NCERT Solutions Class 12

NCERT SolutionsMathematics
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 1)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 2)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 3)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 4)
 Relations and Functions : NCERT Solutions – Class 12 Maths (Ex 5)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 1)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 2)
 Inverse Trigonometric Function : NCERT Solutions – Class 12 Maths (Ex 3)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 1)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 2)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 3)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 4)
 Matrices : NCERT Solutions – Class 12 Maths (Ex 5)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 1)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 2)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 3)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 4)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 5)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 6)
 Determinants : NCERT Solutions – Class 12 Maths (Ex 7)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 1)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 2)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 3)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 4)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 5)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 6)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 7)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 8)
 Continuity and Differentiability : NCERT Solutions – Class 12 Maths (Ex 9)

NCERT SolutionsChemistry
 Aldehydes, Ketones and Carboxylic Acids : NCERT Solutions – Class 12 Chemistry
 Alcohols, Phenols and Ethers : NCERT Solutions – Class 12 Chemistry
 Amines : NCERT Solutions – Class 12 Chemistry
 Biomolecules : NCERT Solutions – Class 12 Chemistry
 Chemical Kinetics : NCERT Solutions – Class 12 Chemistry
 Chemistry in Everyday Life : NCERT Solutions – Class 12 Chemistry
 Coordination Compounds : NCERT Solutions – Class 12 Chemistry
 Electrochemistry : NCERT Solutions – Class 12 Chemistry
 General Principles and Processes of Isolation of Elements : NCERT Solutions – Class 12 Chemistry
 Haloalkanes and Haloarenes : NCERT Solutions – Class 12 Chemistry
 Polymers : NCERT Solutions – Class 12 Chemistry
 Surface Chemistry : NCERT Solutions – Class 12 Chemistry
 The dand fBlock Elements : NCERT Solutions – Class 12 Chemistry
 The pBlock Elements : NCERT Solutions – Class 12 Chemistry
 The Solid State : NCERT Solutions – Class 12 Chemistry
 Solutions : NCERT Solutions – Class 12 Chemistry

NCERT SolutionsBiology

NCERT SolutionsPhysics
 Electrostatic Potential And Capacitance : NCERT Solutions – Class 12 Physics
 Electric Charges And Fields : NCERT Solutions – Class 12 Physics
 Semiconductor Electronics: Materials, Devices And Simple Circuits : NCERT Solutions – Class 12 Physics
 Ray Optics And Optical Instruments : NCERT Solutions – Class 12 Physics
 Nuclei : NCERT Solutions – Class 12 Physics
 Moving Charges And Magnetism : NCERT Solutions – Class 12 Physics
 Magnetism And Matter : NCERT Solutions – Class 12 Physics
 Electromagnetic Induction : NCERT Solutions – Class 12 Physics
 Dual Nature Of Radiation And Matter : NCERT Solutions – Class 12 Physics
 Current Electricity : NCERT Solutions – Class 12 Physics
 Communication Systems : NCERT Solutions – Class 12 Physics
 Atoms : NCERT Solutions – Class 12 Physics
 Alternating Current : NCERT Solutions – Class 12 Physics
Matrices : NCERT Solutions – Class 12 Maths (Ex 5)
Miscellaneous
1. Let A = show that where I is the identity matrix of order 2 and N.
Ans. Using Mathematical Induction, we see the result is true for for
Given: is true, i.e.
To prove:
Proof: L.H.S. = = =
=
=
= = R.H.S.
Thus, is true, therefore, is true.
2. If A = , prove that A’’ = N.
Ans. Given: A = ……….(i)
Let
=
is true for
Now ……….(ii)
Multiplying eq. (ii) by eq. (i),
=
Therefore, is true for all natural numbers by P.M.I.
3. If A = then prove that A’’ = where is any positive integer.
Ans. Given: A’’ =
which is true for
Now, ……….(i)
Again ……….(ii)
[From eq. (i)]
=
Therefore, the result is true for
Hence, by the principal of mathematical induction, the result is true for all positive integers
4. If A and B are symmetric matrices, prove that AB – BA is a skew symmetric matrix.
Ans. A and B are symmetric matrices. A’ = A and B’ = B ……….(i)
Now, (AB – BA)’ = (AB)’ – (BA)’ (AB – BA)’ = B’A’ – A’B’ [Reversal law]
(AB – BA)’ = BA – AB [Using eq. (i)]
(AB – BA)’ = – (AB – BA)
Therefore, (AB – BA) is a skew symmetric.
5. Show that the matrix B’AB is symmetric or skew symmetric according as A is symmetric or skew symmetric.
Ans. (B’AB)’ = [B’(AB]’ = (AB)’ (B’)’ [ (CD)’ = D’C’]
(B’AB)’ = B’A’B ……….(i)
Case I: A is a symmetric matrix, then A’ = A
From eq. (i) (B’AB)’ = B’AB
B’AB is a symmetric matrix.
Case II: A is a skew symmetric matrix. A’ = – A
Putting A’ = – A in eq. (i), (B’AB)’ = B’(– A)B = – B’AB
B’AB is a skew symmetric matrix.
6. Find the values of if the matrix A = satisfies the equation A’A = I.
Ans. Given: A =
A’ =
Now A’A = I
Equating corresponding entries, we have
And
And
, ,
7. For what value of ?
Ans. Given:
Equating corresponding entries, we have
8. If A = show that A^{2} – 5A + 7I = 0.
Ans. Given: A =
A^{2} – 5A + 7I =
= =
= = =
= = 0 = R.H.S. Proved.
9. Find if
Ans. Given:
Equating corresponding entries, we have
10. A manufacturer produces three products, which he sells in two markets. Annual sales are indicated below:
Market  Products  
I. 10,000  2,000  18,000 
II. 6,000  20,000  8,000 
(a) If unit sales prices of and are ` 2.50, ` 1.50 and ` 1.00 respectively, find the total revenue in each market with the help of matrix algebra.
(b) If the unit costs of the above three commodities are ` 2.00, ` 1.00 and 50 paise respectively. Find the gross profit.
Ans. According to question, the matrix A =
(a) Let B be the column matrix representing sale price of each unit of products
Then B =
Now Revenue = Sale price c Number of items sold
= =
Therefore, the revenue collected by sale of all items in Market I = ` 46,000 and the revenue collected by sale of all items in Market II = ` 53,000.
(b) Let C be the column matrix representing cost price of each unit of products
Then C =
Total cost = AC =
= =
The profit collected in two markets is given in matrix form as
Profit matrix = Revenue matrix – Cost matrix
Therefore, the gross profit in both the markets = ` 15000 + ` 17000 = ` 32,000.
11. Find the matrix X so that X
Ans. Given: X ……….(i)
Putting X = in eq. (i),
Equating corresponding entries, we have
…..(ii) …..(iii)
…..(iv) …..(v)
…..(vi) …..(vi)
Solving eq. (ii) and (iii), we have and
Solving eq. (v) and (vi), we have and
Putting these values in X = , X =
12. If A and B are square matrices of the same order such that AB = BA, then prove by induction that AB’’ = B’’A. Further prove that (AB)’’ = A’’B’’ for all N.
Ans. Given: AB = BA …..(i)
Let ……(ii)
For becomes AB = BA
is true for
For
Multiplying both sides by B,
[From eq. (i)]
is also true.
Therefore, is true for all N by P.M.I.
13. If A = is such that A^{2} = I, then:
(A)
(B)
(C)
(D)
Ans. Given: A = and A^{2} = I
Equating corresponding entries, we have
Therefore, option (C) is correct.
14. If the matrix A is both symmetric and skew symmetric, then:
(A) A is a diagonal matrix
(B) A is a zero matrix
(C) A is a square matrix
(D) None of these
Ans. Since, A is symmetric, therefore, A’ = A ……..(i)
And A is skewsymmetric, therefore, A’ = – A
A = – A [From eq. (i)]
A + A = 0 2A = 0 A = 0
Therefore, A is zero matrix.
Therefore, option (B) is correct.
15. If A is a square matrix such that A^{2} = A, then (I + A)^{3} – 7A is equal to:
(A) A
(B) I – A
(C) I
(D) 3A
Ans. Given: A^{2} = A …..(i)
Multiplying both sides by A, A^{3} = A^{2} = A [From eq. (i)] ……(ii)
Also given (I + A)^{3} – 7A = I^{3} + A^{3} + 3I^{2}A + 3IA^{2} – 7A
Putting A^{2} = A [from eq. (i)] and A^{3} = A [from eq. (ii)],
= I + A + 3IA + 3IA – 7A = I + A + 3A + 3A – 7A [ IA = A]
= I + 7A – 7A = I
Therefore, option (C) is correct.