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Rajasthan Board RBSE Class 12 Maths Chapter 4 Determinants Miscellaneous Exercise
Question 1.
Value of determinant
is:
(a) 0
(b) 1
(c) -1
(d) none of these
Solution:
= cos 80° sin 10° – (- cos 109) sin 80°
= cos 80° sin 10° + cos 10° sin 80°
= sin (10° + 80°)
= sin 90° = 1
So, option (b) is correct.
Question 2.
Co-factors of first column of determinant
(a) -1, 3
(b) -1,- 3
(c)- 1, 20
(d) – 1,- 20
Solution:
Co-factor of a11
F11 = (- 1)2 M11
= 1 x (- 1) = – 1
Co-factor of a12
F21 = (-1)11 M21
= (- 1) × 20 = – 20
So, option (d) is correct.
Question 3.
If , then the value of will be :
(a) – 2∆
(b) 8∆
(c) -8∆
(d) -6∆
Solution:
So, option (c) is correct.
Question 4.
Which of the following determinant is identical to determinant
:
Solution:
So, option (c) is correct.
Question 5.
Value of
(a) 0
(b) 1
(c) 1/2
(d) – 1/2
Solution:
= cos 50° cos 10° – sin 50° sin 10°
= cos (50° + 10°)
= cos 60° = \(\frac { 1 }{ 2 }\)
So, option (c) is correct.
Question 6.
Value of
(a) ab + bc + ca
(b) 0
(c) 1
(d) abc
Solution:
Question 7.
If ω is a cube root of unity, then value of
(a) ω2
(b) ω
(c) 1
(d) o
11 04 081
Solution:
Question 8.
If , then x will be:
(a) 6
(b) 7
(c) 8
(d) 0
Solution:
⇒ (4 – 2)2 = (3x – 2) – (x + 6)
⇒ (2)2 = 3x – 2 – x – 6
⇒ 4 = 2x – 8
⇒ 4 + 8 = 2x
x = 6
So, option (a) is correct.
Question 9.
If and F11, F12, F13, …, are the corresponding cofactors of a11, a12, a13, …, then which of the following is true :
(a) a12F12+ a22F22 + a32F32 = 0
(b) a12F12 + a22F22 + a32F32 ≠ ∆
(c) a12F12 + a22F22 + a32F32 = ∆
(d) a12F12 + a22F22 + a32F32 = – ∆.
Solution:
a12F12 + a22F22 + a32F32 = ∆
So, option (c) is correct.
Question 10.
Value of determinant
(a) x + y + z
(b) 2(x + y + z)
(c) 1
(d) 0
Solution:
So, option (d) is correct.
Question 11.
Solve the following equation :
Solution:
⇒ 1(9x – 48) – 2(36 – 42) + 3(32 – 7x) = 0
⇒ 9x – 48 + 12 + 96 – 21x =0
⇒ – 12x + 60 = 0
⇒ – 12x = -60
⇒ x = \(\frac { 60 }{ 12 }\) = -5
So, x = 5.
Question 12.
Find the value of determinant :
Solution:
= 1(27 – 1) – 3(9 – 9) + 9(3 – 81)
= 26 – 0 – 702
= -676.
Question 13.
Find the value of determinant
Solution:
Question 14.
Prove that
Solution:
= a2b2c2 [- 1(1 – 1) – 1(-1 – 1) + 1(1 + 1)]
= a2b2c2 (0 + 2 + 2)
= 4a2b2c2 = R.H.S. Hence Proved.
Question 15.
Prove that x = 2 is a root of following equation. Also find its remaining roots :
Solution:
∵ R1 = R2
∴ Determinant will be zero.
∴ Root of equation is 2.
Question 16.
Prove that:
Solution:
Question 17.
Prove that
.
Solution:
= (a + b + c)[(- b – c – a) (- c – a – b) – 0]
= (a + b + c)(b + c + a)(c + a + b)
= (a + b + c)3 = R.H.S.
Proved.
Question 18.
Prove that :
Solution:
Question 19.
Prove that:
Solution:
= (a – b) (b – c) [(b2 + c2 + bc) – (a2 + b2 + ab)]
= (a – b) (b – c) (b2 + c2 + bc – a2 – b2 – ab)
= (a – b)(b – c) [bc + c2 – a2 – ab]
= (a – b)(b – c) [bc – ab + c2 – a2 ]
= (a – b) (b – c) [b(c – a) + (c2 – a)]
= (a – b)(b – c)(c – a)(b + c + a)
= R.H.S. Proved.
Question 20.
Prove that:
Solution:
Question 21.
If a + b + c = 0, then solve:
Solution:
⇒ (-x) [(c – a) (a – c + x) – (b – c + x) (b – x – a)] = 0
⇒ (-x) [(ac – c2 + cx – a2 + ac – ax)
⇒ (b2 – bx – ab – bc + cx + ac – xb – x2 – ax)] = 0
= (-x) [x2 – (a2 + b2 – ab – bc – bc – ca)]= 0
If – x = 0, then x = 0
Now, if x2 = (a2 + b2 + c2 + bc – ab – ca) = 0 .
Question 22.
Prove that
Solution:
Question 23.
If p + q + r = 0, then prove that
Solution:
Question 24.
Prove that
Solution:
RBSE Solutions for Class 12 Maths
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