREER POINT

So aRb Q 1 – 3 + 3 = 1∉S So b R/ a Q aRb ⇒ b R/ a So R is not symmetric (iii) Transitive : Let a = 3 , b = 1 and c = 2 3 Now aRb ⇒ 3 – 1 + 3 = 2 3 – 1 ∈S bRc ⇒ 1 – 2 3 + 3 = 1 – 3 ∈S and aRc ⇒ 3 – 2 3 + 3 = 0∉S ⇒ a R/ c Q aRb & bRc ⇒ a R/ c Therefore relation R is not transitive hence relation and is reflexive but not symmetric and not transitive.

OR Let (a, b), (c, d), (e, f) be any three elements of A (i) Commutative : Given that (a, b) * (c, d) = (a + c, b + d) = (c + a, d + b) = (c, d) * (a, b) Q (a, b) * (c, d) = (c, d) * (a, b) Therefore * is commutative (ii) Associative : By the given definition (c, d) * (e, f) = (c + e, d + f) then (a, b) * {(c, d) * (e, f)} = (a, b) * (c + e, d + f) ⇒ (a, b) * {(c, d) * (e, f)} = (a + c + e, b + d + f) …(1) Now (a, b) * (c, d) = (a + c, b + d) then {(a, b) * (c, d)} * (e, f) = (a + c, b + d) * (e, f) = (a + c + e, b + d + f) …(2) hence by (1) and (2) (a, b) * {(c, d) * (e, f)} = {(a, b) * (c, d)} * (e, f) So * is associative (iii) Let (x, y) be the identity element in A ∴ (a, b) * (x, y) = (a, b) by definition ⇒ (a + x, b + y) = (a, b) ⇒ a + x = a and b + y = b ⇒ x = 0 and y = 0 hence (0, 0) be the identity element of * Now let (p, q) be the inverse element of (3, –5) in A of operative * then by definition of inverse we say that (3, –5) * (p, q) = (0, 0) (3 + p, –5 +q) = (0, 0) ⇒ 3 + p = 0 and –5 + q = 0 p = – 3 and q = 5 so inverse element is (–3, 5) 23. Tangent to the circle x2 + y2 = 4 at any point on it in the first quadrant makes intercepts OA and OB on x and

y axes respectively, O being the centre of the circle. Find the minimum value of (OA + OB). Sol. a a

24. Show that the differential equation (x – y)dxdy = x + 2y is homogeneous and solve it also.

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CAREER POINT

Sol. a a

OR Find the differential equation of the family of curves (x – h)2 + (y – k)2 = r2, where h and k are arbitrary

constants. Sol. a a 25. Find the equation of a plane passing through the point P(6, 5, 9) and parallel to the plane determined by the

points A(3, –1, 2), B(5, 2, 4) and C(–1, –1, 6). Also find the distance of this plane from the point A.

Sol. P(6, 5, 9) = (x1, y1, z1) = (→a )

A(3, –1, 2)

B(5, 2, 4)

C (–1, –1, 6)

Equation of the plane which is passing through the point P is →→→→

= n.an.r

Where, →n =

→AB ×

→AC (It's a normal of the plane)

Q →AB =

→OB –

→OA =

→b –

→a = k2j3i2 ++

→AC =

→OC –

→OA =

→c –

→a = k4i4 +−

∴ →n =

→AB ×

→AC =

404232kji

−

= i (12 – 0) – j (8 + 8) + k (0 + 12)

= 12 i – 16 j +12 k

∴ Equation of plane is →r .(12 i – 16 j + 12 k ) = (6 i + 5 j + 9 k ).(12 i – 16 j + 12 k )

⇒ →r .(12 i – 16 j + 12 k ) = 72 – 80 + 108 = 100

⇒ 12x – 16y + 12z = 100 ⇒ 3x – 4y + 3z – 25 = 0 …(1) Distance between the point A and the plane (1) is

⇒ 222 3)4(3

25)2(3)1(4)3(3

+−+

−+−− =346

26. If the area bounded by the parabola y2 = 16ax and the line y = 4mx is 12a 2

sq. units, then using integration,

find the value of m. Sol. a a

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