BSEB Bihar Board 12th Maths Important Questions Long Answer Type Part 2 are the best resource for students which helps in revision.
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) \(\left|\begin{array}{lll}
x & a & x+a \\
y & b & y+b \\
z & c & z+c
\end{array}\right|\) = 0
(ii) \(\left|\begin{array}{lll}
a-b & b-c & c-a \\
b-c & c-a & a-b \\
c-a & a-b & b-c
\end{array}\right|\) = 0
(iii) \(\left|\begin{array}{lll}
1 & b c & a(b+c) \\
1 & c a & a(c+a) \\
1 & a b & c(a+b)
\end{array}\right|\) = 0
Solution:
Question 2.
By using determinants property \(\left|\begin{array}{ccc}
1 & 1 & 1 \\
a & b & c \\
a^{3} & b^{3} & c^{3}
\end{array}\right|\) = (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 \(\left|\begin{array}{ccc}
a & a+b & a+b+c \\
2 a & 3 a+2 b & 4 a+3 h+2 c \\
3 a & 6 a+3 b & 10 a+6 b+3 c
\end{array}\right|=a^{2}\)
Solution:
L.H.S = \(\left|\begin{array}{lll}
a & a+b & a+b+c \\
2 a & 3 a+2 b & 4 a+3 b+2 c \\
3 a & 6 a+3 b & 10 a+6 b+3 c
\end{array}\right|\)
R2 → R2 – 2R3, R3 → R3 – 3R1
= \(\left|\begin{array}{rrr}
a & a+b & a+b+c \\
0 & a & 2 a+b \\
0 & 3 a & 7 a+3 b
\end{array}\right|\)
From 1st Column
= a\(\left|\begin{array}{cc}
a & 2 a+b \\
3 a & 7 a+3 b
\end{array}\right|\)
= a(7a2 + 3ab + 6a2 – 3ab)
= a.a2 = a3 = RHS
Question 4.
Without expanding prove that
\(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|\) = 0
Solution:
L.H.S = \(\left|\begin{array}{ccc}
x+y & y+z & z+x \\
z & x & y \\
1 & 1 & 1
\end{array}\right|\)
R1→ R1 + R2
= \(\left|\begin{array}{ccc}
x+y+z & y+z+x & z+x+y \\
z & x & y \\
1 & 1 & 1
\end{array}\right|\)
= (x + y + z)\(\left|\begin{array}{lll}
1 & 1 & 1 \\
z & x & x \\
1 & 1 & 1
\end{array}\right|\)
= (x + y + z) = 0 = 0 R.H.S.
Question 5.
Evaluate : \(\left|\begin{array}{lll}
1 & a & b c \\
1 & b & c a \\
1 & c & a b
\end{array}\right|\)
Solution:
= (a -b)(b – c)(-a + c)
= (a – b) (b – c) (c – a) = R.H.S
Question 6.
Without expanding shw that,
\(\left|\begin{array}{lll}
a+c & q+r & y+z \\
c+a & r+p & z+x \\
a+b & p+q & x+y
\end{array}\right|=2\left|\begin{array}{lll}
a & p & x \\
b & q & y \\
c & r & 2
\end{array}\right|\)
Solution:
Question 7.
Prove that the determinants \(\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|\) is independent of θ
Solution:
Let Δ = \(\left|\begin{array}{ccc}
x & \sin \theta & \cos \theta \\
-\sin \theta & -x & 1 \\
\cos \theta & 1 & x
\end{array}\right|\)
From 1st row→
= \(x \cdot\left|\begin{array}{cc}
-x & 1 \\
1 & x
\end{array}\right|-\sin \theta\left|\begin{array}{cc}
-\sin \theta & 1 \\
\cos \theta & x
\end{array}\right|+\cos \theta\left|\begin{array}{rr}
-\sin \theta & -x \\
\cos \theta & 1
\end{array}\right|\)
= 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 :
\(\left|\begin{array}{ccc}
1 & x & x^{2} \\
x^{2} & 1 & x \\
x & x^{2} & 1
\end{array}\right|\) = (1 – x2)2
Answer:
L.H.S = Δ = \(\left|\begin{array}{ccc}
1 & x & x^{2} \\
x^{2} & 1 & x \\
x & x^{2} & 1
\end{array}\right|\)
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 :
\(\left|\begin{array}{ccc}
1+a^{2}-b^{2} & 2 a b & -2 b \\
2 a b & 1-a^{2}+b^{2} & 2 a \\
2 b & -2 a & 1-a^{2}-b^{2}
\end{array}\right|=\left(1+a^{2}+b^{2}\right)^{3}\)
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 :
\(\left|\begin{array}{ccc}
a^{2}+1 & a b & a c \\
a b & b^{2}+1 & b c \\
c a & c b & c^{2}+1
\end{array}\right|\) = 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(2x\(\sqrt{1-x^{2}}\)) = 2sin-1x
(ii) sin-1\(\frac { 8 }{ 17 }\) + -1\(\frac { 3 }{ 5 }\) = -1\(\frac { 77 }{ 85 }\)
(iii) sin [cot-1(cos(tan-1x)}] = \(\frac{\sqrt{1+x^{2}}}{2+x^{2}}\)
(iv) 2tan-1 \(\frac { 1 }{ 2 }\) + tan-1 \(\frac { 1 }{ 7 }\) = tan-1\(\frac { 31 }{ 17 }\)
(v) \(\tan ^{-1} \sqrt{x} \frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)\) , x ∈ [0,1].
Solution:
(i) Let sin-1 = θ
x = sin θ
L.H.S. = sin-1 (\(2 x \cdot \sqrt{1-x^{2}}\))
= sin-1(2sinθ \(\sqrt{1-\sin ^{2} \theta}\))
= sin-1(2sinθ.cosθ) = sin-1(sin2 θ)
= 2θ = 2sin-1 x = R.H.S.
(ii) L.H.S. = sin-1 \(\frac{8}{17}\) + sin-1 \(\frac{3}{5}\)
(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 = \(\tan ^{-1} \sqrt{x}=\frac{1}{2} \cdot 2 \tan ^{-1} \sqrt{x}\)
= \(\frac{1}{2} \cdot \cos ^{-1}\left(\frac{1-(\sqrt{x})^{2}}{1+(\sqrt{x})^{2}}\right)=\frac{1}{2} \cos ^{-1}\left(\frac{1-x}{1+x}\right)\) = R.H.S
Question 2.
Find the values of each of the following :
(i) \(\tan ^{-1}\left[2 \cos \left(2 \sin ^{-1} \frac{1}{2}\right)\right]\)
(ii) \(\tan \frac{1}{2} \cdot\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right]\), |x| < 1, y > 0, x,y < 1
Solution:
(ii) \(\tan \frac{1}{2} \cdot\left[\sin ^{-1} \frac{2 x}{1+x^{2}}+\cos ^{-1} \frac{1-y^{2}}{1+y^{2}}\right]\)
= tan \(\tan \frac{1}{2}\) [2tan-1 x + 2tan-1 y]
= tan[tan-1x + tan-1 y]
= tan [tan-1( \(\frac{x+y}{1-x y}\) )
= \(\frac{x+y}{1-x y}\)
Question 3.
Show that \(\cot ^{-1}\left(\frac{\sqrt{1+\sin } \bar{x}+\sqrt{1-\sin x}}{\sqrt{1+\sin x}-\sqrt{1-\sin x}}\right)=\frac{x}{2}, x \in\left(0, \frac{\mu}{4}\right)\)
Answer:
Question 4.
Show that \(\frac{9 \pi}{8}-\frac{9}{4} \sin ^{-1} \frac{1}{3}=\frac{9}{4} \sin ^{-1} \frac{2 \sqrt{2}}{3}\)
Answer:
