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 Bihar Board 12th Maths Important Questions Long Answer Type Part 2

Bihar Board 12th Maths Important Questions Long Answer Type Part 2

Determinants

Question 1.
Using the property of determinants and without expending, prove that
(i) ∣∣∣∣xyzabcx+ay+bz+c∣∣∣∣ = 0
(ii) ∣∣∣∣a−bb−cc−ab−cc−aa−bc−aa−bb−c∣∣∣∣ = 0
(iii) ∣∣∣∣∣111bccaaba(b+c)a(c+a)c(a+b)∣∣∣∣∣ = 0
Solution:

Question 2.
By using determinants property ∣∣∣∣1aa31bb31cc3∣∣∣∣ = (a- b)(b – c)(c – a)(a + b + c)
Solution:

= (a-b) (b-c)-(b2 +bc+c2 -a2-ab-b2)
= (a-b)-(b-c)-{(c2-a2) + b-(c-a)}
= (a-b)(b-c)-{(c-a)(c + a) + b-(c-a)}
= (a-b)(b-c)-(c-a)(c + a + b)
= (a-b)(b-c)-(c-a)-(a + b+c) R.H.S.

Question 3.
Prove that ∣∣∣∣a2a3aa+b3a+2b6a+3ba+b+c4a+3h+2c10a+6b+3c∣∣∣∣=a2
Solution:
L.H.S = ∣∣∣∣a2a3aa+b3a+2b6a+3ba+b+c4a+3b+2c10a+6b+3c∣∣∣∣
R2 → R2 – 2R3, R3 → R3 – 3R1
= ∣∣∣∣a00a+ba3aa+b+c2a+b7a+3b∣∣∣∣
From 1st Column
= a∣∣∣a3a2a+b7a+3b∣∣∣
= a(7a2 + 3ab + 6a2 – 3ab)
= a.a2 = a3 = RHS

Question 4.
Without expanding prove that
∣∣∣∣x+yz1y+zx1z+xy1∣∣∣∣ = 0
Solution:
L.H.S = ∣∣∣∣x+yz1y+zx1z+xy1∣∣∣∣
R1→ R1 + R2
= ∣∣∣∣x+y+zz1y+z+xx1z+x+yy1∣∣∣∣
= (x + y + z)∣∣∣∣1z11x11x1∣∣∣∣
= (x + y + z) = 0 = 0 R.H.S.

Question 5.
Evaluate : ∣∣∣∣111abcbccaab∣∣∣∣
Solution:


= (a -b)(b – c)(-a + c)
= (a – b) (b – c) (c – a) = R.H.S

Question 6.
Without expanding shw that,
∣∣∣∣a+cc+aa+bq+rr+pp+qy+zz+xx+y∣∣∣∣=2∣∣∣∣abcpqrxy2∣∣∣∣
Solution:

Question 7.
Prove that the determinants ∣∣∣∣x−sinθcosθsinθ−x1cosθ1x∣∣∣∣ is independent of θ
Solution:
Let Δ = ∣∣∣∣x−sinθcosθsinθ−x1cosθ1x∣∣∣∣
From 1st row→
= x⋅∣∣∣−x11x∣∣∣−sinθ∣∣∣−sinθcosθ1x∣∣∣+cosθ∣∣∣−sinθcosθ−x1∣∣∣
= x(-x2 – 1) – sinθ.(-xsinθ – cos θ) + cos θ(-sin θ + x cos θ)
= -x3 – x + xsin2θ + sin θ.cos θ – sin θ.cos θ + xcos2θ
= -x3.x + x.(sin2θ + cos2θ)
= -x3 – x + x.1
= -x3 – x + x
= -x3 = indepentdent

Question 8.
Evaluate :
∣∣∣∣∣1x2xx1x2x2x1∣∣∣∣∣ = (1 – x2)2
Answer:
L.H.S = Δ = ∣∣∣∣∣1x2xx1x2x2x1∣∣∣∣∣
Operating C1 → C1 + C2 + C3

= (1 – x)2(1 + x + x2)(1) + x(1 + x)
= (1 – x)2 (1 + x + x2)[1 + x + x2]
= (1 – x)2(1 + x + x2 )2
= [(1 – x)(1 + x + x2 )]2 = (1 – x3)2

Question 9.
Evaluate :
∣∣∣∣∣1+a2−b22ab2b2ab1−a2+b2−2a−2b2a1−a2−b2∣∣∣∣∣=(1+a2+b2)3
Answer:

Expanding we get
Δ = (1 + a +b2)2 (1 – a2 – b2) + 2a2 + 2b2
= (1 + a2 + b2 )2 (1 + a2 + b2) = (1 + a2 + b2)3

Question 10.
Evaluate :
∣∣∣∣∣a2+1abcaabb2+1cbacbcc2+1∣∣∣∣∣ = 1 + a2 + b2 + c2
Solution:

This may be expressed as the sum of 8 determinants.

= abc  + (b2c2 – b2c2) + (a2c2 – a2c2) + (a2b2 – a2b2) + a2 +b2 + c2
= a2 + b2 + c2

Inverse Circular Function

Question 1.
Show the following :
(i) sin-1(2x1−x2−−−−−√) = 2sin-1x
(ii) sin-1817 + -135 = -17785
(iii) sin [cot-1(cos(tan-1x)}] = 1+x2√2+x2
(iv) 2tan-1 12 + tan-1 17 = tan-13117
(v) tan−1x−−√12cos−1(1−x1+x) , x ∈ [0,1].
Solution:
(i) Let sin-1 = θ
x = sin θ
L.H.S. = sin-1 (2x⋅1−x2−−−−−√)
= sin-1(2sinθ 1−sin2θ)
= sin-1(2sinθ.cosθ) = sin-1(sin2 θ)
= 2θ = 2sin-1 x = R.H.S.

(ii) L.H.S. = sin-1 817 + sin-1 35

(iii) L.H.S. = sin[cot-1 {cos(tan-1 x)}]
Let tan-1 x = y ⇒ A = tan y
= sin[cot -1(cosy)]

(iv)

(v) L.H.S = tan−1x−−√=12⋅2tan−1x−−√
= 12⋅cos−1(1−(x√)21+(x√)2)=12cos−1(1−x1+x) = R.H.S

Question 2.
Find the values of each of the following :
(i) tan−1[2cos(2sin−112)]
(ii) tan12⋅[sin−12x1+x2+cos−11−y21+y2], |x| < 1, y > 0, x,y < 1
Solution:

(ii) tan12⋅[sin−12x1+x2+cos−11−y21+y2]
= tan tan12 [2tan-1 x + 2tan-1 y]
= tan[tan-1x + tan-1 y]
= tan [tan-1( x+y1−xy )
= x+y1−xy

Question 3.
Show that cot−1(1+sin√x¯+1−sinx√1+sinx√−1−sinx√)=x2,x∈(0,μ4)
Answer:

Question 4.
Show that 9π8−94sin−113=94sin−122√3
Answer: