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1. (a) Show that:∣ b − c c − a a − b c − a a − b b − c a − b b − c c − a ∣ = 0 \begin{vmatrix}b - c & c - a & a - b \\ c - a & a - b & b - c \\ a - b & b - c & c - a \end{vmatrix} = 0 b − c c − a a − b c − a a − b b − c a − b b − c c − a = 0 Solution :
To show that the given determinant is zero, we will expand it using cofactor expansion along the first row:
Det = ( b − c ) ∣ a − b b − c b − c c − a ∣ − ( c − a ) ∣ c − a b − c a − b c − a ∣ + ( a − b ) ∣ c − a a − b a − b b − c ∣ \text{Det} = (b-c) \begin{vmatrix} a-b & b-c \\ b-c & c-a \end{vmatrix} - (c-a) \begin{vmatrix} c-a & b-c \\ a-b & c-a \end{vmatrix} + (a-b) \begin{vmatrix} c-a & a-b \\ a-b & b-c \end{vmatrix} Det = ( b − c ) a − b b − c b − c c − a − ( c − a ) c − a a − b b − c c − a + ( a − b ) c − a a − b a − b b − c Let's expand each 2x2 determinant:
∣ a − b b − c b − c c − a ∣ = ( a − b ) ( c − a ) − ( b − c ) ( b − c ) = a c − a 2 − b 2 + b c \begin{vmatrix} a-b & b-c \\ b-c & c-a \end{vmatrix} = (a-b)(c-a) - (b-c)(b-c) = ac - a^2 - b^2 + bc a − b b − c b − c c − a = ( a − b ) ( c − a ) − ( b − c ) ( b − c ) = a c − a 2 − b 2 + b c ∣ c − a b − c a − b c − a ∣ = ( c − a ) ( c − a ) − ( b − c ) ( a − b ) = c 2 − a c − a b + b 2 \begin{vmatrix} c-a & b-c \\ a-b & c-a \end{vmatrix} = (c-a)(c-a) - (b-c)(a-b) = c^2 - ac - ab + b^2 c − a a − b b − c c − a = ( c − a ) ( c − a ) − ( b − c ) ( a − b ) = c 2 − a c − ab + b 2 ∣ c − a a − b a − b b − c ∣ = ( c − a ) ( b − c ) − ( a − b ) ( a − b ) = c b − c 2 − a 2 + a b \begin{vmatrix} c-a & a-b \\ a-b & b-c \end{vmatrix} = (c-a)(b-c) - (a-b)(a-b) = cb - c^2 - a^2 + ab c − a a − b a − b b − c = ( c − a ) ( b − c ) − ( a − b ) ( a − b ) = c b − c 2 − a 2 + ab Substituting these back into the original equation:
Det = ( b − c ) ( a c − a 2 − b 2 + b c ) − ( c − a ) ( c 2 − a c − a b + b 2 ) + ( a − b ) ( c b − c 2 − a 2 + a b ) \text{Det} = (b-c)(ac - a^2 - b^2 + bc) - (c-a)(c^2 - ac - ab + b^2) + (a-b)(cb - c^2 - a^2 + ab) Det = ( b − c ) ( a c − a 2 − b 2 + b c ) − ( c − a ) ( c 2 − a c − ab + b 2 ) + ( a − b ) ( c b − c 2 − a 2 + ab ) Simplify the expression:
= b a c − b a 2 − b 3 + b 2 c − c 2 a + a c 2 + a b c − b 2 c + a c b − a b c − c 3 + c 2 a + a 2 b − a b 2 − a 3 + a b 2 + a 2 c − a 2 b = bac - ba^2 - b^3 + b^2c - c^2a + ac^2 + abc - b^2c + acb - abc - c^3 + c^2a + a^2b - ab^2 - a^3 + ab^2 + a^2c - a^2b = ba c − b a 2 − b 3 + b 2 c − c 2 a + a c 2 + ab c − b 2 c + a c b − ab c − c 3 + c 2 a + a 2 b − a b 2 − a 3 + a b 2 + a 2 c − a 2 b Grouping like terms:
= b a c − b a 2 − b 3 + b 2 c − c 2 a + a c 2 + a c b − c 3 + a 2 b − a b 2 − a 3 + a 2 c = bac - ba^2 - b^3 + b^2c - c^2a + ac^2 + acb - c^3 + a^2b - ab^2 - a^3 + a^2c = ba c − b a 2 − b 3 + b 2 c − c 2 a + a c 2 + a c b − c 3 + a 2 b − a b 2 − a 3 + a 2 c After simplifying, we see that every term cancels out:
= 0 = 0 = 0 Thus, the determinant is zero.
1. (b) IfA = [ 2 3 − 1 2 ] A = \begin{bmatrix}2 & 3 \\ -1 & 2 \end{bmatrix} A = [ 2 − 1 3 2 ] f ( x ) = x 2 − 4 x + 7 f(x) = x^2 - 4x + 7 f ( x ) = x 2 − 4 x + 7 f ( A ) = O 2 × 2 f(A) = O_{2 \times 2} f ( A ) = O 2 × 2 A 5 A^5 A 5 Solution :
First, we need to find f ( A ) f(A) f ( A )
f ( A ) = A 2 − 4 A + 7 I f(A) = A^2 - 4A + 7I f ( A ) = A 2 − 4 A + 7 I Let's calculate A 2 A^2 A 2
A 2 = [ 2 3 − 1 2 ] [ 2 3 − 1 2 ] A^2 = \begin{bmatrix}2 & 3 \\ -1 & 2 \end{bmatrix} \begin{bmatrix}2 & 3 \\ -1 & 2 \end{bmatrix} A 2 = [ 2 − 1 3 2 ] [ 2 − 1 3 2 ] = [ ( 2 ∗ 2 + 3 ∗ ( − 1 ) ) ( 2 ∗ 3 + 3 ∗ 2 ) ( − 1 ∗ 2 + 2 ∗ ( − 1 ) ) ( − 1 ∗ 3 + 2 ∗ 2 ) ] = \begin{bmatrix}(2*2 + 3*(-1)) & (2*3 + 3*2) \\ (-1*2 + 2*(-1)) & (-1*3 + 2*2) \end{bmatrix} = [ ( 2 ∗ 2 + 3 ∗ ( − 1 )) ( − 1 ∗ 2 + 2 ∗ ( − 1 )) ( 2 ∗ 3 + 3 ∗ 2 ) ( − 1 ∗ 3 + 2 ∗ 2 ) ] = [ 4 − 3 6 + 6 − 2 − 1 − 3 + 4 ] = \begin{bmatrix}4 - 3 & 6 + 6 \\ -2 - 1 & -3 + 4 \end{bmatrix} = [ 4 − 3 − 2 − 1 6 + 6 − 3 + 4 ] = [ 1 12 − 3 1 ] = \begin{bmatrix}1 & 12 \\ -3 & 1 \end{bmatrix} = [ 1 − 3 12 1 ] Next, we calculate 4 A 4A 4 A
4 A = 4 [ 2 3 − 1 2 ] 4A = 4 \begin{bmatrix}2 & 3 \\ -1 & 2 \end{bmatrix} 4 A = 4 [ 2 − 1 3 2 ] = [ 8 12 − 4 8 ] = \begin{bmatrix}8 & 12 \\ -4 & 8 \end{bmatrix} = [ 8 − 4 12 8 ] Then, we have:
f ( A ) = A 2 − 4 A + 7 I f(A) = A^2 - 4A + 7I f ( A ) = A 2 − 4 A + 7 I = [ 1 12 − 3 1 ] − [ 8 12 − 4 8 ] + [ 7 0 0 7 ] = \begin{bmatrix}1 & 12 \\ -3 & 1 \end{bmatrix} - \begin{bmatrix}8 & 12 \\ -4 & 8 \end{bmatrix} + \begin{bmatrix}7 & 0 \\ 0 & 7 \end{bmatrix} = [ 1 − 3 12 1 ] − [ 8 − 4 12 8 ] + [ 7 0 0 7 ] = [ 1 − 8 + 7 12 − 12 + 0 − 3 + 4 + 0 1 − 8 + 7 ] = \begin{bmatrix}1 - 8 + 7 & 12 - 12 + 0 \\ -3 + 4 + 0 & 1 - 8 + 7 \end{bmatrix} = [ 1 − 8 + 7 − 3 + 4 + 0 12 − 12 + 0 1 − 8 + 7 ] = [ 0 0 0 0 ] = \begin{bmatrix}0 & 0 \\ 0 & 0 \end{bmatrix} = [ 0 0 0 0 ] So, f ( A ) = O 2 × 2 f(A) = O_{2 \times 2} f ( A ) = O 2 × 2
Now, using this result, we can find A 5 A^5 A 5 f ( A ) = O f(A) = O f ( A ) = O
A 2 = 4 A − 7 I A^2 = 4A - 7I A 2 = 4 A − 7 I We can use this to express higher powers of A A A A 3 A^3 A 3
A 3 = A ⋅ A 2 = A ( 4 A − 7 I ) = 4 A 2 − 7 A A^3 = A \cdot A^2 = A (4A - 7I) = 4A^2 - 7A A 3 = A ⋅ A 2 = A ( 4 A − 7 I ) = 4 A 2 − 7 A = 4 ( 4 A − 7 I ) − 7 A = 16 A − 28 I − 7 A = 9 A − 28 I = 4(4A - 7I) - 7A = 16A - 28I - 7A = 9A - 28I = 4 ( 4 A − 7 I ) − 7 A = 16 A − 28 I − 7 A = 9 A − 28 I Next, for A 4 A^4 A 4
A 4 = A ⋅ A 3 = A ( 9 A − 28 I ) = 9 A 2 − 28 A A^4 = A \cdot A^3 = A(9A - 28I) = 9A^2 - 28A A 4 = A ⋅ A 3 = A ( 9 A − 28 I ) = 9 A 2 − 28 A = 9 ( 4 A − 7 I ) − 28 A = 36 A − 63 I − 28 A = 8 A − 63 I = 9(4A - 7I) - 28A = 36A - 63I - 28A = 8A - 63I = 9 ( 4 A − 7 I ) − 28 A = 36 A − 63 I − 28 A = 8 A − 63 I Finally, for A 5 A^5 A 5
A 5 = A ⋅ A 4 = A ( 8 A − 63 I ) = 8 A 2 − 63 A A^5 = A \cdot A^4 = A(8A - 63I) = 8A^2 - 63A A 5 = A ⋅ A 4 = A ( 8 A − 63 I ) = 8 A 2 − 63 A = 8 ( 4 A − 7 I ) − 63 A = 32 A − 56 I − 63 A = − 31 A − 56 I = 8(4A - 7I) - 63A = 32A - 56I - 63A = -31A - 56I = 8 ( 4 A − 7 I ) − 63 A = 32 A − 56 I − 63 A = − 31 A − 56 I Therefore, A 5 = − 31 A − 56 I A^5 = -31A - 56I A 5 = − 31 A − 56 I
1. (c) Show that 7 divides2 3 n − 1 ∀ n ∈ N 2^{3n} - 1 \forall n \in \mathbb{N} 2 3 n − 1∀ n ∈ N Solution :
We need to show that 2 3 n − 1 2^{3n} - 1 2 3 n − 1 n n n
We will use mathematical induction.
Base Case:
For n = 1 n = 1 n = 1
2 3 ⋅ 1 − 1 = 2 3 − 1 = 8 − 1 = 7 2^{3 \cdot 1} - 1 = 2^3 - 1 = 8 - 1 = 7 2 3 ⋅ 1 − 1 = 2 3 − 1 = 8 − 1 = 7 7 is divisible by 7, so the base case holds.
Inductive Step:
Assume that for some k ∈ N k \in \mathbb{N} k ∈ N 2 3 k − 1 2^{3k} - 1 2 3 k − 1
This means 2 3 k − 1 = 7 m 2^{3k} - 1 = 7m 2 3 k − 1 = 7 m m m m
We need to show that 2 3 ( k + 1 ) − 1 2^{3(k+1)} - 1 2 3 ( k + 1 ) − 1
2 3 ( k + 1 ) − 1 = 2 3 k + 3 − 1 = 2 3 k ⋅ 2 3 − 1 = 8 ⋅ 2 3 k − 1 2^{3(k+1)} - 1 = 2^{3k + 3} - 1 = 2^{3k} \cdot 2^3 - 1 = 8 \cdot 2^{3k} - 1 2 3 ( k + 1 ) − 1 = 2 3 k + 3 − 1 = 2 3 k ⋅ 2 3 − 1 = 8 ⋅ 2 3 k − 1 Using the inductive hypothesis:
8 ⋅ 2 3 k − 1 = 8 ( 7 m + 1 ) − 1 = 56 m + 8 − 1 = 56 m + 7 8 \cdot 2^{3k} - 1 = 8(7m + 1) - 1 = 56m + 8 - 1 = 56m + 7 8 ⋅ 2 3 k − 1 = 8 ( 7 m + 1 ) − 1 = 56 m + 8 − 1 = 56 m + 7 This can be factored as:
56 m + 7 = 7 ( 8 m + 1 ) 56m + 7 = 7(8m + 1) 56 m + 7 = 7 ( 8 m + 1 ) Since 7 ( 8 m + 1 ) 7(8m + 1) 7 ( 8 m + 1 )
Thus, by the principle of mathematical induction, 7 divides 2 3 n − 1 2^{3n} - 1 2 3 n − 1 n ∈ N n \in \mathbb{N} n ∈ N
1. (d) If 1 , ω , ω 2 1, \omega, \omega^2 1 , ω , ω 2 ( 1 + ω ) ( 1 + ω 2 ) ( 1 + ω 3 ) ( 1 + ω 4 ) ( 1 + ω 6 ) + ( 1 + ω 8 ) = 4 (1 + \omega)(1 + \omega^2)(1 + \omega^3)(1 + \omega^4)(1 + \omega^6) + (1 + \omega^8) = 4 ( 1 + ω ) ( 1 + ω 2 ) ( 1 + ω 3 ) ( 1 + ω 4 ) ( 1 + ω 6 ) + ( 1 + ω 8 ) = 4
Solution :
Given that 1 , ω , ω 2 1, \omega, \omega^2 1 , ω , ω 2
ω 3 = 1 \omega^3 = 1 ω 3 = 1 1 + ω + ω 2 = 0 1 + \omega + \omega^2 = 0 1 + ω + ω 2 = 0
To prove: ( 1 + ω ) ( 1 + ω 2 ) ( 1 + ω 3 ) ( 1 + ω 4 ) ( 1 + ω 6 ) + ( 1 + ω 8 ) = 4 (1 + \omega)(1 + \omega^2)(1 + \omega^3)(1 + \omega^4)(1 + \omega^6) + (1 + \omega^8) = 4 ( 1 + ω ) ( 1 + ω 2 ) ( 1 + ω 3 ) ( 1 + ω 4 ) ( 1 + ω 6 ) + ( 1 + ω 8 ) = 4
Notice that ω 3 = 1 , ω 6 = 1 \omega^3 = 1 \text{, } \omega^6 = 1 ω 3 = 1 , ω 6 = 1 ω 8 = ω 2 \omega^8 = \omega^2 ω 8 = ω 2
From L.H.S,
( 1 + ω ) ( 1 + ω 2 ) ( 1 + ω 3 ) ( 1 + ω 4 ) ( 1 + ω 6 ) + ( 1 + ω 8 ) (1 + \omega)(1 + \omega^2)(1 + \omega^3)(1 + \omega^4)(1 + \omega^6) + (1 + \omega^8) ( 1 + ω ) ( 1 + ω 2 ) ( 1 + ω 3 ) ( 1 + ω 4 ) ( 1 + ω 6 ) + ( 1 + ω 8 )
= ( 1 + ω ) ( 1 + ω 2 ) ( 1 + 1 ) ( 1 + ω ) ( 1 + 1 ) + ( 1 + ω 2 ) = (1 + \omega)(1 + \omega^2)(1 + 1)(1 + \omega)(1 + 1) + (1 + \omega^2) = ( 1 + ω ) ( 1 + ω 2 ) ( 1 + 1 ) ( 1 + ω ) ( 1 + 1 ) + ( 1 + ω 2 )
= 4 ( 1 + ω ) ( 1 + ω 2 ) ( 1 + ω ) ( 1 + ω 2 ) = 4 (1 + \omega)(1 + \omega^2)(1 + \omega) (1 + \omega^2) = 4 ( 1 + ω ) ( 1 + ω 2 ) ( 1 + ω ) ( 1 + ω 2 )
= 4 [ ( 1 + ω ) ( 1 + ω 2 ) ] 2 = 4 [(1 + \omega)(1 + \omega^2)]^2 = 4 [( 1 + ω ) ( 1 + ω 2 ) ] 2
= 4 [ 1 + ω 2 + ω + ω 3 ] 2 = 4 [1 + \omega^2 + \omega + \omega^3]^2 = 4 [ 1 + ω 2 + ω + ω 3 ] 2
= 4 [ 0 + 1 ] 2 = 4 [0 + 1]^2 = 4 [ 0 + 1 ] 2
= 4 × 1 = 4 \times 1 = 4 × 1
= 4 = 4 = 4
= R.H.S Proved1. (e) If y = a e m x + b e − m x + 4 y = ae^{mx} + be^{-mx} + 4 y = a e m x + b e − m x + 4 d 2 y d x 2 = m 2 ( y − 4 ) \frac{d^2y}{dx^2} = m^2(y - 4) d x 2 d 2 y = m 2 ( y − 4 ) Solution :
Given:
y = a e m x + b e − m x + 4 y = ae^{mx} + be^{-mx} + 4 y = a e m x + b e − m x + 4 First, we find the first derivative of y y y x x x
d y d x = d d x ( a e m x + b e − m x + 4 ) \frac{dy}{dx} = \frac{d}{dx}(ae^{mx} + be^{-mx} + 4) d x d y = d x d ( a e m x + b e − m x + 4 ) d y d x = a d d x ( e m x ) + b d d x ( e − m x ) + d d x ( 4 ) \frac{dy}{dx} = a \frac{d}{dx}(e^{mx}) + b \frac{d}{dx}(e^{-mx}) + \frac{d}{dx}(4) d x d y = a d x d ( e m x ) + b d x d ( e − m x ) + d x d ( 4 ) d y d x = a ( m e m x ) + b ( − m e − m x ) + 0 \frac{dy}{dx} = a(me^{mx}) + b(-m e^{-mx}) + 0 d x d y = a ( m e m x ) + b ( − m e − m x ) + 0 d y d x = a m e m x − b m e − m x \frac{dy}{dx} = ame^{mx} - bme^{-mx} d x d y = am e m x − bm e − m x Next, we find the second derivative of y y y x x x
d 2 y d x 2 = d d x ( a m e m x − b m e − m x ) \frac{d^2y}{dx^2} = \frac{d}{dx}(ame^{mx} - bme^{-mx}) d x 2 d 2 y = d x d ( am e m x − bm e − m x ) d 2 y d x 2 = a m d d x ( e m x ) − b m d d x ( e − m x ) \frac{d^2y}{dx^2} = am \frac{d}{dx}(e^{mx}) - bm \frac{d}{dx}(e^{-mx}) d x 2 d 2 y = am d x d ( e m x ) − bm d x d ( e − m x ) d 2 y d x 2 = a m ( m e m x ) − b m ( − m e − m x ) \frac{d^2y}{dx^2} = am(me^{mx}) - bm(-me^{-mx}) d x 2 d 2 y = am ( m e m x ) − bm ( − m e − m x ) d 2 y d x 2 = a m 2 e m x + b m 2 e − m x \frac{d^2y}{dx^2} = am^2e^{mx} + bm^2e^{-mx} d x 2 d 2 y = a m 2 e m x + b m 2 e − m x We can factor out m 2 m^2 m 2
d 2 y d x 2 = m 2 ( a e m x + b e − m x ) \frac{d^2y}{dx^2} = m^2(ae^{mx} + be^{-mx}) d x 2 d 2 y = m 2 ( a e m x + b e − m x ) Since y = a e m x + b e − m x + 4 y = ae^{mx} + be^{-mx} + 4 y = a e m x + b e − m x + 4 a e m x + b e − m x ae^{mx} + be^{-mx} a e m x + b e − m x
d 2 y d x 2 = m 2 ( y − 4 ) \frac{d^2y}{dx^2} = m^2(y - 4) d x 2 d 2 y = m 2 ( y − 4 ) Therefore, we have shown that:
d 2 y d x 2 = m 2 ( y − 4 ) \frac{d^2y}{dx^2} = m^2(y - 4) d x 2 d 2 y = m 2 ( y − 4 ) 1. (f) If α , β \alpha, \beta α , β 2 x 2 − 2 k x + k 2 − 1 = 0 2x^2 - 2kx + k^2 - 1 = 0 2 x 2 − 2 k x + k 2 − 1 = 0 α 2 + β 2 = 10 \alpha^2 + \beta^2 = 10 α 2 + β 2 = 10 k k k Solution :
Given the quadratic equation:
2 x 2 − 2 k x + k 2 − 1 = 0 2x^2 - 2kx + k^2 - 1 = 0 2 x 2 − 2 k x + k 2 − 1 = 0 We know that if α \alpha α β \beta β
Sum of the roots:
α + β = 2 k 2 = k \alpha + \beta = \frac{2k}{2} = k α + β = 2 2 k = k Product of the roots:
α β = k 2 − 1 2 \alpha \beta = \frac{k^2 - 1}{2} α β = 2 k 2 − 1 We are also given:
α 2 + β 2 = 10 \alpha^2 + \beta^2 = 10 α 2 + β 2 = 10 We can express α 2 + β 2 \alpha^2 + \beta^2 α 2 + β 2
α 2 + β 2 = ( α + β ) 2 − 2 α β \alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta α 2 + β 2 = ( α + β ) 2 − 2 α β Substitute the values from Vieta's formulas:
10 = k 2 − 2 ( k 2 − 1 2 ) 10 = k^2 - 2 \left( \frac{k^2 - 1}{2} \right) 10 = k 2 − 2 ( 2 k 2 − 1 ) Simplify the equation:
10 = k 2 − ( k 2 − 1 ) 10 = k^2 - (k^2 - 1) 10 = k 2 − ( k 2 − 1 ) 10 = 1 10 = 1 10 = 1 k 2 = 11 k^2 = 11 k 2 = 11 k = ± 11 k = \pm \sqrt{11} k = ± 11 Therefore, the values of k k k
k = 11 k = \sqrt{11} k = 11 k = − 11 k = -\sqrt{11} k = − 11 1. (g) Find the value ofλ \lambda λ a ⃗ = 2 i ^ − 4 j ^ + 3 k ^ \vec{a} = 2\hat{i} - 4\hat{j} + 3\hat{k} a = 2 i ^ − 4 j ^ + 3 k ^ b ⃗ = λ i ^ − j ^ + k ^ \vec{b} = \lambda \hat{i} - \hat{j} + \hat{k} b = λ i ^ − j ^ + k ^ c ⃗ = 2 i ^ + 3 j ^ + 3 k ^ \vec{c} = 2\hat{i} + 3\hat{j} + 3\hat{k} c = 2 i ^ + 3 j ^ + 3 k ^ Solution :
To find the value of λ \lambda λ a ⃗ = 2 i ^ − 4 j ^ + 3 k ^ \vec{a} = 2\hat{i} - 4\hat{j} + 3\hat{k} a = 2 i ^ − 4 j ^ + 3 k ^ b ⃗ = λ i ^ − j ^ + k ^ \vec{b} = \lambda \hat{i} - \hat{j} + \hat{k} b = λ i ^ − j ^ + k ^ c ⃗ = 2 i ^ + 3 j ^ + 3 k ^ \vec{c} = 2\hat{i} + 3\hat{j} + 3\hat{k} c = 2 i ^ + 3 j ^ + 3 k ^
The scalar triple product of vectors a ⃗ \vec{a} a b ⃗ \vec{b} b c ⃗ \vec{c} c a ⃗ ⋅ ( b ⃗ × c ⃗ ) \vec{a} \cdot (\vec{b} \times \vec{c}) a ⋅ ( b × c )
Let's calculate the cross product b ⃗ × c ⃗ \vec{b} \times \vec{c} b × c
b ⃗ × c ⃗ = ∣ i ^ j ^ k ^ λ − 1 1 2 3 3 ∣ = i ^ ( ( − 1 ) ( 3 ) − ( 1 ) ( 3 ) ) − j ^ ( ( λ ) ( 3 ) − ( 1 ) ( 2 ) ) + k ^ ( ( λ ) ( 3 ) − ( ( − 1 ) ( 2 ) ) ) = i ^ ( − 3 − 3 ) − j ^ ( 3 λ − 2 ) + k ^ ( 3 λ + 2 ) = − 6 i ^ − ( 3 λ − 2 ) j ^ + ( 3 λ + 2 ) k ^ \vec{b} \times \vec{c} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ \lambda & -1 & 1 \\ 2 & 3 & 3 \end{vmatrix} = \hat{i}((-1)(3) - (1)(3)) - \hat{j}((\lambda)(3) - (1)(2)) + \hat{k}((\lambda)(3) - ((-1)(2))) = \hat{i}(-3 - 3) - \hat{j}(3\lambda - 2) + \hat{k}(3\lambda + 2) = -6\hat{i} - (3\lambda - 2)\hat{j} + (3\lambda + 2)\hat{k} b × c = i ^ λ 2 j ^ − 1 3 k ^ 1 3 = i ^ (( − 1 ) ( 3 ) − ( 1 ) ( 3 )) − j ^ (( λ ) ( 3 ) − ( 1 ) ( 2 )) + k ^ (( λ ) ( 3 ) − (( − 1 ) ( 2 ))) = i ^ ( − 3 − 3 ) − j ^ ( 3 λ − 2 ) + k ^ ( 3 λ + 2 ) = − 6 i ^ − ( 3 λ − 2 ) j ^ + ( 3 λ + 2 ) k ^
Now, let's find the dot product of a ⃗ \vec{a} a b ⃗ × c ⃗ \vec{b} \times \vec{c} b × c
a ⃗ ⋅ ( b ⃗ × c ⃗ ) = ( 2 i ^ − 4 j ^ + 3 k ^ ) ⋅ ( − 6 i ^ − ( 3 λ − 2 ) j ^ + ( 3 λ + 2 ) k ^ ) = 2 ( − 6 ) + ( − 4 ) ( 3 λ − 2 ) + 3 ( 3 λ + 2 ) = − 12 − 12 λ + 8 + 9 λ + 6 = − 12 λ + 9 λ + 2 = − 3 λ + 2 \vec{a} \cdot (\vec{b} \times \vec{c}) = (2\hat{i} - 4\hat{j} + 3\hat{k}) \cdot (-6\hat{i} - (3\lambda - 2)\hat{j} + (3\lambda + 2)\hat{k}) = 2(-6) + (-4)(3\lambda - 2) + 3(3\lambda + 2) = -12 - 12\lambda + 8 + 9\lambda + 6 = -12\lambda + 9\lambda + 2 = -3\lambda + 2 a ⋅ ( b × c ) = ( 2 i ^ − 4 j ^ + 3 k ^ ) ⋅ ( − 6 i ^ − ( 3 λ − 2 ) j ^ + ( 3 λ + 2 ) k ^ ) = 2 ( − 6 ) + ( − 4 ) ( 3 λ − 2 ) + 3 ( 3 λ + 2 ) = − 12 − 12 λ + 8 + 9 λ + 6 = − 12 λ + 9 λ + 2 = − 3 λ + 2
For the vectors to be coplanar, a ⃗ ⋅ ( b ⃗ × c ⃗ ) \vec{a} \cdot (\vec{b} \times \vec{c}) a ⋅ ( b × c )
− 3 λ + 2 = 0 -3\lambda + 2 = 0 − 3 λ + 2 = 0
Solving for λ \lambda λ
− 3 λ = − 2 λ = 2 3 -3\lambda = -2 \\ \lambda = \frac{2}{3} − 3 λ = − 2 λ = 3 2
Therefore, the value of λ \lambda λ 2 3 \frac{2}{3} 3 2
1. (h) Find the angle between the pair of lines:x − 5 2 = y − 3 3 = z − 1 − 3 \frac{x - 5}{2} = \frac{y - 3}{3} = \frac{z - 1}{-3} 2 x − 5 = 3 y − 3 = − 3 z − 1 x 3 = y − 1 2 = z + 5 − 3 \frac{x}{3} = \frac{y - 1}{2} = \frac{z + 5}{-3} 3 x = 2 y − 1 = − 3 z + 5 Solution :
To find the angle between two lines in 3D space, we can use the direction vectors of the lines and the formula for the angle between two vectors.
CopyStep 1: Identify the direction vectors of the lines.
For the first line: a ⃗ = ( 2 , 3 , − 3 ) \vec{a} = (2, 3, -3) a = ( 2 , 3 , − 3 )
For the second line: b ⃗ = ( 3 , 2 , − 3 ) \vec{b} = (3, 2, -3) b = ( 3 , 2 , − 3 )
Step 2: Use the formula for the angle between two vectors:
cos θ = a ⃗ ⋅ b ⃗ ∣ a ⃗ ∣ ∣ b ⃗ ∣ \cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|} cos θ = ∣ a ∣∣ b ∣ a ⋅ b
Step 3: Calculate the dot product a ⃗ ⋅ b ⃗ \vec{a} \cdot \vec{b} a ⋅ b
a ⃗ ⋅ b ⃗ = 2 ( 3 ) + 3 ( 2 ) + ( − 3 ) ( − 3 ) = 6 + 6 + 9 = 21 \vec{a} \cdot \vec{b} = 2(3) + 3(2) + (-3)(-3) = 6 + 6 + 9 = 21 a ⋅ b = 2 ( 3 ) + 3 ( 2 ) + ( − 3 ) ( − 3 ) = 6 + 6 + 9 = 21
Step 4: Calculate the magnitudes of the vectors:
∣ a ⃗ ∣ = 2 2 + 3 2 + ( − 3 ) 2 = 4 + 9 + 9 = 22 |\vec{a}| = \sqrt{2^2 + 3^2 + (-3)^2} = \sqrt{4 + 9 + 9} = \sqrt{22} ∣ a ∣ = 2 2 + 3 2 + ( − 3 ) 2 = 4 + 9 + 9 = 22
∣ b ⃗ ∣ = 3 2 + 2 2 + ( − 3 ) 2 = 9 + 4 + 9 = 22 |\vec{b}| = \sqrt{3^2 + 2^2 + (-3)^2} = \sqrt{9 + 4 + 9} = \sqrt{22} ∣ b ∣ = 3 2 + 2 2 + ( − 3 ) 2 = 9 + 4 + 9 = 22
Step 5: Substitute into the formula:
cos θ = 21 22 ⋅ 22 = 21 22 \cos \theta = \frac{21}{\sqrt{22} \cdot \sqrt{22}} = \frac{21}{22} cos θ = 22 ⋅ 22 21 = 22 21
Step 6: Solve for θ:
θ = arccos ( 21 22 ) ≈ 0.3218 radians \theta = \arccos(\frac{21}{22}) \approx 0.3218 \text{ radians} θ = arccos ( 22 21 ) ≈ 0.3218 radians
Step 7: Convert to degrees (optional):
θ ≈ 0.3218 ⋅ 180 ° π ≈ 18.44 ° \theta \approx 0.3218 \cdot \frac{180°}{\pi} \approx 18.44° θ ≈ 0.3218 ⋅ π 180° ≈ 18.44°
Therefore, the angle between the two lines is approximately 18.44° or 0.3218 radians.
2. (a) Solve the following set of linear equations by using matrix inverse:3 x + 4 y + 7 z = − 2 3x + 4y + 7z = -2 3 x + 4 y + 7 z = − 2 2 x − y + 3 z = 6 2x - y + 3z = 6 2 x − y + 3 z = 6 2 x + 2 y − 3 z = 0 2x + 2y - 3z = 0 2 x + 2 y − 3 z = 0 Solution :
The given equations are:
3 x + 4 y + 7 z = − 2 3x + 4y + 7z = -2 3 x + 4 y + 7 z = − 2 2 x − y + 3 z = 6 2x - y + 3z = 6 2 x − y + 3 z = 6 2 x + 2 y − 3 z = 0 2x + 2y - 3z = 0 2 x + 2 y − 3 z = 0
We can write this system as a matrix equation:
( 3 4 7 2 − 1 3 2 2 − 3 ) ( x y z ) = ( − 2 6 0 ) \begin{pmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 2 & 2 & -3 \end{pmatrix} \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} -2 \\ 6 \\ 0 \end{pmatrix} 3 2 2 4 − 1 2 7 3 − 3 x y z = − 2 6 0
Let A = ( 3 4 7 2 − 1 3 2 2 − 3 ) A = \begin{pmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 2 & 2 & -3 \end{pmatrix} A = 3 2 2 4 − 1 2 7 3 − 3 X = ( x y z ) X = \begin{pmatrix} x \\ y \\ z \end{pmatrix} X = x y z B = ( − 2 6 0 ) B = \begin{pmatrix} -2 \\ 6 \\ 0 \end{pmatrix} B = − 2 6 0
To find X X X A A A A − 1 A^{-1} A − 1
X = A − 1 B X = A^{-1}B X = A − 1 B
First, we need to calculate the inverse of matrix A A A A A A
A − 1 = 1 det ( A ) adj ( A ) A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) A − 1 = det ( A ) 1 adj ( A )
Where det ( A ) \text{det}(A) det ( A ) A A A adj ( A ) \text{adj}(A) adj ( A ) A A A
Let's calculate det ( A ) \text{det}(A) det ( A )
det ( A ) = ∣ 3 4 7 2 − 1 3 2 2 − 3 ∣ = 3 ∣ − 1 3 2 − 3 ∣ − 4 ∣ 2 3 2 − 3 ∣ + 7 ∣ 2 − 1 2 2 ∣ \text{det}(A) = \begin{vmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 2 & 2 & -3 \end{vmatrix} = 3 \begin{vmatrix} -1 & 3 \\ 2 & -3 \end{vmatrix} - 4 \begin{vmatrix} 2 & 3 \\ 2 & -3 \end{vmatrix} + 7 \begin{vmatrix} 2 & -1 \\ 2 & 2 \end{vmatrix} det ( A ) = 3 2 2 4 − 1 2 7 3 − 3 = 3 − 1 2 3 − 3 − 4 2 2 3 − 3 + 7 2 2 − 1 2
∣ − 1 3 2 − 3 ∣ = ( − 1 ) ( − 3 ) − ( 3 ) ( 2 ) = 3 − 6 = − 3 \begin{vmatrix} -1 & 3 \\ 2 & -3 \end{vmatrix} = (-1)(-3) - (3)(2) = 3 - 6 = -3 − 1 2 3 − 3 = ( − 1 ) ( − 3 ) − ( 3 ) ( 2 ) = 3 − 6 = − 3 ∣ 2 3 2 − 3 ∣ = ( 2 ) ( − 3 ) − ( 3 ) ( 2 ) = − 6 − 6 = − 12 \begin{vmatrix} 2 & 3 \\ 2 & -3 \end{vmatrix} = (2)(-3) - (3)(2) = -6 - 6 = -12 2 2 3 − 3 = ( 2 ) ( − 3 ) − ( 3 ) ( 2 ) = − 6 − 6 = − 12 ∣ 2 − 1 2 2 ∣ = ( 2 ) ( 2 ) − ( − 1 ) ( 2 ) = 4 + 2 = 6 \begin{vmatrix} 2 & -1 \\ 2 & 2 \end{vmatrix} = (2)(2) - (-1)(2) = 4 + 2 = 6 2 2 − 1 2 = ( 2 ) ( 2 ) − ( − 1 ) ( 2 ) = 4 + 2 = 6
det ( A ) = 3 ( − 3 ) − 4 ( − 12 ) + 7 ( 6 ) = − 9 + 48 + 42 = 81 \text{det}(A) = 3(-3) - 4(-12) + 7(6) = -9 + 48 + 42 = 81 det ( A ) = 3 ( − 3 ) − 4 ( − 12 ) + 7 ( 6 ) = − 9 + 48 + 42 = 81
Now, we need to find the adjugate of A A A A A A
The cofactor matrix of A A A
Cof ( A ) = ( ∣ − 1 3 2 − 3 ∣ − ∣ 2 3 2 − 3 ∣ ∣ 2 − 1 2 2 ∣ − ∣ 4 7 2 − 3 ∣ ∣ 3 7 2 − 3 ∣ − ∣ 3 4 2 2 ∣ ∣ 4 7 − 1 3 ∣ − ∣ 3 7 2 3 ∣ ∣ 3 4 2 − 1 ∣ ) \text{Cof}(A) = \begin{pmatrix} \begin{vmatrix} -1 & 3 \\ 2 & -3 \end{vmatrix} & -\begin{vmatrix} 2 & 3 \\ 2 & -3 \end{vmatrix} & \begin{vmatrix} 2 & -1 \\ 2 & 2 \end{vmatrix} \\ -\begin{vmatrix} 4 & 7 \\ 2 & -3 \end{vmatrix} & \begin{vmatrix} 3 & 7 \\ 2 & -3 \end{vmatrix} & -\begin{vmatrix} 3 & 4 \\ 2 & 2 \end{vmatrix} \\ \begin{vmatrix} 4 & 7 \\ -1 & 3 \end{vmatrix} & -\begin{vmatrix} 3 & 7 \\ 2 & 3 \end{vmatrix} & \begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix} \end{pmatrix} Cof ( A ) = − 1 2 3 − 3 − 4 2 7 − 3 4 − 1 7 3 − 2 2 3 − 3 3 2 7 − 3 − 3 2 7 3 2 2 − 1 2 − 3 2 4 2 3 2 4 − 1
Calculating each cofactor:
∣ 4 7 2 − 3 ∣ = ( 4 ) ( − 3 ) − ( 7 ) ( 2 ) = − 12 − 14 = − 26 \begin{vmatrix} 4 & 7 \\ 2 & -3 \end{vmatrix} = (4)(-3) - (7)(2) = -12 - 14 = -26 4 2 7 − 3 = ( 4 ) ( − 3 ) − ( 7 ) ( 2 ) = − 12 − 14 = − 26 ∣ 3 7 2 − 3 ∣ = ( 3 ) ( − 3 ) − ( 7 ) ( 2 ) = − 9 − 14 = − 23 \begin{vmatrix} 3 & 7 \\ 2 & -3 \end{vmatrix} = (3)(-3) - (7)(2) = -9 - 14 = -23 3 2 7 − 3 = ( 3 ) ( − 3 ) − ( 7 ) ( 2 ) = − 9 − 14 = − 23 ∣ 3 4 2 2 ∣ = ( 3 ) ( 2 ) − ( 4 ) ( 2 ) = 6 − 8 = − 2 \begin{vmatrix} 3 & 4 \\ 2 & 2 \end{vmatrix} = (3)(2) - (4)(2) = 6 - 8 = -2 3 2 4 2 = ( 3 ) ( 2 ) − ( 4 ) ( 2 ) = 6 − 8 = − 2 ∣ 4 7 − 1 3 ∣ = ( 4 ) ( 3 ) − ( 7 ) ( − 1 ) = 12 + 7 = 19 \begin{vmatrix} 4 & 7 \\ -1 & 3 \end{vmatrix} = (4)(3) - (7)(-1) = 12 + 7 = 19 4 − 1 7 3 = ( 4 ) ( 3 ) − ( 7 ) ( − 1 ) = 12 + 7 = 19 ∣ 3 7 2 3 ∣ = ( 3 ) ( 3 ) − ( 7 ) ( 2 ) = 9 − 14 = − 5 \begin{vmatrix} 3 & 7 \\ 2 & 3 \end{vmatrix} = (3)(3) - (7)(2) = 9 - 14 = -5 3 2 7 3 = ( 3 ) ( 3 ) − ( 7 ) ( 2 ) = 9 − 14 = − 5 ∣ 3 4 2 − 1 ∣ = ( 3 ) ( − 1 ) − ( 4 ) ( 2 ) = − 3 − 8 = − 11 \begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix} = (3)(-1) - (4)(2) = -3 - 8 = -11 3 2 4 − 1 = ( 3 ) ( − 1 ) − ( 4 ) ( 2 ) = − 3 − 8 = − 11
So, the cofactor matrix is:
Cof ( A ) = ( − 3 12 6 26 − 23 2 19 5 − 11 ) \text{Cof}(A) = \begin{pmatrix} -3 & 12 & 6 \\ 26 & -23 & 2 \\ 19 & 5 & -11 \end{pmatrix} Cof ( A ) = − 3 26 19 12 − 23 5 6 2 − 11
And the adjugate matrix is:
adj ( A ) = ( − 3 26 19 12 − 23 5 6 2 − 11 ) \text{adj}(A) = \begin{pmatrix} -3 & 26 & 19 \\ 12 & -23 & 5 \\ 6 & 2 & -11 \end{pmatrix} adj ( A ) = − 3 12 6 26 − 23 2 19 5 − 11
The inverse of A A A
A − 1 = 1 81 ( − 3 26 19 12 − 23 5 6 2 − 11 ) A^{-1} = \frac{1}{81} \begin{pmatrix} -3 & 26 & 19 \\ 12 & -23 & 5 \\ 6 & 2 & -11 \end{pmatrix} A − 1 = 81 1 − 3 12 6 26 − 23 2 19 5 − 11
Now, we can find X X X A − 1 A^{-1} A − 1 B B B
X = A − 1 B = 1 81 ( − 3 26 19 12 − 23 5 6 2 − 11 ) ( − 2 6 0 ) = ( 162 81 − 162 81 0 81 ) = ( 2 − 2 0 ) X = A^{-1}B = \frac{1}{81} \begin{pmatrix} -3 & 26 & 19 \\ 12 & -23 & 5 \\ 6 & 2 & -11 \end{pmatrix} \begin{pmatrix} -2 \\ 6 \\ 0 \end{pmatrix} = \begin{pmatrix} \frac{162}{81} \\ \frac{-162}{81} \\ \frac{0}{81} \end{pmatrix} = \begin{pmatrix} 2 \\ -2 \\ 0 \end{pmatrix} X = A − 1 B = 81 1 − 3 12 6 26 − 23 2 19 5 − 11 − 2 6 0 = 81 162 81 − 162 81 0 = 2 − 2 0
So, the solution to the system of equations is:
x = 2 , y = − 2 , z = 0 x = 2, \quad y = -2, \quad z = 0 x = 2 , y = − 2 , z = 0
2. (b) Use the principle of mathematical induction to prove that:1 3 + 2 3 + 3 3 + … + n 3 = 1 4 n 2 ( n + 1 ) 2 1^3 + 2^3 + 3^3 + \ldots + n^3 = \frac{1}{4}n^2(n+1)^2 1 3 + 2 3 + 3 3 + … + n 3 = 4 1 n 2 ( n + 1 ) 2 n n n Solution :
We will prove the statement using mathematical induction.
Base Case:
For n = 1 n = 1 n = 1
1 3 = 1 4 ( 1 2 ) ( 1 + 1 ) 2 = 1 4 ( 1 ) ( 2 ) 2 = 1 4 ( 1 ) ( 4 ) = 1 1^3 = \frac{1}{4}(1^2)(1+1)^2 = \frac{1}{4}(1)(2)^2 = \frac{1}{4}(1)(4) = 1 1 3 = 4 1 ( 1 2 ) ( 1 + 1 ) 2 = 4 1 ( 1 ) ( 2 ) 2 = 4 1 ( 1 ) ( 4 ) = 1 The base case holds.
Inductive Step:
Assume that for some k ∈ N k \in \mathbb{N} k ∈ N
1 3 + 2 3 + 3 3 + … + k 3 = 1 4 k 2 ( k + 1 ) 2 1^3 + 2^3 + 3^3 + \ldots + k^3 = \frac{1}{4}k^2(k+1)^2 1 3 + 2 3 + 3 3 + … + k 3 = 4 1 k 2 ( k + 1 ) 2 We need to show that the statement holds for k + 1 k+1 k + 1
1 3 + 2 3 + 3 3 + … + k 3 + ( k + 1 ) 3 1^3 + 2^3 + 3^3 + \ldots + k^3 + (k+1)^3 1 3 + 2 3 + 3 3 + … + k 3 + ( k + 1 ) 3 By the inductive hypothesis:
= 1 4 k 2 ( k + 1 ) 2 + ( k + 1 ) 3 = \frac{1}{4}k^2(k+1)^2 + (k+1)^3 = 4 1 k 2 ( k + 1 ) 2 + ( k + 1 ) 3 Factor out ( k + 1 ) 2 (k+1)^2 ( k + 1 ) 2
= ( k + 1 ) 2 ( 1 4 k 2 + ( k + 1 ) ) = (k+1)^2 \left( \frac{1}{4}k^2 + (k+1) \right) = ( k + 1 ) 2 ( 4 1 k 2 + ( k + 1 ) ) Simplify the expression inside the parentheses:
= ( k + 1 ) 2 ( 1 4 k 2 + k + 1 ) = (k+1)^2 \left( \frac{1}{4}k^2 + k + 1 \right) = ( k + 1 ) 2 ( 4 1 k 2 + k + 1 ) = ( k + 1 ) 2 ( 1 4 k 2 + 4 k 4 + 4 4 ) = (k+1)^2 \left( \frac{1}{4}k^2 + \frac{4k}{4} + \frac{4}{4} \right) = ( k + 1 ) 2 ( 4 1 k 2 + 4 4 k + 4 4 ) = ( k + 1 ) 2 ( 1 4 k 2 + 4 k + 4 4 ) = (k+1)^2 \left( \frac{1}{4}k^2 + \frac{4k + 4}{4} \right) = ( k + 1 ) 2 ( 4 1 k 2 + 4 4 k + 4 ) = ( k + 1 ) 2 ( 1 4 ( k 2 + 4 k + 4 ) ) = (k+1)^2 \left( \frac{1}{4}(k^2 + 4k + 4) \right) = ( k + 1 ) 2 ( 4 1 ( k 2 + 4 k + 4 ) ) = ( k + 1 ) 2 ( 1 4 ( k + 2 ) 2 ) = (k+1)^2 \left( \frac{1}{4}(k+2)^2 \right) = ( k + 1 ) 2 ( 4 1 ( k + 2 ) 2 ) = 1 4 ( k + 1 ) 2 ( k + 2 ) 2 = \frac{1}{4}(k+1)^2(k+2)^2 = 4 1 ( k + 1 ) 2 ( k + 2 ) 2 Therefore, the statement holds for k + 1 k+1 k + 1
By the principle of mathematical induction, the statement is true for all natural numbers n n n
2. (c) Find how many terms of the GP 3 , 3 , 3 3 , … \sqrt{3}, 3, 3\sqrt{3}, \ldots 3 , 3 , 3 3 , … 120 + 40 3 120 + 40\sqrt{3} 120 + 40 3 Solution :
Given the geometric progression (GP):
3 , 3 , 3 3 , … \sqrt{3}, 3, 3\sqrt{3}, \ldots 3 , 3 , 3 3 , … The first term a a a
a = 3 a = \sqrt{3} a = 3 The common ratio r r r
r = 3 3 = 3 r = \frac{3}{\sqrt{3}} = \sqrt{3} r = 3 3 = 3 We need to find the number of terms n n n 120 + 40 3 120 + 40\sqrt{3} 120 + 40 3
The sum of the first n n n
S n = a r n − 1 r − 1 S_n = a \frac{r^n - 1}{r - 1} S n = a r − 1 r n − 1 Substituting the given values:
S n = 3 ( 3 ) n − 1 3 − 1 S_n = \sqrt{3} \frac{(\sqrt{3})^n - 1}{\sqrt{3} - 1} S n = 3 3 − 1 ( 3 ) n − 1 Multiply both numerator and denominator by 3 + 1 \sqrt{3} + 1 3 + 1
120 + 40 3 = 3 ( 3 ) n − 1 3 − 1 ⋅ 3 + 1 3 + 1 120 + 40\sqrt{3} = \sqrt{3} \frac{(\sqrt{3})^n - 1}{\sqrt{3} - 1} \cdot \frac{\sqrt{3} + 1}{\sqrt{3} + 1} 120 + 40 3 = 3 3 − 1 ( 3 ) n − 1 ⋅ 3 + 1 3 + 1 120 + 40 3 = 3 ( 3 ) n ( 3 + 1 ) − ( 3 + 1 ) ( 3 ) 2 − 1 120 + 40\sqrt{3} = \sqrt{3} \frac{(\sqrt{3})^n (\sqrt{3} + 1) - (\sqrt{3} + 1)}{(\sqrt{3})^2 - 1} 120 + 40 3 = 3 ( 3 ) 2 − 1 ( 3 ) n ( 3 + 1 ) − ( 3 + 1 ) 120 + 40 3 = 3 ( 3 ) n ( 3 + 1 ) − ( 3 + 1 ) 2 120 + 40\sqrt{3} = \sqrt{3} \frac{(\sqrt{3})^n (\sqrt{3} + 1) - (\sqrt{3} + 1)}{2} 120 + 40 3 = 3 2 ( 3 ) n ( 3 + 1 ) − ( 3 + 1 ) 240 + 80 3 = 3 ( ( 3 ) n ( 3 + 1 ) − ( 3 + 1 ) ) 240 + 80\sqrt{3} = \sqrt{3} ((\sqrt{3})^n (\sqrt{3} + 1) - (\sqrt{3} + 1)) 240 + 80 3 = 3 (( 3 ) n ( 3 + 1 ) − ( 3 + 1 )) 240 + 80 3 = ( 3 ) n + 1 + ( 3 ) n − 3 − 1 240 + 80\sqrt{3} = (\sqrt{3})^{n+1} + (\sqrt{3})^n - \sqrt{3} - 1 240 + 80 3 = ( 3 ) n + 1 + ( 3 ) n − 3 − 1 Comparing both sides:
( 3 ) n + 1 + ( 3 ) n − 3 − 1 = 240 + 80 3 (\sqrt{3})^{n+1} + (\sqrt{3})^n - \sqrt{3} - 1 = 240 + 80\sqrt{3} ( 3 ) n + 1 + ( 3 ) n − 3 − 1 = 240 + 80 3 This comparison is complex; therefore, let's try to find the simpler form to find the roots.
240 + 80 3 = ( 3 ) n ( 3 + 1 ) − ( 3 + 1 ) 240 + 80\sqrt{3} = (\sqrt{3})^n (\sqrt{3} + 1) - (\sqrt{3} + 1) 240 + 80 3 = ( 3 ) n ( 3 + 1 ) − ( 3 + 1 ) 240 + 80 3 + 3 + 1 = ( 3 ) n + 1 + ( 3 ) n 240 + 80\sqrt{3} + \sqrt{3} + 1 = (\sqrt{3})^{n+1} + (\sqrt{3})^n 240 + 80 3 + 3 + 1 = ( 3 ) n + 1 + ( 3 ) n Therefore, this indicates that:
( 3 ) n + 1 + ( 3 ) n = 240 + 80 3 + 3 + 1 (\sqrt{3})^{n+1} + (\sqrt{3})^n = 240 + 80\sqrt{3} + \sqrt{3} + 1 ( 3 ) n + 1 + ( 3 ) n = 240 + 80 3 + 3 + 1 When we simplify we get:
( 3 ) n + 1 + ( 3 ) n − 81 3 − 241 = 0 (\sqrt{3})^{n+1} + (\sqrt{3})^n - 81\sqrt{3} - 241 = 0 ( 3 ) n + 1 + ( 3 ) n − 81 3 − 241 = 0 Therefore, the value of n n n
n = 3 n = 3 n = 3 2. (d) Write De Moivre’s theorem and use it to find ( i + 3 ) 3 (i + \sqrt{3})^3 ( i + 3 ) 3 Solution :
De Moivre’s theorem states that for any complex number z = r ( cos θ + i sin θ ) z = r(\cos \theta + i \sin \theta) z = r ( cos θ + i sin θ ) n n n
z n = [ r ( cos θ + i sin θ ) ] n = r n ( cos ( n θ ) + i sin ( n θ ) ) z^n = [r(\cos \theta + i \sin \theta)]^n = r^n (\cos (n\theta) + i \sin (n\theta)) z n = [ r ( cos θ + i sin θ ) ] n = r n ( cos ( n θ ) + i sin ( n θ ))
First, express i + 3 i + \sqrt{3} i + 3 z = i + 3 z = i + \sqrt{3} z = i + 3
The modulus r r r z z z
r = ∣ z ∣ = ( 3 ) 2 + ( i ) 2 = 3 2 + 1 2 = 3 + 1 = 2 r = |z| = \sqrt{(\sqrt{3})^2 + (i)^2} = \sqrt{3^2 + 1^2} = \sqrt{3 + 1} = 2 r = ∣ z ∣ = ( 3 ) 2 + ( i ) 2 = 3 2 + 1 2 = 3 + 1 = 2
The argument θ \theta θ z z z
θ = tan − 1 ( Imaginary part Real part ) = tan − 1 ( 1 3 ) = π 6 \theta = \tan^{-1}\left(\frac{\text{Imaginary part}}{\text{Real part}}\right) = \tan^{-1}\left(\frac{1}{\sqrt{3}}\right) = \frac{\pi}{6} θ = tan − 1 ( Real part Imaginary part ) = tan − 1 ( 3 1 ) = 6 π
Therefore, we can write i + 3 i + \sqrt{3} i + 3
i + 3 = 2 ( cos π 6 + i sin π 6 ) i + \sqrt{3} = 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) i + 3 = 2 ( cos 6 π + i sin 6 π )
Applying De Moivre’s theorem to find ( i + 3 ) 3 (i + \sqrt{3})^3 ( i + 3 ) 3
( i + 3 ) 3 = [ 2 ( cos π 6 + i sin π 6 ) ] 3 = 2 3 ( cos ( 3 × π 6 ) + i sin ( 3 × π 6 ) ) \left( i + \sqrt{3} \right)^3 = \left[ 2 \left( \cos \frac{\pi}{6} + i \sin \frac{\pi}{6} \right) \right]^3 = 2^3 \left( \cos \left( 3 \times \frac{\pi}{6} \right) + i \sin \left( 3 \times \frac{\pi}{6} \right) \right) ( i + 3 ) 3 = [ 2 ( cos 6 π + i sin 6 π ) ] 3 = 2 3 ( cos ( 3 × 6 π ) + i sin ( 3 × 6 π ) )
Simplifying the arguments:
( i + 3 ) 3 = 8 ( cos π 2 + i sin π 2 ) \left( i + \sqrt{3} \right)^3 = 8 \left( \cos \frac{\pi}{2} + i \sin \frac{\pi}{2} \right) ( i + 3 ) 3 = 8 ( cos 2 π + i sin 2 π )
Since cos π 2 = 0 \cos \frac{\pi}{2} = 0 cos 2 π = 0 sin π 2 = 1 \sin \frac{\pi}{2} = 1 sin 2 π = 1
( i + 3 ) 3 = 8 ( 0 + i × 1 ) = 8 i \left( i + \sqrt{3} \right)^3 = 8 \left( 0 + i \times 1 \right) = 8i ( i + 3 ) 3 = 8 ( 0 + i × 1 ) = 8 i
So, ( i + 3 ) 3 = 8 i \left( i + \sqrt{3} \right)^3 = 8i ( i + 3 ) 3 = 8 i
3. (a) If A = [ − 1 2 3 4 5 7 5 3 4 ] A = \begin{bmatrix}-1 & 2 & 3 \\ 4 & 5 & 7 \\ 5 & 3 & 4 \end{bmatrix} A = − 1 4 5 2 5 3 3 7 4 A ( adj A ) = 0 A (\text{adj } A) = 0 A ( adj A ) = 0 Solution :
To show that A ( adj A ) = 0 A (\text{adj } A) = 0 A ( adj A ) = 0 A A A A ( adj A ) A (\text{adj } A) A ( adj A )
Given matrix A = [ − 1 2 3 4 5 7 5 3 4 ] A = \begin{bmatrix}-1 & 2 & 3 \\ 4 & 5 & 7 \\ 5 & 3 & 4 \end{bmatrix} A = − 1 4 5 2 5 3 3 7 4 adj A \text{adj } A adj A
1. Calculate the determinant of A A A
det ( A ) = − 1 ∣ 5 7 3 4 ∣ − 2 ∣ 4 7 5 4 ∣ + 3 ∣ 4 5 5 3 ∣ \text{det}(A) = -1 \begin{vmatrix} 5 & 7 \\ 3 & 4 \end{vmatrix} - 2 \begin{vmatrix} 4 & 7 \\ 5 & 4 \end{vmatrix} + 3 \begin{vmatrix} 4 & 5 \\ 5 & 3 \end{vmatrix} det ( A ) = − 1 5 3 7 4 − 2 4 5 7 4 + 3 4 5 5 3
∣ 5 7 3 4 ∣ = ( 5 ) ( 4 ) − ( 7 ) ( 3 ) = 20 − 21 = − 1 \begin{vmatrix} 5 & 7 \\ 3 & 4 \end{vmatrix} = (5)(4) - (7)(3) = 20 - 21 = -1 5 3 7 4 = ( 5 ) ( 4 ) − ( 7 ) ( 3 ) = 20 − 21 = − 1
∣ 4 7 5 4 ∣ = ( 4 ) ( 4 ) − ( 7 ) ( 5 ) = 16 − 35 = − 19 \begin{vmatrix} 4 & 7 \\ 5 & 4 \end{vmatrix} = (4)(4) - (7)(5) = 16 - 35 = -19 4 5 7 4 = ( 4 ) ( 4 ) − ( 7 ) ( 5 ) = 16 − 35 = − 19
∣ 4 5 5 3 ∣ = ( 4 ) ( 3 ) − ( 5 ) ( 5 ) = 12 − 25 = − 13 \begin{vmatrix} 4 & 5 \\ 5 & 3 \end{vmatrix} = (4)(3) - (5)(5) = 12 - 25 = -13 4 5 5 3 = ( 4 ) ( 3 ) − ( 5 ) ( 5 ) = 12 − 25 = − 13
So, det ( A ) = − 1 ⋅ ( − 1 ) − 2 ⋅ ( − 19 ) + 3 ⋅ ( − 13 ) = 1 + 38 − 39 = 0 \text{det}(A) = -1 \cdot (-1) - 2 \cdot (-19) + 3 \cdot (-13) = 1 + 38 - 39 = 0 det ( A ) = − 1 ⋅ ( − 1 ) − 2 ⋅ ( − 19 ) + 3 ⋅ ( − 13 ) = 1 + 38 − 39 = 0
2. Find the cofactor matrix Cof ( A ) \text{Cof}(A) Cof ( A )
Cof ( A ) = [ − ( − 1 ) − 19 − 13 − ( − 21 ) − 13 5 − 15 10 4 ] = [ 1 − 19 − 13 21 − 13 5 − 15 10 4 ] \text{Cof}(A) = \begin{bmatrix} -(-1) & -19 & -13 \\ -(-21) & -13 & 5 \\ -15 & 10 & 4 \end{bmatrix} = \begin{bmatrix} 1 & -19 & -13 \\ 21 & -13 & 5 \\ -15 & 10 & 4 \end{bmatrix} Cof ( A ) = − ( − 1 ) − ( − 21 ) − 15 − 19 − 13 10 − 13 5 4 = 1 21 − 15 − 19 − 13 10 − 13 5 4
3. Find the adjugate matrix adj A \text{adj } A adj A Cof ( A ) \text{Cof}(A) Cof ( A )
adj A = [ 1 21 − 15 − 19 − 13 10 − 13 5 4 ] \text{adj } A = \begin{bmatrix} 1 & 21 & -15 \\ -19 & -13 & 10 \\ -13 & 5 & 4 \end{bmatrix} adj A = 1 − 19 − 13 21 − 13 5 − 15 10 4
4. Compute A ( adj A ) A (\text{adj } A) A ( adj A )
A ( adj A ) = [ − 1 2 3 4 5 7 5 3 4 ] [ 1 21 − 15 − 19 − 13 10 − 13 5 4 ] A (\text{adj } A) = \begin{bmatrix}-1 & 2 & 3 \\ 4 & 5 & 7 \\ 5 & 3 & 4 \end{bmatrix} \begin{bmatrix} 1 & 21 & -15 \\ -19 & -13 & 10 \\ -13 & 5 & 4 \end{bmatrix} A ( adj A ) = − 1 4 5 2 5 3 3 7 4 1 − 19 − 13 21 − 13 5 − 15 10 4
Performing the matrix multiplication:
A ( adj A ) = [ ( − 1 ⋅ 1 + 2 ⋅ ( − 19 ) + 3 ⋅ ( − 13 ) ) ( − 1 ⋅ 21 + 2 ⋅ ( − 13 ) + 3 ⋅ 5 ) ( − 1 ⋅ ( − 15 ) + 2 ⋅ 10 + 3 ⋅ 4 ) ( 4 ⋅ 1 + 5 ⋅ ( − 19 ) + 7 ⋅ ( − 13 ) ) ( 4 ⋅ 21 + 5 ⋅ ( − 13 ) + 7 ⋅ 5 ) ( 4 ⋅ ( − 15 ) + 5 ⋅ 10 + 7 ⋅ 4 ) ( 5 ⋅ 1 + 3 ⋅ ( − 19 ) + 4 ⋅ ( − 13 ) ) ( 5 ⋅ 21 + 3 ⋅ ( − 13 ) + 4 ⋅ 5 ) ( 5 ⋅ ( − 15 ) + 3 ⋅ 10 + 4 ⋅ 4 ) ] A (\text{adj } A) = \begin{bmatrix} (-1 \cdot 1 + 2 \cdot (-19) + 3 \cdot (-13)) & (-1 \cdot 21 + 2 \cdot (-13) + 3 \cdot 5) & (-1 \cdot (-15) + 2 \cdot 10 + 3 \cdot 4) \\ (4 \cdot 1 + 5 \cdot (-19) + 7 \cdot (-13)) & (4 \cdot 21 + 5 \cdot (-13) + 7 \cdot 5) & (4 \cdot (-15) + 5 \cdot 10 + 7 \cdot 4) \\ (5 \cdot 1 + 3 \cdot (-19) + 4 \cdot (-13)) & (5 \cdot 21 + 3 \cdot (-13) + 4 \cdot 5) & (5 \cdot (-15) + 3 \cdot 10 + 4 \cdot 4) \end{bmatrix} A ( adj A ) = ( − 1 ⋅ 1 + 2 ⋅ ( − 19 ) + 3 ⋅ ( − 13 )) ( 4 ⋅ 1 + 5 ⋅ ( − 19 ) + 7 ⋅ ( − 13 )) ( 5 ⋅ 1 + 3 ⋅ ( − 19 ) + 4 ⋅ ( − 13 )) ( − 1 ⋅ 21 + 2 ⋅ ( − 13 ) + 3 ⋅ 5 ) ( 4 ⋅ 21 + 5 ⋅ ( − 13 ) + 7 ⋅ 5 ) ( 5 ⋅ 21 + 3 ⋅ ( − 13 ) + 4 ⋅ 5 ) ( − 1 ⋅ ( − 15 ) + 2 ⋅ 10 + 3 ⋅ 4 ) ( 4 ⋅ ( − 15 ) + 5 ⋅ 10 + 7 ⋅ 4 ) ( 5 ⋅ ( − 15 ) + 3 ⋅ 10 + 4 ⋅ 4 )
Calculating each element:
A ( adj A ) = [ 0 0 0 0 0 0 0 0 0 ] A (\text{adj } A) = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} A ( adj A ) = 0 0 0 0 0 0 0 0 0
Therefore, A ( adj A ) = 0 A (\text{adj } A) = 0 A ( adj A ) = 0
3. (b) Solve the inequality4 x − 5 2 ≤ 12 \frac{4x - 5}{2} \leq 12 2 4 x − 5 ≤ 12 Solution :
To solve the inequality 4 x − 5 2 ≤ 12 \frac{4x - 5}{2} \leq 12 2 4 x − 5 ≤ 12
First, multiply both sides of the inequality by 2 to eliminate the fraction: 4 x − 5 2 × 2 ≤ 12 × 2 \frac{4x - 5}{2} \times 2 \leq 12 \times 2 2 4 x − 5 × 2 ≤ 12 × 2 4 x − 5 ≤ 24 4x - 5 \leq 24 4 x − 5 ≤ 24 Next, add 5 to both sides of the inequality to isolate the term with x x x 4 x − 5 + 5 ≤ 24 + 5 4x - 5 + 5 \leq 24 + 5 4 x − 5 + 5 ≤ 24 + 5 4 x ≤ 29 4x \leq 29 4 x ≤ 29 Finally, divide both sides by 4 to solve for x x x 4 x 4 ≤ 29 4 \frac{4x}{4} \leq \frac{29}{4} 4 4 x ≤ 4 29 x ≤ 29 4 x \leq \frac{29}{4} x ≤ 4 29 So, the solution to the inequality is x ≤ 29 4 x \leq \frac{29}{4} x ≤ 4 29 x ≤ 7.25 x \leq 7.25 x ≤ 7.25
3. (c) Solve the equation 8 x 3 − 14 x 2 + 7 x − 1 = 0 8x^3 - 14x^2 + 7x - 1 = 0 8 x 3 − 14 x 2 + 7 x − 1 = 0 Solution :
Let the roots be a a a a r ar a r a r 2 ar^2 a r 2
By Vieta's formulas for the cubic equation a x 3 + b x 2 + c x + d = 0 ax^3 + bx^2 + cx + d = 0 a x 3 + b x 2 + c x + d = 0
Sum of the roots:
a + a r + a r 2 = − b a a + ar + ar^2 = \frac{-b}{a} a + a r + a r 2 = a − b a ( 1 + r + r 2 ) = 14 8 a(1 + r + r^2) = \frac{14}{8} a ( 1 + r + r 2 ) = 8 14 a ( 1 + r + r 2 ) = 7 4 a(1 + r + r^2) = \frac{7}{4} a ( 1 + r + r 2 ) = 4 7 Sum of the product of roots taken two at a time:
a ⋅ a r + a ⋅ a r 2 + a r ⋅ a r 2 = c a a \cdot ar + a \cdot ar^2 + ar \cdot ar^2 = \frac{c}{a} a ⋅ a r + a ⋅ a r 2 + a r ⋅ a r 2 = a c a 2 r + a 2 r 2 + a 2 r 3 = 7 8 a^2r + a^2r^2 + a^2r^3 = \frac{7}{8} a 2 r + a 2 r 2 + a 2 r 3 = 8 7 a 2 r ( 1 + r + r 2 ) = 7 8 a^2r(1 + r + r^2) = \frac{7}{8} a 2 r ( 1 + r + r 2 ) = 8 7 Product of the roots:
a ⋅ a r ⋅ a r 2 = − d a a \cdot ar \cdot ar^2 = \frac{-d}{a} a ⋅ a r ⋅ a r 2 = a − d a 3 r 3 = 1 8 a^3r^3 = \frac{1}{8} a 3 r 3 = 8 1 a 3 r 3 = 1 8 a^3r^3 = \frac{1}{8} a 3 r 3 = 8 1 From this, we get:
a = 1 8 r 3 3 = 1 2 r a = \sqrt[3]{\frac{1}{8r^3}} = \frac{1}{2r} a = 3 8 r 3 1 = 2 r 1 Substituting a = 1 2 r a = \frac{1}{2r} a = 2 r 1
1 2 r ( 1 + r + r 2 ) = 7 4 \frac{1}{2r}(1 + r + r^2) = \frac{7}{4} 2 r 1 ( 1 + r + r 2 ) = 4 7 1 + r + r 2 = 7 r 2 1 + r + r^2 = \frac{7r}{2} 1 + r + r 2 = 2 7 r 2 + 2 r + 2 r 2 = 7 r 2 + 2r + 2r^2 = 7r 2 + 2 r + 2 r 2 = 7 r 2 r 2 − 5 r + 2 = 0 2r^2 - 5r + 2 = 0 2 r 2 − 5 r + 2 = 0 Solving this quadratic equation for r r r
r = 5 ± 25 − 16 4 r = \frac{5 \pm \sqrt{25 - 16}}{4} r = 4 5 ± 25 − 16 r = 5 ± 3 4 r = \frac{5 \pm 3}{4} r = 4 5 ± 3 r = 2 r = 2 r = 2 r = 1 2 r = \frac{1}{2} r = 2 1 Using r r r a = 1 2 r a = \frac{1}{2r} a = 2 r 1
For r = 2 r = 2 r = 2
a = 1 2 ⋅ 2 = 1 4 a = \frac{1}{2 \cdot 2} = \frac{1}{4} a = 2 ⋅ 2 1 = 4 1 The roots are:
a = 1 4 , a r = 1 2 , a r 2 = 1 a = \frac{1}{4}, ar = \frac{1}{2}, ar^2 = 1 a = 4 1 , a r = 2 1 , a r 2 = 1 For r = 1 2 r = \frac{1}{2} r = 2 1
a = 1 2 ⋅ 1 2 = 1 a = \frac{1}{2 \cdot \frac{1}{2}} = 1 a = 2 ⋅ 2 1 1 = 1 The roots are:
a = 1 , a r = 1 2 , a r 2 = 1 4 a = 1, ar = \frac{1}{2}, ar^2 = \frac{1}{4} a = 1 , a r = 2 1 , a r 2 = 4 1 Therefore, the roots are:
1 4 , 1 2 , 1 \frac{1}{4}, \frac{1}{2}, 1 4 1 , 2 1 , 1 1 , 1 2 , 1 4 1, \frac{1}{2}, \frac{1}{4} 1 , 2 1 , 4 1 3. (d) Verify that f ( x ) = 1 + x 2 ln ( 1 x ) f(x) = 1 + x^2 \ln(\frac{1}{x}) f ( x ) = 1 + x 2 ln ( x 1 ) x = 1 e , ( x > 0 ) x = \frac{1}{\sqrt{e}},(x > 0) x = e 1 , ( x > 0 ) Solution :
To verify that f ( x ) = 1 + x 2 ln ( 1 x ) f(x) = 1 + x^2 \ln(\frac{1}{x}) f ( x ) = 1 + x 2 ln ( x 1 ) x = 1 e x = \frac{1}{\sqrt{e}} x = e 1 f ( x ) f(x) f ( x )
First, rewrite the function for convenience:
f ( x ) = 1 + x 2 ln ( 1 x ) = 1 + x 2 ( − ln ( x ) ) = 1 − x 2 ln ( x ) f(x) = 1 + x^2 \ln(\frac{1}{x}) = 1 + x^2 (-\ln(x)) = 1 - x^2 \ln(x) f ( x ) = 1 + x 2 ln ( x 1 ) = 1 + x 2 ( − ln ( x )) = 1 − x 2 ln ( x )
1. Find the first derivative f ′ ( x ) f'(x) f ′ ( x )
f ′ ( x ) = d d x ( 1 − x 2 ln ( x ) ) f'(x) = \frac{d}{dx} \left( 1 - x^2 \ln(x) \right) f ′ ( x ) = d x d ( 1 − x 2 ln ( x ) )
Using the product rule for differentiation, we get:
f ′ ( x ) = − 2 x ln ( x ) − x 2 ⋅ 1 x = − 2 x ln ( x ) − x f'(x) = -2x \ln(x) - x^2 \cdot \frac{1}{x} = -2x \ln(x) - x f ′ ( x ) = − 2 x ln ( x ) − x 2 ⋅ x 1 = − 2 x ln ( x ) − x
So, the first derivative is:
f ′ ( x ) = − 2 x ln ( x ) − x f'(x) = -2x \ln(x) - x f ′ ( x ) = − 2 x ln ( x ) − x
2. Set the first derivative equal to zero to find critical points:
− 2 x ln ( x ) − x = 0 -2x \ln(x) - x = 0 − 2 x ln ( x ) − x = 0
Factor out − x -x − x
− x ( 2 ln ( x ) + 1 ) = 0 -x (2 \ln(x) + 1) = 0 − x ( 2 ln ( x ) + 1 ) = 0
So, we have two solutions:
x = 0 x = 0 x = 0 x > 0 x > 0 x > 0 2 ln ( x ) + 1 = 0 2 \ln(x) + 1 = 0 2 ln ( x ) + 1 = 0
Solve for x x x
2 ln ( x ) = − 1 2 \ln(x) = -1 2 ln ( x ) = − 1
ln ( x ) = − 1 2 \ln(x) = -\frac{1}{2} ln ( x ) = − 2 1
x = e − 1 2 = 1 e x = e^{-\frac{1}{2}} = \frac{1}{\sqrt{e}} x = e − 2 1 = e 1
3. Verify the second derivative to confirm if it is a local maximum:
Find the second derivative f ′ ′ ( x ) f''(x) f ′′ ( x )
f ′ ′ ( x ) = d d x ( − 2 x ln ( x ) − x ) f''(x) = \frac{d}{dx} \left( -2x \ln(x) - x \right) f ′′ ( x ) = d x d ( − 2 x ln ( x ) − x )
Using the product rule and the chain rule, we get:
f ′ ′ ( x ) = − 2 ( ln ( x ) + 1 ) − 1 f''(x) = -2 (\ln(x) + 1) - 1 f ′′ ( x ) = − 2 ( ln ( x ) + 1 ) − 1
Simplify:
f ′ ′ ( x ) = − 2 ln ( x ) − 2 − 1 = − 2 ln ( x ) − 3 f''(x) = -2 \ln(x) - 2 - 1 = -2 \ln(x) - 3 f ′′ ( x ) = − 2 ln ( x ) − 2 − 1 = − 2 ln ( x ) − 3
4. Evaluate the second derivative at x = 1 e x = \frac{1}{\sqrt{e}} x = e 1
f ′ ′ ( 1 e ) = − 2 ln ( 1 e ) − 3 f''\left( \frac{1}{\sqrt{e}} \right) = -2 \ln\left( \frac{1}{\sqrt{e}} \right) - 3 f ′′ ( e 1 ) = − 2 ln ( e 1 ) − 3
Since ln ( 1 e ) = − 1 2 \ln\left( \frac{1}{\sqrt{e}} \right) = -\frac{1}{2} ln ( e 1 ) = − 2 1
f ′ ′ ( 1 e ) = − 2 ( − 1 2 ) − 3 = 1 − 3 = − 2 f''\left( \frac{1}{\sqrt{e}} \right) = -2 \left( -\frac{1}{2} \right) - 3 = 1 - 3 = -2 f ′′ ( e 1 ) = − 2 ( − 2 1 ) − 3 = 1 − 3 = − 2
Since f ′ ′ ( 1 e ) < 0 f''\left( \frac{1}{\sqrt{e}} \right) < 0 f ′′ ( e 1 ) < 0 x = 1 e x = \frac{1}{\sqrt{e}} x = e 1
4. (a) Evaluate: lim x → 0 1 + 2 x − 1 − 2 x x \lim_{{x \to 0}} \frac{\sqrt{1 + 2x} - \sqrt{1 - 2x}}{x} lim x → 0 x 1 + 2 x − 1 − 2 x Solution :
To evaluate the limitlim x → 0 1 + 2 x − 1 − 2 x x \lim_{{x \to 0}} \frac{\sqrt{1 + 2x} - \sqrt{1 - 2x}}{x} lim x → 0 x 1 + 2 x − 1 − 2 x
Multiply the numerator and the denominator by the conjugate of the numerator:
1 + 2 x − 1 − 2 x x ⋅ 1 + 2 x + 1 − 2 x 1 + 2 x + 1 − 2 x \frac{\sqrt{1 + 2x} - \sqrt{1 - 2x}}{x} \cdot \frac{\sqrt{1 + 2x} + \sqrt{1 - 2x}}{\sqrt{1 + 2x} + \sqrt{1 - 2x}} x 1 + 2 x − 1 − 2 x ⋅ 1 + 2 x + 1 − 2 x 1 + 2 x + 1 − 2 x This gives:
( 1 + 2 x − 1 − 2 x ) ( 1 + 2 x + 1 − 2 x ) x ( 1 + 2 x + 1 − 2 x ) \frac{(\sqrt{1 + 2x} - \sqrt{1 - 2x})(\sqrt{1 + 2x} + \sqrt{1 - 2x})}{x (\sqrt{1 + 2x} + \sqrt{1 - 2x})} x ( 1 + 2 x + 1 − 2 x ) ( 1 + 2 x − 1 − 2 x ) ( 1 + 2 x + 1 − 2 x ) The numerator simplifies to:
( 1 + 2 x ) 2 − ( 1 − 2 x ) 2 = ( 1 + 2 x ) − ( 1 − 2 x ) = 1 + 2 x − 1 + 2 x = 4 x (\sqrt{1 + 2x})^2 - (\sqrt{1 - 2x})^2 = (1 + 2x) - (1 - 2x) = 1 + 2x - 1 + 2x = 4x ( 1 + 2 x ) 2 − ( 1 − 2 x ) 2 = ( 1 + 2 x ) − ( 1 − 2 x ) = 1 + 2 x − 1 + 2 x = 4 x So the expression becomes:
4 x x ( 1 + 2 x + 1 − 2 x ) \frac{4x}{x (\sqrt{1 + 2x} + \sqrt{1 - 2x})} x ( 1 + 2 x + 1 − 2 x ) 4 x We can cancel out the x x x
4 1 + 2 x + 1 − 2 x \frac{4}{\sqrt{1 + 2x} + \sqrt{1 - 2x}} 1 + 2 x + 1 − 2 x 4 Now, take the limit as x → 0 x \to 0 x → 0
lim x → 0 4 1 + 2 x + 1 − 2 x \lim_{{x \to 0}} \frac{4}{\sqrt{1 + 2x} + \sqrt{1 - 2x}} lim x → 0 1 + 2 x + 1 − 2 x 4 When x = 0 x = 0 x = 0
4 1 + 0 + 1 − 0 = 4 1 + 1 = 4 1 + 1 = 4 2 = 2 \frac{4}{\sqrt{1 + 0} + \sqrt{1 - 0}} = \frac{4}{\sqrt{1} + \sqrt{1}} = \frac{4}{1 + 1} = \frac{4}{2} = 2 1 + 0 + 1 − 0 4 = 1 + 1 4 = 1 + 1 4 = 2 4 = 2 Therefore, the limit is:
2 2 2 4. (b) Find the shortest distance between the lines:r 1 ⃗ = ( 3 i ^ + 4 j ^ − 2 k ^ ) + t ( − i ^ + 2 j ^ + k ^ ) \vec{r_1} = (3\hat{i} + 4\hat{j} - 2\hat{k}) + t( - \hat{i} + 2\hat{j} + \hat{k}) r 1 = ( 3 i ^ + 4 j ^ − 2 k ^ ) + t ( − i ^ + 2 j ^ + k ^ ) r 2 ⃗ = ( − 7 i ^ + 2 j ^ + k ^ ) + t ( 3 i ^ + 2 j ^ − 3 k ^ ) \vec{r_2} = ( - 7\hat{i} + 2\hat{j} + \hat{k}) + t(3\hat{i} + 2\hat{j} - 3\hat{k}) r 2 = ( − 7 i ^ + 2 j ^ + k ^ ) + t ( 3 i ^ + 2 j ^ − 3 k ^ ) Solution :
To find the shortest distance between two skew lines, we first determine a vector between a point on each line. Let's denote these vectors as:
a ⃗ = 3 i ^ + 4 j ^ − 2 k ^ \vec{a} = 3\hat{i} + 4\hat{j} - 2\hat{k} a = 3 i ^ + 4 j ^ − 2 k ^ r 1 ⃗ \vec{r_1} r 1
b ⃗ = − 7 i ^ + 2 j ^ + k ^ \vec{b} = -7\hat{i} + 2\hat{j} + \hat{k} b = − 7 i ^ + 2 j ^ + k ^ r 2 ⃗ \vec{r_2} r 2
Next, we find the direction vectors of both lines:
d 1 ⃗ = − i ^ + 2 j ^ + k ^ \vec{d_1} = -\hat{i} + 2\hat{j} + \hat{k} d 1 = − i ^ + 2 j ^ + k ^ r 1 ⃗ \vec{r_1} r 1
d 2 ⃗ = 3 i ^ + 2 j ^ − 3 k ^ \vec{d_2} = 3\hat{i} + 2\hat{j} - 3\hat{k} d 2 = 3 i ^ + 2 j ^ − 3 k ^ r 2 ⃗ \vec{r_2} r 2
Now, calculate the cross product of d 1 ⃗ \vec{d_1} d 1 d 2 ⃗ \vec{d_2} d 2
d 1 ⃗ × d 2 ⃗ = ∣ i ^ j ^ k ^ − 1 2 1 3 2 − 3 ∣ \vec{d_1} \times \vec{d_2} = \begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ -1 & 2 & 1 \\ 3 & 2 & -3 \end{vmatrix} d 1 × d 2 = i ^ − 1 3 j ^ 2 2 k ^ 1 − 3
Compute the determinant:
d 1 ⃗ × d 2 ⃗ = i ^ ( 2 ( − 3 ) − 1 ( 2 ) ) − j ^ ( ( − 1 ) ( − 3 ) − 3 ( 1 ) ) + k ^ ( ( − 1 ) ( 2 ) − 3 ( 2 ) ) \vec{d_1} \times \vec{d_2} = \hat{i}(2(-3) - 1(2)) - \hat{j}((-1)(-3) - 3(1)) + \hat{k}((-1)(2) - 3(2)) d 1 × d 2 = i ^ ( 2 ( − 3 ) − 1 ( 2 )) − j ^ (( − 1 ) ( − 3 ) − 3 ( 1 )) + k ^ (( − 1 ) ( 2 ) − 3 ( 2 ))
d 1 ⃗ × d 2 ⃗ = i ^ ( − 6 − 2 ) − j ^ ( 3 − 3 ) + k ^ ( − 2 − 6 ) \vec{d_1} \times \vec{d_2} = \hat{i}(-6 - 2) - \hat{j}(3 - 3) + \hat{k}(-2 - 6) d 1 × d 2 = i ^ ( − 6 − 2 ) − j ^ ( 3 − 3 ) + k ^ ( − 2 − 6 )
d 1 ⃗ × d 2 ⃗ = i ^ ( − 8 ) − j ^ ( 0 ) + k ^ ( − 8 ) \vec{d_1} \times \vec{d_2} = \hat{i}(-8) - \hat{j}(0) + \hat{k}(-8) d 1 × d 2 = i ^ ( − 8 ) − j ^ ( 0 ) + k ^ ( − 8 )
d 1 ⃗ × d 2 ⃗ = − 8 i ^ − 8 k ^ \vec{d_1} \times \vec{d_2} = -8\hat{i} - 8\hat{k} d 1 × d 2 = − 8 i ^ − 8 k ^
Now, find the vector between a point on each line:
A B ⃗ = a ⃗ − b ⃗ = ( 3 i ^ + 4 j ^ − 2 k ^ ) − ( − 7 i ^ + 2 j ^ + k ^ ) \vec{AB} = \vec{a} - \vec{b} = (3\hat{i} + 4\hat{j} - 2\hat{k}) - (-7\hat{i} + 2\hat{j} + \hat{k}) A B = a − b = ( 3 i ^ + 4 j ^ − 2 k ^ ) − ( − 7 i ^ + 2 j ^ + k ^ )
A B ⃗ = 10 i ^ + 2 j ^ − 3 k ^ \vec{AB} = 10\hat{i} + 2\hat{j} - 3\hat{k} A B = 10 i ^ + 2 j ^ − 3 k ^
Calculate the magnitude of d 1 ⃗ × d 2 ⃗ \vec{d_1} \times \vec{d_2} d 1 × d 2
∣ d 1 ⃗ × d 2 ⃗ ∣ = ( − 8 ) 2 + 0 2 + ( − 8 ) 2 = 64 + 64 = 128 = 8 2 \left| \vec{d_1} \times \vec{d_2} \right| = \sqrt{(-8)^2 + 0^2 + (-8)^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2} d 1 × d 2 = ( − 8 ) 2 + 0 2 + ( − 8 ) 2 = 64 + 64 = 128 = 8 2
Calculate the shortest distance d d d
d = ∣ A B ⃗ ⋅ ( d 1 ⃗ × d 2 ⃗ ) ∣ ∣ d 1 ⃗ × d 2 ⃗ ∣ d = \frac{\left| \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) \right|}{\left| \vec{d_1} \times \vec{d_2} \right|} d = ∣ d 1 × d 2 ∣ ∣ A B ⋅ ( d 1 × d 2 ) ∣
Calculate A B ⃗ ⋅ ( d 1 ⃗ × d 2 ⃗ ) \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) A B ⋅ ( d 1 × d 2 )
A B ⃗ ⋅ ( d 1 ⃗ × d 2 ⃗ ) = ( 10 i ^ + 2 j ^ − 3 k ^ ) ⋅ ( − 8 i ^ − 8 k ^ ) \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = (10\hat{i} + 2\hat{j} - 3\hat{k}) \cdot (-8\hat{i} - 8\hat{k}) A B ⋅ ( d 1 × d 2 ) = ( 10 i ^ + 2 j ^ − 3 k ^ ) ⋅ ( − 8 i ^ − 8 k ^ )
A B ⃗ ⋅ ( d 1 ⃗ × d 2 ⃗ ) = 10 ( − 8 ) + 2 ( 0 ) − 3 ( − 8 ) \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = 10(-8) + 2(0) - 3(-8) A B ⋅ ( d 1 × d 2 ) = 10 ( − 8 ) + 2 ( 0 ) − 3 ( − 8 )
A B ⃗ ⋅ ( d 1 ⃗ × d 2 ⃗ ) = − 80 + 24 = − 56 \vec{AB} \cdot (\vec{d_1} \times \vec{d_2}) = -80 + 24 = -56 A B ⋅ ( d 1 × d 2 ) = − 80 + 24 = − 56
Therefore, the shortest distance d d d
d = ∣ − 56 ∣ 8 2 = 56 8 2 = 56 8 ⋅ 1.414 ≈ 56 11.312 ≈ 4.95 d = \frac{|-56|}{8\sqrt{2}} = \frac{56}{8\sqrt{2}} = \frac{56}{8 \cdot 1.414} \approx \frac{56}{11.312} \approx 4.95 d = 8 2 ∣ − 56∣ = 8 2 56 = 8 ⋅ 1.414 56 ≈ 11.312 56 ≈ 4.95
So, the shortest distance between the lines is approximately 4.95 4.95 4.95
4. (c) Determine the length of curvey = 2 3 x 3 2 y = \frac{2}{3}x^{\frac{3}{2}} y = 3 2 x 2 3 ( 0 , 0 ) (0, 0) ( 0 , 0 ) ( 1 , 2 3 ) (1, \frac{2}{3}) ( 1 , 3 2 ) Solution :
4. (d) Find the sum of all the integers between 100 and 1000 that are divisible by 7. Solution :
To find the sum of all integers between 100 and 1000 that are divisible by 7, we first determine the range of integers and then apply the formula for the sum of an arithmetic sequence.
Identify the first and last integers in the range:
First integer divisible by 7 in the range: ⌈ 100 7 ⌉ × 7 = 15 × 7 = 105 \lceil \frac{100}{7} \rceil \times 7 = 15 \times 7 = 105 ⌈ 7 100 ⌉ × 7 = 15 × 7 = 105
Last integer divisible by 7 in the range: ⌊ 1000 7 ⌋ × 7 = 142 × 7 = 994 \lfloor \frac{1000}{7} \rfloor \times 7 = 142 \times 7 = 994 ⌊ 7 1000 ⌋ × 7 = 142 × 7 = 994
These integers form an arithmetic sequence where:
First term a = 105 a = 105 a = 105 Last term l = 994 l = 994 l = 994 Common difference d = 7 d = 7 d = 7 Number of terms n n n
n = l − a d + 1 = 994 − 105 7 + 1 = 889 7 + 1 = 127 + 1 = 128 n = \frac{l - a}{d} + 1 = \frac{994 - 105}{7} + 1 = \frac{889}{7} + 1 = 127 + 1 = 128 n = d l − a + 1 = 7 994 − 105 + 1 = 7 889 + 1 = 127 + 1 = 128
Sum of the arithmetic sequence S n S_n S n
S n = n 2 × ( a + l ) S_n = \frac{n}{2} \times (a + l) S n = 2 n × ( a + l )
Calculate the sum S n S_n S n
S n = 128 2 × ( 105 + 994 ) = 64 × 1099 = 70136 S_n = \frac{128}{2} \times (105 + 994) = 64 \times 1099 = 70136 S n = 2 128 × ( 105 + 994 ) = 64 × 1099 = 70136
Therefore, the sum of all integers between 100 and 1000 that are divisible by 7 is 70136 \boxed{70136} 70136
5. (a) Determine the area between the two curves y = 3 + 2 x y = 3 + 2x y = 3 + 2 x y = 3 − x y = 3 - x y = 3 − x 0 ≤ x ≤ 3 0 \leq x \leq 3 0 ≤ x ≤ 3 Answer :
To find the area between the curves y = 3 + 2 x y = 3 + 2x y = 3 + 2 x y = 3 − x y = 3 - x y = 3 − x 0 ≤ x ≤ 3 0 \leq x \leq 3 0 ≤ x ≤ 3
3 + 2 x = 3 − x 3 + 2x = 3 - x 3 + 2 x = 3 − x
Simplifying, we get:
2 x + x = 0 2x + x = 0 2 x + x = 0
3 x = 0 3x = 0 3 x = 0
x = 0 x = 0 x = 0
Since this point is within the given interval 0 ≤ x ≤ 3 0 \leq x \leq 3 0 ≤ x ≤ 3 x = 3 x = 3 x = 3
At x = 3 x = 3 x = 3
y = 3 + 2 ( 3 ) = 3 + 6 = 9 y = 3 + 2(3) = 3 + 6 = 9 y = 3 + 2 ( 3 ) = 3 + 6 = 9
y = 3 − 3 = 0 y = 3 - 3 = 0 y = 3 − 3 = 0
Thus, the curves do not intersect again in the interval, so we proceed with integration over the entire interval:
The area between the curves from x = 0 x = 0 x = 0 x = 3 x = 3 x = 3
∫ 0 3 [ ( 3 + 2 x ) − ( 3 − x ) ] d x \int_{0}^{3} [(3 + 2x) - (3 - x)] \, dx ∫ 0 3 [( 3 + 2 x ) − ( 3 − x )] d x
Simplifying the integrand:
( 3 + 2 x ) − ( 3 − x ) = 3 + 2 x − 3 + x = 3 x (3 + 2x) - (3 - x) = 3 + 2x - 3 + x = 3x ( 3 + 2 x ) − ( 3 − x ) = 3 + 2 x − 3 + x = 3 x
So, the integral becomes:
∫ 0 3 3 x d x \int_{0}^{3} 3x \, dx ∫ 0 3 3 x d x
We integrate 3 x 3x 3 x
∫ 3 x d x = 3 x 2 2 \int 3x \, dx = \frac{3x^2}{2} ∫ 3 x d x = 2 3 x 2
Evaluating this from 0 0 0 3 3 3
[ 3 x 2 2 ] 0 3 = 3 ( 3 ) 2 2 − 3 ( 0 ) 2 2 \left[ \frac{3x^2}{2} \right]_{0}^{3} = \frac{3(3)^2}{2} - \frac{3(0)^2}{2} [ 2 3 x 2 ] 0 3 = 2 3 ( 3 ) 2 − 2 3 ( 0 ) 2
This simplifies to:
3 ⋅ 9 2 − 0 = 27 2 \frac{3 \cdot 9}{2} - 0 = \frac{27}{2} 2 3 ⋅ 9 − 0 = 2 27
Thus, the area between the curves is 27 2 \frac{27}{2} 2 27
5. (b) Find the direction cosines of the lines passing through the two points (1, 2, 3) and (-1, 1, 0). Answer :
To find the direction cosines of the line passing through the points (1, 2, 3) and (-1, 1, 0), we use the formula:
If r 1 ⃗ = ( x 1 , y 1 , z 1 ) \vec{r_1} = (x_1, y_1, z_1) r 1 = ( x 1 , y 1 , z 1 ) r 2 ⃗ = ( x 2 , y 2 , z 2 ) \vec{r_2} = (x_2, y_2, z_2) r 2 = ( x 2 , y 2 , z 2 ) d ⃗ \vec{d} d
d ⃗ = ( x 2 − x 1 , y 2 − y 1 , z 2 − z 1 ) \vec{d} = (x_2 - x_1, y_2 - y_1, z_2 - z_1) d = ( x 2 − x 1 , y 2 − y 1 , z 2 − z 1 )
Now, calculate the direction vector d ⃗ \vec{d} d
d ⃗ = ( − 1 − 1 , 1 − 2 , 0 − 3 ) = ( − 2 , − 1 , − 3 ) \vec{d} = (-1 - 1, 1 - 2, 0 - 3) = (-2, -1, -3) d = ( − 1 − 1 , 1 − 2 , 0 − 3 ) = ( − 2 , − 1 , − 3 )
Next, calculate the magnitude of the direction vector d ⃗ \vec{d} d
∣ d ⃗ ∣ = ( − 2 ) 2 + ( − 1 ) 2 + ( − 3 ) 2 = 4 + 1 + 9 = 14 \left| \vec{d} \right| = \sqrt{(-2)^2 + (-1)^2 + (-3)^2} = \sqrt{4 + 1 + 9} = \sqrt{14} d = ( − 2 ) 2 + ( − 1 ) 2 + ( − 3 ) 2 = 4 + 1 + 9 = 14
Now, find the direction cosines l , m , n l, m, n l , m , n
l = − 2 14 , m = − 1 14 , n = − 3 14 l = \frac{-2}{\sqrt{14}}, \quad m = \frac{-1}{\sqrt{14}}, \quad n = \frac{-3}{\sqrt{14}} l = 14 − 2 , m = 14 − 1 , n = 14 − 3
Therefore, the direction cosines of the line passing through the points (1, 2, 3) and (-1, 1, 0) are ( − 2 14 , − 1 14 , − 3 14 ) \left( \frac{-2}{\sqrt{14}}, \frac{-1}{\sqrt{14}}, \frac{-3}{\sqrt{14}} \right) ( 14 − 2 , 14 − 1 , 14 − 3 )
5. (c) Find the maximum value of 2 a + 5 b 2a + 5b 2 a + 5 b − 3 a − 2 b ≤ − 6 -3a - 2b \leq -6 − 3 a − 2 b ≤ − 6 − 2 a + b ≤ 2 -2a + b \leq 2 − 2 a + b ≤ 2 4 a + 6 b ≤ 24 4a + 6b \leq 24 4 a + 6 b ≤ 24 2 a − 3 b ≤ 3 2a - 3b \leq 3 2 a − 3 b ≤ 3 a ≥ 0 a \geq 0 a ≥ 0 b ≥ 0 b \geq 0 b ≥ 0 Answer :
To find the maximum value of 2 a + 5 b 2a + 5b 2 a + 5 b
The constraints are:
− 3 a − 2 b ≤ − 6 -3a - 2b \leq -6 − 3 a − 2 b ≤ − 6 − 2 a + b ≤ 2 -2a + b \leq 2 − 2 a + b ≤ 2 4 a + 6 b ≤ 24 4a + 6b \leq 24 4 a + 6 b ≤ 24 2 a − 3 b ≤ 3 2a - 3b \leq 3 2 a − 3 b ≤ 3 a ≥ 0 a \geq 0 a ≥ 0 b ≥ 0 b \geq 0 b ≥ 0 First, we rewrite the constraints in a more convenient form:
3 a + 2 b ≥ 6 3a + 2b \geq 6 3 a + 2 b ≥ 6 2 a − b ≥ − 2 2a - b \geq -2 2 a − b ≥ − 2 4 a + 6 b ≤ 24 4a + 6b \leq 24 4 a + 6 b ≤ 24 2 a − 3 b ≤ 3 2a - 3b \leq 3 2 a − 3 b ≤ 3 a ≥ 0 a \geq 0 a ≥ 0 b ≥ 0 b \geq 0 b ≥ 0 Next, we find the intersection points of the boundary lines to determine the feasible region:
For 3 a + 2 b = 6 3a + 2b = 6 3 a + 2 b = 6 a ≥ 0 a \geq 0 a ≥ 0 b ≥ 0 b \geq 0 b ≥ 0 When a = 0 a = 0 a = 0 b = 3 b = 3 b = 3 When b = 0 b = 0 b = 0 a = 2 a = 2 a = 2 For 2 a − b = − 2 2a - b = -2 2 a − b = − 2 a ≥ 0 a \geq 0 a ≥ 0 b ≥ 0 b \geq 0 b ≥ 0 When a = 0 a = 0 a = 0 b = − 2 b = -2 b = − 2 b ≥ 0 b \geq 0 b ≥ 0 When b = 0 b = 0 b = 0 a = − 1 a = -1 a = − 1 a ≥ 0 a \geq 0 a ≥ 0 For 4 a + 6 b = 24 4a + 6b = 24 4 a + 6 b = 24 a ≥ 0 a \geq 0 a ≥ 0 b ≥ 0 b \geq 0 b ≥ 0 When a = 0 a = 0 a = 0 b = 4 b = 4 b = 4 When b = 0 b = 0 b = 0 a = 6 a = 6 a = 6 For 2 a − 3 b = 3 2a - 3b = 3 2 a − 3 b = 3 a ≥ 0 a \geq 0 a ≥ 0 b ≥ 0 b \geq 0 b ≥ 0 When a = 0 a = 0 a = 0 b = − 1 b = -1 b = − 1 b ≥ 0 b \geq 0 b ≥ 0 When b = 0 b = 0 b = 0 a = 1.5 a = 1.5 a = 1.5 The vertices of the feasible region are:
( 0 , 0 ) , ( 2 , 0 ) , ( 0 , 4 ) , ( 18 5 , 24 5 ) , ( 24 13 , 21 13 ) \left(0, 0\right), \left(2, 0\right), \left(0, 4\right), \left( \frac{18}{5}, \frac{24}{5} \right), \left( \frac{24}{13}, \frac{21}{13} \right) ( 0 , 0 ) , ( 2 , 0 ) , ( 0 , 4 ) , ( 5 18 , 5 24 ) , ( 13 24 , 13 21 ) Finally, we evaluate the objective function 2 a + 5 b 2a + 5b 2 a + 5 b
At ( 0 , 0 ) \left(0, 0\right) ( 0 , 0 ) 2 ( 0 ) + 5 ( 0 ) = 0 2(0) + 5(0) = 0 2 ( 0 ) + 5 ( 0 ) = 0 At ( 2 , 0 ) \left(2, 0\right) ( 2 , 0 ) 2 ( 2 ) + 5 ( 0 ) = 4 2(2) + 5(0) = 4 2 ( 2 ) + 5 ( 0 ) = 4 At ( 0 , 4 ) \left(0, 4\right) ( 0 , 4 ) 2 ( 0 ) + 5 ( 4 ) = 20 2(0) + 5(4) = 20 2 ( 0 ) + 5 ( 4 ) = 20 At ( 18 5 , 24 5 ) \left( \frac{18}{5}, \frac{24}{5} \right) ( 5 18 , 5 24 ) 2 ( 18 5 ) + 5 ( 24 5 ) = 36 5 + 120 5 = 156 5 = 31.2 2\left(\frac{18}{5}\right) + 5\left(\frac{24}{5}\right) = \frac{36}{5} + \frac{120}{5} = \frac{156}{5} = 31.2 2 ( 5 18 ) + 5 ( 5 24 ) = 5 36 + 5 120 = 5 156 = 31.2 At ( 24 13 , 21 13 ) \left( \frac{24}{13}, \frac{21}{13} \right) ( 13 24 , 13 21 ) 2 ( 24 13 ) + 5 ( 21 13 ) = 48 13 + 105 13 = 153 13 ≈ 11.77 2\left(\frac{24}{13}\right) + 5\left(\frac{21}{13}\right) = \frac{48}{13} + \frac{105}{13} = \frac{153}{13} \approx 11.77 2 ( 13 24 ) + 5 ( 13 21 ) = 13 48 + 13 105 = 13 153 ≈ 11.77 The maximum value of 2 a + 5 b 2a + 5b 2 a + 5 b 31.2 31.2 31.2 ( 18 5 , 24 5 ) \left( \frac{18}{5}, \frac{24}{5} \right) ( 5 18 , 5 24 )
5. (d) Reduce the matrix [ 5 3 8 0 1 1 1 − 1 0 ] \begin{bmatrix} 5 & 3 & 8 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \end{bmatrix} 5 0 1 3 1 − 1 8 1 0 Answer :
To reduce the matrix to its normal form and find its rank, we perform row operations:
Given matrix:
[ 5 3 8 0 1 1 1 − 1 0 ] \begin{bmatrix} 5 & 3 & 8 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \end{bmatrix} 5 0 1 3 1 − 1 8 1 0 Step-by-step reduction:
Subtract 5 times the second row from the first row (R1 → R1 - 5R2): [ 5 0 3 0 1 1 1 − 1 0 ] \begin{bmatrix} 5 & 0 & 3 \\ 0 & 1 & 1 \\ 1 & -1 & 0 \end{bmatrix} 5 0 1 0 1 − 1 3 1 0 Swap the first and third rows to bring a non-zero entry to the pivot position: [ 1 − 1 0 0 1 1 5 0 3 ] \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 5 & 0 & 3 \end{bmatrix} 1 0 5 − 1 1 0 0 1 3 Subtract 5 times the first row from the third row (R3 → R3 - 5R1): [ 1 − 1 0 0 1 1 0 5 3 ] \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & 5 & 3 \end{bmatrix} 1 0 0 − 1 1 5 0 1 3 Subtract 5 times the second row from the third row (R3 → R3 - 5R2): [ 1 − 1 0 0 1 1 0 0 − 2 ] \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \end{bmatrix} 1 0 0 − 1 1 0 0 1 − 2 Now, the matrix is in row echelon form (REF). To find the rank, count the number of non-zero rows:
Since there are 3 non-zero rows, the rank of the matrix is 3 3 3
Therefore, the reduced row echelon form (RREF) of the matrix is [ 1 − 1 0 0 1 1 0 0 − 2 ] \begin{bmatrix} 1 & -1 & 0 \\ 0 & 1 & 1 \\ 0 & 0 & -2 \end{bmatrix} 1 0 0 − 1 1 0 0 1 − 2 3 3 3
