CBSE Board-XII Maths

2 / 17 MATHEMATICS CBSE-XII-2015 EXAMINATION CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com...

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MATHEMATICS

CBSE-XII-2015 EXAMINATION

CAREER POINT

MATHEMATICS Paper & Solution Time : 3 Hrs. 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.

Code : 65/3/A Max. Marks : 100

Section A 1.

Find the sum of the order and the degree of the following differential equation : 3

2  dy  d y y=x   + 2  dx  dx 3

Sol.

2

 dy  d y y=x   + 2  dx  dx order of equation = 2 degree of equation = 1 hence sum of order and degree = 3

2.

Find the solution of the following differential equation :

Sol.

x (1 + y 2 ) dx + y (1 + x 2 ) dy = 0 By data x (1 + y 2 ) dx + y (1 + x 2 ) dy = 0 ∴

x 1+ x

2

dx +

Integrating, we get

y 1 + y2



dy = 0

x 1+ x2

dx +



y 1 + y2

dy = 0

1 + x 2 + 1 + y2 = C 3. Sol.

 cos θ sin θ  n If A =   , then for any natural number n, find the value of Det (A ). sin cos − θ θ    cos θ sin θ  QA=   − sin θ cos θ Now A2 = A.A =  cos θ sin θ   cos θ sin θ  − sin θ cos θ ⋅ − sin θ cos θ     cos 2 θ − sin 2 θ 2 sin θ cos θ  A2 =  2 2   − 2 sin θ cos θ cos θ sin θ  cos 2θ sin 2θ  A2 =   − sin 2θ cos 2θ Similarly we can say that  cos nθ sin nθ  An =   − sin nθ cos nθ CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

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 cos nθ sin nθ  Now det (An) =   − sin nθ cos nθ = cos2nθ + sin2nθ = 1 4.



171 which is perpendicular to both of the vectors a = ˆi + 2ˆj − 3kˆ and

Find a vector of magnitude →

b = 3ˆi − ˆj + 2kˆ .

Sol.













Let c be any vector perpendicular to a and b then c = ( a × b ) →

c = {( ˆi + 2ˆj − 3kˆ ) × ( 3ˆi − ˆj + 2kˆ )} i j k → c = 1 2 − 3 = i – 11j – 7k 3 −1 2 →

Now vector of magnitude 171 in the direction of c is given by = magnitude. cˆ (i − 11j − 7k ) = 171 1 + 121 + 49 = i – 11 j – 7k 5. Sol.

Find the angle between the lines 2x = 3y = –z and 6x = –y = –4z. The equation of straight lines are x −0 y−0 z−0 x−0 y−0 z−0 = = = = and 1 1 1 1 −1 −1 − 2 3 6 4 The given lines are parallel to the vectors → → 1 1 1 1 and a = i+ j–k b= i–j– k 2 3 6 4 → → 1 1 (2i – 12 j – 3k) a = (3i + 2j – 6k) and b= 6 12 Now angle between these two vectors is given by → →

cos θ = cos θ =

a .b | a || b | (3i + 2 j − 6k ).(2i − 12 j − 3k )

9 + 4 + 36 4 + 144 + 9 6 − 24 + 18 =0 cos θ = 49 157 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 CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

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MATHEMATICS

CBSE-XII-2015 EXAMINATION  →

CAREER POINT

 →

OA + OB OC = 2

 →

then





a+ b OC = 2

 →

Section B π/ 2

7.



Evaluate :

0

π/ 2

Sol.

I=

2



0 π/ 2

I=

cos x dx dx 2 x

∫ 1 + 3 sin

0 π/ 2

I=

cos 2 x dx 1 + 3 sin 2 x

1 − sin 2 x dx 1 + 3 sin 2 x  1

4

1

∫ − 3 + 3 . (1 + 3 sin 0

I=–

1 3

π/ 2



4 3

dx +

0

π/ 2

1 π 4 I=– . + 3 2 3

∫ 0

2

  dx x) 

π/ 2

1

∫ 1 + 3 sin

2

0

x

dx

sec 2 x dx 1 + 4 tan 2 x

Let 2 tan x = t 1 sec 2 x dx = dt 2 π 4 1 I=– + . 6 3 2



dt

∫ 1+ t

2

0

π 2 + . (tan–1 t ) ∞0 6 3 π 2 π I=– + × 6 3 2 π π π ⇒ I= I=– + 6 3 6

I=–

π/ 4

8.

Evaluate :

0

π/ 4

Sol.

I=

sin x + cos x

∫ 3 + 1 − (sin x − cos x)

0 π/ 4

I=

 sin x + cos x   dx 3 + sin 2 x 

∫ 

∫ 0

2

sin x + cos x dx 4 − (sin x − cos x ) 2

dx …(1)

Let sin x – cos x = t (cos x + sin x) dx = dt at x = 0, t = 0 – 1 = –1 CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

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CBSE-XII-2015 EXAMINATION

at x =

π , 4

t=0 we know that  1 1 a+x   a 2 − x 2 dx = 2a ln  a − x  + c   

0

dt I= 4 − t2 −1





0

1   2 + t  I=   ln  2 × 2   2 − t  −1   2 − 1    ln 1 − l n   2 + 1   1 1 1 I = – ln ⇒ I = + ln3 4 3 4

I=

9.

1 4









Let a = ˆi + 4ˆj + 2kˆ , b = 3ˆi − 2ˆj + 7 kˆ and c = 2ˆi − ˆj + 4kˆ . 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 ) i j k → d=λ 1 4 2

…(1)

3 −2 7 →

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



d = λ{32i – j – 14k}

…(2)

→ →

again given that c . d = 27 λ( 2ˆi − ˆj + 4kˆ ).( 32ˆi − ˆj − 14kˆ ) = 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 = ( ˆi + 2ˆj + 3kˆ ) + λ( 2ˆi + 3ˆj + 4kˆ )



r = ( 2ˆi + 4ˆj + 5kˆ ) + µ( 4ˆi + 6ˆj + 8kˆ )

Sol.

OR Find the equation of the plane passing through the line of intersection of the planes 2x + y – z = 3 and x −1 y − 3 5 − z 5x – 3y + 4z + 9 = 0 and is parallel to the line = = . 2 4 −5 From the given equations, we observe that →

first line passes through a = ˆi + 2ˆj + 3kˆ →

and parallel to α = 2 ˆi + 3ˆj + 4kˆ →

and second line passes through b = 2ˆi + 4ˆj + 5kˆ CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

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CBSE-XII-2015 EXAMINATION

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and parallel to β = 4 ˆi + 6ˆj + 8kˆ Because both lines are parallel then shortest distance between both lines are given by →

SD =





(b− a) × α

…(1)



|α| →



where b – a = 2ˆi + 4ˆj + 5kˆ – ˆi − 2ˆj − 3kˆ = ˆi + 2ˆj + 2kˆ …(2)  →

SD =

= =

(i + 2 j + 2k ) × (2i + 3 j + 4k ) 4 + 9 + 16

2i − k 29 5 29

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 x −1 y − 3 5 − z as per data this plane is parallel to the straight line = = 2 4 −5 x −1 y − 3 z − 5 Then normal of plane is perpendicular to straight line = = 2 4 5 ⇒ a1a2 + b1b2 + c1c2 = 0 2(2 + 5λ) + 4(1– 3λ) + 5(– 1+ 4λ) = 0 4 + 10λ + 4 – 12λ – 5 + 20λ = 0 3 18λ + 3 = 0 ⇒ λ =− 18 1 λ=– 6 1 Now from equation (1) the equation of plane is 2x + y – z – 3 – (5x – 3y + 4z + 9) = 0 6 Multiply by 6 we get 12x + 6y – 6z – 18 – 5x + 3y – 4z – 9 = 0 7x + 9y – 10z – 27 = 0 11.

Sol.

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

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

MATHEMATICS

CBSE-XII-2015 EXAMINATION

CAREER POINT

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

E1 E2 E1 is the event that on throwing a dice 1 or 2 comes 2 1 So P(E1 ) = = 6 3 E2 is the event that on throwing a dice 3, 4, 5, 6 comes 4 2 So P(E 2 ) = = 6 3 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(E 2 ).P(A / E 2 ) P(E2/A) = P(E1 ).P(A / E1 ) + P(E 2 ).P(A / E 2 ) 1 3 P(A/E2) = , P(A/E1) = 2 8 1 2 1 . 1 24 8 3 2 = 3 = × P(A/E2) = = 2 1 1 3 1 1 3 11 11 + . + . 3 2 3 8 3 8 12.

Sol.

Using elementary row operations (transformations), find the inverse of the following matrix :  0 1 2   1 2 3  3 1 0   OR 6 7 0 1 1  2 0 If A = − 6 0 8  , B = 1 0 2 , C = − 2 , then calculate AC, BC and (A + B) C. also verify that  7 − 8 0 1 2 0  3  (A + B) C = AC + BC. 0 1 2 Given that A = 1 2 3 3 1 0 we write A = I A 0 1 2 1 0 0 So 1 2 3 = 0 1 0 A 3 1 0 0 0 1 operate R1 ↔ R2 we get 1 2 3 0 1 0 0 1 2 = 1 0 0 A     3 1 0 0 0 1 CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

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operate R3 → R3 – 3R1, we get 3  0 1 0  1 2 0 1 2  = 1 0 0 A  0 − 5 − 9 0 − 3 1 operate R1 →R1 – 2R2 and R3 →R3 + 5R2 we get 1 0 − 1 − 2 1 0 0 1 2  =  1 0 0 A    0 0 1   5 − 3 1 operate R1 → R1 + R3 and R2 → R2 – 2R3 1 0 0  3 − 2 1  0 1 0  =  − 9 6 − 2 A     0 0 1  5 − 3 1   3 −2 1  –1 Hence A = − 9 6 − 2  5 − 3 1  OR

6 7 0  A = − 6 0 8 , B =  7 − 8 0 2 C = − 2  3 

0 1 1  1 0 2,   1 2 0

6 7  2  0  Now AC = − 6 0 8  − 2  7 − 8 0  3  3×3 3×1 0 12 21 9 − +     …(1) = − 12 + 0 + 24 = 12   14 + 16 + 0  30 0 1 1   2  BC = 1 0 2 − 2 1 2 0  3  3×3 3×1

 0 − 2 + 3  1  …(2) = 2 + 0 + 6 =  8  2 − 4 + 0 − 2 Now (A+B) C =  0 6 7  0 1 1    2         − 6 0 8  + 1 0 2  − 2   7 − 8 0 1 2 0   3        7 8  2  0  = − 5 0 10 − 2  8 − 6 0   3  CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

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CBSE-XII-2015 EXAMINATION

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− 14 + 24 10   0  = − 10 + 0 + 30 = 20 …(3)  16 + 12 + 0  28 Now add (1) and (2)  9   1  10  AC + BC = 12  +  8  = 20 …(4) 30 − 2 28 = (A + B) C Now from (3) & (4) (A + B) C = AC + BC Hence proved 13. Sol.

Discuss the continuity an differentiability of the function f(x) = |x| + |x – 1| in the interval (–1, 2). f(x) = |x| + |x – 1| Defined this function we get x<0  − 2x + 1  f(x) =  1 0 ≤ x ≤1  2x − 1 x >1  Now test continuity at x = 0 and 1 at x = 0 f(0) = 1 LHL at x = 0 f(0 – 0) = lim – 2(0 – h) + 1 = +1 h →0

RHL at x = 0 f(0 + 0) = lim 1 = 1 h →0

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 lim f(1 – h) = lim 1 = 1 h →0

RHL at x =1

h →0

lim f(1 + h) = lim 2(1 + h) – 1

h →0

h →0

=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) x<0  − 2x + 1  Differentiability → Q f(x) =  1 0 ≤ x ≤1  2x − 1 x >1  f(0) = 1 & f(1) = 1 at x = 0 f ( 0 − h ) − f ( 0) L f ′(0) = lim h →0 −h −2(0 − h ) + 1 − 1 = lim = –2 h →0 −h f (0 + h ) − f (0) Rf ′(0) = lim h →0 h 1−1 = lim =0 h →0 h Lf ′(0) ≠ Rf ′ (0) CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

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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

Sol.

Given that x = a(cos 2t + 2t sin 2t) Diff. w.r. to t dx = a(–2sin 2t + 4t cos 2t + 2sin 2t) dt dx = 4at cos2t ...(1) dt Again y = a(sin 2t – 2tcos 2t) D.w.r. to t dy = a[2cos2t – 2cos2t + 4t sin 2t] dt dy = 4at sin 2t ...(2) dt dy dy / dt then = Now from (1) & (2) dx dx / dt dy 4at sin 2 t = dx 4at cos 2 t dy = tan 2t dx Again diff. w.r. to x dt d2y = 2sec22t . 2 dx dx dt from (1) Put dx 1 d2y 2 × = 2 2 4 at cos 2t cos 2 t dx 2 1 d y . sec32t = 2 2at dx

15.

If (ax + b) ey/x = x, then show that 2  d 2 y   dy  x3  2  =  x − y   dx   dx y/x Given that (ax + b)e = x x ⇒ ey/x = …(1) ax + b Diff. w.r. to x  dy   x dx − y  ax + b − ax y/x ⇒ e =  = 2 (ax + b) 2  x    From (1)

Sol.

d2y . dx 2

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

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dy −y b dx = x2 (ax + b) 2 dy bx x –y= …(2) dx ax + b Again differentiate w.r. to x d 2 y dy dy (ax + b).b − abx x 2 + – = dx dx dx (ax + b) 2

x × ax + b

x

x

d2y b2 = dx 2 (ax + b) 2

Now from equ. (2)

 dy  −y x d 2 y  dx  x 2 = x   dx    

2

d 2 y  dy  =x − y dx 2  dx  Hence proved

2

⇒ x3

16.

Evaluate :



sin x − x cos x dx x ( x + sin x ) OR

Evaluate : Sol.

x

3

∫ (x − 1)(x

2

+ 1)

dx

sin x − x cos x dx x ( x + sin x ) sin x − x cos x I= dx  sin x  x 2 1 +  x   sin x =t Let 1 + x x cos x − sin x dx = dt x2 dt I=– t I = – ln |t| + c sin x I = – ln 1 + +c x I=

∫ ∫



I = ln

x +c x + sin x OR

I= I=

x

3

∫ (x − 1)(x ∫

2

+ 1)

dx

x3 −1 + 1 dx ( x − 1)( x 2 + 1) CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

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MATHEMATICS I=



CBSE-XII-2015 EXAMINATION

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( x − 1)( x 2 + x + 1) 1 dx + dx 2 ( x − 1)( x + 1) ( x − 1)( x 2 + 1)



x2 +1 x 1 dx + dx + dx 2 2 x +1 x +1 ( x − 1)( x 2 + 1) x 1 dx + dx I=x+ 2 x +1 ( x − 1)( x 2 + 1) …(1) I = x + I1 + I2 x dx where I1 = x2 +1 Let x2 + 1 = t 1 1 dt 2x dx = dt I1 = 2 t 1 1 x dx = dt I1 = ln |t| + C1 2 2 1 2 …(2) I1 = ln (x + 1) + C1 2 1 …(3) Now I2 = dx ( x − 1)( x 2 + 1) A 1 Bx + C + 2 Let = …(4) 2 x − 1 ( x − 1)( x + 1) x +1 I=

















A ( x 2 + 1) + B( x − 1) x + C( x − 1) ( x − 1)( x 2 + 1) Compare numerate A(x2 + 1) + Bx(x – 1) + C(x – 1) = 1 1 at x = 1 2A = 1 ⇒ A= 2 =

⇒C=A–1

at x = 0

A–C=1

at x = –1

2A + 2B – 2C = 1

⇒ C=−

1 2 1 1 1 x +1 1 From (4) – = . 2 ( x − 1)( x + 1) 2 x − 1 2 x 2 + 1 1 1 1 1 1 x dx – dx = dx – I2 = 2 2 2 x −1 2 2 ( x − 1)( x + 1) x +1 1 1 1 1 I2 = ln(x – 1) – ln(x2 + 1) – tan–1x + …(5) 2 4 2 2 Now from equ (1), (4) and (5) 1 1 1 I = x + ln(x2 + 1) + ln|x – 1| – ln(x2 + 1) 2 2 4 1 1 1 I = x + ln(x2 + 1) + ln|x – 1| – tan–1 x + C 4 2 2 1 + 2B + 1 = 1



17.



1 2

⇒ B=−



∫x

1 dx +1

2

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

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MATHEMATICS

Sol.

CBSE-XII-2015 EXAMINATION

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? The members of the two families can be represented by the 2 × 3 matrix m w c A

 4 6 2 F=  2 2 4 B and the recommended daily allowance of calories and proteins for each member can be represented by 3 × 2 calories proteins m 2400 45 R = w 1900 55 c 1800 33

The total requirement of calories and proteins for each of the two families is given by the matrix multiplication 2400 45 4 6 2  FR =   1900 55 2 2 4 1800 33 A 9600 + 11400 + 3600 180 + 330 + 66 =  4800 + 3800 + 7200 90 + 110 + 132  B A

24600 576 15800 332  B Hence family A required 24600 calories and 576 gm proteins and family B requires 15800 calories and 332 gm proteins.

=

18. Sol.

  1  π Evaluate : tan 2 tan −1   +   5 4    1  π Q tan 2 tan −1   +   5 4 

Q 2 tan–1 x = tan–1

2x 1− x2

 1 So tan 2 tan −1   + 5 

1    −1 2 × 5 π  + tan −1 1  = tan tan 1 4   1− 25  

2 25   + tan −1 1 = tan tan −1 × 5 24   5   = tan tan −1 + tan −1 1 12   5    −1 12 + 1  = tan tan  5  1 − × 1 12    −1 17 12  ×  = tan tan 12 7   17   = tan tan −1  7  CAREER POINT, CP Tower, Road No.1, IPIA, Kota (Raj.), Ph: 0744-3040000 Website : www.careerpointgroup.com, Email: [email protected]

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MATHEMATICS = 19.

CAREER POINT

17 7

Using properties of determinants, prove that a3 2 a b3 c3

Sol.

CBSE-XII-2015 EXAMINATION

2 b = 2(a – b)(b – c)(c – a)(a + b + c). 2 c

a3 Let ∆ = b 3 c3

2 a 2 a a3 2 b = 2 b b3 2 c

2 c

c3

1 a a3 ∆ = 2 1 b b3 1 c

c3

operate R2 → R2 – R3

R3 → R3 – R1 we get

3

1 a a 3 ∆ = 2 0 b − a b − a3 0 c−a

c3 − a 3

1 a a3 ∆ = 2 0 b − a (b − a )(b 2 + a 2 + ab) 0 c−a

(c − a )(c 2 + a 2 + ac)

1 a a3 ∆ = 2 (b – a) (c – a) 0 1 a 2 + b 2 + ab 0 1 a 2 + c 2 + ac

operate R2 → R2 – R3 1 a a3 ∆ = 2 (b – a) (c – a) 0 0 b 2 − c 2 + ab − ac 0 1

a 2 + c 2 + ac

1 a a3 ∆ = 2 (b – a) (c – a) 0 0 (b − c)(a + b + c) 0 1 a 2 + c 2 + ac

∆ = 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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MATHEMATICS

5 Red 2 Black urn. No.of black ballx = x P( x )

CBSE-XII-2015 EXAMINATION

1

CAREER POINT

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

2

0

1

2

20 42

20 42

2 42

 R  5 4 20 (a) P(0) = P(R1) × P  2  = × =  R 1  7 6 42 B  R  (b) P(1) = P(R1) × P  2  + P(B1) × P  2   R1   B1  5 2 2 5 20 × + × = 7 6 7 6 42 B  2 1 2 (c) P(2) = P(B1) × P  2  = × =  B1  7 6 42 (iii) Mean ( x ) = x1P1 + x2P2 + x3P3 20   20   2   =  0 ×  + 1 ×  +  2 ×  42   42   42   24 12 = = 42 21 =

2

2

(iv) Variance =

12   P( x )( x − x ) 2 = P(0).  0 −  + 42   x =0



2

12   12   P(1). 1 −  + P(2)  2 −  42   42   2 20 144 20 900 2 (72) + + = = × × × 2 2 42 (42) 42 (42) 42 (42) 2 (42) 3 21.

Sol.

2

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 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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MATHEMATICS

CBSE-XII-2015 EXAMINATION

A 10

4

8

(0, 8) B

6

CAREER POINT

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

4

5

2

E(10, 0) 2

4

6

8

10 2

1

3 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  26   26  and this passes through L  0,  and M  ,0   5   3  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.

Sol.

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. 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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MATHEMATICS So Q So Q (iii) Transitive : Let

Now

aRb ⇒

CBSE-XII-2015 EXAMINATION

CAREER POINT

aRb 1 – 3 + 3 = 1∉S b R/ a aRb ⇒ b R/ a So R is not symmetric

a = 3 , b = 1 and

c=2 3

3 – 1 + 3 = 2 3 – 1 ∈S

bRc ⇒ 1 – 2 3 + 3 = 1 –

3 ∈S

3 – 2 3 + 3 = 0∉S ⇒ a R/ c and aRc ⇒ 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. Sol.

24.

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). a a Show that the differential equation (x – y)

dy = x + 2y is homogeneous and solve it also. dx

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MATHEMATICS Sol.

Sol.

CAREER POINT

CBSE-XII-2015 EXAMINATION

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. 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 )



C (–1, –1, 6)

A (3, –1, 2)

B (5, 2, 4)

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

 →

→ →

→ →

r .n = a .n

 →

Where, n = AB × AC (It's a normal of the plane) Q



 →

 →

 →

 →

 →

 →





AB = OB – OA = b – a = 2ˆi + 3ˆj + 2kˆ →



AC = OC – OA = c – a = − 4ˆi + 4kˆ ˆi ˆj kˆ →  →  → n = AB × AC = 2 3 2

−4 0 4 = ˆi (12 – 0) – ˆj (8 + 8) + kˆ (0 + 12) = 12 ˆi – 16 ˆj +12 kˆ →

∴ Equation of plane is r .(12 ˆi – 16 ˆj + 12 kˆ ) = (6 ˆi + 5 ˆj + 9 kˆ ).(12 ˆi – 16 ˆj + 12 kˆ ) →

⇒ r .(12 ˆi – 16 ˆj + 12 kˆ ) = 72 – 80 + 108 = 100 ⇒ 12x – 16y + 12z = 100 ⇒ 3x – 4y + 3z – 25 = 0 …(1) Distance between the point A and the plane (1) is ⇒

3(3) − 4(−1) + 3(2) − 25 32 + (−4) 2 + 32

=

6 34

26.

If the area bounded by the parabola y2 = 16ax and the line y = 4mx is

Sol.

find the value of m. a a

a2 sq. units, then using integration, 12

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