class xii practice questions
TRANSCRIPT
CLASS XII (ASSIGNMENTS) Matrices and Determinants
Properties of determinants :-
1. The value of a determinant remains unchanged if its rows and columns are interchanged.
2. The sign of value of a determinant is changed if its any two rows or columns are interchanged.
3. The value of a determinant is zero if its any two rows or columns are identical.
4. If a determinant is multiplied by a scalar (number) , its only one row or column gets multiplied by that constant.
5. If any two row or column of a determinant are proportional, its value becomes zero.
6. If all elements of a row or column are expressed as sum of two or more elements ,the whole of the determinant can be expressed in sum of two or more determinants.
7. If some multiple of one row or column is added or subtracted to another row or column (elementwise), its value remains unchanged.
Tips to solve properties based problems:-
1. If a determinant is of order ,we can apply only n-1 propertis at a time to it.
2. The format of application of properties is :-Row affected Row affected n (Row used)
Ex.
3. The format for interchanging Rows or columns :- 4. You can never multiply a number to Row affected, it is always
multiplied to Row used. 5. First always try to make elements of any one row or column
identical so that you could take out common from that row or column. It makes all the elements of that Row or column unity(1) and then you make at the most two elements of that row or columns zero (0).Now expand that the determinant by that row or column.
EXAMPLE:
We shall apply
and
Taking out common b from
Expand by =
Important Questions (ASSIGNMENT)
Q. 1. Construct a 3X4 matrix, whose elements are given by
aij=
Q. 2. Construct a 2x3 matrix A = [aij] whose elements are given by aij
= , i ≠ j = |i + 2j|, i = j
Q.3. Construct a 3 × 3 matrix whose elements aij are given by
Q.4. How many number of matrices are possible of order 3 × 3 with each entry 0 or 1. Q. 5. Express the matrix
A= as the sum of symmetric and skew-symmetric matrix Q. 6. Obtain the inverse of the matrix
A =
using elementary transformations.
Q. 7. If
f(x)= Prove that f(x). f(y) = f(x + y)
8. Find x,y,z if
[x 3 2] =[0 0 1]
Q. 9. If A and B are invertible matrices of the same order, then prove that (AB)-1 = B-1A-1
10. Express as a sum of a symmetric and a skew – symmetric matrix.
Q. 11. If
A =
Show that A2 -5A + 7I = 0, Use this to find A-1.
Q. 12. Express the matrix
A =
as the sum of a symmetric and a skew-symmetric matrix. Q. 13. Find the values of x, y, z if the matrix
A =
satisfy the equation A'A = I3.
Q. 14. Show that :
=
Q. 15. Find the inverse of
, if a2 + b2 + c2 + d2 = 1.
Q. 16.(i) Find x such that:
(ii) find x, p, y &q
(iii) Find the value of x, if
Q. 17. For what value of k the matrix A = has no inverse.
Q. 18. Prove that the product of matrices
and
is the null matrix, when and differ by an odd multiple of . Q. 19. A matrix X has a + b rows and a + 2 columns while the matrix Y has b + 1 rows and a + 3 columns. Both matrices XY and YX exist. Find a and b. Can you say XY and YX are of the same type? Are they equal.
Q. 20. Find the matrix A satisfying the matrix equation
A =
21. If A = , B = , c = Find a matrix D such that CD - AB = 0.
22. Let A = , Verify that.
23. If A = ,find k so that.
24.Find B if .
25. Find A = , find a and b such that such that where I is unit matrix of order 2.
Q. 26.Prove that every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric. Hence represent
as above.
27. If , prove that ,n N
28. If A = ,Using principle of mathematically induction prove
that
29. If A = , B = , c = Find a matrix D such that CD - AB = 0.
30. Let A = , Verify that.
Q. 31.(i)Prove that every square matrix can be uniquely expressed as the sum of a symmetric matrix and a skew-symmetric. Hence represent
as above.
(ii) Express the matrix as the sum of symmetric and skew symmetric matrix.
Q.32(i).Using elementary row transformation find the inverse of the matrix
(ii) Using the method of reduction (i.e elementary row transformations),find the inverse of
(iii)If A =
; show that A2 = A-1 . (Without using elementary transformations)
(iv) Using elementary transformation , find the inverse of the
matrix.
Q. 33.(i) Show that the matrix A = satisfies the equation x2 – 4x –5=0. Hence find A-1
(ii) Solve the equations by matrix method
(iii) Solve using matrices :
(iv) Solve by matrix method:
2x + y + z = 1 x - 2y – z = 3/2 3y - 5z = 9
34.(i) If A = and B = find the product AB and
use this result to solve the following system of equations:
(ii) If find A-1 and use it solve the system of
equations:
(iii) Determine the product
and use it solve the system of equations : x-y+z=4 x-2y-2z=9 2x+y+3z=1
(iv) If
A =
, find A-1, using A solve the following system of linear equations.
2x – y + z + 3 = 0 3x – z + 8 = 0 2x + 6y – 2 = 0
35.(i) For what value of a and b ,the following system of equations is
consistent?
(ii) Find whether the following system of equations is consistent or not, find the solution of the system also. 3x-y+2z=3, x-2y-z = 1, 2x+y + 3z = 5.
(iii) Using matrix methed, solve the following system of equations:
(iv) Solve the following system of linear equation
(v)The sum of three numbers is 6. If we multiply third number by 3 and add second number to it, we get 11. By adding first and third numbers, we get double of the second number. Represent it algebraically and find the numbers using matrix method.
(vi) The management committee of a residential colony decided to award
some of its members for honesty, some for helping others and some others
for supervising the workers to keep the colony neat and clean. The sum of all
the awardees is 12. Three times the sum of awardees for cooperation and
supervision added to two times the number of awardees for honesty is 33. If
the sum of the number awardees for honesty and supervision is twice the
number of awardees for helping others, using matrix method, find the
number of awardees of each category. Apart from above values, suggest one
more value which the management of the colony must include for awards.
(vii) Show that the following system of equations is consistent 2x – y + 3z = 5, 3x + 2y – z = 7, 4x + 5y – 5z = 9. Also, find the solution.
(viii)Using matrix method, solve the following system of equations for x, y and z :
Determinants
Q. 1. (i)Find the equation of the line joining A(1,3) and B(0,0) using determinants and find if D (K, 0) is a point such that area of a triangle ABD is 3 square units.
(ii) if |𝒙 + 𝟏 𝒙 − 𝟏𝒙 − 𝟑 𝒙 + 𝟐
| = |𝟒 −𝟏𝟏 𝟑
|, then write the value of x.
Q. 2. Using determinants, find the area of the triangle whose vertices are (1, 4), (2, 3), (-5, 3). Are the given points collinear.
Q. 3. If the points (a1, b1), (a2, b2) and (a1 + a2,(b1 + b2) are collinear, Show that a1b2 = a2b1.
Q. 4. If a, b and c are real numbers, and
, show that either a + b + c = 0 or a = b = c.
Q.5. Using properties of determinants, prove
that = 0.
Q. 6. Without expanding show that
Q. 7. Prove that :
Q. 8. Solve :
Q. 9. If a, b, c are all positive and are pth , qth and rth terms of G.P., then show that
Q. 10. If
= 0, then Prove that a, b, c are in G.P or x, y, z are in G.P
Q. 11. If x, y, z are different and
Q. 12. Show that points A (a, b + c), B (b, c + a), C (c, a + b) are collinear.
Q.13. Using properties of determinants,prove that:
Q.14. Using properties ,prove that:
Q. 15. Using properties of determinants solve for x:
Q16. Using properties prove that :
Q.17. Prove using properties of determinants.
Q. 18. Show that :
Q. 19. Prove that :
Q. 20.
ASSIGNMENT
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.]
Question.2 Use the product [−𝟒 𝟒 𝟒−𝟕 𝟏 𝟑𝟓 −𝟑 −𝟏
] [𝟏 −𝟏 𝟏𝟏 −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 𝟏𝟏 −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]
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
]
(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
],
Question. If A is singular matrix then under what condition set of equations AX = B may be consistent. [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.]
Q. The management committee of a residential colony decided to award
some of its members for honesty, some for helping others and some others
for supervising the workers to keep the colony neat and clean. The sum of
all the awardees is 12. Three times the sum of awardees for cooperation
and supervision added to two times the number of awardees for honesty is
33. If the sum of the number awardees for honesty and supervision is
twice the number of awardees for helping others, using matrix method,
find the number of awardees of each category. Apart from above values,
suggest one more value which the management of the colony must include
for awards.
[HINT: x+y+z = 12, 3(y+z) + 2x = 33, x – 2y +z = 0, |A|= 3, Adj A = 𝟗 −𝟑 𝟎𝟏 𝟎 −𝟏
−𝟕 𝟑 𝟏
X=3, Y=4 &Z=5. Regularity, sincerity.]
ASSIGNMENT ( WITH HINTS)(determinant)
Question: (i) Let ∆ = 𝑨𝒙 𝒙² 𝟏𝑩𝒚 𝒚² 𝟏
𝑪𝒛 𝒛² 𝟏
and ∆₁ = 𝑨 𝑩 𝑪𝒙 𝒚 𝒛𝒛𝒚 𝒛𝒙 𝒙𝒚
, then ∆ - ∆′₁
= 0
[Hint ∆₁ = 𝒙𝒚𝒛
𝒙𝒚𝒛
𝑨𝒙 𝑩𝒚 𝑪𝒛
𝒙² 𝒚² 𝒛²𝟏 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 𝟏𝟐𝒔𝒊𝒏𝒕 𝒕 𝟐𝒕𝒔𝒊𝒏𝒕 𝒕 𝒕
, then 𝐥𝐢𝐦𝒕→𝟎
𝒇(𝒕)
𝒕² is equal to 0,1,2,3.
[Hint 0, 𝒇(𝒕)
𝒕² =
𝒄𝒐𝒔𝒕 𝒕 𝟏𝟐𝒔𝒊𝒏𝒕
𝒕𝟏 𝟐
𝒔𝒊𝒏𝒕
𝒕𝟏 𝟏
→ 𝟏 𝟎 𝟏𝟐 𝟏 𝟐𝟏 𝟏 𝟏
as t→∞ ].
(iv) There are two values of a which makes determinant ∆ = 𝟏 −𝟐 𝟓𝟐 𝒂 −𝟏 𝟎 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 𝟏 𝒃 𝒄 + 𝒂 𝟏
𝑪 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
(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) 𝒃²𝒄² 𝒃𝒄 𝒃+ 𝒄𝒄²𝒂² 𝒄𝒂 𝒄+ 𝒂𝒂²𝒃² 𝒂𝒃 𝒂+ 𝒃
= 0 ⇨ 𝒂𝒃𝒄
𝒂𝒃𝒄
𝒃²𝒄² 𝒃𝒄 𝒃 + 𝒄𝒄²𝒂² 𝒄𝒂 𝒄 + 𝒂𝒂²𝒃² 𝒂𝒃 𝒂 + 𝒃
= 𝟏
𝒂𝒃𝒄 𝒂𝒃²𝒄² 𝒂𝒃𝒄 𝒂𝒃+ 𝒂𝒄𝒃𝒄²𝒂² 𝒃𝒄𝒂 𝒃𝒄+ 𝒃𝒂𝒄𝒂²𝒃² 𝒄𝒂𝒃 𝒄𝒂+ 𝒄𝒃
𝒂𝒃𝒄.𝒂𝒃𝒄
𝒂𝒃𝒄 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 𝒂² − 𝒃𝒄𝟏 𝒃 𝒃² − 𝒄𝒂𝟏 𝒄 𝒄² − 𝒂𝒃
= 0 ⇨ 𝟏 𝒂 𝒂²𝟏 𝒃 𝒃²𝟏 𝒄 𝒄²
- 𝟏 𝒂 𝒃𝒄𝟏 𝒃 𝒄𝒂𝟏 𝒄 𝒂𝒃
= 𝟏 𝒂 𝒂²𝟏 𝒃 𝒃²𝟏 𝒄 𝒄²
- 𝟏
𝒂𝒃𝒄 𝒂 𝒂² 𝒂𝒃𝒄𝒃 𝒃² 𝒃𝒄𝒂𝒄 𝒄𝟐 𝒄𝒂𝒃
𝟏 𝒂 𝒂²𝟏 𝒃 𝒃²𝟏 𝒄 𝒄²
- 𝒂 𝒂² 𝟏𝒃 𝒃² 𝟏𝒄 𝒄𝟐 𝟏
= 0 (PassC3overthefirsttwocolumns.)
(v) 𝟏 𝒂 𝒂𝟐𝟏 𝒃 𝒃𝟐𝟏 𝒄 𝒄𝟐
= 𝟏 𝒃𝒄 𝒃+ 𝒄𝟏 𝒄𝒂 𝒄 + 𝒂𝟏 𝒂𝒃 𝒂 + 𝒃
R.H.S. 𝟏
𝒂𝒃𝒄
𝒂 𝒂𝒃𝒄 𝒂𝒃 + 𝒂𝒄𝒃 𝒂𝒃𝒄 𝒃𝒄+ 𝒃𝒂𝒄 𝒂𝒃𝒄 𝒄𝒂+ 𝒄𝒃
= 𝟏 𝒂 𝒂𝒃+ 𝒂𝒄𝟏 𝒃 𝒃𝒄+ 𝒃𝒂𝟏 𝒄 𝒄𝒂+ 𝒄𝒃
( applying C1 ↔ C2)
= - 𝟏 𝒂 −𝒃𝒄𝟏 𝒃 −𝒄𝒂𝟏 𝒄 −𝒂𝒃
(apply C3 → C3 – (ab+bc+ca))C1)
= 𝟏
𝒂𝒃𝒄
𝒂 𝒂² 𝒂𝒃𝒄𝒃 𝒃² 𝒂𝒃𝒄𝒄 𝒄𝟐 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)
𝐥𝐨𝐠𝒂 𝒑 𝟏𝐥𝐨𝐠𝐛 𝒒 𝟏𝐥𝐨𝐠𝐜 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 𝒂²𝟏 𝒃 𝒃²𝟏 𝒄 𝒄²
(same method as given below) (viii)
𝒂 𝒂² 𝒃𝒄𝒃 𝒃² 𝒄𝒂𝒄 𝒄𝟐 𝒂𝒃
= 𝟏 𝒂² 𝒂³𝟏 𝒃² 𝒃³𝟏 𝒄² 𝒄³
( Multiply by abc as R1 with a,R2 by b and R3 by c then divide with abc )
Find the values of: (ix)
𝟏 𝒂𝒃𝟏
𝒂+
𝟏
𝒃
𝟏 𝒃𝒄𝟏
𝒃+
𝟏
𝒄
𝟏 𝒄𝒂𝟏
𝒂+
𝟏
𝒄
(Operate C2 → 𝟏
𝒂𝒃𝒄 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 + 𝒂² − 𝒃² 𝟐𝒂𝒃 −𝟐𝒃
𝟐𝒂𝒃 𝟏 − 𝒂² + 𝒃² 𝟐𝒂𝟐𝒃 −𝟐𝒂 𝟏 − 𝒂² − 𝒃²
= (1+a2+b2)3 ( Apply C1 → (C1 - bC3) and C2 → (C2+aC3)
(b)
𝒄𝒐𝒔𝜶𝒄𝒐𝒔𝜷 𝒄𝒐𝒔𝜶𝒔𝒊𝒏𝜷 −𝒔𝒊𝒏𝜶−𝒔𝒊𝒏𝜷 𝒄𝒐𝒔𝜷 𝟎
𝒔𝒊𝒏𝜶𝒄𝒐𝒔𝜷 𝒔𝒊𝒏𝜶𝒔𝒊𝒏𝜷 𝒄𝒐𝒔𝜶 = 1 (Apply R3 → sin𝛂R3 + cos𝛂R1)
(C)
(𝒚 + 𝒛)² 𝒙𝒚 𝒛𝒙
𝒙𝒚 (𝒙+ 𝒛)² 𝒚𝒛
𝒙𝒛 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
𝒄+ 𝒂)² 𝒃𝟐 𝒄𝒂
𝒂 + 𝒃)² 𝒄𝟐 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 (𝟏
𝒂 +
𝟏
𝒃 +
𝟏
𝒄+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 + 𝒂² − 𝒃² 𝟐𝒂𝒃 −𝟐𝒃
𝟐𝒂𝒃 𝟏 − 𝒂² + 𝒃² 𝟐𝒂𝟐𝒃 −𝟐𝒂 𝟏 − 𝒂² − 𝒃²
= (𝟏 + 𝒂² + 𝒃²)3
Apply C1→C1-b C3, C2→C2+a C3, we get 𝟏 + 𝒂² + 𝒃² 𝟎 −𝟐𝒃
𝟎 𝟏 + 𝒂² + 𝒃² 𝟐𝒂𝒃(𝟏+ 𝒂𝟐 + 𝒃𝟐) −𝒂(𝟏+ 𝒂𝟐 + 𝒃𝟐) 𝟏 − 𝒂² − 𝒃²
= (𝟏 + 𝒂² + 𝒃²)² 𝟏 𝟎 −𝟐𝒃𝟎 𝟏 𝟐𝒂𝒃 −𝒂 𝟏 − 𝒂² − 𝒃²
expand along C1, We get (𝟏 +
𝒂² + 𝒃²)³ .
**(h) Evaluate
(𝑿𝟏) (𝑿
𝟐) (𝑿
𝟑)
(𝒀𝟏) (𝒀
𝟐) (𝒀
𝟑)
(𝒁𝟏) (𝒁
𝟐) (𝒁
𝟑)
where (𝑿𝟏) =C(x,1) ( binomial
coefficient)
Solution:
𝒙
𝟏!
𝒙(𝒙−𝟏)
𝟐!
𝒙(𝒙−𝟏)(𝒙−𝟐)
𝟑!𝒚
𝟏!
𝒚(𝒚−𝟏)
𝟐!
𝒚(𝒚−𝟏)(𝒚−𝟐)
𝟑!𝒛
𝟏!
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
𝒙𝒚𝒛
𝟏𝟐 𝟏 𝒂 𝒂²𝟏 𝒃 𝒃²𝟏 𝒄 𝒄²
= 𝒙𝒚𝒛
𝟏𝟐 (a-b)(b-c)(c-a) =
𝒙𝒚𝒛
𝟏𝟐 (x-y)(y-z)(z-x).
Question: If x,y,z are all different and if 𝒙 𝒙² 𝟏 + 𝒙³𝒚 𝒚² 𝟏 + 𝒚³
𝒛 𝒛² 𝟏 + 𝒛³
= 0 , prove
that xyz = -1
Solution: 𝒙 𝒙² 𝟏 + 𝒙³𝒚 𝒚² 𝟏 + 𝒚³
𝒛 𝒛² 𝟏 + 𝒛³
= 𝒙 𝒙² 𝟏𝒚 𝒚² 𝟏
𝒛 𝒛² 𝟏
+ 𝒙 𝒙² 𝒙³𝒚 𝒚² 𝒚³
𝒛 𝒛² 𝒛³
=
𝒙 𝒙² 𝟏𝒚 𝒚² 𝟏
𝒛 𝒛² 𝟏
+ xyz 𝟏 𝒙 𝒙²𝟏 𝒚 𝒚²
𝟏 𝒛 𝒛²
= 0
𝟏 𝒙 𝒙²𝟏 𝒚 𝒚²
𝟏 𝒛 𝒛²
(1+xyz) = 0 ⇨ (x-y)(y-z)(z-x)(1+xyz) = 0 ⇨ xyz=-1 ∵
x ≠y≠ z. Question: By using properties of determinant,show that
𝒂² + 𝟏 𝒂𝒃 𝒂𝒄
𝒂𝒃 𝒃² + 𝟏 𝒃𝒄𝒄𝒂 𝒄𝒃 𝒄² + 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 respectively R1→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:
𝟐
𝒙 +
𝟑
𝒚 +
𝟏𝟎
𝒛 = 4,
𝟒
𝒙 -
𝟔
𝒚 +
𝟓
𝒛 = 1,
𝟔
𝒙 +
𝟗
𝒚 -
𝟐𝟎
𝒛 = 2; x, y, z ≠ 0.
[X=2,Y=3,Z=5,|A|=1200,adjA =[𝟕𝟓 𝟏𝟓𝟎 𝟕𝟓𝟏𝟏𝟎 −𝟏𝟎𝟎 𝟑𝟎𝟕𝟐 𝟎 −𝟐𝟒
] ]
(ii) 𝟏
𝒙 -
𝟏
𝒚 +
𝟏
𝒛 =4,
𝟐
𝒙 +
𝟏
𝒚 -
𝟑
𝒛 = 0,
𝟏
𝒙 +
𝟏
𝒚 +
𝟏
𝒛 = 2
[ x=1/2,y=-1,z=1 adjA= [𝟒 𝟐 𝟐
−𝟓 𝟎 𝟓𝟏 −𝟐 𝟑
] |A| = 10]
(iii) 𝟐
𝒙 -
𝟑
𝒚 +
𝟑
𝒛 = 10,
𝟏
𝒙 +
𝟏
𝒚 +
𝟏
𝒛 = 10,
𝟑
𝒙 -
𝟏
𝒚 +
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
] ]
Relations & functions
Q.1. Show that the relation R defined by (a, b) R (c, d) ⟹ a + d = b + c on the set N×N is an equivalence relation.
Q.Show that ƒ: N N defined by ƒ (x) = , if n is odd
, if n is even is many-one onto function.
Composition of Function and Invertible Function.
Q.1. If f(x) = x + 7 and g(x) = x – 7, x ε R, find (fog)(7).
Solution :
We have, f(x) = x + 7 and g(x) = x – 7. Then (fog)(x) = fo(g(x)) = f(x – 7) = (x – 7) + 7 = x. Therefore, (fog)(7) = 7. [Ans.]
Q.2. If f : R→ R and g : R→ R are defined respectively as f(x) = x2 + 3x + 1 and g(x) = 2x – 3, find (a) fog , (b) gof.
Solution : [ 4x2 – 6x + 1. [Ans.] , 2x2 + 6x – 1. [Ans.]
Q.3. If f : R→ R defined as f(x) = (2x – 7)/4 is an invertible function, find f–1. [f –1(x) = (4x + 7)/2. [Ans.]
Q.4. If f : R→ R defined by f(x) = (3x + 5)/2 is an invertible function, find f–1.Solution : [ f–1(x) = (2x – 5)/3. [Ans.]
1.4 Binary Operations.
Q.1.
i. Is the binary operation *, defined on set N, given by a*b = (a + b)/2, for all a, b ε N, commutative ?
ii. Is the above binary operation * associative ?
Q.2. Let * be a binary operation defined by a*b = 2a + b – 3. Find 3*4.
ASSIGNMENT( Relations & Functions)
Q. 1 Show that the function f : N →N, defined by f(x) = x2 + x +1 is one-one but not onto.
Q. 2. Let f: N be a function defined as f ( x ) = 4x2 + 12x + 15 .Show
that f: N where S is the range of f, is invertible .Find the inverse of f .
Q. 3. Show that the relation R on the set R of all real numbers, defined as R = { (a, b): a2 + b2 =1 } is neither reflexive nor transitive but symmetric
Q. 4. Show that ƒ : R –{0} R–{0} given by ƒ (x) = 3/x is invertible and it is inverse of itself.
Q. 5 If R1 & R2 are equivalence relations on set A. Show that R1 U R2 is reflexive, symmetric but not transitive.
Q. 6. Let ƒ (x) = [x] and g(x) = |x|, find goƒ (-5/3) – ƒog (-5/3)
Q. 7. Show that the function ƒ: R R defined by ƒ (x) = 3x3 + 5 for x R is a bijection.
Q. 8. Show that the relation R on the set R of all real numbers, defined as R = { (a, b): |a| ≤ b } is neither reflexive nor symmetric but transitive.
Q. 9. Show that the function ƒ : N → N given by ƒ (1) = ƒ (2) =1 and ƒ (x) = x-1,for every x > 2 is onto but not one-one.
Q. 10. If * is a binary operation on R defined by a * b = a/4 + b/7 for a, b R, find the value of ( 2 * 5 ) * 7
Q. 11 Let R be a relation on NXN, defined by (a,b) R (c,d) ⇔ ad(b+c)) = bc(a+d), ∀(a,b) , (c,d) є NXN ,Show that R is an equivalence relation on NXN.
Same for addition
Q. 12 Let R be a relation on NXN, defined by (a,b) R (c,d) ⇔ ad = bc, ∀(a,b) , (c,d) є NXN ,Show that R is an equivalence relation on NXN.
Q.13 Let A = {1,2,3}. Find the number of relations on A containing (1,2) & (2,3) which are reflexive, transitive but not symmetric.
Q.14 Show that the function ƒ : N → N given by f(x) =x – (-1)x, is bijection. ( same as ques.2 of misc.)
Q.15 Let f: [-1,∞) →[-1,∞) is given by f(x)= (x+1)2 – 1. Show that f is invertible and find the set = { x : f(x) = f-1 (x) }.
Relations & functions for class—XII Level—2 Q.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) = 𝒙
𝟐+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 1-1, as
consider any x1, x2 Є [-1, 1] such that f(x1) = f(x2) ⇨ 𝒙₁
𝟐+𝑿₁=
𝑿₂
𝟐+𝑿₂⇨ x1x2+2x1 =
x1x2 +2x2 ⇨ x1 = x2 For the range of f
Let y = f(x) ⇨ y = 𝒙
𝟐+𝒙 ⇨ xy +2y =x ⇨ (y – 1) x= -2y ⇨ x =
−𝟐𝒚
𝒚−𝟏
As x Є [-1, 1], so -1 ≤ −𝟐𝒚
𝒚−𝟏 ≤1 , but (y – 1)2 >0 , y ≠ 1⇨ -(y – 1)2 ≤
−𝟐𝒚
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
𝒙
𝟐+𝒙= y ⇨ xy +2y =x ⇨ 2y = x (1 – y) ⇨ x=
𝟐𝒚
𝟏−𝒚 f-1(y) = x =
𝟐𝒚
𝟏−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.
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. *.
Q.1 Let f(x) = 𝒙+𝟑,𝒊𝒇 𝒙<1𝟒𝒙−𝟐,𝒊𝒇 𝟏≤𝒙≤𝟒.𝒙²+𝟓,𝒊𝒇 𝒙>4 Find f(-1) ,f(4)
and f(5). Q.2 If f(x) = x2 – 𝟏/𝒙² , then find the value of f(x) + f (𝟏/𝒙²).
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) 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. Level
Inverse Trigonometric Functions
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+𝒀
𝟏−𝑿𝒀 ; xy<1
tan-1 x – tan-1 y= 𝐭𝐚𝐧−𝟏 𝑿−𝒀
𝟏+𝑿𝒀 ; xy>-1
2tan-1 x= 𝐭𝐚𝐧−𝟏 𝟐𝒙
𝟏−𝒙² = 𝐬𝐢𝐧−𝟏 𝟐𝒙
𝟏+𝒙² = 𝐜𝐨𝐬−𝟏 𝟏−𝒙²
𝟏+𝒙²
Q. 1. Find the value of : tan-1 (1) + cos -1 (-1/2) + sin-1 (-1/2).
Q.2 Prove that 𝐭𝐚𝐧−𝟏 𝒙 + 𝐭𝐚𝐧−𝟏 𝟐𝒙
𝟏−𝒙² = 𝐭𝐚𝐧−𝟏 𝟑𝒙−𝒙³
𝟏−𝟑𝒙² , x<
𝟏
√𝟑
Q. 3. Solve : tan-12x + tan-13x = 𝝅
𝟒
Q. 4. Prove :
Q. 5. Solve : sin-1 ( 1 –x) – 2sin-1x = .
Q. 6. Evaluate: tan-1 - sec-1 (-2) + cosec-1 .
Q. 7. Prove :
tan-1 =
Q. 8. Simplify :
sin-1 ,
Q. 9. Prove: sec2 (tan-12) + cosec2 ( cot-13) = 15.
Q. 10. Simplify :
tan-1
Q. 11. Prove :
tan-1 =
Q. 12. If sin(sin-1 , then find the value of x.
Q. 13. Prove that :
2tan-1 = cos-1
Q. 14. Find the principal value of Sec-1
Q. 15. Find value of
Sin
Q. 1. If 2 tan-1 (Cos q) = tan-1 (2 cosec q), find q.
Q.2
Question.3 if 𝐜𝐨𝐬−𝟏 𝒙
𝒂 + 𝐜𝐨𝐬−𝟏 𝒚
𝒃 = 𝜽, then prove that
𝒙²
𝒂² -
𝟐𝒙𝒚
𝒂𝒃 cos𝜽
+ 𝒚²
𝒂² = sin2𝜽 [Hint: 𝐜𝐨𝐬−𝟏 𝒙
𝒂 + 𝐜𝐨𝐬−𝟏 𝒚
𝒃 = 𝐜𝐨𝐬−𝟏 [
𝒙𝒚
𝒂𝒃 - √𝟏 −
𝒙²
𝒂²√𝟏 −
𝒚²
𝒃² ] =
𝜽 ⇨ (𝒙𝒚
𝒂𝒃− cos𝜽 )2 = (√𝟏 −
𝒙²
𝒂²√𝟏 −
𝒚²
𝒃²)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 − 𝒙²)(𝟏− 𝒚²) - xy = - √𝟏 − 𝒛² and simply it.
[Hint: for (ii) tan-1 x + tan-1y = 𝐭𝐚𝐧−𝟏 𝑿+𝒀
𝟏−𝑿𝒀 ]
Question.5 Write the following functions in the simplest
form:(i)𝐭𝐚𝐧−𝟏 (𝒄𝒐𝒔𝒙
𝟏+𝒔𝒊𝒏𝒙 ) (ii) 𝐭𝐚𝐧−𝟏 (
𝒄𝒐𝒔𝒙
𝟏−𝒔𝒊𝒏𝒙) (iii) 𝐭𝐚𝐧−𝟏 √
𝒂−𝒙
𝒂+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 𝒙
𝒂
]Question.6 If y = 𝐜𝐨𝐭−𝟏(√𝒄𝒐𝒔𝒙) - 𝐭𝐚𝐧−𝟏(√𝒄𝒐𝒔𝒙), prove that siny =
tan2(x/2). [Hint: y = 𝝅
𝟐 - 2 𝐭𝐚𝐧−𝟏(√𝒄𝒐𝒔𝒙) , use formula 2𝐭𝐚𝐧−𝟏 𝒙 =
𝐜𝐨𝐬−𝟏(𝟏−𝒙²
𝟏+𝒙²)]
Question.7 (i) Prove that 𝐭𝐚𝐧−𝟏 𝟏 + 𝐭𝐚𝐧−𝟏 𝟐 + 𝐭𝐚𝐧−𝟏 𝟑 = π.
(ii) Prove that 𝐜𝐨𝐭−𝟏(𝒂𝒃+𝟏
𝒂−𝒃) +𝐜𝐨𝐭−𝟏(
𝒄𝒃+𝟏
𝒃−𝒄) + 𝐜𝐨𝐭−𝟏(
𝒂𝒄+𝟏
𝒄−𝒂) = 0.
[Hint: for (i) 𝐭𝐚𝐧−𝟏 𝟐 = 𝝅
𝟐 - 𝐜𝐨𝐭−𝟏 𝟐 =
𝝅
𝟐− 𝐭𝐚𝐧−𝟏 𝟏
𝟐 , then use
formula of tan-1 x + tan-1y = 𝐭𝐚𝐧−𝟏 𝑿+𝒀
𝟏−𝑿𝒀 ]
(ii) [Hint: write 𝐜𝐨𝐭−𝟏 𝒙 = 𝐭𝐚𝐧−𝟏 𝟏
𝒙 ]
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 √𝒙² + 𝒙 + 𝟏 = 𝝅
𝟐 .
(i) [Hint: write 𝐜𝐨𝐬−𝟏 𝒙 = 𝝅
𝟐 - 𝐬𝐢𝐧−𝟏 𝒙, put 𝐬𝐢𝐧−𝟏 𝒙 = y]
(ii) [Hint: use 𝐭𝐚𝐧−𝟏 𝒙 = 𝐜𝐨𝐬−𝟏 𝟏
√𝟏+𝒙² ] Indu thakur
Question.9 Using principal values, evaluate 𝐜𝐨𝐬−𝟏(𝒄𝒐𝒔𝟐𝝅
𝟑) +
𝐬𝐢𝐧−𝟏( 𝒔𝒊𝒏𝟐𝝅
𝟑). [answer is π]
Question.10 Show that tan(𝟏
𝟐𝐬𝐢𝐧−𝟏 𝟑
𝟒) =
𝟒− √𝟕
𝟑 and justify why the
other value is ignored?
[ Hint: put 𝟏
𝟐𝐬𝐢𝐧−𝟏 𝟑
𝟒 =∅ ⇨ ¾ = sin2∅ = 2tan∅/(1+tan2∅), find
tan∅]
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).
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 - 𝟏
𝟕𝟐𝟗 =
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---2
Q.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 = 𝟏𝒐𝒓𝟐
𝒌(𝟓− 𝒙)𝒊𝒇 𝒙 = 𝟑𝒐𝒓𝟒𝒐, 𝒐𝒕𝒉𝒆𝒓𝒘𝒊𝒔𝒆
(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—2
Ans.1 A and B are possible events ⟹A ≠ ф⇨ P(A)≠ 0 , P(B) ≠0 But 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 4
P(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 150. P(A/ E1)= 5/100, P(A/ E2)= 10/100 By Baye’s
theorem P(E1/A) = 3/7.
Probability
Q. 1. A die is thrown twice and the sum of the numbers appearing is observed to be 6. What is the conditional probability that the number 4 has appeared at least once. [2/5]
Q. 2. Assume that each born child is equally likely to be a boy or a girl. If a family has two children, what is the conditional probability that both are girls given that at least one is a girl. [1/3]
Q. 3. If A and B are two independent events, show that the probability of occurrence of at least one of A and B is given by : 1 – P(A').P(B')
Q. 4. An urn contains 5 red and 5 black balls. A ball is drawn at random, its colour is noted and is returned to the urn. Moreover, 2 additional balls of the colour drawn are put in the urn and then a ball is drawn at random. What is the probability that the second ball is red.[1/2]
Q. 5. Two cards are drawn simultaneously (or successively without replacement) from a well shuffled pack of 52 cards. Find the mean and standard deviation of the number of kings.[34/221, 0.37]
Q. 6.
Q. 7. A doctor is to visit a patient .From the past experiences , it is known that the probabilities that he will come by train , bus , scooter
or by other means of transport are respectively . The
probabilities that he will be late are , if he comes by train , bus, and scooter respectively , but if he come by other means of transport , then he will not be late .When he arrives he is late .What is the probability that he comes by train ? which life skill is the doctor lacking?, ans. Lacks responsibility & dedication to his work.
13.1. Conditional Probability.
Q.1. A and B toss a coin alternately till one of them gets a head and wins the game. If A starts first, find the probability that B will win the game. [1/3.]
Q.2. A and B throw two dice simultaneously turn by turn. A will win if he throws a total of 5, B will win if he throws a doublet. Find the probability that B will win the game, though A started it. [4/7]
Q.3. Two dice are rolled once. Find the probability that :
i. the numbers on two dices are different. ii. the total of numbers on the two dice is at least 4.
[11/12]
Q.4. Two unbiased dice are tossed simultaneously. Find the probability that the sum of the numbers will be a multiple of 3 or 5.
[19/36]
Q.5. Two unbiased dice are thrown. Find the probability that the sum of the numbers obtained on the two dice is neither a multiple of 3 nor a multiple of 4.
[4/9]
13.2. Multiplication Theorem on Probability/Independent Events.
Q.1. There are two bags. The first bag contains 4 white and 2 black balls, while the second bag contains 3 white and 4 black balls. A ball is picked up at random and a ball is drawn out. Find the probability that it is a white ball.
[23/42]
Q.2. In a group of 9 students, there are 5 boys and 4 girls. A team of 4 students is to be selected for a quiz competition. Find the probability that there will be 2 boys and 2 girls in that team.[10/21]
Q.3. 12 cards numbered 1 to 12, are placed in a box, mixed up thoroughly and then a card is drawn at random from the box. If it is known that the number on the drawn card is more than 3, find the probability that it is an even number. [5/9]
13.3. Bayes’ Theorem.
Q.1. In a bolt factory, machines A, B, C manufacture 25%, 35% and 40% respectively of the total bolts. Of their output 5%, 4% and 2% respectively are defective bolts. A bolt is drawn at random and is found to be defective. Find the probability that it is manufactured by machine B.
28/69.[Ans.]
Q.2. In a bulb factory, machines A, B and C manufacture 60%, 30% and 10% bulbs respectively. 1%, 2% and 3% of the bulbs produced respectively by A, B and C are found to be defective. A bulb is picked up at random from the total production and found to be defective. Find the probability that this bulb was produced by the machine A.
Ans. = 2/5.]
Q.3. A class consists of 50 students out of which there are 10 girls. In the class 2 girls and 5 boys are rank holders in an examination. If a student is selected at random from the class and is found to be a rank holder, what is the probability that the student selected is a girl ?
2/7 . [Ans.]
Q.4. A man is known to speak the truth 3 out of 4 times. He throws a
die and reports that it is a six. Find the probability that it is actually a
six. How is truthfulness helpful in life?----- Ans.--- truthfulness is essential life
skill. Everybody trusts a truthful person.
3/8. [Ans.]
Q.5. A company has two plants which manufacture scooters. Plant I manufactures 80% of the scooters while Plant II manufactures 20% of the scooters. At Plant I, 85 out of 100 scooters are rated as being of standard quality, while at Plant II only 65 out of 100 scooters are rated as being of standard quality. If a scooter is of standard quality , what is the probability that it come from Plant I.
0.84 .[Ans.]
Q.6. A manufacturing firm produces steel pipes in three plants A, B and C with daily production of 500, 1000 and 2000 units respectively. The fractions of defective steel pipes output produced by the plant A, B and C are respectively 0.005, 0.008 and 0.010. If a pipe is selected from a day’s total production and found to be defective, find out the probability that it came from the first plant.
5/61[Ans.]
Q.7. An insurance company insured 6000 scooter drivers, 3000 car drivers and 9000 truck drivers. The probability of an accident involving a scooter, a car and a truck is 0.02, 0.06 and 0.30 respectively. One of the insured persons meets with an accident. Find the probability that he is a car driver.
0.06.[Ans.]
Q.8. An insurance company insured 4000 doctors, 8000 teachers and 12000 engineers. The probabilities of a doctor, a teacher and an engineer dying before the age of 58 years are 0.01, 0.03 and 0.05 respectively. If one of the insured person dies before the age of 58 years, find the probability that he is a doctor.
Solution : Do yourself. [Ans. = 1/22]
Q.9. A car manufacturing factory has two plants X and Y. Plant X manufactures 70% of the cars and plant Y manufactures 30%. At plant X, 80% of the cars are rated of standard quality and at plant Y, 90% are rated of standard quality. A car is picked up at random and is found to be of standard quality. Find the probability that it has come from plant X.
56/83. [Ans.]
Q. 10 in an examination , an examinee either guesses or copies or
knows the answer of MCQs with four choices. The prob. That he makes
a guess is 1/3 and the prob. That he copies answer is 1/6, the prob.
That his answer is correct, given that he copied it , is 1/8. Find the
prob. That he copies the answer, given that he correctly answered it.if
a student copies an answer, what value is he violating?
[hint P(A)=1/3 , P(B)= 1/6 , P(C) = 1 – 1/3 – 1/6 =1/2 , P(E/A)=1/4,
P(E/B)=1/8, P(E/C)=1 & P(B/E)= 1/29]
13.4. Probability Distribution.
Q.1. Two cards are drawn successively with replacement from a well shuffled pack of 52 cards. Find the probability distribution of number of jacks.
Hence required probability distribution is :
x 0 1 2
P(X) 144/169 24/169 1/169
Q.2. Two cards are drawn successively with replacement from a well shuffled deck of 52 cards. Find the probability distribution of number of aces.
Solution :Do yourself. [Ans. of Q.1. above]
Q.3. A pair of dice is tossed twice. If the random variable X is defined as the number of doublets, find the probability distribution of X. Hence, the required probability distribution is
x 0 1 2
P(X) 25/36 10/36 1/36
Q.4. Find the probability distribution of the number of successes in two tosses of a die, where a success is defined as a number less than 3. Also find the mean and the variance of the distribution.
Mean = Σpixi = 4/9×0 + 4/9×1 + 1/9×2 = 6/9 = 2/3.[Ans.] Variance = Σpixi2 – (Mean)2 = (4/9×0 + 4/9×12 + 1/9×22) – (2/3)2 = 8/9 – 4/9 = 4/9.[Ans.]
Q.5. An urn contains 5 white and 3 red balls. Find the probability distribution of the number of red balls, with replacements, in three draws.
Hence the required probability distribution is
x 0 1 2 3
P(X) 125/512 225/512 135/512 27/512
Q.6. A pair of dice is thrown 4 times. If getting a doublet is considered a success, find the probability distribution of number of success. Therefore, required probability distribution is
x 0 1 2 3 4
P(X) 625/1296 500/1296 150/1296 20/1296 1/1296
13.5. Binomial Distribution.
Q.1. The mean and variance of a binomial distributions are 4 and 4/3 respectively. Find the distribution and P(X ≥ 1).
728/729.[Ans.]
Q. 2 A drunk man takes a step forward with prob. 0.4 &backward with
prob. 0.6. find the prob. That at the end of 11 steps, he is just one step
away from the starting point. Is drinking alcohol a good habit?
[ hint required prob. = P(X=5) + P(X=6) = 462(0.24)5 ]
ASSIGNMENT(3-DIMENTIAL GEOMETRY)
Direction Ratio & Direction Cosines of a Line.
Q.1. The equation of a line is given by (4 – x)/2 = (y + 3)/3 = (z + 2)/6. Write the direction cosines of a line parallel to the above line.
l = – 2/7, m = 3/7, n = 6/7 [Ans.]
Q.2. The equation of a line is (2x – 5)/4 = (y + 4)/3 = (6 – z)/6. Find the direction cosines of a line parallel to this line. [Ans. = 2/7, 3/7, – 6/7]
11.2. Equation of a Line in Space.
Q.1. Find the foot of the perpendicular drawn from the point P(1, 6, 3) on the line x/1 = (y – 1)/2 = (z – 2)/3. Also find the distance from P.
PQ = √13 units. [Ans.]
Q.3. Find the point on the line : (x + 2)/3 = (y + 1)/2 = (z – 3)/2 at a distance 3√2 from the point (1, 2, 3).
P(56/17, 43/17, 111/17).
Q.4. Find the equation of the perpendicular drawn from the point (2, 4, – 1) to the line (x + 5)/1 = (y + 3)/4 = (z – 6)/–9 .
(x – 2)/6 = (y – 4)/3 = (z + 1)/2 [Ans.]
Q.2. Find the length and the foot of the perpendicular drawn from the point (2, – 1 , 5) to the line (x – 11)/10 = (y + 2)/– 4 = (z + 8)/ –11.
[Ans. = Point (1, 2, 3); distance = √14 units.]
11.3. Angle Between Two Lines.
Q.1. Find the angle between the pair of lines given by r→ = (2i – 5j + k) + λ(3i + 2j + 6k) and r→ = 7i – 6j – 6k + μ(i + 2j + 2k).
θ = cos –1(19/21). [Ans.]
11.4. Shortest Distance Between Two Lines.
Q.1. The vector equations of two lines are : r→ = i + 2j + 3k + λ(i – 3j + 2k ) and r→ = 4i + 5j + 6k + μ(2i + 3j + k). Find the shortest distance between the above lines.
3/√19 units. [Ans.]
Q.2. Find the shortest distance between the following lines : (x – 3)/1 = (y – 5)/–2 = (z – 7)/1 and (x + 1)/7 = (y + 1)/–6 = (z + 1)/1.
2√29. [Ans.]
Equation of a Plane.
Q.1. Find the equation of the plane passing through the points (1, 2, 3) and (0, – 1, 0) and parallel to the line (x – 1)/2 = (y + 2)/3 = z/–3.
6x – 3y + z = 3. [Ans.]
Q.2. Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4). 5x – 7y + 11z + 4 = 0. [Ans.]
Q.3. Find the equation of the plane passing through the point (–1, –1, 2) and perpendicular to each of the following planes : 2x + 3y – 3z = 2 and 5x – 4y + z = 6.
11x + 17y + 23z – 18 = 0 [Ans.]
Q.4.Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line (x + 3)/2 = (y – 3)/7 = (z – 2)/5.
8x – 13y + 15z + 13 = 0. [Ans.]
11.6. Distance of a Point from a Plane.
Q.1. Find the image of the point (1, 2, 3) in the plane x + 2y + 4z = 38.
(3, 6, 11). [Ans.]
Q.2. Find the co-ordinates of the image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0. [Ans. = (–3, 5, 2) ]
Q.3. From the point P(1, 2, 4), a perpendicular is drawn on the plane 2x + y – 2z + 3 = 0. Find the equation, the length and the co-ordinates of the foot of the perpendicular. (11/9, 19/9, 34/9) [Ans.] Length of perpendicular from (1, 2, 4) is PM = |{2(1) + 2 – 2(4) + 3}/√(4 + 1 + 4)| = 1/3 unit. [Ans.]
Q.4. Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, – 1 , 2), B(5, 2, 4) and C(– 1, – 1 , 6).
6/√(34). [Ans.]
Q.5. Find the co-ordinates of the point where the line (x + 1)/2 = (y + 2)/3 = (z + 3)/4 meets the plane x + y + 4z = 6. P(1, 1, 1) [Ans.]
Q.6. Find the distance of the point (– 2, 3 – 4) from the line (x + 2)/3 = (2y + 3)/4 = (3z + 4)/5 measured parallel to the plane 4x + 12y – 3z + 1 = 0. 17/2.[Ans.]
Class – XII Subject – Mathematics (Three Dimensional Geometry)
1. Find the d.c’s of X, Y and Z-axis.[example 1,0,0(x-axis)0,1,0(y-axis]
2. The equation of a line is given by (4 – x)/2 = (y + 3)/3 = (z + 2)/6. Write the direction cosines of a line parallel to the above line. [Ans: – 2/7, 3/7, 6/7]
3. The equation of a line is (2x – 5)/4 = (y + 4)/3 = (6 – z)/6. Find the direction cosines of a line parallel to this line. [ Ans : 2/7, 3/7, – 6/7]
4. Find coordinates of the foot of the per. drawn from the origin to the plane 2x – 3y + 4z -6 = 0. [example ans. 12/29, -18/29, 24/29]
5. Find the angle between the line (x + 1)/2 = y/3 = (z-3)/6 and the
plane 10x + 2y – 11z = 3.[example ans. sinφ = |�� .��
|�� ||�� || = 8/21]
*6. Find the image of the point (1, 6, 3) in the line x = (y – 1)/2 = (z – 2)/3. [ Hint: it will be found that foot of per. from P to the line is N (1,3,5). If Q(𝜶, 𝜷, 𝜸) is the image of P then N is the mid point of PQ
⇨ 𝜶+𝟏
𝟐= 𝟏,
𝜷+𝟔
𝟐= 𝟑,
𝜸+𝟑
𝟐= 𝟓 ⇨ Q(1, 0, 7)]
7. Find the foot of the perpendicular drawn from the point P(1, 6, 3) on the line x/1 = (y – 1)/2 = (z – 2)/3. Also find the distance from P. [ Ans : √13 units., eqn. Of line is (x-1)/0=(y-6)/-3=(z+1)/2]
8. Find the length and the foot of the perpendicular drawn from the point (2, – 1 , 5) to the line (x – 11)/10 = (y + 2)/– 4 = (z + 8)/ –11. [ Ans. = Point (1, 2, 3); distance = √14 units.]
*9. Show that the angles between the diagonals of a cube is cos-
1(1/3).
Ans. (same as example. 26 of misc.) Let the length of cube be a , in fig. Of example : A,B,C are on x-axis, y-axis, z-axis resp.& F,G,D are in yz, xz ,xy –axis , O is origin , d.r’s of OE (E (a,a,a)) & AF are a,a,a & -a,a,a resp. ∴ cosѲ =(-a2+a2+a2)/√3a.√3a = 1/3(angle b/w two lines)
10. Find the length of the perpendicular drawn from the point (2, 3, 7) to the plane 3x – y – z = 7 . Also find the coordinates of the foot of the perpendicular. [ same as Q. 8 ]
*11. Find the point on the line (x + 2)/3 = (y + 1)/2 = (z – 3)/2 at a distance 3√2 from the point (1, 2, 3).
Ans. (x + 2)/3 = (y + 1)/2 = (z – 3)/2 =k , any point on the line A(3k-2, 2k-1, 2k+3) , if its distance from B(1,2,3) is 3√2, then AB² = (3√2)²
⇨ K= 0 OR 30/17 , ∴ point A(-2,-1,3) OR A(56/17,43/17,111/17) [by putting value of k in A]
12. Find the equation of the perpendicular drawn from the point (2, 4, – 1) to the line (x + 5)/1 = (y + 3)/4 = (z – 6)/–9 .
[ Ans : (x – 2)/6 = (y – 4)/3 = (z + 1)/2]
*13. Find the equation of the line passing through the point (-1, 3, -2) and perpendicular to the lines x = y/2 = z/3 and (x + 2)/(-3) = (y – 1)/2 = (z + 1)/5.
Ans. eqn. Of line (x+1)/a=(y-3)/b=(z+2)/c ........(i) ,where a,b,c are d.r’s , since (i) is per. to lines ∴ a+2b+3c=0 & -3a+2b+5c=0, after solving , we get a/4=b/-14=c/8 =k(say) , so eqn. Of line (by putting the values of a=4k, b=-14k,c=8k) is (x+1)/4 = (y-3)/-14 = (z+2)/8
14. Find the foot of the per. drawn from the point (0, 2, 3) on the line (x + 3)/5 = (y – 1)/2 = (z + 4)/3 Also, find the length of the perpendicular. [ans..(2,3,-1), use distance formula, length= √21]
*15. Find the equation of the plane passing through the line intersection of the planes x – 2y + z = 1 and 2x + y + z = 8 and parallel to the line with direction ratios (1, 2, 1) Also, find the perpendicular distance of the point P(3, 1, 2) from this plane.
Ans. equation of the plane passing through the line intersection of the planes x – 2y + z -1 +k(2x + y + z – 8)=0 or (1+2k)x+(-2+k)y+(1+k)z-1-8k=0 ........(i) is parallel to line ∴ (1+2k).1+(-2+k).2+(1+k).1=0 ⇨ k= 2/5(normal to the plane must be per. to the line), put k in (i) ⇨ 9x-8y+7z-21=0, per. dis. From P Is 22/√(194)
*19. Find the shortest distance between the following lines : (x – 3)/1 = (y – 5)/–2 = (z – 7)/1 and (x + 1)/7 = (y + 1)/–6 = (z + 1)/1.
[ncert Ans : 2√29 , use formula ]
20. Find the equation of the plane passing through the points (1, 2, 3) and (0, – 1, 0) and parallel to the line (x – 1)/2 = (y + 2)/3 = z/–3. [Ans :same as Q. 26 , 6x – 3y + z = 3]
21. Find the shortest distance between the following pairs of lines whose cartesian equations are : (x -1)/2 = (y + 1)/3 = z and (x + 1)/3 = (y – 2) , z = 2. [ SAME as Q. 19 ]
*22. Find the distance of the point P(2, 3, 4) from the plane 3x + 2y +2 z + 5=0 measured parallel to the line (x+3)/3 = (y-2)/6 = z/2
Ans. Any point on line (x+3)/3 = (y-2)/6 = z/2=k is Q(3k+2,6k+3,2k+4) It lies on plane ⇨ k=-1 ∴ point Q( -1,-3,2),PQ=7
23. Find the equation of the plane passing through the points (0, – 1, – 1), (4, 5, 1) and (3, 9, 4). [ Ans : 5x – 7y + 11z + 4 = 0]
*24. Find the equation of the plane passing through the point (–1, –1, 2) and perpendicular to each of the following planes : 2x + 3y – 3z = 2 and 5x – 4y + z = 6.
[ Ans : 9x + 17y + 23z – 20 = 0, same as example of misc. Eqn. Of plane a(x+1)+b(y+1)+c(z-2)=0........(i) By condition of
perpendicularity to the plane (i) with the planes 2a+3b-3c=0 & 5a-4b+c =0 , after solving , we get a=9c/23 & b=17c/23, put in (i) ]
25. Find the equation of the plane passing through the point (1, 1, -1) and perpendicular to the planes x + 2y + 3z – 7=0 and 2x – 3y + 4z = 0. [ SAME as Q. 24]
26. Find the equation of the plane passing through the points (3, 4, 1) and (0, 1, 0) and parallel to the line (x + 3)/2 = (y – 3)/7 = (z – 2)/5.
Ans : 8x – 13y + 15z + 13 = 0, eqn. Of plane a(x-3)+b(y-4)+c(z-1)=0.......(i), It passes through (0,1,0) ,then -3a-3b-c=0 // to line ∴ 2a+7b+5c=0, after solving a=8c/15, b=-13c/15, put in (i)
*27. Find the image of the point P(1,2,3) in the plane x+2y+4Z=38
[ Ans :Let M is foot of per. (mid point of PQ) , if Q (𝜶, 𝜷, 𝜸 ) , d.r’s of a normal to the plane are 1,2,4, eqn. Of line PM is (x-1)/1=(y-2)/2=(z-3)/4 = k ⇨ any point on a line Q(1+K,2+2K,3+4K) ∴ M{(2+k)/2,(4+2k)/2,(6+4k)/2} put in plane ⇨ k=2 ∴ Q(3,6,11) ]
28. Find the co-ordinates of the image of the point (1, 3, 4) in the plane 2x – y + z + 3 = 0. [Ans : (–3, 5, 2)]
*29. Find the distance between the point P(6, 5, 9) and the plane determined by the points A(3, – 1 , 2), B(5, 2, 4) and C(– 1, – 1 , 6).
Ans. [example28 of misc. Ans: 6/√(34),we can find eqn. Of plane through A,B,C and find distance of P from plane or can be found as
PD = 𝑨𝑷 . 𝑨𝑩 × 𝑨𝑪 ( proj. Of 𝑨𝑷 on 𝑨𝑩 × 𝑨𝑪 , D is foot of per. from P) or find eqn. of plane passes through A,B,C , then find dist. ]
30. Find the co-ordinates of the point where the line (x + 1)/2 = (y + 2)/3 = (z + 3)/4 meets the plane x + y + 4z = 6. [Ans: P(1, 1, 1)]
*31. Find the distance of the point A(– 2, 3 – 4) from the line (x + 2)/3 = (2y + 3)/4 = (3z + 4)/5 measured parallel to the plane 4x + 12y – 3z + 1 = 0.
[Ans: |AP|= 17/2, eqn. Can be written as (x + 2)/3 = (y + 3/2)/2 = (z + 4/3)/5/3 = t ⇨ P{ -2+3t, (-3/2)+2t, (-4/3)+5t/3} be any point on line D.R’S of AP {3t, 2t-(9/2), (5t/3)+(8/3)} , since AP is parallel to plane 4.3t+ (12).[ 2t-(9/2)]+ (-3). (5t/3)+(8/3) = 0 ⇨ t=2 ∴ P (4,5/2,2) ]
Ques. 32 Find the distance of the point P(-2, 3, -4) from the line (x+2)/3 = (2y+3)/4 = (3z+4)/6 measured // to the plane 4x + 12y -3 z + 1=0.
Ans. same as Q. 22, point on lineQ(3k-2,2k-(3/2),(5k/3)-(4/3)), d.r’s
of PQ are (3k,2k-(9/2),(5k/3)+(8/3)) , since PQ is // to plane ∴ PQ is per. to normal to the plane ⇨ (3k).4+{2k-(9/2)}.12+{(5k/3)+(8/3)}.(-3)=0 ∴ k= 2 ⇨ Q(4,5/2,2), PQ = 17/2
Question 33 find whether the lines �� = (i-j-k) +λ (2i+j) and �� =
(2i-j) +μ (i+j-k) intersect or not . if intersecting, find their point of
intersection.
Ans. 𝒂𝟐 -𝒂𝟏 = I +k , 𝒃𝟏 × 𝒃𝟐
=
𝒊 𝒋 𝒌𝟐 𝟏 𝟎𝟏 𝟏 −𝟏
= -i+2j+k , its
magnitude is √6
Shortest distance = = 0/√6 = 0
Hence lines intersect. Point of intersection is given by
(i-j-k) +λ (2i+j) = (2i-j) +μ (i+j-k) ⇨ (1+2λ) = 2+μ , -1+λ - -1+μ
and -1 = -μ ⇨ λ = 1 and μ =1 satisfy μ = 2λ -1 , put the values of
λ and μ in given lines , we obtain the position vector of point of
intersection of the two given lines as 3i- k i.e., the point of
intersection is (3,o,-1).
Ques. 34 Show that the four points (0,-1,-1), (4,5,1), (3,9,4) and (-4,4,4) are coplanar. Also find the eqn. of the plane containing them.
Ans. Eqn. of plane passes through (0,-1,-1) is a(x-0)+b(y+1)+(z+1)=0
It passes through (4,5,1), (3,9,4), we get 4a+6b+2c=0, 3a+10b+5c=0
⇨ eqn. of plane is 5x-7y+11z+4=0 and (-4,4,4) will satisff the eqn. of plane to be coplanar.
Q. 35 Show that lines (x+3)/-3 = (y-1)/1 = (z-5)/5 & (x+1)/-1=(y-2)/2 =(z-5)/5 are coplanar . find the eqn. of plane.
Ans.(as example 21), The given lines are coplanar if
(𝒙𝟐 − 𝒙𝟏) (𝒚𝟐 − 𝒚𝟏) (𝒛𝟐 − 𝒛𝟏)𝒍𝟏 𝒎𝟏 𝒏𝟏𝒍𝟐 𝒎𝟐 𝒏𝟐
= (−𝟏 + 𝟑) (𝟐 − 𝟏) 𝟎
−𝟑 𝟏 𝟓−𝟏 𝟐 𝟓
= 0 , so lines are coplanar. Eqn of plane
= (𝒙 − 𝒙𝟏) (𝒚 − 𝒚𝟏) (𝒛 − 𝒛𝟏)
𝒍𝟏 𝒎𝟏 𝒏𝟏𝒍𝟐 𝒎𝟐 𝒏𝟐
= (𝒙 + 𝟑) (𝒚 − 𝟏) (𝒛 − 𝟓)
−𝟑 𝟏 𝟓
−𝟏 𝟐 𝟓
= x-2y+z=0.
Q.36 Find the shortest distance and eqn. Of shortest distance b/w the lines (x-6)/3 = (y-7)/-1 = (z-4)/1 & x/-3 = (y+9)/2 = (z-2)/4 .
S (3p+6,-p+7,p+4) ......(i)
D (-3q,2q-9,4q+2) .......(ii)
Ans. let SD be the shortest dist. , then co-ordinates of eqn.(i) are ( 3p+6, -p+7, p+4) & eqn.(ii) are (-3q, 2q-9, 4q+2)
d.c’s of SD are proportional to -3q-3p-6, 2q+p-16, 4q-p-2 and if SD is at right angles to both lines , we get 3.( -3q-3p-6)+(-1).( 2q+p-16)+ 1. (4q-p-2)=0 and -3.(-3q-3p-6)+2.( 2q+p-16)+4.( 4q-p-2) =0 ⇨ 7q+11p+4=0 & 29q+7p-22=0 ⇨ p=-1 , q=1 and SD=3√(30) , eqn. Will be (x-3)/-6 = (y-8)/-15 = (z-3)/3
Q.37 Find the shortest distance and eqn. b/w the lines �� = (8+3λ)i-(9+16λ)j+(10+7λ)k & �� = (15i+29j+5k)+μ(3i+8j-5k).
Ans. let SD be the shortest dist. b/w two lines , let position vector of S be (8i-9j+10k)+λ(3i-16j+7k) and position vector of D is
(15i+29j+5k+μ(3i+8j-5k), 𝑺𝑫 = �� − �� = (3μ-3λ+7)i+(8μ+16λ+38)j+(-7λ-5μ-5)k since SD is perpendicular on line (i) & (ii) then
[(3μ-3λ+7)i+(8μ+16λ+38)j+(-7λ-5μ-5)k]. (3i-16j+7k)=0 ⇨ 157λ+77μ+311=0...(iii) & 77λ+49μ+175=0...(iv), by solving λ = -1,μ=-
2 ∴ 𝑺𝑫 = �� − �� = 4i+6j+12k ⇨| 𝑺𝑫 |= 14 & eqn. Is �� =
(5i+7j+3k)+t[(9i+13j+15k)-(5i+7j+3k)] = (5i+7j+3k)+t(4i+6j+12k)
where �� =5i+7j+3k, �� =9i+13j+15k & use �� =�� +t(�� − �� ).
Vector Algebra
Q. 1. Find a vector in the direction of vector that has magnitude 7 units.
Q. 2. Show that the points A, B and C with position
vectors, respectively, form the vertices of a right angled triangle.
Q. 3. Find
, if two vectors are such that .
Q. 4. Find the area of the parallelogram whose adjacent sides are
determined by the vectors
Q. 5. If a unit vector makes angles with , and acute
angle θ with , then find θ and hence the components of .
Q. 6. Let , and be three vectors such that and each one of them being perpendicular to the sum of other two,
Find
Q. 7. Find the value of
Q. 8. The scalar product of the vector with a unit vector along
the sum of vectors Find the value of .
Q. 9. If the sum of two unit vectors is a unit vector, Prove that the
magnitude of their difference is .
Q. 10. If are position vectors of points A and B respectively, then find the position vector of points of trisection of AB.
Q. 11. Prove that the line segment joining the mid-points of two sides of a triangle is parallel to the third side and equal to half of it.
Q. 12. ABCD is a parallelogram. If the coordinates of A, B, C are (-2, -1), (3, 0) and (1, -2) respectively, Find the co-ordinate of D.
Q. 13. Show that the points A, B, C with position
vectors
Q. 14. If a vector makes with OX, OY and OZ respectively, prove
that
Q. 15. If inclined at an angle , then prove that
sin = .
Q. 16. If .
Q. 17. If .
Q. 18. Consider two points P and Q with position vectors
. Find the position vector of a point R which divides the line joining P and Q in the ratio 2:1, (i) internally, and (ii) externally.
Q. 19. Find a unit vector perpendicular to each of the
vectors
Q. 20. Show that the vectors 2 i – j + k , i– 3 j – 5 k and 3 i – 4 j – 4 k form the vertices of a right angled triangle
Q. 21.
ASSIGNMENT(LINEAR PROGRAMMING)
Q.1. David wants to invest at most Rs12,000 in Bonds A and B. According to the rule, he has to invest at least Rs2,000 in Bond A and at least Rs4,000 in Bond B. If the rate of interest in bonds A and B respectively are 8% and 10% per annum, formulate the problem as L.P.P. and solve it graphically for maximum interest. Also determine the maximum interest received in a year. Do you think your investment in bond A &B help in the well being of the nation?Do you think that a person should start saving at an early age for his retirement?
[ANS:Therefore, the interest I is maximum at C(2000, 10000) and the maximum interest is Rs1160.]
Q.2. A farmer has a supply of chemical fertilizer of type A which contains 10% of nitrogen and 6% of phosphoric acid and of type B which contains 5% of nitrogen and 10% of phosphoric acid. After soil testing it is found that at least 7 kg of nitrogen and the same quantity of phosphoric acid is required for a good crop. The fertilizer of type A
costs Rs5.00 per kg and the type B costs Rs8.00 per kg. Using linear programming find how many kgs of each type of the fertilizer should be bought to meet the requirement and the cost be minimum. Solve the problem graphically.
[ANS: Therefore, cost is minimum at P(50, 40). That is 50 kg of type A and 40 kg of type B is mixed to meet the requirement.
Q.3. A dietician wishes to mix two types of foods in such a way that vitamin contents of the mixture contain at least 8 units of vitamin A and 10 units of vitamin C. Food I contains 2 units/kg of vitamin A and 1 unit/kg of vitamin C. Food II contains 1 unit/kg of vitamin A and 2 units/kg of vitamin C. It costs Rs50 per kg to purchase Food I and Rs70 per kg to purchase Food II. Formulate this problem as a linear programming problem to minimize the cost of such a mixture.
[ANS:Hence, the minimum value of Z is Rs380.]
Q.4. Two tailors A and B are paid Rs150 and Rs200 per day respectively. A can stich 6 shirts and 4 paints while B can stich 10 shirts and 4 paints per day. Form a linear programming problem to minimize the labour cost to produce at least 60 shirts and 32 paints. Solve the problem graphically. These shirts & paints have been specially designed for a ‘VAN MAHOTSAVA’ programme. Would you like to participate in this programme? Why?
[Ans. = Tailor A works for 5 days and B for 3 days; Minimum cost = Rs1,350]
Q.5. A cooperative society of farmers has 50 hectare of land to grow two crops X and Y. The profit from crops X and Y per hectare are estimated as Rs10,500 and Rs9,000 respectively. To control weeds, a liquid herbicide has to be used for crops X and Y at rates of 20 litres and 10 litres per hectare. Further, not more than 800 litres of herbicide should be used in order to protect fish and wild life using a pond which collects drainage from this land. How much land should be allocated to each crop so as to maximize the total profit of the society ? In what ways does a cooperative society help the farmers?
[ANS:Hence , society will get the maximum profit of Rs4,95,000 by allocating 30 hectare for crop X and 20 hectare for crop Y.]
Q.6. Kellogg is a new cereal formed of a mixture of bran and rice that contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing this new cereal if bran costs Rs5 per kilogram and rice costs Rs4 per kilogram.
[ANS: Therefore, minimum cost of producing this cereal is Rs4.60 per kg.]
Q.7. A new cereal, formed of a mixture of bran and rice, contains at least 88 grams of protein and at least 36 milligrams of iron. Knowing that bran contains 80 grams of protein and 40 milligrams of iron per kilogram, and that rice contains 100 grams of protein and 30 milligrams of iron per kilogram, find the minimum cost of producing a kilogram of this new cereal if bran costs Rs. 28 per kilogram and rice costs Rs. 25 per kilogram.
Solution : – Do yourself. [Ans. = Rs. 26.8]
Q.8. A dealer wishes to purchase a number of fans and CFLs. He has only Rs5,760 to invest and has space for at most 20 items. A fan cost him Rs360 and a CFL Rs240. He expects to sell a fan at a profit of Rs22 and a CFL at a profit of Rs18. Assuming that he can sell all the atoms that he buys, how should he invest his money to maximize the profit ? Solve graphically and find the maximum profit.What is the full form of CFL? Which one isw more energy efficient: CFL or ordinary incandescent bulb?
[ANS: Therefore, for maximum profit he should buy and sell 8 fans and 12 CFLs. His maximum profit is Rs392.]
Q.9. A man has Rs1,500 for purchasing rice and wheat. A bag of rice and a bag of wheat cost Rs180 and Rs120 respectively. He has the storage capacity of at most 10 bags. He earns a profit of Rs11 and Rs9 per bag of rice and wheat respectively. Formulate the above problem as an LPP to maximize the profit and solve it graphically.
Solution :
Do yourself. [Ans. Max. Profit = Rs100, No. of rice bags = 5 = No. of wheat bags]
Q.10. A factory owner purchases two types of machines, A and B for his factory. The requirements and the limitations for the machines are as follows:
Machine Area occupied
Labour force
Daily output (in units)
A 1000 m2 12 men
60
B 1200 m2 8 men 40
He has maximum area of 9000 m2 available, and 72 skilled labourers who can operate both the machines. How many machines of each type should he buy to maximize the daily output?
Solution : [ANS: The maximum output is at B and C. But the number of machines cannot be a fraction. Hence no. of machines of type A = 6 and no. of machines of type B = 0 ]
Q.11 If a 19 years old girl drives her car at 25 km/hr, she has to spend Rs. 2/km on petrol. If she drives it at a faster speed of 40 km/hr, the petrol cost increases to Rs.5/km. She has Rs. 100 to spend on petrol and wishes to find the max. Distance she can travel within one hour . express it as a L.P.P. and then solve it . Is the girl eligible for a driving licence? What is the benefit of driving at economic speed?
Solution: [ans: max distance is 30 km.at (50/3,40/3) .yes ∵ minimum age require for licence is 18 years. Reduces fuel consumption and causes less pollution.]
