BSEB Bihar Board 12th Maths Important Questions Long Answer Type Part 1 are the best resource for students which helps in revision.

## Bihar Board 12th Maths Important Questions Long Answer Type Part 1

Relations & functions

Question 1.

Show that the relation R in the set {1, 2, 3} Jut, 3} given by R = {(1, 1), (2,2), (3,3), (l, 2), (2,3)} is reflexive but neither symmetric nor transitive,

Solution:

Given Set (A) ={1,2, 3}

Given relation R = { (1,1), (2, 2), (3,3), (1, 2), (2, 3)}

∵ (1, 1), (2,2), (3, 3) ∈ R and 1,2, 3, ∈ A

∵ R is Reflexive.

∵ (1,2) ∈ R but (2, 1) ∉ R

∵ R is not symmetric. .

∵ (1, 2) ∈ R, (2, 3) ∈ R but (1.3) ∉ R

∵ R is not transitive.

Question 2.

Show that the relation R in the set Z of integers given by R = {(a. b} 2 divides a – b} is an equivalence relation.

Solution:

Given set = Z (set of integers)

Given relation R = { (a, b): 2 divides a-b)

For equivalence relation.

(i) R be reflexive.

(ii) R be symmetric.

(iii) R be transitive

∵ \(\frac{a-a}{2}=\frac{0}{2}\)= 0 (Integer)

∵\(\frac{b-b}{2}=\frac{0}{2}\) = 0 (Integer)

∴ R is reflexive.

For symmetric if a Rb

⇒ b Ra

∵ aRb ⇒ \(\frac{a-b}{2}\) = Integer

⇒ \(\frac{-(b-a)}{2}\) = Integer

∴ \(\frac{b-a}{2}\) = Integer

∴ bRa

∴ R is symmetric

For transitive, If a Rb, b Rc ⇒ aRC

∵ bRb ⇒ \(\frac{a-b}{2}\) = Integer

bRc\(\frac{b-c}{2}\) = Integer.

Eqn. (i) + Eqn. (ii)

∴ a Rc

∴ R in transitive.

Hence R is an equivalence relation.

Question 3.

Show that each of the relation R in the set A = { x ∈ Z/0 ≤ x ≤ 12} given by

(i) R – {(a, b): |a-b| is a multiple of 4.}

(ii) R – {(a, b) : a = b} is an equivalence relation. Find the set of .all elements related to 1 in each case.

Solution:

Given set A = {x ∈ R/o ≤ x ≤ 12}

(i) Relation R = {(a, b):|a- b| is a multiple of 4}.

For equivalence relation R be reflexive.

R be symmetric. .

R be transitive. .

∵ a-a = 0 = 0 x 4 = multiple of 4

b – b = 0 = 0 x 4 = multiple of 4 and so on.

∴ R is reflexive .

For symmetric, If a Rb

⇒ b Ra

Given a Rb

means |a – b| is a multiple of 4.

⇒ |(b – a)| is a multiple of 4.

|b – a| is a multiple of 4.

∴ b Ra

∴ R is symmetric.

For transitive, if a Rb, bRc ⇒ a Rc ,

Given a Rb & b Rc

=> |a – b| = multiple of 4. …………(i)

and |b-c| = multiple of 4 …(ii)

eqn. (i) + eqn. (ii) .

|a -b| + |b-c|= multiple 4.

⇒ |a – c| = multiple of 4.

⇒ a Rc

∴ R is transitive.

∴ R is equivalence relation.

(ii) R = {(a,b):a = b)

For equivalence relation R be reflexive.

R be symmetric

R be transitive.

∵ Each element is equal to itself,

∵ R is reflexive.

For symmetric, if a Rb ⇒ b Ra

Given aRb

⇒ a = b

⇒ b = a

⇒ b Ra

∴ R is symmetric

For transitive, if a Rb, b Rc

⇒ aRc

Given, aRb, bRc

⇒ a = b, b = c

⇒ a = c

⇒ aRc

∴ R is transitive

∴ R is an equivalence relation.

When 1st components of each pair = 1 then

(i) R = {(1,1), (1,5),(1,9)}

∴ A={1,5, 9,}

(ii) R = {(1,1)}

∴ A = {1}

Question 4.

Prove that the greatest integer function f : R → R given by fix) – (x} is neither one-one nor on to where (x) denotes the greatest integer less than or equal to x.

Solution:

Given f: R → R.f(x) = [x].

Putx = 1.5 ∴ f(1 – 5) = [1 – 5] = 1

Putx = 1.8 – ∴ f(1 – 8) = [1 – 8] = l

∴ f is not one-one.

∵ Co-domain = R.

Rage = y =f(x) = [x] = Integral value,

∵ Co-domain ≠ Rage

∴ f is not on to.

Question 5.

Show that if f : R – – [latex]\frac{3}{5}[/latex] is defined by f(x) = \(\frac{3 x}{5 x} \cdot \frac{4}{7}\) and g:R –[latex]\frac{3}{5}[/latex] → R [latex]\frac{7}{5}[/latex] is defined by f(x) \(\frac{7 x+4}{5 x-3}\) then fog = I_{A} and gof = I_{B}.

Where A = R – [latex]\frac{3}{5}[/latex], B = R – \(\frac{7}{5}\), I_{A}(x) = x ∀ x ∈ A.

I_{B}(x) = x ∀ x ∈ B.

Solution:

gof(x) = g(f(x)) = \(g\left(\frac{3 x+4}{5 x-7}\right)\)

Question 6.

Let f:N → R foe a function defined as f(x) = 4x^{2} + 12x +15 show that f: N → S, where S is the Rage of/is invertiahte fined f^{-1}.

Solution:

Given f(x) = 4x^{2} + 12x + 15

or, y = (2x)^{2} + 2.2x.3 + 9 + 6

or, y = (2x + 3)^{2} + 6

or, (2x + 3)^{2} = y – 6

or, 2x + 3 = \(\sqrt{y-6}\)

⇒ x = \(\frac{\sqrt{y-6}-3}{2}\), y ≥ 6

⇒ g(y) = \(\frac{\sqrt{y-6}-3}{2}\)

∵ gof(x) = g(f(x)) = g( 4x^{2} + 12x + 15)

= g(12x + 3)^{2}) + 6)

= y – 6 + 6 = y

Hence gof = I_{n}fog = I_{s}

⇒ f is invertiable with f^{-1} = g.

Question 7.

Let A = N x N and * be the binary operation on A defined by (a, b) * (c, d) = (a + c, b + d). Show that * is commutative and associative,. find the identity element for * on A if any.

Solution:

Given (a, b) * (c,d) = ( a + c, b + d)

For commutative, (a, b) * (c,d) = (c,d) (a,b)

L.H.S. = (a,b) * (c,d) = (a + c,b + d)

R.H.S = (c, d) * (a, b) = ( c + a, d + b) = ( a + c, b + d)

∴ L.H.S. = R.H.S.

∴ * is commutative binary operation.

For associative, (a, b) *((c, d)*(e,f)) = ((a,b) * (c, d)) * (e,/)

L.H.S. =(a,b)*((c,d)*(e,f))

= (a, b) *(c + e, d+f) = (a + c + e, b + d + f)

R.aS. = ((a, b)*(c,d) *(e, f )

= (a + c, b + d) *(e, f) = (a + c + e, b + d +f)

∴ L.H.S = R.H. S

∴ * is associative binary operation.

Algebra Matrix

Question 1.

If A = \(\left[\begin{array}{rrr}

2 & 0 & 1 \\

2 & 1 & 3 \\

1 & -1 & 0

\end{array}\right]\) find A^{2} – 5A + 6I ?

Solution:

Question 2.

If A = \(\left[\begin{array}{ll}

3 & -2 \\

4 & -2

\end{array}\right]\) and I = \(\left[\begin{array}{cc}

1 & 0 \\

0 & 1

\end{array}\right]\) k so that A^{2} = k.A – 2I

Solution:

∴ 3k – 2 = 1

⇒ 3k = 3

⇒ k = 1

Question 3.

A trust fund has Rs. 30,000 that must be invested in two different types of bonds, the first bond pays 5% Interest per year and the second band pays 7% interest per year. Using, matrix nmltiplicatHm determine how to divide Rs. 30,006 among the two types at bonds. It the* trust fund must obtained on arrival total interest of (a) Rs. 1800.

Solution:

Let invest in 1st bond = xRs.

∴ Invest in 2nd bond = 30,000 – x Rs.

Interest in 1st bond = 5%

Interest in 2nd bond = 7%

(a) When total annual interest of Rs. 1800 then,

⇒ -2x + 21,0000 = 18,0000

⇒ 21,0000 – 18,0000 = 2x

⇒ 2x = 3,0000

x = 15,000

∴ Invest in 1st bond = 15,000 Rs.

Invest in 2nd bond = 30,000 – 15,000 = 15,000 Rs.

Question 4.

Express the matrix A = \(\left[\begin{array}{rrr}

2 & -2 & -4 \\

-1 & 3 & 4 \\

1 & -2 & -3

\end{array}\right]\) as the sum of a symmetric and skew symmetric matrix.

Solution:

∴ Sum of symmetric and skew symmetric

Question 5.

By using elementary operatives find the inverse of the matrix A = \(\left[\begin{array}{rr}

1 & 2 \\

2 & -1

\end{array}\right]\)

Solution:

given A = \(\left[\begin{array}{rr}

1 & 2 \\

2 & -1

\end{array}\right]\)

∴ |A| = \(\left|\begin{array}{rr}

1 & 2 \\

2 & -1

\end{array}\right|\) = -1 – 4 = -5

From 1st row, co-factor of I = -1

co-factor of 2 = – 2 .

From 2nd row, co-factor of 2 = -2

co-factor of -1 = 1