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MATRIX: Definition: A matrix is defined as an ordered rectangular array of numbers. Elementary Transformation 3. R i → R i + kR j means multiply each element of j th row by k and add it to the corresponding elements of i th row. 4. In applying one or more row operations while finding A -1 by elementary row operations, we obtain all zeros in one or more, then A -1 does not exist. NOTE If A is symm. As well as skew- symm., then A is a null matrix.( if A = A T then A is Symm. And if A = - A T then A is skew- symm.)

Assignments for class XII

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Page 1: Assignments for class XII

MATRIX:

Definition: A matrix is defined as an ordered rectangular array of numbers.

Elementary Transformation

3. Ri → Ri + kRj means multiply each element of jth

row by k and add it to the corresponding elements of ith

row.

4. In applying one or more row operations while finding A-1 by elementary row operations, we obtain all zeros in one or more, then A-1 does not exist.

NOTE If A is symm. As well as skew-symm., then A is a null matrix.( if A = AT then A is Symm. And if A = - AT then A is skew- symm.)

A =[1 0 00 2 00 0 4] is symmetric and B[ 0 −5 8

5 0 12−8 −12 0 ] is skew-

symmetric.

NOTE: (i) If A and B are symmetric matrices, then BA-2AB is neither symm. nor skew-symm.

(ii) If A is symm. matrix then BTAB is symm. (iii) If A and B are symmetric matrices of same order, then AB is symm. iff AB=BA (iv) Zero matrix is both symm. and skew-

Page 2: Assignments for class XII

symm. (v) Sum of two skew-symm. matrices is always skew-symm (vi) If A is a symm., then A3 is a symm. and if A is skew-symm., then A2 is a symm.

The Determinant of a Matrix

DEFINITION: Determinants play an important role in finding the inverse of a matrix and also in solving systems of linear equations. In the following we assume we have a square matrix (m = n). The determinant of a matrix A will be denoted by det(A) or |A|. Firstly the determinant of a 2×2 and 3×3 matrix will be introduced, then the n×n case will be shown.

Consistent and Inconsistent SolutionsConsistent system : A system of equation is said to be consistent if its solution ( one or more ) exists.Inconsistent system : A system of equation is said to be inconsistent if its solution does not exist. Working rule to check consistency : Case I When A0 System is consistence and has uniquesolution. Case II When A=0 .Find Adj(A) and then find Adj(A) .B. If Adj(A) .B 0 then system is inconsistence .Case III If Adj(A).B=0 Then it may have infinite solutions then it is consistence or have no solution then it is inconsistence.Properties of Determinants:Property 1. If each element of a row (column)ofdeterminant is zero , then value of determinant is zero.Property 2. Value of a determinant is not changed bychanging the rows into columns and columns into rows.

Property 3. If two adjacent rows (columns)of a determinant are interchanged , then the sign of the determinant is changed but its numerical value is unchanged.

Page 3: Assignments for class XII

Property 4. If two rows (columns) are identical, then the value of the determinant is zero.Property 5. If every element of a row (column) is multiplied by some constant k,the value of the determinant is multiplied by k.Property 6 .If each element in any row (column) consist of two terms , then the determinant can be expressed as the sum of the determinants of same order. Property 7 . The value of a determinant remain unchanged if to each element of a row (column) be add ( or subtracted) equimultiplies of the corresponding elements of one or more rows (columns) of the determinant.Property 8. The value of the determinant of a diagonal matrix is equal to the product of the diagonal elements.Property 9. The value of the determinant of a skew-symmetric matrix of odd order is always zero.Property 10. The determinant of a symmetric matrix of even order is always a perfect square.ASSIGNMENT(matrices)

Qoestion.1 Using matrices, solve the following system of equations

(i) x+2y+z = 1 , 2x – y+z = 5 , 3x+y – z = 0.

[Hint use AX = B ⇨ X = A-1 B, |A|=15≠0 means A is invertible. Adj(A) =

[0 3 35 −4 15 5 −5] ,A-1 = adj(A)

¿ A∨¿¿Ans. x =1, y=-1, z=2]

(ii) 2x+y – 3z = 13, x + y – z = 6 , 2x – y+4z = -12.

[ Ans. |A| = 9, adj(A) = [ 3 −1 2−6 14 −1−3 4 1 ] , x=1, y=2, z=-3.]

(iii) 2x+y+z = 1 , x – 2y – z = 3/2 , 3y – 5z = 9.

[Hint |A| = 34, adj (A) = [13 8 15 −10 33 −6 −5], x=1, y= 1/2., z=-3/2.]

Page 4: Assignments for class XII

Question.2 Use the product [−4 4 4−7 1 35 −3 −1] [1 −1 1

1 −2 −22 1 3 ] to solve the

equations x – y+z = 4, x – 2y – 2z = 9, 2x+y+3z = 1.

[Hint take product of above two matrices, we get identity matrix, then use AB=BA = I means B is the inverse of A

Or A is the inverse of B.

⇨ [−4 4 4−7 1 35 −3 −1] [1 −1 1

1 −2 −22 1 3 ] = 8I3 ,

according to above equation let A

let B (1/8) B is the inverse of A. Ans. x=3, y=-2, z=-1.]

Question.3 Solve the following system of homogenous equations: 2x+3y – z = 0, x – y – 2z = 0, 3x+y+3z = 0.

Solution: system of homogenous equations can be written as AX = O

, A = [2 3 −11 −1 −23 1 3 ] ,|A| = -33

So, the system has only the trivial solution given by x=y=z=0. If |A| = 0 then system has non-trivial solution.]

Question.4 Show that system of equations x+y – z = 0, x – 2y+z = 0, 3x+6y – 5z = 0 has non-trivial solution. Find sol.Answer: |A| = 0, it has infinitely many solutions ∴ let z = k a arbitrary x+y = z = k , x – 2y = -z = -k , 3y = 2k i.e, y = 2k/3 ⇨ x = k/3 from first equation by putting the values of x, y & z in third equation, we get 0 which is true. The required solution is z = k, y = 2k/3, x = k/3 where k is arbitrary.

Question.5 Show that system of equations3x+2y +7 z = 0, 4x – 3y - 2z = 0, 5x+9y +23z = 0 has non-trivial solution. Find the solution. [Hint x = -k, y = -2k, z = k]

Page 5: Assignments for class XII

Question.6 The system of equations 2x+3y = 7 , 14x+21y = 49 has (a) only one solution (b) finitely many solution (c) no solution (d) infinitely many solution . [give reason]

Question.6 Find the inverse (using elementary transformations) of

following matrices: (i) A = [0 1 21 2 33 1 1]

[Hint: A-1 = [1/2 −1 /2 1/2−4 3 −15/2 −3 /2 1/2] ,R1↔R2,R3→R3 –

3R1,R3→R3+5R2,R1→R1 – 2R2,R2→R2 – R3, R1→R1+(1/2) R3,R3→(1/2)R3]

(ii) A = [3 4 20 2 −31 −2 6 ] [Hint: A-1 = [ 3 −14 −8

−3/2 8 9/2−1 5 3 ] ,R1↔R3,R3→R3 –

3R1,R1→R1+R2,R2→1/2R2,R3→R3 – 10R2, R3→-R3,R1→R1 – 3R3,R2→R2+3/2R3]

(iii) A = [ 1 3 −2−3 0 −52 5 0 ] [Hint: A-1 = [ 1 −2/5 −3/5

−2/5 4/5 11 /25−3/5 1/25 9/25 ]

,R2→R2+R1+R3,R3→R3 – 2R1,R1→R1+3R3,R3→8R3,R3→R3+R2

R3→1/25R3,R1→R1 – 10R3,R2→R2+7R3,R2→1/8R2]

Question. If A is singular matrix then underwhatcondition set of equations AX = B may beconsistent. [answer if (adjA)B = O ,then eqns. Will have infinitly many sols. Hence consistent.]

Question. If A is a square matrix of order 3 such that |adjA| = 289, find |A|. [ |A| = ±17 ∵ |adjA| = |A|n-1.]

ASSIGNMENT ( WITH HINTS)(determinant)

Question: (i) Let ∆ = Ax x ² 1By y ² 1Cz z ² 1

and ∆₁ = A B Cx y zzy zx xy

, then ∆ - ∆₁ = 0

Page 6: Assignments for class XII

[Hint ∆₁ = xyzxyz

Ax By Czx ² y ² z ²1 1 1

]

(ii) If f(x) = 0 x−a x−b

x+a 0 x−cx+b x+c 0

, then which is correctf(a)=0 , f(b)=0,

f(0)=0 and f(1)=0 [ Hint f(0)=0 det.(skewsymm.matrix)=0].

**(iii) Let f(t) = cost t 1

2 sint t 2 tsint t t

, then limt →0¿

f ( t)t ² is equal to 0,1,2,3.

[Hint 0, f ( t)t ² =

cost t 12 sintt

1 2

sintt

1 1

→ 1 0 12 1 21 1 1

as t→∞ ].

(iv) There are two values of a which makes determinant ∆ = 1 −2 52 a −10 4 2a

= 86, then sum of these numbers is 4,5,-4,9. [Hint a=-4, operate R2

– 2R1]

Question.1 Prove that the points P (a, b+c), Q(b, c+a), R(c, a+b) are collinear.

Answer : If P,Q and R are collinear then a b+c 1b c+a 1C a+b 1

= 0

By applying C2 → C2+C1 a a+b+c 1b a+b+c 1c a+b+c 1

= (a+b+c) a 1 1b 1 1c 1 1

=0 ( ∵ C2, C3 are identical)

Question.2 Find the value of k if the area of the triangle with vertices (-2,0),(0,4) and (0,k) is 4 square units.

Answer: Area of ∆ = ½ −2 0 10 4 10 k 1

= 4

⇨ the absolute value of ½ (-2)(4 – k) = 4 ⇨ the absolute value of (k – 4) = 4 ⇨ |k – 4| = 4 ⇨ k – 4 = 4, -4 ⇨ k = 8, 0.Question.3 Without expanding, show that

Page 7: Assignments for class XII

(i) b−c c−a a−bc−a a−b b−ca−b b−c c−a

= 0

Operating C1 → C1+C2+C3, we get 0 c−a a−b0 a−b b−c0 b−c c−a

= 0.

(ii) 0 a −b

−a 0 −cb c o

= 0 Taking out (-1) from C1,C2 and C3, we get,

∆ = (-1)(-1)(-1) 0 −a ba 0 c

−b −c o = -1

0 a −b−a 0 −cb c o

=

- ∆ (by interchanging rows and columns) 2∆ = 0 ⇨ ∆ = 0

(iii) b ² c ² bc b+cc ²a ² ca c+aa ²b ² ab a+b

= 0 ⇨ abcabc

b ² c ² bc b+cc ²a ² ca c+aa ²b ² ab a+b

= 1abc

ab ² c ² abc ab+acbc ²a ² bca bc+baca ²b ² cab ca+cb

abc .abcabc

bc 1 ab+acca 1 bc+baab 1 ca+cb

= abc

bc 1 ab+bc+acca 1 ac+bc+baab 1 ab+ca+cb

( Operating C3 → C3+C1)

abc(ab+bc+ac) x 0 = 0 ( two cols. Are identical)

(iv) 1 a a ²−bc1 b b ²−ca1 c c ²−ab

= 0 ⇨ 1 a a ²1 b b ²1 c c ²

- 1 a bc1 b ca1 c ab

= 1 a a ²1 b b ²1 c c ²

-

1abc

a a ² abcb b ² bcac c 2 cab

1 a a ²1 b b ²1 c c ²

- a a ² 1b b ² 1c c 2 1

= 0 (PassC3overthefirsttwocolumns.) (v) 1 a a2

1 b b21 c c2

= 1 bc b+c1 ca c+a1 ab a+b

Page 8: Assignments for class XII

R.H.S. 1abc a abc ab+ac

b abc bc+bac abc ca+cb

= 1 a ab+ac1 b bc+ba1 c ca+cb

( applying C1 ↔ C2) = - 1 a −bc

1 b −ca1 c −ab

(apply C3 → C3 – (ab+bc+ca))C1) = 1

abc a a ² abcb b ² abcc c 2 abc

= 1 a a21 b b21 c c2

(apply C2 ↔ C3 and C1 ↔ C2)Q. If a,b,c are +ve and are the pth,qth and rth terms resp. Of a G.P.,show without expanding that

** (vi) log a p 1logb q 1log c r 1

= 0 (put a=xyp-1 ,b=xyq-1 ,c=xyr-1 , apply

C1 →C1-logx.C3 ,C1→C1+C3) (vii) 1 a bc1 b ca1 c ab

= 1 a a ²1 b b ²1 c c ²

(same method as given below) (viii) a a ² bc

b b ² cac c 2 ab

= 1 a ² a ³1 b ² b ³1 c ² c ³

( Multiply by abc as R1 with a,R2 by b and R3 by c then divide with abc ) Find the values of: (ix)

1 ab1a+ 1b

1 bc1b+

1c

1 ca1a+ 1c

(Operate C2 → 1abc C2.

and value is 0

(x) sinα cosα cos (α+δ)sinβ cosβ cos (β+δ)sinγ cosγ cos (γ+δ)

(Operating C3 → C3 – cosδ.C2+ sin δ .C1 and value is 0) Q. Prove that :

(a) 1+a ²−b ² 2ab −2b

2ab 1−a ²+b ² 2a2b −2a 1−a ²−b ²

= (1+a2+b2)3 ( Apply C1 → (C1 - bC3) and C2 → (C2+aC3)

Page 9: Assignments for class XII

(b) cosαcosβ cosαsinβ −sinα−sinβ cosβ 0sinαcosβ sinαsinβ cosα

= 1 (Apply R3 → sin𝛂R3 + cos𝛂R1)

(C) ( y+z ) ² xy zxxy (x+z ) ² yzxz zy (x+ y ) ²

= xyz(x+y+z)3 (Apply R1 → x R1, R2

→ y R2, R3 → z R3 and take x,y,z common from C1,C2,C3 resp.)

(d (b+c )² a2 bc

¿ ² ¿ca¿a+b¿ ² ¿c 2¿ab¿ =(a-b)(b-c)(c-a)(a+b+c)

(a2+b2+c2 )( Apply C1 → (C1+C2 – 2C3)

(e) a+b+c −c −b

−c a+b+c −a−b −a a+b+c

= 2(a+b)(b+c)(c+a) (Apply C1

→(C1+C3 )and C2 →(C2+C3))

(f) a b−c c+b

a+c b c−aa−b a+b c

= (a+b+c)(a2+b2+c2)

(g) b+c c+a a+bq+r r+ p p+qy+z z+x x+ y

= 2 a b cp q rx y z

(apply C1→C1-C2-C3, C2→C2-

C1,C3→C3-C1, C2↔C3)

(h) 1+a 1 1

1 1+b 11 1 1+c

= abc (1a +

1b +

1c+1(ab+bc+ca+abc).

(Hint taking a,b,c common from each row , apply R1→R1+R2+R3 then expand along first row).

(i) 1+a ²−b ² 2ab −2b

2ab 1−a ²+b ² 2a2b −2a 1−a ²−b ²

= ¿)3

Apply C1→C1-b C3, C2→C2+a C3, we get 1+a ²+b ² 0 −2b

0 1+a ²+b ² 2ab (1+a2+b2) −a(1+a2+b2) 1−a ²−b ²

Page 10: Assignments for class XII

= (1+a ²+b ² ¿ ² 1 0 −2b0 1 2ab −a 1−a ²−b ²

expand along C1, We get (

1+a ²+b ² ¿ ³ .

**(h) Evaluate

(X1 ) (X2 ) (X3 )(Y1 ) (Y2 ) (Y3 )(Z1 ) (Z2 ) (Z3 )

where (X1 ) =C(x,1) ( binomial

coefficient)

Solution:

x1!

x (x−1)2!

x ( x−1 )( x−2)3 !

y1!

y ( y−1)2!

y ( y−1 )( y−2)3 !

z1!

z (z−1)2!

z ( z−1 )(z−2)3 !

= xyz

2!3 !

1 x−1 ( x−1 )(x−2)1 y−1 ( y−1)( y−2)1 z−1 ( z−1 )(z−2)

( taking x,y,z common from

R1,R2,R3 resp. and ½!,1/3! From C2,C3 resp.) ( by formula of C(n,r) =

n!(n−r )!n ! )

Apply C3→C3 + C2 and put a= x-1, b=y-1, c=z-1

xyz12

1 a a ²1 b b ²1 c c ²

= xyz12 (a-b)(b-c)(c-a) =

xyz12 (x-y)(y-z)(z-x).

Question: If x,y,z are all different and if x x ² 1+ x ³y y ² 1+ y ³z z ² 1+z ³

= 0 ,

prove that xyz = -1

Solution: x x ² 1+ x ³y y ² 1+ y ³z z ² 1+z ³

= x x ² 1y y ² 1z z ² 1

+ x x ² x ³y y ² y ³z z ² z ³

= x x ² 1y y ² 1z z ² 1

+ xyz 1 x x ²1 y y ²1 z z ²

= 0

1 x x ²1 y y ²1 z z ²

(1+xyz) = 0 ⇨ (x-y)(y-z)(z-x)(1+xyz) = 0 ⇨

xyz=-1 ∵ x ≠y≠ z.

Page 11: Assignments for class XII

Question: By using properties of determinant,show that

a ²+1 ab acab b ²+1 bcca cb c ²+1

= 1+a2+b2+c2

[Hint: multiply and divide by a,b,c with R1,R2,R3 respectively,taking a,b,c common from C1,C2,C3 respectivelyR1→R1+R2+R3]Question: show that

a−b−c 2a 2a

2b b−c−a 2b2c 2c c−a−b

=(a+b+c)3.[Hint:R1→R1+R2+R3]

Question(i) Using matrix method, solve the following system of equations:

2x + 3

y + 10z = 4, 4

x - 6y + 5

z = 1, 6x + 9

y - 20z = 2; x, y, z ≠ 0.

[X=2,Y=3,Z=5,|A|=1200,adjA =[ 75 150 75110 −100 3072 0 −24 ] ]

(ii) 1x - 1

y + 1z =4, 2

x + 1y - 3

z = 0, 1x + 1

y + 1z = 2

[ x=1/2,y=-1,z=1 adjA= [ 4 2 2−5 0 51 −2 3] |A| = 10]

(iii) 2

x - 3y + 3

z = 10, 1x + 1

y + 1z = 10, 3

x - 1y + 2

z =13; X, Y, Z ≠ 0.

[X=1/2,Y=1/3,Z=1/5,|A|=-9, adjA = [ 3 3 −61 −5 1

−4 −7 5 ] ]

Inverse Trigonometric Functions

Page 12: Assignments for class XII

Table of domain and range of inverse trigonometric function

Inverse-forward identities are

Forward-inverse identities are

Page 13: Assignments for class XII

Relation between inverse functions:

Page 14: Assignments for class XII
Page 15: Assignments for class XII

  

 

For suitable values of x and y

sin-1 x + sin-1 y= sin-1 (x√1-y2+ y√1-x2)

sin-1 x - sin-1 y=sin-1 (x√1-y2- y√1-x2)

cos-1 x + cos-1 y= cos-1 (xy- √1-x2√1-y2)

cos-1 x - cos-1 y= cos-1 (xy+√1-x2√1-y2)

tan-1 x + tan-1y = tan−1 X+Y1−XY         ; xy<1

tan-1 x – tan-1 y= tan−1 X−Y1+XY ; xy>-1

Page 16: Assignments for class XII

2tan-1 x= tan−1 2 x1−x ² = sin−1 2x

1+ x ² = cos−1 1−x ²1+x ²

Trigonometry examples

Example 1:

Solve the following equation:

Suggested answer:

Page 17: Assignments for class XII

Problems – Solve Inverse Trigonometric Functions Problems:

Problem 1:

Prove that tan−1 x + tan−1 2 x1−x ² = tan−1 3 x−x ³

1−3 x ² , x<1√3

Solution:

Let x = tan θ , then θ = tan-1 x. we have

You will take R.H.S to prove the given expression

Page 18: Assignments for class XII

R.H.S =     tan−1 3 x−x ³1−3 x ²

           = tan-1 (tan 3θ)            = 3 θ   = 3 tan-1 x

           = tan-1 x + 2 tan-1 x

    =   tan−1 x + tan−1 2 x1−x ²         = L.H.S

Hence, the given expression will be proved.

OR

Wecantake L.H.S. = tan−1 x + tan−1 2 x1−x ² = tan−1(

x+ 2x1−x ²

1−[ 2 x2

1−x 2])

By using

tan-1 x + tan-1y = tan−1 X+Y1−XY         =R.H.S.

Example problem 2:

Solve tan-1 2x + tan-1 3x = π/4

Solution:

Given: tan-1 2x + tan-1 3x = π/4

Or

 tan-1 ((2x + 3x)/(1 - 2x . 3x)) = π/4

tan-1 ((5x)/(1 - 6x2)) = π/4

∴ (5x)/(1 - 6x2) = tan π/4 = 1

or

Page 19: Assignments for class XII

6x2 + 5x – 1 = 0

That means, (6x – 1)(x + 1) = 0

Which gives

             x = 1/6 or x = -1

since x = -1 does not satisfy the equation ,the equation of the L.H.S is negative, so  x = 1/6 is the only solution of the given equation.

ASSIGNMENT:

Question.1 Evaluate: (i) sin-1(sin10) (ii) cos-1 (cos10) (iii) tan-1(tan(-6))

Answer: (i) if –π/2 ≤x ≤π/2, then sin-1(sinx)=x but x= 10 radians does not lie between –π/2 and π/2

3π – 10 lies between –π/2 and π/2 ∴ sin-

1(sin(3π-10)) = 3π-10. Similarly for (ii) cos-1 (cos10) = cos-1 (cos(4π-10)) = 4π-10. [10 radians does not lie between 0 and π. ∴ 0≤4π-10≤π]

For (iii) tan-1(tan(-6)) = tan-1(tan(2π-6)) = 2π-6 . { -6 radians does not lie in [ –π/2 , π/2]}

Question.2 If x = cos-1(cos4) and y = sin-1(sin3), then which holds? (give reason)

(i) x=y=1 (ii) x+y+1=0 (iii) x+2y=2 (iv) tan(x+y) = -tan7.

Question.3 if cos−1 xa + cos−1 y

b = θ, then prove that x ²a ² -

2xyab cosθ

+ y ²a ² = sin2θ

Page 20: Assignments for class XII

[Hint: cos−1 xa + cos−1 y

b = cos−1 ¿ - √1− x ²a ² √1− y ²

b ² ] = θ ⇨ ¿cosθ)2 = ¿)2

Simplify it]

Question.4 *(i) sin-1x + sin-1y + sin-1z = π, then prove that

X4+y4+z4+4x2y2z2 = 2(x2y2+y2z2+z2x2)

(ii) If tan−1 x + tan−1 y + tan−1 z = π/2 ; prove that xy+yz+xz = 1.

(iii) If tan−1 x + tan−1 y + tan−1 z = π , prove that x+y+z = xyz. [Hint: for (i) sin-1x + sin-1y = π - sin-1z ⇨ cos(sin-1x + sin-

1y) =cos( π - sin-1z)

Use cos(A-B) = cosAcosB – sinAsinB and cos(π – 𝛂)= -cos𝛂

It becomes √(1−x ²)(1− y ²) - xy = - √1−z ² and simply it.

[Hint: for (ii) tan-1 x + tan-1y = tan−1 X+Y1−XY ]

Question.5 Write the following functions in the simplest form:

(i)tan−1 ¿ ) (ii) tan−1 ¿) (iii) tan−1 √ a−xa+ x , -a<x<a

[Hint: for (i) write cosx = cos2x/2 – sin2x/2 and 1+sinx =(cosx/2 +sinx/2)2 , then use tan(A-B), answer is π/4 – x/2 ]

[ Hint: for (ii) write cosx = sin(π/2 – x) and sinx = cos(π/2 – x), then use formula of 1-cos(π/2 – x)= 2sin2(π/4 – x/2) and sin(π/2 – x) = 2 sin(π/4 – x/2) cos(π/4 – x/2)

Same method can be applied for (i) part also. Answer is π/4 + x/2]

[ for (iii) put x=a cos𝛂, then answer will be ½ cos−1 xa ]

Page 21: Assignments for class XII

Question.6 If y = cot−1¿¿) - tan−1(√cosx ), prove that siny =

tan2(x/2). [Hint: y = π2 - 2 tan−1 ¿¿ , use formula 2tan−1 x = cos−1 ¿)]

Question.7 (i) Prove that tan−11 +tan−1 2 + tan−13 = π.

(ii) Prove that cot−1¿) +cot−1¿) + cot−1¿) = 0.

[Hint: for (i) tan−1 2 = π2 - cot−12 =

π2−tan−1 1

2 , then use formula of

tan-1 x + tan-1y = tan−1 X+Y1−XY    ]  

(ii) [Hint: write cot−1 x  = tan−1 1x ]

Question.8 Solve the following equations:

(i) sin−1 x + sin−1(1−x) = cos−1 x .

(ii) tan−1 √x (x+1) + sin−1√ x ²+ x+1 = π2 .

(i) [Hint: write cos−1 x = π2 - sin−1 x, put sin−1 x = y]

(ii) [Hint: use tan−1 x=¿ cos−1 1

√1+x ² ]

Question.9 Using principal values, evaluate cos−1 ¿¿) + sin−1¿). [answer is π]

Question.10 Show that tan(12

sin−1 34 ) =

4−√ 73 and justify why

the other value is ignored?

[ Hint: put 12

sin−1 34 =∅ ⇨ ¾ = sin2∅ = 2tan∅/(1+tan2∅), find tan∅]

** SOME HOT QUESTIONS:

Page 22: Assignments for class XII

1. Which is greater tan 1 or tan-11?

2. Find the value of sin(2tan−1 23) + cos( tan−1√3¿¿ .

3. Find the value of x which satisfies the equation sin−1 x + sin−1(1−x) = cos−1 x .

4. Solve the equation: sin−16 x + sin−1¿) = -π/2.

5. Show that tan−1 ¿tanx2) =

12

cos−1¿) .

6. If 𝛂 = tan−1(2 tan ²α ) - 12

sin−1 ¿¿), then find the general value of 𝛂.ANSWERS WITH HINTS:1. Since 1> π/4 ⇨ tan1> 1> tan-11.

2. sin(2tan−1 23) + cos( tan−1√3¿¿= sin2x + cosy ⇨

2 tanx1+ tan ² x +

1

√1+ tan ² y = 2.( 2

3)

1+(4/9)+

12 =

3726 .

3. sin ¿ + sin−1(1−x)¿ = sin(cos−1 x) by using sin(A+B)=sinA cosB + cosA sinB

⇨ x√1−(1−x ) ² + (1-x)√1−x ² = √1−x ² ∵ sin(cos−1 x) = √1−cos ²(¿cos−1 x )¿

⇨ 2x – x2 = 1 ⇨ x = 0 or ½.

4. sin−16 x = - π2 - sin−1¿¿) ⇨ 6x = sin[-

π2 - sin−1¿¿)]

= -cos[sin−1¿¿ ] = -cos[cos−1 √1−108x ² ] = - √1−108 x ² etc.

Page 23: Assignments for class XII

5. 1/2(2tan−1 ¿tanx2 )) , use formula 2tan−1 A = cos−1 1−A ²

1+A ² and

tan2x/2 = 1−cosx1+cosx .

6. Put tan𝛂 = t and use sin2𝛂 = 2 tanα

1+ tan ²α and cos2𝛂 = 1−tan ²α1+tan ² α

then put t/3 = T,answer is 𝛂 = nπ, nπ+π/4. tan−1 2t ² - tan−1 t = ½ sin−1 6 t

9+t ² ⇨ tan−1 2 t ²−t1+2 t ³ = ½ sin−1 2 t /3

1+(t /3)² = ½ sin−1 2T

1+(T ) ²

tan−1 2 t ²−t1+2 t ³ = (2T), then tan𝛂 = 0 ,1.½

ASSESSMENT OF Relations & functions for class—XII Level—1

Q.1 Let f(x) = { x+3 , if x<14 x−2, if 1≤x ≤4.x ²+5 , if x>4

Find f(-1) ,f(4) and f(5).

Q.2 If f(x) = x2 - 1x ² , then find the value of f(x) + f ( 1

x ²¿.

Q.3 Let Q be the set all rational numbers and relation on Q defined by R = {(X, Y): 1+XY > 0}. Prove then R is reflexive and symmetric but not transitive.

Q.4 Write the identity element for the binary operation *defined on set R by a*b = 3ab/8 ∀ a, b ЄR.

Q.5 Show that the function f: R → R defined by f(x) = sin x is neither 1-1 nor onto.

Answers (Level—1)

Page 24: Assignments for class XII

Ans.1 f (-1) = 2, f (4) = 14, f(5)= 30. Ans.2 0. Ans.3 Consider any x, y Є Q, since 1+x.x =1+x2 ≥ 1 ⇨ (x,x)Є R ⇨ reflexive

Let (x,y) Є R ⇨ 1+xy > 0 ⇨ 1+yx > 0 ⇨ (y,x) ЄR ⇨ symmetric.

But not transitive . Since (-1, 0) and (0, 2) ЄR, because 1 > 0 by putting values. But (-1, 2) ∉ R because -1<0. Ans.4 Let e be the identity element in R. Then a *e =a =e*a ∀ aЄR ⇨ a*e =a ∀ a ЄR ⇨ e = 8/3 in R. Ans.5 f is not 1-1 because sin 0 = 0 =sin π,so the different elements o, π have same images. f is not onto because -1 ≤ sin x ≤ 1 for all x ЄR ∴ the range of f =[-1,1], which is a proper subset of R.

LevelQ.1 If f: R→ R is given by f(x) = (3 – x3)1/3 show that fof =Ig where Ig is the identity map on R.

Q.2 Show that the function f: [-1, 1] →R defined by f(x) = x2+ x is 1-1 .

Find the range of f. Also find the inverse of the function f: [-1, 1] → range of f.

Q.3 Show that the function f: R → R defined by f(x) = cos (5x+2) is neither 1-1 nor onto?

Q.4 If f: R → R be given by f(x) = sin2x +sin2(x+π/3) +cosx .cos(x+π/3) ∀ x Є R, and g: R → R be a function such that g(5/4) =1 , then prove that (gof) : R → R is a constant function.

Q.5 Let R1=R – {-1} and an operation * is defined on R1 by a*b = a + b + ab ∀ a, b Є R1 .

Find the identity element and inverse of an element. ANSWERS OF Level—2

Ans.1 As f: R → R, fof exists and fof : R → R is given by (fof) (x) = f(f(x)) = f(3 – x3)1/3 = (3 – ((3 – x3)1/3 )3 )1/3 = (3 – (3 – x3))1/3 =x ∀ x ЄR Ans.2 f is 1-

Page 25: Assignments for class XII

1, as consider any x1, x2 Є [-1, 1] such that f(x1) = f(x2) ⇨ x₁2+X ₁= X ₂

2+X ₂⇨ x1x2+2x1 = x1x2 +2x2 ⇨ x1 = x2 For the range of f

Let y = f(x) ⇨ y = x2+ x ⇨ xy +2y =x ⇨ (y – 1) x= -2y ⇨ x = −2 y

y−1

As x Є [-1, 1], so -1 ≤ −2 yy−1 ≤1 , but (y – 1)2 >0 , y ≠ 1⇨ -(y – 1)2 ≤ −2 y

y−1

(y-1)2≤ (y – 1)2 ⇨ -(y2 – 2y +1) ≤ -2y2+2y ≤ y2 – 2y +1 ,y≠ 1⇨ Y2 – 1 ≤ 0 and 0 ≤ 3y2 – 4y +1 ⇨ y ε [-1,1] and (y – 1/3) (y – 1) ≥ 0 ,y ≠ 1 ⇨y Є [-1,1] and y ε (-∞ ,1/3] U [1,∞) , y ≠ 1 ⇨ y Є [-1,1] and y ε (-∞ ,1/3] U (1,∞)⇨ y Є[-1,1/3].∴ its inverse exists as f is 1-1 and onto, to find f-1

x2+ x= y ⇨ xy +2y =x ⇨ 2y = x (1 – y) ⇨ x= 2 y

1− y f-1(y) = x = 2 y1− y .

Ans. 3 For f is not 1-1, 5x+2 = π/2 ⇨ x = (π – 4)/10 ∴ 5x+2 =π/2, again 5x+2 = 3π/2 ⇨ x = (3π – 4)/10, Now f ((π – 4)/10)) = cos[5((π – 4)/10) +2] = cosπ/2 =0

f ((3π – 4)/10) = cos[5((3π – 4)/10) +2] = cos3π/2 = 0.

For f is not onto, as -1 ≤ cos (5x+2) ≤1, then -1≤ y ≤1, range of f = [-1, 1] = {y : -1≤ y ≤1 } ≠ co-domain R.

Ans. 4 ½[ 2sin2x +2sin2(x+π/3) +2cosx cos(x+π/3)]

f (x)= ½[ 1 – cos2x +1 – cos (2x+2π/3)+ cos (2x+π/3)+cosπ/3]

( As we know that 2sin2x= 1 – cos2x and 2cosA cosB= Cos(A+B) + cos(A-B).) ½[5/2 – {cos2x + cos(2x+2π/3)} + cos(2x+π/3) ⇨ ½[5/2 – 2cos(2x+π/3) cos π/3 + cos(2x+π/3)] = 5/4 ∀ x ЄR∴ for any x Є R , we have (gof)(x) = g(f(x)) = g(5/4)=1 ,so it is constant function.

Page 26: Assignments for class XII

Ans. 5 * can be shown to be a binary operation on R1 as let a ≠ -1, b ≠ -1 . a*b = a+b+ab Є R – {-1} ⇨ a+b+ab ≠ -1⇨ a(1+b)+(b+1) ≠0 ⇨ (a+1) (1+b) ≠0 ⇨ a ≠ -1 and b ≠ -1 which is true.Now if e is the identity element, then a*e =a ⇨ a+e+ae =a ⇨ e (1+a) = 0 ⇨ e =0 or a = -1 ⇨e =0 , 0 is the identity w.r.t. *

Let a’ be inverse of a, then a*a’ =0 ⇨ a+a’+aa’ = 0 ⇨ a’(1+a) = - a ∴ a’ = - a/(1+a) , is the inverse of a w.r.t. *.

   

ASSIGNMENT(continuity & differentiability) (XII)

**Question 1 Determine a and b so that the function f given by

f(x) = 1−sin ² x3 cos ² x , x<п/2

=a, x=п/2

= b(1−sinx)

(п−2x ) ² , x>п/2 Is continuous at x=п/2.

Answer [a = 1/3 , b = 8/3] **Question 2 Find k such that following functions are continuous at indicated point

(i) f(x) ={1−cos4 x8x ²

, x ≠0

k , x=0 at x=0

(ii) f(x) = (2x+2 - 16)/(4x – 16) , x≠2

= k, x = 0 at x=2. Answer [ (i) k=1,(ii) k=1/2]

**Question 3 The function f is defined as {x ²+ax+b ,0≤x<23 x+2 ,2≤ x≤4

2ax+5b ,4< x≤8 If

f(x) is continuous on [0,8], find the values of a and b. Answer [a=3,b=-2]

  

Page 27: Assignments for class XII

** Question 4 If f(x) = {√1+ px−√1−pxx

,−1≤ x<0

2 x+1x−1

,0≤ x ≤1 is continuous in the [-

1,1], find p. Answer [p=-1]

**Question 5 Find the value of a and b such that the f(x) defined as

f(x) = { x+a√2 sinx ,0≤ x<п/42 xcotx+b ,п /4≤ x≤п/2

acos2 x−bsinx ,п2<x≤п

is continuous for all values of x in [0,п].

ANSWER [a=п/6 , b=-п/12]

** Question 6 Prove that limx→π /4

tan3 x−tanx

cos (x+ √π4

) = -4

[ Hint: Nr. Can be written as tanx(tanx-1)(tanx+1) =- [tanx(cosx-sinx)(tanx+1)]/cosx Cosx-sinx = √ 2 cos¿) ]

**Question 7 Prove that (i) limx→ 1

√ 2

x−cos (sin−1 x )1−tan (sin−1 x ) = −1

√2 [ Hint: put x=

sin ]Ѳ

(ii) limx→∞x ¿¿ ) = -3/2. [Hint: π4 = tan−11 & use formula of tan−1 x−tan−1 y

]

Question 8 f(x) = a x2+b

x2+1 , limx→0

f (x ) =1 & limx→∞f (x) =1, then p.t. f(-

2)=f(2)=1. [ Hint: limx→∞

1

x2 =0]

Question 9 limx→0

ex−1√1−cosx

[Dr. = √2|sinx/2| &limx→0

e x−1x

=1

|sinx/2| =+ve & -ve as x→0+ & x→0- , ⇨ limit does not exist] Question 10 Show that the function

Page 28: Assignments for class XII

f(x)¿ {sin 3 xtan 3 x

, x<0

32, x=0

log(1+3 x)e2x−1

, x>0

is continuous at x=0.

[Hint: use limx→0

e x−1x

=1 , limx→0

log (1+x)x =1]

Question11 Show that f(x) = |x-3|,x∊R is cts. But not diff. at x=3.

[Hint:show L.H.lt=R.H.lt by |x-3| = x-3, if x ≥3 and –x+3, if x<3, L.hd=-

1≠1(R.h.d) Question 12 Discuss the continuity of the fn. f(x) = |x+1|

+|x+2|, at x = -1 & -2 [Hint:f(x) = {−2x−3 ,when x←21 ,when−2≤x←1

2 x+3 ,when x ≥−1

yes cts. At x=-1,-2

Question 13 Find the values of p and q so that f(x) ={x ²+3x+ p , if x≤12x+2 , if x>1 is

diff. at x = 1. [ answer is p=3 , q=5]

Question 14 For what choice of a, b, c if any , does the function

F(x)={ax ²+bx+c ,0≤ x≤1bx−c ,1<x ≤2

c , x>2becomes diff at x=1,2 & show that a=b=c=0.

Question15For what values a,b f(x)={ e2x−1 ,when x ≤0

ax+ bx ²2

,when x>0 is diff.at x=0

[Hint: L.H.d= 2by using limx→ 0

ex−1x

=1& R.H.d=a, since f‘(x)=0exists, a=2,b∊R]

ASSIGMENT OF DIFFERENTITION

Question 1 Show that y = aex and y = be –x cut at right angles

aab=1 [ by equating , we get ex = √ ba ⇨ x= ½ log ( b/a) , find

Page 29: Assignments for class XII

slopes(dy/dx) at pt. of intersection is (½ log ( b/a , √ab).

Question 2 (i) If y√1−x ² + x√1− y ² = 1, prove that

dy/dx= (-1)√ 1− y ²1−x ²

[Hint: put y=sinѲ & x= sinφ , use formula of sin(θ+φ ¿¿

(ii) If cos-1¿ ) = tan-1a , find dy/dx. [let cos(tan-1a )= k(constant), then assume c= 1-k/1+k , dy/dx= y/x]

(iii) If y x = e y−x, prove that dy/dx = (1+ logy ) ²logy

(iv) If xm.yn = (x+y)m+n, then find dy/dx. [ y/x]Question 3 Differentiate w.r.t. x : **(i) Using logarithmic differentiation, differentiate:

Solution:

(ii) x tanx + √ x ²+1x

(iii) (iogx)x + xlogx

Question 4 (i) If y x = e y−x, prove that dy/dx = (1+ logy ) ²logy

(ii) If f(1)= 4,f’(1)=2,find d/dx{logf(ex)} at the point x =0.[1/2](iii)If y = √ x+√x+√ x+…….∞ ,show that (2y – 1)dy/dx =1.(iv) If x = (t+1/t)a , y= a(t+1/t) where a>0,a≠1,t≠0, find dy/dx.[Hint: take dy/dt & dx/dt , then find dy/dx = ylogy/ax. ]Question5(i)differentiate: Sec-1(1/(2x2 – 1)),w.r.t.sin-1(3x –4x3).[Hint: let u=1st fn. & v= 2nd fn. , find du/dv = 1]

(ii)differentiate: tan-1 (√1+x ²−1x ),w.r.t. sin-1 (

2x1+ x ² ) if 1<x<1;x≠0

Page 30: Assignments for class XII

[ du/dv= ¼, put x=tan ⇨ u= /2, v=2 , u&v as assumed Ѳ Ѳ Ѳabove](iii) If y = e(msin-1x) , show that (1-x2)y2 – xy1 – m2y= 0.Question 6 Water is driping out from a conical funnel, at the uniform rate of 2cm3/sec. through a tiny hole at the vertex at the bottom. When the slant height of the water is 4cm.,find the rate of decrease of the slant height of the water given that the vertical angle of the funnel is 1200 .[Hint: Let l is slant height ,V = 1/3.π .l(√ 3/2)2.l/2=πl3/8(vertical angle will be 600 (half cone), take dv/dt=-2cm3/sec. l=-1/3⇨ π cm/s.]**Question 7(i) Let f be differentiable for all x. If f(1)=-2 and if f `(x) ≥2 ∀ x∊[1, 6], then prove f(6) ≥8.[ use L.M.V.Thm.,f`(c)≥2,c∊[1, 6]](ii) If the function f(x)= x3 – 6x2+ax+b defined on [1, 3] satisfies the rolle’s theorem for c = (2√ 3 +i)/√ 3 , then p.t. a = 11 & b∊R.[Hint: Take f(1)=f(3) , use rolle’s thm. f`(c)=0⇨ a=11]Question 8 (i) Show that f(x)= x/sinx is increasing in (0, п/2)[HINT: f’(x)>0 , tanx >x](ii) Find the intervals of increase and decrease for f(x) = x3 + 2x2 – 1.[Answer is increasing in (-∞, -4/3)U(0, ∞) & decreasing in (-4/3, 0)](iii) Find the interval of increase&decrease for f(x) =log(1+x)-(x/1+x) ORProve that x/1+x < log(1+x) < x for x > 0.[ Hint: f(x)strictly ↑ in [0, ∞) , x>0 ⇨f(x)>f(0), let g(x)=x-log(1+x)g(x)>0 ↑ in [0,∞) & f(x) ↓ in (-∞, 0].](iv) For which value of a , f(x)=a(x+sinx)+a is increasing. [Hint: f’(x) a(1+cosx) ≥0 ⇨ a>0 ∵ -1≤cosx≤1]**Question 9 Problem: Using differentials, approximate the expression

Solution: We let

Hence, x = 0.05 and y = /4.

Page 31: Assignments for class XII

Differentiating, we obtain

Substituting, we get

Question 10 For the curve y = 4x3 − 2x5, find all the points at which the tangents passes through the origin. [Hint: eqn. Of tangent at (x0,y0) , put x,y=0,(x0,y0)lies on given curve]Question 11 Find the stationary points of the function f(x) = 3x4 – 8x3+6x2 and distinguish b/w them. Also find the local max. And local mini. Values, if they exist.[ f’(x)=0⇨ x=0,1 f has local mini. At x=0∵f’’>0 & f’’(1)=0, f has point of inflexion at x=1,f(1)=1] Question 12 Show that the semi – vertical angle of right circular cone of given total surface area and max. Volume is sin-1 1/3.[Hint: take S=Пr(l+r) ⇨ l= S/пr – r , take derivative of V² OR can use trigonometric functions for l & h]Question 13 A window has the shape of a rectangle surmounted by an equilateral ∆. If the perimeter of the window is 12 m., find the dimensions of the rectangle so that it may produce the largest area of the window.[Hint: let x=length, y=breadth, then y=6 – 3y/2, A= XY+√3X2 /4, take derivative of A & it is max. ,x=4(6+√3)/11 ,y=6(5−√3)/11]

ASSIGNMENT OF INTEGRATION

Page 32: Assignments for class XII

Question 1 Evaluate: (i)** Integrate .[ Use the power substitution

Put ]

** (iii) Integrate . [ Use the power substitution Put ]

(iii) ∫0

π /4

secx tan3 x dx [answer is (2 - √2)/3 ]

(iv) ∫ dx [multiply&divide by sin(a-b)] (v)∫ √ 1−√ x

1+√x dx

[multiply & divide by √1−√ x ] (Vi)∫ x

x3−1dx [by partial fraction]

(v)∫ (x−4)ex

(x−2)3 dx [ use ∫ex(f(x)+f’(x))dx] (vi)∫o

π /2dx

3+2 sinx+cosx dx [put sinx= 2 tanx /2

1+ tan ² x /2, cosx =1−tan ² x /2

1+tan ² x /2 , then put t=tanx/2. Answer is tan−1 2– п/2]

(vii) ∫0

3 /2

¿ xcosπx∨¿¿ dx [∫0

1 /2

¿ xcosπx∨¿¿+ ∫1 /2

3 /2

¿ xcosπx∨¿¿ = ∫+ve dx+∫ -ve dx , answer is

5/2п- 1/п2] (viii) [ write sin2x = 1-cos2x answer is п/6] (ix)

∫0

π /2

√ tanx + √cotx dx [ answer is √2π] (x) ∫0

a

sin−1 √ xa+x

dx [ put x=atan2 , answer Ѳ

is a/2( -2) ] п (xi) ∫0

πx

1+sin ² x dx [ use property ∫

0

a

f ( x )dx = ∫0

a

f (a− x )dx , ∫0

2a

f ( x )dx =2∫0

a

f ( x )dx ∵f(2a-x) = f(x) , then put t=tanx, answer is /2√2п² ] (xii) ∫

−5

0

f ( x ) dx , where f(x) =|x|+|x+2|+|x+5|. [∫−5

−2

(−x+3)dx + ∫−2

0

( x+7) dx , answer is 31.5 ] (xiii) Evaluate ∫ ex (1−x ) ²

( x ²+1 ) ² dx [use ∫ ex(f(x)+f’(x))dx

Question 2 Using integration, find the area of the regions: (i) { (x,y): |x-1| ≤y ≤√5−x ² } (ii) {(x,y):0≤y≤x2+3; 0≤y≤2x+3; 0≤x≤3}

Page 33: Assignments for class XII

[(i) A= ∫−1

2

√5−x ²dx- ∫−1

1

(−x+1 )dx - ∫1

2

(x−1) dx = 5/2 [ sin−1( 2

√5) +sin−1( 1

√5) ] –

] [(ii) ½ A=∫0

2

(x ²+3¿)¿ dx +∫2

3

(2 x+3)dx , answer is 50/3](iii) Find the area bounded by the curve x 2 = 4y & the line x = 4y – 2.[A = ∫

−1

2x+24

dx - ∫−1

2x ²4

dx = 9/8 sq. Unit.]**(iv) Sketch the graph of f(x) = {|x−2|+2 , x≤2

x ²−2 , x>2 ,evaluate∫0

4

f (x ) dx[hint: ∫

0

4

f (x )dx = ∫0

2

(4−x )dx + ∫2

4

(x ²−2) dx = 62/3.]**Question 3 evaluate ∫ √1+ x

√ x dx [ mult. & divide by √1+x , put 1+x =A.(d/dx)(x2+x)+B ,find A=B=1/2, integrate]

Definite integral as the limit of a sum , use formula : ∫a

b

f ( x )dx limh→0

∑r=1

n

f (a+rh), where

nh=b-a & n→∞ Question 4 Evaluate ( i )∫0

4

¿¿) dx (ii) ∫0

3

(x ²−2 x+2) dx[ use lim

h→ 0

eh−1h

= 1 for part (i) , use formulas of special sequences, answer is 6]

Page 34: Assignments for class XII

Some special case :

(1) Evaluate: ∫ dx(x−3)√ x+1

[ put x+1=t²] (2) ∫ dx(x ²−4)√ x+1

[ put x+1 = t² ]

(3) Evaluate: ∫ dx(x+1)√x ²−1

(4) Evaluate: ∫ dx

x ²√x ²+1 [ put x=1/t for both]

(5) Evaluate: ∫ (x ²+1)dxx4+1

[ divide Nr. & Dr. By x2 , then write x²+1/x²=(x-1/x)² +2

according to Nr. , let x-1/x=t]

(6) Evaluate ∫ x √1+x−x ² dx [ let x=A(d/dx) ( 1+x-x²) +B]

(7) Integrating by parts evaluate ∫ x ²( xsinx+cosx ) ² = ∫ ( xsecx ) . xcosx

( xsinx+cosx )²

(8) Evaluate ∫ 11+cotxdx =∫ sinx

sinx+cosxdx [ put sinx=Ad/dx(sinx+cosx)+B(sinx+cosx)

+C

Page 35: Assignments for class XII

If Nr. Is constant term then use formulas of sinx,cosx as Ques. No. 1 (vi) part]

ASSESSMENT OF DIFFERENTIAL EQUATIONS FOR CLASS—

XI Level--1 Q.1 Find the order and degree of the following differential equations. State also whether they are linear or non-linear.

(i) X2( d ² ydx ² )3 + y ( dydx )4 y4 =0. (ii) d ² y

dx ² = 3√1+( dydx

¿)² ¿ .

Q.2 Form the differential equation corresponding to y2 = a (b – x)(b+ x) by eliminating parameters a and b. Q.3 Solve the differential equation (1+e2x) dy + (1+y2) ex dx = 0, when x= 0, y =1.

Q.4 Solve the differential equation: dydx = 1−cosx1+cosx .

Q.5 Verify that y = A cosx – Bsinx is a solution of the differential

equation d ² ydx ² + y = 0.

Answers of Level—11. (i) order =2 , degree=3 , non linear( because degree is more than 1 ) , (ii) order 2 , degree 3 , non-linear .

2. y2 = a(b – x)(b+ x) = a (b2 – x2), 2y dydx =-2ax ⇨ ydydx = -ax, again

differentiate Y d ² ydx ² + ¿ )2 = -a , by using the value of a from above step ,

we will get , x{ Y d ² ydx ² + ¿ )2 } = ydydx .

3. dy1+ y ² = - e

xdx1+e2 x , Integrating both sides, we get

tan−1 y = - ∫ exdx

1+e2 x , put ex = t⇨ tan−1 y = - tan−1 t +c

Using x=0, y=1, we have y = 1/ex.

4. y = 2 tan(x/2) – x +c , put tan(x/2) = 1−cosx1+cosx

5. dydx = - A sinx – B cosx , d ² ydx ² = - A cosx + B sinx = -y.

Level---2

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Q.1 Solve: y dx + x log ( yx ) dy – 2x dy = 0 .

Q. 2 Which of following transformations reduce the differential

equation dzdx + zx log z = zx ² ( log z ) ² into the form

dudx + P(x) u = Q(x) ?

(i) u = log x (ii) u = ex (iii) u = ¿-1 (iv) u = ( log z )2

Q.3 Solve: dydx +xy = xy3

Q.4 Solve: dydx = cos (x+ y) + sin( x+ y )

Q.5 Solve: dydx + x sin 2 y = x3 cos2y

Answers of Level ---2 1. put x = vy , answer = 1+ log (yx ) = ky .

2. (iii) differentiate w.r.t. x dudx = - 1¿¿ .1

z dzdx , put the value of dzdx in the

given differential equation. 3. put 1y ² = t, answer is 1

y ² = 1 + cex ². 4.

put x+y = v, answer is log (1+tan ( x+ y )

2) = x + c.

5. put tan y = v, I.F. = ex ² , also use sin 2 y = 2siny cosy.

ASSESSMENT OF PROBABILITY FOR CLASS –XII Level—1 Q. 1 If the mean and variance of a binomial distribution are 4 and 4/3 respectively, find P(X≥1). Q. 2 If P(A) = 3/8 , P(B) = ½ and P(A∩B) = ¼ , find P(A/B). Q.3 A bag contains 4 white and 2 black balls. Another bag contains 3 white and 5 black balls.

If one ball is drawn from each bag, find the probability that (i) Both are white balls. (ii) One is white and one is black.

Q.4 If A and B are independent events and P(A∩B) = 1/8, P( A’ ∩ B’) =3/8 , find P(A) and P(B).

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Q.5 The probabilities of P, Q and R solving a problem are ½, 1/3 and ¼ respectively. If the problem is attempted by

all simultaneously, find the probability of exactly one of them solving it.

Answers of Level—1 Ans.1 np = 4, npq = 4/3 ⇨ q=1/3 ⇨ p = 1 - 1/3

= 2/3 ⇨ n=6 ⇨ P(X≥1) =1 – C(6,0) (2/3)0 (1/3)6 = 1 - 1729 = 728

729 .

Ans.2 P(A/B) = P(A ∩ B ) /P(B) = 1−P(AUB)

1−P(B) = 1−[ 3

8+ 1

2−1

4]

1−1 /2 = ¾.

Ans.3 (i) P(A ∩ B) = P(A).P(B) = (2/3).(3/8) [A,B are independent events] (ii) P(A’ ∩ B) + P(A ∩ B’) = P(A’).P(B)+P(A).P(B’) =( 1/3).(3/8)+ (2/3).(5/8)=13/24. [A’, B are indep. Events, B’ A are indep. events], where A = drawing a white ball from first bag. B= drawing a same ball from second bag.A’ = drawing a black ball from first bag and B’ =drawing from second bag. Ans.4 P(A∩B) = P(A).P(B) = 1/8 let x=P(A), y= P(B), P( A’ ∩ B’) =3/8 = P( A’) .P( B’) =(1- X)(1 – Y) ⇨ X+Y – XY = 5/8 ⇨ X=1/2 , Y= ¼.Ans. 5 P(A’)=1/2 ,P(B’) = 1-1/3=2/3 , P(C’)=3/4 ∴ Req. Prob. = P(A)P(B’)P(C’) = P(A’)P(B)P(C’)+ P(A’)P(B’)P(C) [A,B,C are indep. events] = (1/2)(2/3)(3/4)+(1/2)(1/3)(3/4)+(1/2)(2/3)(1/4)=11/24.

Level---2Q.1 If A ∩ B = ф, show that P(A/B) =0, where A and B are possible events.Q.2 A pair of dice is thrown if the sum is even, find the probability that at least one of the dice Shows three. Q.3 Let X denotes the number of hours you study during a randomly selected school day.The probability that X can take the value x, has the following form, where k is some unknown constant P(X=0)=0.1 and P(X=x) = { kx if x=1∨2

k (5−x ) if x=3∨4o ,otherwise

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(i) Find k. (ii) What is the probability that you study at least two hour? Exactly two hour? At most two hours?Q.4 Six dice are thrown 729 times. How many do you expect at least three dice to show a 5 or 6? Q.5 In a class; 5% of the boys and 10% of the girls have an I.Q. of more than 150. In this class 60% of the students are boys. If a student is selected at random and is formed and is found To have an I.Q. of more than 150, find the probability that the student is a boy. Answers of Level—2Ans.1 A and B are possible events ⇨A ≠ ⇨ P(A)≠ 0 , P(B) ≠0 фBut A∩B = ф ⇨ P(A∩B) = P( ) = ф P(A/B) = P (A ∩B)

P(B) =0. Ans. 2 n(S)=36, n(A)=18 Out of these 18, the cases which at least one die shows up 3 are (1, 3),(3,1),(3,3),(3,5),(5,3) Required probability=5/18. Ans.3 X 0 1 2 3 4P(X) 0.1 K 2K 2K K(i) k=0.15 (ii) 0.75, 0.3, o.55. Ans.4 P(success)= 2/6=1/3 ∴ q=2/3 P(x success i.e., getting a 5 or 6)= C(6, x) Px q6-x P(at least three successes in six trials) = P(x≥3)=1 – [p(0)+p(1)+p(2)]By using above result we get 1 – (16/81)(31/9) = 233/729 ∴ required answer is 233/729x729=233. Ans.5 Let E1: The student chosen is a boy. P(E1)=60/100 ∴ E2: ........................................girl. P(E2) = 40/100 E1, E2 are mutually exclusive. A: a student has an I.Q. of more than

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150. P(A/ E1)= 5/100, P(A/ E2)= 10/100 By Baye’s theorem P(E1/A) = 3/7.