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Mathematics – Class 12 – Chapterwise Question Bank with Answers – For 2025 Board Examination

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Mathematics - Class 12

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Mathematics - Class 12

Mathematics – Class 12 – Chapterwise Question Bank with Answers – For 2025 Board Examination – 63 papers - 475 Pages – Complex Numbers & De Moivres Theorem, Quadratic Expressions, Theory of Equations, Functions, Mathematical Induction, Partial Fractions, Binomial Theorem, Permutations & C...

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Uploaded on August 29, 2024
Number of pages 475
Written in 2024/2025
Type Exam (elaborations)
Contains Questions & answers

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,Contents
1. Complex Numbers and De-Moivre's Theorem 1-19
2. Quadratic Expressions 20-24
3. Theory of Equations 25-38
4. Functions 39-49
5. Mathematical Induction 50-53
6. Partial Fractions 54-57
7. Binomial Theorem 58-69
8. Permutations and Combinations 70-80
9. Matrices and Determinants 81-101
10. Measures of Dispersion 102-110
11. Probability 111-132
12. Trigonometric Functions and Identities 133-148
13. Trigonometric Equations 149-153
14. Properties of Triangles 154-168
15. Inverse Trigonometric Functions 169-173
16. Hyperbolic Functions 174-178
17. Rectangular Cartesian Coordinates 179-190
18. Straight Line and Pair of Straight Lines 191-221
19. Circle and System of Circles 222-252
20. Conic Sections 253-276
21. Vector Algebra 277-303
22. Three Dimensional Geometry 304-320
23. Limits and Continuity 321-335
24. Differentiation 336-352
25. Applications of Derivatives 353-370
26. Indefinite Integrals 371-394
27. Definite Integrals and Its Applications 395-412
28. Differential Equations 413-426
29. Miscellaneous 427-428

Practice Sets (1-3) 429-472

, 1
Complex Numbers and
De-Moivre’s Theorem
1. If 1, a , a 2 , ...., a n − 1 are the nth roots of unity. Sol. (d)
n −1 2α = −1 − i 3, 2 β = −1 + i 3
1
Then ∑ − ai
is equal to (5α 4 + 5β 4 + 7α −1β −1)
i = 12 [17 Sep. 2020, Shift-I] 7
= 5(α 4 + β 4) +
(n − 2 ) 2 n −1 + 1 αβ
(a) (n − 2 )2 n (b)
2n − 1 7
= 5[(α 2 + β 2)2 − 2α 2β 2] +
(n − 2 ) 2 n − 1 1 α ⋅β
(c) (d)
2n − 1 (n − 2 ) 2 n  1 2 
= 5  (2α + 2 β)2 − 2αβ  − 2 α 2 β 2  +
7
Sol. (b)   4   α ⋅β
n −1 1
∑ 1, α , α 2 , ......, α n−1 are the nth root  1 2 
i =1 2− αi = 5  × 4 − 2 − 2 + 7
4 
of unity  
Q x n −1 = 0 = 5 [1 − 2] + 7 = 2
⇒ x n − 1 = (x − 1)(x − α)(x − α 2)......(x − α n −1) i
3. If a + bi = , then (a , b) =
⇒ log(x n − 1) = log(x − 1) + log(x − α) 1− i [17 Sep. 2020, Shift-I]
+ ......log(x − α n −1) −1 −1
(a)  ,  (b)  , 
1 1
Differentiating w.r.t ‘x’, we get 2 2 2 2
1 −1 −1 1
nx n −1 1 1 1 (c)  ,  (d)  , 
⇒ = + + ...... + 2 2   2 2
x −1
n
( x − 1) ( x − α) (x − α n −1)
At (x = 2) Sol. (d)
n ⋅ 2n −1 1 1 1 a + bi =
i
⇒ =1+ + + ...... + 1− i
2n − 1 2− α 2− α2 2 − α n −1
i(1 + i) 1 i 1 1
⇒ a + bi = =− + ⇒ a=− ,b=
n −1 1  n ⋅ 2n −1  (n − 2)2n −1 + 1 2 2 2 2 2
⇒ ∑ = n − 1 =
i =1 2− α  2 −1
i
 2n − 1 4. Let z1, z 2 be two complex numbers such that
3π
2. If 2α = − 1 − i 3 and 2β = − 1 + i 3, then z1 − iz 2 = 0 and arg(z1 z 2) = , then arg(z1) =
4
5α 4 + 5β 4 + 7 α −1β −1 is equal to [17 Sep. 2020, Shift-II]
[17 Sep. 2020, Shift-I] π − π π π
(a) (b) (c) (d)
(a) −1 (b) −2 (c) 0 (d) 2 4 8 8 3

, 2 AP EAMCET Chapterwise Mathematics

Sol. (c)  1
7. Let the complex numbers α and   lie on
Given z1 − i z2 = 0 α
⇒ z1 = i z2 ⇒ z1 = iz2 ⇒ z1 = − iz2 circles (x − x 0)2 + (y − y 0)2 = r 2 and
π
Clearly argument of z1 = argument of z2 − (x − x 0)2 + (y − y 0)2 = 4 r 2 respectively.
2
π If z 0 = x 0 + iy 0 satisfies the equation 2 | z 0 |2 =
or argument z1 = argument z2 −
2 r 2 + 2, then| α | = [18 Sep. 2020, Shift-I]
π
⇒ argument z2 = argument z1 + (a)
1
(b)
1
(c)
1
(d)
1
2 2 2 7 3
Let argument z1 = α
3π Sol. (c)
Then, given argument ( z1 z2 ) = As point α lies on the circle
4
3π (x − x 0)2 + (y − y0)2 = r 2
⇒ argument z1 + argument z2 =
4 ∴ |α − z0|2 = r 2, where z0 = x 0 + iy0
π 3π ⇒ |α|2 + |z0|2 −(α z0 + α z0) = r 2
α+α+ = …(i)
2 4 1
3π π Q lies on the circle (x − x 0)2 + (y − y0)2 = 4r 2
2α = − α
4 2 1
2

π ∴ − z0 = 4r 2
⇒ α= α
8
1  αz αz 
(3 + 2 i) (4 − 7 i) (12 + 13 i) ⇒ + |z0|2 −  02 + 02  = 4r 2
5. If x + iy = , then |α|2  |α| |α| 
(13 − 12 i) (2 − 3 i) (11 + 3 i)
⇒ 1 + |z0|2|α|2 − (αz0 + α z0) = 4r 2|α|2 …(ii)
x 2 + y2 = [17 Sep. 2020, Shift-II]
By subtracting Eqs. (i) and (ii), we get
1
(a) 1 (b) 2 (c) (d) 3 1−|α|2 −|z0|2 (1−|α|2) = r 2(4|α|2 −1)
2
⇒ (|α|2 −1) (|z0|2 −1) = r 2(4|α|2 −1)
Sol. (c)
Given, r2 + 2
Q |z0|2 = , we get
(3 + 2i) (4 − 7i)(12 + 13i) 2
z = x + iy =
(13 − 12i) (2 − 3i) (11 + 3i) r2
(|α|2 −1) = r 2(4|α|2 −1)
|3 + 2i|⋅ |4 − 7i|⋅ |12 + 13i| 2
|z| =
|13 − 12i|⋅ |2 − 3i|⋅ |11 + 3i| ⇒ |α|2 −1 = 8|α|2 −2
1
( 32 + 22) ( 42 + 72) ( (122 + 132) ⇒ 7|α|2 = 1 ⇒ |α|=
= 7
( 132 + 122) ⋅ ( 22 + 32) ( 112 + 32)
8. If α and β are non-real roots of
42 + 72 65 1
= = = x 3 − x 2 − x − 2 = 0, then
112 + 32 130 2
α 2020 + β 2020 + α 2020 ⋅ β 2020 =
1 1
⇒ x + y =
2 2
or x + y =
2 2
[18 Sep. 2020, Shift-I]
2 2
(a) 1 (b) 2020
6. What is the modulus of the complex number (c) 1 + α + β (d) − 1
(1 + 2 i) (− 2 + i)? [17 Sep. 2020, Shift-II]
Sol. (c)
(a) 5 (b) 5 (c) 5 5 (d) 35 Given equation, x 3 − x 2 − x − 2 = 0
Sol. (b) ⇒ (x − 2)(x 2 + x + 1) = 0
|($i + 2$i) ⋅ (−2 + i$)| = |$i + 2i||
⋅ −2 + i| −1 ± 3i
∴ α and β are or we can say
= ( 12 + 22) ⋅ ( (−2)2 + 12) 2
α and β are non-real complex roots of unity.
= 5× 5 = 5.

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