BCS-012 Solved Assignment
Q1: For what value of ‘
k
k
k k k ‘ the points
(
−
k
+
1
,
2
k
)
,
(
k
,
2
−
2
k
)
(
−
k
+
1
,
2
k
)
,
(
k
,
2
−
2
k
)
(-k+1,2k),(k,2-2k) (-k+1,2 k),(k, 2-2 k) ( − k + 1 , 2 k ) , ( k , 2 − 2 k ) and
(
−
4
−
k
,
6
−
2
k
)
(
−
4
−
k
,
6
−
2
k
)
(-4-k,6-2k) (-4-k, 6-2 k) ( − 4 − k , 6 − 2 k ) are collinear.
Q2: Solve the following system of equations by using Matrix Inverse Method.
3
x
+
4
y
+
7
z
=
14
2
x
−
y
+
3
z
=
4
2
x
+
2
y
−
3
z
=
0
3
x
+
4
y
+
7
z
=
14
2
x
−
y
+
3
z
=
4
2
x
+
2
y
−
3
z
=
0
{:[3x+4y+7z=14],[2x-y+3z=4],[2x+2y-3z=0]:} \begin{aligned}
& 3 x+4 y+7 z=14 \\
& 2 x-y+3 z=4 \\
& 2 x+2 y-3 z=0
\end{aligned} 3 x + 4 y + 7 z = 14 2 x − y + 3 z = 4 2 x + 2 y − 3 z = 0 Q3: Use principle of Mathematical Induction to prove that:
1
1
∗
2
+
1
2
∗
3
+
…
…
…
…
+
1
n
(
n
+
1
)
=
n
n
+
1
1
1
∗
2
+
1
2
∗
3
+
…
…
…
…
+
1
n
(
n
+
1
)
=
n
n
+
1
(1)/(1**2)+(1)/(2**3)+dots dots dots dots+(1)/(n(n+1))=(n)/(n+1) \frac{1}{1 * 2}+\frac{1}{2 * 3}+\ldots \ldots \ldots \ldots+\frac{1}{n(n+1)}=\frac{n}{n+1} 1 1 ∗ 2 + 1 2 ∗ 3 + … … … … + 1 n ( n + 1 ) = n n + 1 Q4: How many terms of the sequence
3
,
3
,
3
3
,
…
3
,
3
,
3
3
,
…
sqrt3,3,3sqrt3,dots \sqrt{3}, 3,3 \sqrt{3}, \ldots 3 , 3 , 3 3 , … must be taken to get the sum
39
+
13
3
39
+
13
3
39+13sqrt3 39+13 \sqrt{3} 39 + 13 3 ?
Q5: If
y
=
a
e
m
x
+
b
e
−
m
x
y
=
a
e
m
x
+
b
e
−
m
x
y=ae^(mx)+be^(-mx) y=a e^{m x}+b e^{-m x} y = a e m x + b e − m x , Prove that
d
2
y
/
d
x
2
=
m
2
y
d
2
y
/
d
x
2
=
m
2
y
d^(2)y//dx^(2)=m^(2)y d^2 y / d x^2=m^2 y d 2 y / d x 2 = m 2 y
Q6: Integrate function
f
(
x
)
=
x
/
[
(
x
+
1
)
(
2
x
−
1
)
]
f
(
x
)
=
x
/
[
(
x
+
1
)
(
2
x
−
1
)
]
f(x)=x//[(x+1)(2x-1)] f(x)=x /[(x+1)(2 x-1)] f ( x ) = x / [ ( x + 1 ) ( 2 x − 1 ) ] w.r.t
x
x
x x x
Q7: If 1,
w
,
w
2
w
,
w
2
w,w^(2) w, w^2 w , w 2 are Cube Roots of unity show that
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
0
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
0
(1+w)^(2)-(1+w)^(3)+w^(2)=0 (1+w)^2-(1+w)^3+w^2=0 ( 1 + w ) 2 − ( 1 + w ) 3 + w 2 = 0 .
Q8: If
α
,
β
α
,
β
alpha,beta \alpha, \beta α , β are roots of equation
2
x
2
−
3
x
−
5
=
0
2
x
2
−
3
x
−
5
=
0
2x^(2)-3x-5=0 2 x^2-3 x-5=0 2 x 2 − 3 x − 5 = 0 , them find a Quadratic equation whose roots are
α
2
,
β
2
α
2
,
β
2
alpha^(2),beta^(2) \alpha^2, \beta^2 α 2 , β 2
Q9: Solve the inequality
3
5
(
x
−
2
)
≤
5
3
(
2
−
x
)
3
5
(
x
−
2
)
≤
5
3
(
2
−
x
)
(3)/(5)(x-2) <= (5)/(3)(2-x) \frac{3}{5}(x-2) \leq \frac{5}{3}(2-x) 3 5 ( x − 2 ) ≤ 5 3 ( 2 − x ) and graph the solution set.
Q10: If a positive number exceeds its positive square root by 12 , then find the number.
Q11: Find the area bounded by the curves
x
2
=
y
x
2
=
y
x^(2)=y \mathrm{x}^2=\mathrm{y} x 2 = y and
y
=
x
y
=
x
y=x y=x y = x . Q12: Find the inverse of the matrix
A
=
(
1
6
4
2
4
−
1
−
1
2
5
)
A
=
1
6
4
2
4
−
1
−
1
2
5
A=([1,6,4],[2,4,-1],[-1,2,5]) A=\left(\begin{array}{ccc}1 & 6 & 4 \\ 2 & 4 & -1 \\ -1 & 2 & 5\end{array}\right) A = ( 1 6 4 2 4 − 1 − 1 2 5 ) , if it exists,
Q13: If
m
m
m m m times the
m
th
m
th
m^(“th “) m^{\text {th }} m th term of an A.P. is
n
n
n n n times its
n
th
n
th
n^(“th “) n^{\text {th }} n th term, show that
(
m
+
n
)
th
(
m
+
n
)
th
(m+n)^(“th “) (m+n)^{\text {th }} ( m + n ) th term of the A.P. is zero.
Q14: Show that
i)
lim
n
→
0
|
x
|
x
lim
n
→
0
|
x
|
x
lim_(n rarr0)(|x|)/(x) \lim _{n \rightarrow 0} \frac{|x|}{x} lim n → 0 | x | x does not exist
ii)
f
(
x
)
=
|
x
|
f
(
x
)
=
|
x
|
f(x)=|x| \mathrm{f}(x)=|x| f ( x ) = | x | is continuous at
x
=
0
x
=
0
x=0 x=0 x = 0 .
Q15: Suriti wants to Invest at most 12000 in saving certificates and National Saving Bonds. She has to invest at least 2000 in Saving certificates and at least 4000 in National Saving Bonds. If Rate of
Interest on saving certificates is
8
%
8
%
8% 8 \% 8 % per annum and rate of interest on national saving bond is
10
%
10
%
10% 10 \% 10 % per annum. How much money should she invest to earn maximum yearly income? Find also the maximum yearly income.
Q16: A spherical balloon is being Inflated at the rate of
900
cm
3
/
sec
900
cm
3
/
sec
900cm^(3)//sec 900 \mathrm{~cm}^3 / \mathrm{sec} 900 cm 3 / sec . How fast is the Radius of the balloon Increasing when the Radius is 15 cm .
Expert Answer
BCS-012 Solved Assignment
Question:-01
For what value of ‘
k
k
k k k
(
−
k
+
1
,
2
k
)
,
(
k
,
2
−
2
k
)
(
−
k
+
1
,
2
k
)
,
(
k
,
2
−
2
k
)
(-k+1,2k),(k,2-2k) (-k+1,2 k),(k, 2-2 k) ( − k + 1 , 2 k ) , ( k , 2 − 2 k )
(
−
4
−
k
,
6
−
2
k
)
(
−
4
−
k
,
6
−
2
k
)
(-4-k,6-2k) (-4-k, 6-2 k) ( − 4 − k , 6 − 2 k )
Answer:
To determine the value of
k
k
k k k such that the points
(
−
k
+
1
,
2
k
)
(
−
k
+
1
,
2
k
)
(-k+1,2k) (-k+1, 2k) ( − k + 1 , 2 k ) ,
(
k
,
2
−
2
k
)
(
k
,
2
−
2
k
)
(k,2-2k) (k, 2 – 2k) ( k , 2 − 2 k ) , and
(
−
4
−
k
,
6
−
2
k
)
(
−
4
−
k
,
6
−
2
k
)
(-4-k,6-2k) (-4 – k, 6 – 2k) ( − 4 − k , 6 − 2 k ) are collinear, we need to use the condition that the area of the triangle formed by three collinear points is zero.
The formula for the area of a triangle formed by three points
(
x
1
,
y
1
)
(
x
1
,
y
1
)
(x_(1),y_(1)) (x_1, y_1) ( x 1 , y 1 ) ,
(
x
2
,
y
2
)
(
x
2
,
y
2
)
(x_(2),y_(2)) (x_2, y_2) ( x 2 , y 2 ) , and
(
x
3
,
y
3
)
(
x
3
,
y
3
)
(x_(3),y_(3)) (x_3, y_3) ( x 3 , y 3 ) is:
Area
=
1
2
|
x
1
(
y
2
−
y
3
)
+
x
2
(
y
3
−
y
1
)
+
x
3
(
y
1
−
y
2
)
|
Area
=
1
2
x
1
(
y
2
−
y
3
)
+
x
2
(
y
3
−
y
1
)
+
x
3
(
y
1
−
y
2
)
“Area”=(1)/(2)|x_(1)(y_(2)-y_(3))+x_(2)(y_(3)-y_(1))+x_(3)(y_(1)-y_(2))| \text{Area} = \frac{1}{2} \left| x_1(y_2 – y_3) + x_2(y_3 – y_1) + x_3(y_1 – y_2) \right| Area = 1 2 | x 1 ( y 2 − y 3 ) + x 2 ( y 3 − y 1 ) + x 3 ( y 1 − y 2 ) | For the given points
(
−
k
+
1
,
2
k
)
(
−
k
+
1
,
2
k
)
(-k+1,2k) (-k + 1, 2k) ( − k + 1 , 2 k ) ,
(
k
,
2
−
2
k
)
(
k
,
2
−
2
k
)
(k,2-2k) (k, 2 – 2k) ( k , 2 − 2 k ) , and
(
−
4
−
k
,
6
−
2
k
)
(
−
4
−
k
,
6
−
2
k
)
(-4-k,6-2k) (-4 – k, 6 – 2k) ( − 4 − k , 6 − 2 k ) , we can substitute their coordinates into this formula and set the area equal to zero.
Step 1: Write the coordinates
Let the three points be:
(
x
1
,
y
1
)
=
(
−
k
+
1
,
2
k
)
(
x
1
,
y
1
)
=
(
−
k
+
1
,
2
k
)
(x_(1),y_(1))=(-k+1,2k) (x_1, y_1) = (-k + 1, 2k) ( x 1 , y 1 ) = ( − k + 1 , 2 k )
(
x
2
,
y
2
)
=
(
k
,
2
−
2
k
)
(
x
2
,
y
2
)
=
(
k
,
2
−
2
k
)
(x_(2),y_(2))=(k,2-2k) (x_2, y_2) = (k, 2 – 2k) ( x 2 , y 2 ) = ( k , 2 − 2 k )
(
x
3
,
y
3
)
=
(
−
4
−
k
,
6
−
2
k
)
(
x
3
,
y
3
)
=
(
−
4
−
k
,
6
−
2
k
)
(x_(3),y_(3))=(-4-k,6-2k) (x_3, y_3) = (-4 – k, 6 – 2k) ( x 3 , y 3 ) = ( − 4 − k , 6 − 2 k )
The area formula becomes:
Area
=
1
2
|
(
−
k
+
1
)
(
(
2
−
2
k
)
−
(
6
−
2
k
)
)
+
k
(
(
6
−
2
k
)
−
2
k
)
+
(
−
4
−
k
)
(
2
k
−
(
2
−
2
k
)
)
|
=
0
Area
=
1
2
(
−
k
+
1
)
(
(
2
−
2
k
)
−
(
6
−
2
k
)
)
+
k
(
(
6
−
2
k
)
−
2
k
)
+
(
−
4
−
k
)
(
2
k
−
(
2
−
2
k
)
)
=
0
“Area”=(1)/(2)|(-k+1)((2-2k)-(6-2k))+k((6-2k)-2k)+(-4-k)(2k-(2-2k))|=0 \text{Area} = \frac{1}{2} \left| (-k+1)((2-2k) – (6-2k)) + k((6-2k) – 2k) + (-4-k)(2k – (2 – 2k)) \right| = 0 Area = 1 2 | ( − k + 1 ) ( ( 2 − 2 k ) − ( 6 − 2 k ) ) + k ( ( 6 − 2 k ) − 2 k ) + ( − 4 − k ) ( 2 k − ( 2 − 2 k ) ) | = 0 Simplify the expression step by step.
Step 3: Simplify each term
(
2
−
2
k
)
−
(
6
−
2
k
)
=
2
−
2
k
−
6
+
2
k
=
−
4
(
2
−
2
k
)
−
(
6
−
2
k
)
=
2
−
2
k
−
6
+
2
k
=
−
4
(2-2k)-(6-2k)=2-2k-6+2k=-4 (2 – 2k) – (6 – 2k) = 2 – 2k – 6 + 2k = -4 ( 2 − 2 k ) − ( 6 − 2 k ) = 2 − 2 k − 6 + 2 k = − 4 So, the first term becomes
(
−
k
+
1
)
(
−
4
)
=
4
(
k
−
1
)
(
−
k
+
1
)
(
−
4
)
=
4
(
k
−
1
)
(-k+1)(-4)=4(k-1) (-k + 1)(-4) = 4(k – 1) ( − k + 1 ) ( − 4 ) = 4 ( k − 1 ) .
(
6
−
2
k
)
−
2
k
=
6
−
2
k
−
2
k
=
6
−
4
k
(
6
−
2
k
)
−
2
k
=
6
−
2
k
−
2
k
=
6
−
4
k
(6-2k)-2k=6-2k-2k=6-4k (6 – 2k) – 2k = 6 – 2k – 2k = 6 – 4k ( 6 − 2 k ) − 2 k = 6 − 2 k − 2 k = 6 − 4 k So, the second term becomes
k
(
6
−
4
k
)
=
6
k
−
4
k
2
k
(
6
−
4
k
)
=
6
k
−
4
k
2
k(6-4k)=6k-4k^(2) k(6 – 4k) = 6k – 4k^2 k ( 6 − 4 k ) = 6 k − 4 k 2 .
2
k
−
(
2
−
2
k
)
=
2
k
−
2
+
2
k
=
4
k
−
2
2
k
−
(
2
−
2
k
)
=
2
k
−
2
+
2
k
=
4
k
−
2
2k-(2-2k)=2k-2+2k=4k-2 2k – (2 – 2k) = 2k – 2 + 2k = 4k – 2 2 k − ( 2 − 2 k ) = 2 k − 2 + 2 k = 4 k − 2 So, the third term becomes
(
−
4
−
k
)
(
4
k
−
2
)
=
(
−
4
)
(
4
k
−
2
)
+
(
−
k
)
(
4
k
−
2
)
=
−
16
k
+
8
−
4
k
2
+
2
k
(
−
4
−
k
)
(
4
k
−
2
)
=
(
−
4
)
(
4
k
−
2
)
+
(
−
k
)
(
4
k
−
2
)
=
−
16
k
+
8
−
4
k
2
+
2
k
(-4-k)(4k-2)=(-4)(4k-2)+(-k)(4k-2)=-16 k+8-4k^(2)+2k (-4 – k)(4k – 2) = (-4)(4k – 2) + (-k)(4k – 2) = -16k + 8 – 4k^2 + 2k ( − 4 − k ) ( 4 k − 2 ) = ( − 4 ) ( 4 k − 2 ) + ( − k ) ( 4 k − 2 ) = − 16 k + 8 − 4 k 2 + 2 k .
Simplifying the third term gives
−
14
k
+
8
−
4
k
2
−
14
k
+
8
−
4
k
2
-14 k+8-4k^(2) -14k + 8 – 4k^2 − 14 k + 8 − 4 k 2 .
Step 4: Combine all terms
4
(
k
−
1
)
+
(
6
k
−
4
k
2
)
+
(
−
14
k
+
8
−
4
k
2
)
=
0
4
(
k
−
1
)
+
(
6
k
−
4
k
2
)
+
(
−
14
k
+
8
−
4
k
2
)
=
0
4(k-1)+(6k-4k^(2))+(-14 k+8-4k^(2))=0 4(k – 1) + (6k – 4k^2) + (-14k + 8 – 4k^2) = 0 4 ( k − 1 ) + ( 6 k − 4 k 2 ) + ( − 14 k + 8 − 4 k 2 ) = 0 Simplifying:
4
k
−
4
+
6
k
−
4
k
2
−
14
k
+
8
−
4
k
2
=
0
4
k
−
4
+
6
k
−
4
k
2
−
14
k
+
8
−
4
k
2
=
0
4k-4+6k-4k^(2)-14 k+8-4k^(2)=0 4k – 4 + 6k – 4k^2 – 14k + 8 – 4k^2 = 0 4 k − 4 + 6 k − 4 k 2 − 14 k + 8 − 4 k 2 = 0
(
4
k
+
6
k
−
14
k
)
+
(
−
4
−
4
k
2
−
4
k
2
)
+
8
=
0
(
4
k
+
6
k
−
14
k
)
+
(
−
4
−
4
k
2
−
4
k
2
)
+
8
=
0
(4k+6k-14 k)+(-4-4k^(2)-4k^(2))+8=0 (4k + 6k – 14k) + (-4 – 4k^2 – 4k^2) + 8 = 0 ( 4 k + 6 k − 14 k ) + ( − 4 − 4 k 2 − 4 k 2 ) + 8 = 0
−
4
k
−
8
k
2
+
4
=
0
−
4
k
−
8
k
2
+
4
=
0
-4k-8k^(2)+4=0 -4k – 8k^2 + 4 = 0 − 4 k − 8 k 2 + 4 = 0 Step 5: Solve for
k
k
k k k
This simplifies to:
−
8
k
2
−
4
k
+
4
=
0
−
8
k
2
−
4
k
+
4
=
0
-8k^(2)-4k+4=0 -8k^2 – 4k + 4 = 0 − 8 k 2 − 4 k + 4 = 0 Dividing the entire equation by -4:
2
k
2
+
k
−
1
=
0
2
k
2
+
k
−
1
=
0
2k^(2)+k-1=0 2k^2 + k – 1 = 0 2 k 2 + k − 1 = 0 This is a quadratic equation. Solve it using the quadratic formula:
k
=
−
1
±
1
2
−
4
(
2
)
(
−
1
)
2
(
2
)
=
−
1
±
1
+
8
4
=
−
1
±
9
4
k
=
−
1
±
1
2
−
4
(
2
)
(
−
1
)
2
(
2
)
=
−
1
±
1
+
8
4
=
−
1
±
9
4
k=(-1+-sqrt(1^(2)-4(2)(-1)))/(2(2))=(-1+-sqrt(1+8))/(4)=(-1+-sqrt9)/(4) k = \frac{-1 \pm \sqrt{1^2 – 4(2)(-1)}}{2(2)} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm \sqrt{9}}{4} k = − 1 ± 1 2 − 4 ( 2 ) ( − 1 ) 2 ( 2 ) = − 1 ± 1 + 8 4 = − 1 ± 9 4
k
=
−
1
±
3
4
k
=
−
1
±
3
4
k=(-1+-3)/(4) k = \frac{-1 \pm 3}{4} k = − 1 ± 3 4 Thus,
k
=
−
1
+
3
4
=
2
4
=
1
2
k
=
−
1
+
3
4
=
2
4
=
1
2
k=(-1+3)/(4)=(2)/(4)=(1)/(2) k = \frac{-1 + 3}{4} = \frac{2}{4} = \frac{1}{2} k = − 1 + 3 4 = 2 4 = 1 2 or
k
=
−
1
−
3
4
=
−
4
4
=
−
1
k
=
−
1
−
3
4
=
−
4
4
=
−
1
k=(-1-3)/(4)=(-4)/(4)=-1 k = \frac{-1 – 3}{4} = \frac{-4}{4} = -1 k = − 1 − 3 4 = − 4 4 = − 1 .
Step 6: Conclusion
The value of
k
k
k k k can be either
1
2
1
2
(1)/(2) \frac{1}{2} 1 2 or
−
1
−
1
-1 -1 − 1 .
Question:-02
Solve the following system of equations by using Matrix Inverse Method.
3
x
+
4
y
+
7
z
=
14
2
x
−
y
+
3
z
=
4
2
x
+
2
y
−
3
z
=
0
3
x
+
4
y
+
7
z
=
14
2
x
−
y
+
3
z
=
4
2
x
+
2
y
−
3
z
=
0
{:[3x+4y+7z=14],[2x-y+3z=4],[2x+2y-3z=0]:} \begin{aligned}
& 3x + 4y + 7z = 14 \\
& 2x – y + 3z = 4 \\
& 2x + 2y – 3z = 0
\end{aligned} 3 x + 4 y + 7 z = 14 2 x − y + 3 z = 4 2 x + 2 y − 3 z = 0 Answer:
To solve the system of equations using the matrix inverse method, we represent the system in matrix form as
A
x
=
b
A
x
=
b
Ax=b A \mathbf{x} = \mathbf{b} A x = b , where:
A
=
[
3
4
7
2
−
1
3
2
2
−
3
]
,
x
=
[
x
y
z
]
,
b
=
[
14
4
0
]
A
=
3
4
7
2
−
1
3
2
2
−
3
,
x
=
x
y
z
,
b
=
14
4
0
A=[[3,4,7],[2,-1,3],[2,2,-3]],quadx=[[x],[y],[z]],quadb=[[14],[4],[0]] A = \begin{bmatrix} 3 & 4 & 7 \\ 2 & -1 & 3 \\ 2 & 2 & -3 \end{bmatrix}, \quad \mathbf{x} = \begin{bmatrix} x \\ y \\ z \end{bmatrix}, \quad \mathbf{b} = \begin{bmatrix} 14 \\ 4 \\ 0 \end{bmatrix} A = [ 3 4 7 2 − 1 3 2 2 − 3 ] , x = [ x y z ] , b = [ 14 4 0 ] Step 1: Find the inverse of matrix
A
A
A A A
To find the inverse of
A
A
A A A , we first calculate its determinant.
det
(
A
)
=
3
(
(
−
1
)
(
−
3
)
−
2
(
3
)
)
−
4
(
2
(
−
3
)
−
2
(
3
)
)
+
7
(
2
(
2
)
−
2
(
−
1
)
)
det
(
A
)
=
3
(
−
1
)
(
−
3
)
−
2
(
3
)
−
4
2
(
−
3
)
−
2
(
3
)
+
7
2
(
2
)
−
2
(
−
1
)
“det”(A)=3((-1)(-3)-2(3))-4(2(-3)-2(3))+7(2(2)-2(-1)) \text{det}(A) = 3 \left( (-1)(-3) – 2(3) \right) – 4 \left( 2(-3) – 2(3) \right) + 7 \left( 2(2) – 2(-1) \right) det ( A ) = 3 ( ( − 1 ) ( − 3 ) − 2 ( 3 ) ) − 4 ( 2 ( − 3 ) − 2 ( 3 ) ) + 7 ( 2 ( 2 ) − 2 ( − 1 ) ) Simplifying:
det
(
A
)
=
3
(
3
−
6
)
−
4
(
−
6
−
6
)
+
7
(
4
+
2
)
det
(
A
)
=
3
3
−
6
−
4
−
6
−
6
+
7
4
+
2
“det”(A)=3(3-6)-4(-6-6)+7(4+2) \text{det}(A) = 3 \left( 3 – 6 \right) – 4 \left( -6 – 6 \right) + 7 \left( 4 + 2 \right) det ( A ) = 3 ( 3 − 6 ) − 4 ( − 6 − 6 ) + 7 ( 4 + 2 )
det
(
A
)
=
3
(
−
3
)
−
4
(
−
12
)
+
7
(
6
)
det
(
A
)
=
3
(
−
3
)
−
4
(
−
12
)
+
7
(
6
)
“det”(A)=3(-3)-4(-12)+7(6) \text{det}(A) = 3(-3) – 4(-12) + 7(6) det ( A ) = 3 ( − 3 ) − 4 ( − 12 ) + 7 ( 6 )
det
(
A
)
=
−
9
+
48
+
42
=
81
det
(
A
)
=
−
9
+
48
+
42
=
81
“det”(A)=-9+48+42=81 \text{det}(A) = -9 + 48 + 42 = 81 det ( A ) = − 9 + 48 + 42 = 81 Since
det
(
A
)
=
81
≠
0
det
(
A
)
=
81
≠
0
“det”(A)=81!=0 \text{det}(A) = 81 \neq 0 det ( A ) = 81 ≠ 0 , matrix
A
A
A A A is invertible.
Step 2: Find the adjoint of
A
A
A A A
The adjoint of
A
A
A A A , denoted as
adj
(
A
)
adj
(
A
)
“adj”(A) \text{adj}(A) adj ( A ) , is the transpose of the cofactor matrix of
A
A
A A A . Let’s compute the cofactors of each element of
A
A
A A A :
The cofactor of
A
11
A
11
A_(11) A_{11} A 11 (3) is:
Cofactor
(
A
11
)
=
(
−
1
)
1
⋅
|
−
1
3
2
−
3
|
=
(
−
1
)
⋅
(
(
−
1
)
(
−
3
)
−
3
(
2
)
)
=
−
3
−
6
=
−
9
Cofactor
(
A
11
)
=
(
−
1
)
1
⋅
−
1
3
2
−
3
=
(
−
1
)
⋅
(
−
1
)
(
−
3
)
−
3
(
2
)
=
−
3
−
6
=
−
9
“Cofactor”(A_(11))=(-1)^(1)*|[-1,3],[2,-3]|=(-1)*((-1)(-3)-3(2))=-3-6=-9 \text{Cofactor}(A_{11}) = (-1)^1 \cdot \begin{vmatrix} -1 & 3 \\ 2 & -3 \end{vmatrix} = (-1) \cdot \left( (-1)(-3) – 3(2) \right) = -3 – 6 = -9 Cofactor ( A 11 ) = ( − 1 ) 1 ⋅ | − 1 3 2 − 3 | = ( − 1 ) ⋅ ( ( − 1 ) ( − 3 ) − 3 ( 2 ) ) = − 3 − 6 = − 9
The cofactor of
A
12
A
12
A_(12) A_{12} A 12 (4) is:
Cofactor
(
A
12
)
=
(
−
1
)
2
⋅
|
2
3
2
−
3
|
=
(
2
(
−
3
)
−
3
(
2
)
)
=
−
6
−
6
=
−
12
Cofactor
(
A
12
)
=
(
−
1
)
2
⋅
2
3
2
−
3
=
2
(
−
3
)
−
3
(
2
)
=
−
6
−
6
=
−
12
“Cofactor”(A_(12))=(-1)^(2)*|[2,3],[2,-3]|=(2(-3)-3(2))=-6-6=-12 \text{Cofactor}(A_{12}) = (-1)^2 \cdot \begin{vmatrix} 2 & 3 \\ 2 & -3 \end{vmatrix} = \left( 2(-3) – 3(2) \right) = -6 – 6 = -12 Cofactor ( A 12 ) = ( − 1 ) 2 ⋅ | 2 3 2 − 3 | = ( 2 ( − 3 ) − 3 ( 2 ) ) = − 6 − 6 = − 12
The cofactor of
A
13
A
13
A_(13) A_{13} A 13 (7) is:
Cofactor
(
A
13
)
=
(
−
1
)
3
⋅
|
2
−
1
2
2
|
=
(
−
1
)
⋅
(
2
(
2
)
−
(
−
1
)
(
2
)
)
=
−
(
4
+
2
)
=
−
6
Cofactor
(
A
13
)
=
(
−
1
)
3
⋅
2
−
1
2
2
=
(
−
1
)
⋅
2
(
2
)
−
(
−
1
)
(
2
)
=
−
4
+
2
=
−
6
“Cofactor”(A_(13))=(-1)^(3)*|[2,-1],[2,2]|=(-1)*(2(2)-(-1)(2))=-(4+2)=-6 \text{Cofactor}(A_{13}) = (-1)^3 \cdot \begin{vmatrix} 2 & -1 \\ 2 & 2 \end{vmatrix} = (-1) \cdot \left( 2(2) – (-1)(2) \right) = – \left( 4 + 2 \right) = -6 Cofactor ( A 13 ) = ( − 1 ) 3 ⋅ | 2 − 1 2 2 | = ( − 1 ) ⋅ ( 2 ( 2 ) − ( − 1 ) ( 2 ) ) = − ( 4 + 2 ) = − 6
The cofactor of
A
21
A
21
A_(21) A_{21} A 21 (2) is:
Cofactor
(
A
21
)
=
(
−
1
)
2
⋅
|
4
7
2
−
3
|
=
(
4
(
−
3
)
−
7
(
2
)
)
=
−
12
−
14
=
−
26
Cofactor
(
A
21
)
=
(
−
1
)
2
⋅
4
7
2
−
3
=
4
(
−
3
)
−
7
(
2
)
=
−
12
−
14
=
−
26
“Cofactor”(A_(21))=(-1)^(2)*|[4,7],[2,-3]|=(4(-3)-7(2))=-12-14=-26 \text{Cofactor}(A_{21}) = (-1)^2 \cdot \begin{vmatrix} 4 & 7 \\ 2 & -3 \end{vmatrix} = \left( 4(-3) – 7(2) \right) = -12 – 14 = -26 Cofactor ( A 21 ) = ( − 1 ) 2 ⋅ | 4 7 2 − 3 | = ( 4 ( − 3 ) − 7 ( 2 ) ) = − 12 − 14 = − 26
The cofactor of
A
22
A
22
A_(22) A_{22} A 22 (-1) is:
Cofactor
(
A
22
)
=
(
−
1
)
3
⋅
|
3
7
2
−
3
|
=
(
−
1
)
⋅
(
3
(
−
3
)
−
7
(
2
)
)
=
−
(
−
9
−
14
)
=
23
Cofactor
(
A
22
)
=
(
−
1
)
3
⋅
3
7
2
−
3
=
(
−
1
)
⋅
3
(
−
3
)
−
7
(
2
)
=
−
(
−
9
−
14
)
=
23
“Cofactor”(A_(22))=(-1)^(3)*|[3,7],[2,-3]|=(-1)*(3(-3)-7(2))=-(-9-14)=23 \text{Cofactor}(A_{22}) = (-1)^3 \cdot \begin{vmatrix} 3 & 7 \\ 2 & -3 \end{vmatrix} = (-1) \cdot \left( 3(-3) – 7(2) \right) = -(-9 – 14) = 23 Cofactor ( A 22 ) = ( − 1 ) 3 ⋅ | 3 7 2 − 3 | = ( − 1 ) ⋅ ( 3 ( − 3 ) − 7 ( 2 ) ) = − ( − 9 − 14 ) = 23
The cofactor of
A
23
A
23
A_(23) A_{23} A 23 (3) is:
Cofactor
(
A
23
)
=
(
−
1
)
4
⋅
|
3
4
2
2
|
=
(
3
(
2
)
−
4
(
2
)
)
=
6
−
8
=
−
2
Cofactor
(
A
23
)
=
(
−
1
)
4
⋅
3
4
2
2
=
3
(
2
)
−
4
(
2
)
=
6
−
8
=
−
2
“Cofactor”(A_(23))=(-1)^(4)*|[3,4],[2,2]|=(3(2)-4(2))=6-8=-2 \text{Cofactor}(A_{23}) = (-1)^4 \cdot \begin{vmatrix} 3 & 4 \\ 2 & 2 \end{vmatrix} = \left( 3(2) – 4(2) \right) = 6 – 8 = -2 Cofactor ( A 23 ) = ( − 1 ) 4 ⋅ | 3 4 2 2 | = ( 3 ( 2 ) − 4 ( 2 ) ) = 6 − 8 = − 2
The cofactor of
A
31
A
31
A_(31) A_{31} A 31 (2) is:
Cofactor
(
A
31
)
=
(
−
1
)
3
⋅
|
4
7
−
1
3
|
=
(
−
1
)
⋅
(
4
(
3
)
−
7
(
−
1
)
)
=
−
(
12
+
7
)
=
−
19
Cofactor
(
A
31
)
=
(
−
1
)
3
⋅
4
7
−
1
3
=
(
−
1
)
⋅
4
(
3
)
−
7
(
−
1
)
=
−
(
12
+
7
)
=
−
19
“Cofactor”(A_(31))=(-1)^(3)*|[4,7],[-1,3]|=(-1)*(4(3)-7(-1))=-(12+7)=-19 \text{Cofactor}(A_{31}) = (-1)^3 \cdot \begin{vmatrix} 4 & 7 \\ -1 & 3 \end{vmatrix} = (-1) \cdot \left( 4(3) – 7(-1) \right) = -(12 + 7) = -19 Cofactor ( A 31 ) = ( − 1 ) 3 ⋅ | 4 7 − 1 3 | = ( − 1 ) ⋅ ( 4 ( 3 ) − 7 ( − 1 ) ) = − ( 12 + 7 ) = − 19
The cofactor of
A
32
A
32
A_(32) A_{32} A 32 (2) is:
Cofactor
(
A
32
)
=
(
−
1
)
4
⋅
|
3
7
2
3
|
=
(
3
(
3
)
−
7
(
2
)
)
=
9
−
14
=
−
5
Cofactor
(
A
32
)
=
(
−
1
)
4
⋅
3
7
2
3
=
3
(
3
)
−
7
(
2
)
=
9
−
14
=
−
5
“Cofactor”(A_(32))=(-1)^(4)*|[3,7],[2,3]|=(3(3)-7(2))=9-14=-5 \text{Cofactor}(A_{32}) = (-1)^4 \cdot \begin{vmatrix} 3 & 7 \\ 2 & 3 \end{vmatrix} = \left( 3(3) – 7(2) \right) = 9 – 14 = -5 Cofactor ( A 32 ) = ( − 1 ) 4 ⋅ | 3 7 2 3 | = ( 3 ( 3 ) − 7 ( 2 ) ) = 9 − 14 = − 5
The cofactor of
A
33
A
33
A_(33) A_{33} A 33 (-3) is:
Cofactor
(
A
33
)
=
(
−
1
)
5
⋅
|
3
4
2
−
1
|
=
(
−
1
)
⋅
(
3
(
−
1
)
−
4
(
2
)
)
=
−
(
−
3
−
8
)
=
11
Cofactor
(
A
33
)
=
(
−
1
)
5
⋅
3
4
2
−
1
=
(
−
1
)
⋅
3
(
−
1
)
−
4
(
2
)
=
−
(
−
3
−
8
)
=
11
“Cofactor”(A_(33))=(-1)^(5)*|[3,4],[2,-1]|=(-1)*(3(-1)-4(2))=-(-3-8)=11 \text{Cofactor}(A_{33}) = (-1)^5 \cdot \begin{vmatrix} 3 & 4 \\ 2 & -1 \end{vmatrix} = (-1) \cdot \left( 3(-1) – 4(2) \right) = -(-3 – 8) = 11 Cofactor ( A 33 ) = ( − 1 ) 5 ⋅ | 3 4 2 − 1 | = ( − 1 ) ⋅ ( 3 ( − 1 ) − 4 ( 2 ) ) = − ( − 3 − 8 ) = 11
Thus, the cofactor matrix is:
Cofactor
(
A
)
=
[
−
9
−
12
−
6
−
26
23
−
2
−
19
−
5
11
]
Cofactor
(
A
)
=
−
9
−
12
−
6
−
26
23
−
2
−
19
−
5
11
“Cofactor”(A)=[[-9,-12,-6],[-26,23,-2],[-19,-5,11]] \text{Cofactor}(A) = \begin{bmatrix} -9 & -12 & -6 \\ -26 & 23 & -2 \\ -19 & -5 & 11 \end{bmatrix} Cofactor ( A ) = [ − 9 − 12 − 6 − 26 23 − 2 − 19 − 5 11 ] Now, take the transpose of the cofactor matrix to get the adjugate matrix
adj
(
A
)
adj
(
A
)
“adj”(A) \text{adj}(A) adj ( A ) :
adj
(
A
)
=
[
−
9
−
26
−
19
−
12
23
−
5
−
6
−
2
11
]
adj
(
A
)
=
−
9
−
26
−
19
−
12
23
−
5
−
6
−
2
11
“adj”(A)=[[-9,-26,-19],[-12,23,-5],[-6,-2,11]] \text{adj}(A) = \begin{bmatrix} -9 & -26 & -19 \\ -12 & 23 & -5 \\ -6 & -2 & 11 \end{bmatrix} adj ( A ) = [ − 9 − 26 − 19 − 12 23 − 5 − 6 − 2 11 ] Step 3: Find the inverse of
A
A
A A A
The inverse of
A
A
A A A is given by:
A
−
1
=
1
det
(
A
)
adj
(
A
)
=
1
81
[
−
9
−
26
−
19
−
12
23
−
5
−
6
−
2
11
]
A
−
1
=
1
det
(
A
)
adj
(
A
)
=
1
81
−
9
−
26
−
19
−
12
23
−
5
−
6
−
2
11
A^(-1)=(1)/(“det”(A))”adj”(A)=(1)/(81)[[-9,-26,-19],[-12,23,-5],[-6,-2,11]] A^{-1} = \frac{1}{\text{det}(A)} \text{adj}(A) = \frac{1}{81} \begin{bmatrix} -9 & -26 & -19 \\ -12 & 23 & -5 \\ -6 & -2 & 11 \end{bmatrix} A − 1 = 1 det ( A ) adj ( A ) = 1 81 [ − 9 − 26 − 19 − 12 23 − 5 − 6 − 2 11 ] Step 4: Solve for
x
x
x \mathbf{x} x
Now, use the inverse matrix to solve for
x
x
x \mathbf{x} x by multiplying both sides of
A
x
=
b
A
x
=
b
Ax=b A \mathbf{x} = \mathbf{b} A x = b by
A
−
1
A
−
1
A^(-1) A^{-1} A − 1 :
x
=
A
−
1
b
=
1
81
[
−
9
−
26
−
19
−
12
23
−
5
−
6
−
2
11
]
[
14
4
0
]
x
=
A
−
1
b
=
1
81
−
9
−
26
−
19
−
12
23
−
5
−
6
−
2
11
14
4
0
x=A^(-1)b=(1)/(81)[[-9,-26,-19],[-12,23,-5],[-6,-2,11]][[14],[4],[0]] \mathbf{x} = A^{-1} \mathbf{b} = \frac{1}{81} \begin{bmatrix} -9 & -26 & -19 \\ -12 & 23 & -5 \\ -6 & -2 & 11 \end{bmatrix} \begin{bmatrix} 14 \\ 4 \\ 0 \end{bmatrix} x = A − 1 b = 1 81 [ − 9 − 26 − 19 − 12 23 − 5 − 6 − 2 11 ] [ 14 4 0 ] Multiply the matrices:
x
=
1
81
[
(
−
9
)
(
14
)
+
(
−
26
)
(
4
)
+
(
−
19
)
(
0
)
(
−
12
)
(
14
)
+
(
23
)
(
4
)
+
(
−
5
)
(
0
)
(
−
6
)
(
14
)
+
(
−
2
)
(
4
)
+
(
11
)
(
0
)
]
x
=
1
81
(
−
9
)
(
14
)
+
(
−
26
)
(
4
)
+
(
−
19
)
(
0
)
(
−
12
)
(
14
)
+
(
23
)
(
4
)
+
(
−
5
)
(
0
)
(
−
6
)
(
14
)
+
(
−
2
)
(
4
)
+
(
11
)
(
0
)
x=(1)/(81)[[(-9)(14)+(-26)(4)+(-19)(0)],[(-12)(14)+(23)(4)+(-5)(0)],[(-6)(14)+(-2)(4)+(11)(0)]] \mathbf{x} = \frac{1}{81} \begin{bmatrix} (-9)(14) + (-26)(4) + (-19)(0) \\ (-12)(14) + (23)(4) + (-5)(0) \\ (-6)(14) + (-2)(4) + (11)(0) \end{bmatrix} x = 1 81 [ ( − 9 ) ( 14 ) + ( − 26 ) ( 4 ) + ( − 19 ) ( 0 ) ( − 12 ) ( 14 ) + ( 23 ) ( 4 ) + ( − 5 ) ( 0 ) ( − 6 ) ( 14 ) + ( − 2 ) ( 4 ) + ( 11 ) ( 0 ) ] Simplifying:
x
=
1
81
[
−
126
−
104
+
0
−
168
+
92
+
0
−
84
−
8
+
0
]
=
1
81
[
−
230
−
76
−
92
]
x
=
1
81
−
126
−
104
+
0
−
168
+
92
+
0
−
84
−
8
+
0
=
1
81
−
230
−
76
−
92
x=(1)/(81)[[-126-104+0],[-168+92+0],[-84-8+0]]=(1)/(81)[[-230],[-76],[-92]] \mathbf{x} = \frac{1}{81} \begin{bmatrix} -126 – 104 + 0 \\ -168 + 92 + 0 \\ -84 – 8 + 0 \end{bmatrix} = \frac{1}{81} \begin{bmatrix} -230 \\ -76 \\ -92 \end{bmatrix} x = 1 81 [ − 126 − 104 + 0 − 168 + 92 + 0 − 84 − 8 + 0 ] = 1 81 [ − 230 − 76 − 92 ] Finally, divide each element by 81:
x
=
[
−
230
81
−
76
81
−
92
81
]
x
=
−
230
81
−
76
81
−
92
81
x=[[(-230)/(81)],[(-76)/(81)],[(-92)/(81)]] \mathbf{x} = \begin{bmatrix} \frac{-230}{81} \\ \frac{-76}{81} \\ \frac{-92}{81} \end{bmatrix} x = [ − 230 81 − 76 81 − 92 81 ] Thus, the solution to the system is:
x
=
−
230
81
,
y
=
−
76
81
,
z
=
−
92
81
x
=
−
230
81
,
y
=
−
76
81
,
z
=
−
92
81
x=(-230)/(81),quad y=(-76)/(81),quad z=(-92)/(81) x = \frac{-230}{81}, \quad y = \frac{-76}{81}, \quad z = \frac{-92}{81} x = − 230 81 , y = − 76 81 , z = − 92 81
Question:-03
Use principle of Mathematical Induction to prove that:
1
1
∗
2
+
1
2
∗
3
+
…
…
…
+
1
n
(
n
+
1
)
=
n
n
+
1
1
1
∗
2
+
1
2
∗
3
+
…
…
…
+
1
n
(
n
+
1
)
=
n
n
+
1
(1)/(1**2)+(1)/(2**3)+dots dots dots+(1)/(n(n+1))=(n)/(n+1) \frac{1}{1 * 2} + \frac{1}{2 * 3} + \ldots \ldots \ldots + \frac{1}{n(n+1)} = \frac{n}{n+1} 1 1 ∗ 2 + 1 2 ∗ 3 + … … … + 1 n ( n + 1 ) = n n + 1 Answer:
We will use the principle of Mathematical Induction to prove the statement:
P
(
n
)
:
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
n
(
n
+
1
)
=
n
n
+
1
P
(
n
)
:
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
n
(
n
+
1
)
=
n
n
+
1
P(n):(1)/(1*2)+(1)/(2*3)+cdots+(1)/(n(n+1))=(n)/(n+1) P(n): \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1} P ( n ) : 1 1 ⋅ 2 + 1 2 ⋅ 3 + ⋯ + 1 n ( n + 1 ) = n n + 1 Step 1: Base Case
We first check the base case
n
=
1
n
=
1
n=1 n = 1 n = 1 .
For
n
=
1
n
=
1
n=1 n = 1 n = 1 , the left-hand side is:
1
1
⋅
2
=
1
2
1
1
⋅
2
=
1
2
(1)/(1*2)=(1)/(2) \frac{1}{1 \cdot 2} = \frac{1}{2} 1 1 ⋅ 2 = 1 2 The right-hand side for
n
=
1
n
=
1
n=1 n = 1 n = 1 is:
1
1
+
1
=
1
2
1
1
+
1
=
1
2
(1)/(1+1)=(1)/(2) \frac{1}{1 + 1} = \frac{1}{2} 1 1 + 1 = 1 2 Since both sides are equal, the base case holds true.
Step 2: Inductive Hypothesis
Assume that the statement is true for some arbitrary positive integer
n
=
k
n
=
k
n=k n = k n = k . That is, we assume:
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
k
(
k
+
1
)
=
k
k
+
1
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
k
(
k
+
1
)
=
k
k
+
1
(1)/(1*2)+(1)/(2*3)+cdots+(1)/(k(k+1))=(k)/(k+1) \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1} 1 1 ⋅ 2 + 1 2 ⋅ 3 + ⋯ + 1 k ( k + 1 ) = k k + 1 Step 3: Inductive Step
We need to prove that the statement is true for
n
=
k
+
1
n
=
k
+
1
n=k+1 n = k + 1 n = k + 1 . That is, we need to prove:
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
(
k
+
1
)
(
k
+
2
)
=
k
+
1
k
+
2
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
(
k
+
1
)
(
k
+
2
)
=
k
+
1
k
+
2
(1)/(1*2)+(1)/(2*3)+cdots+(1)/((k+1)(k+2))=(k+1)/(k+2) \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2} 1 1 ⋅ 2 + 1 2 ⋅ 3 + ⋯ + 1 ( k + 1 ) ( k + 2 ) = k + 1 k + 2 Start with the left-hand side for
n
=
k
+
1
n
=
k
+
1
n=k+1 n = k+1 n = k + 1 :
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
k
(
k
+
1
)
+
1
(
k
+
1
)
(
k
+
2
)
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
k
(
k
+
1
)
+
1
(
k
+
1
)
(
k
+
2
)
(1)/(1*2)+(1)/(2*3)+cdots+(1)/(k(k+1))+(1)/((k+1)(k+2)) \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} + \frac{1}{(k+1)(k+2)} 1 1 ⋅ 2 + 1 2 ⋅ 3 + ⋯ + 1 k ( k + 1 ) + 1 ( k + 1 ) ( k + 2 ) By the inductive hypothesis, we know that:
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
k
(
k
+
1
)
=
k
k
+
1
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
k
(
k
+
1
)
=
k
k
+
1
(1)/(1*2)+(1)/(2*3)+cdots+(1)/(k(k+1))=(k)/(k+1) \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{k(k+1)} = \frac{k}{k+1} 1 1 ⋅ 2 + 1 2 ⋅ 3 + ⋯ + 1 k ( k + 1 ) = k k + 1 So, the left-hand side becomes:
k
k
+
1
+
1
(
k
+
1
)
(
k
+
2
)
k
k
+
1
+
1
(
k
+
1
)
(
k
+
2
)
(k)/(k+1)+(1)/((k+1)(k+2)) \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} k k + 1 + 1 ( k + 1 ) ( k + 2 ) Now, combine these two terms:
k
k
+
1
+
1
(
k
+
1
)
(
k
+
2
)
=
k
(
k
+
2
)
+
1
(
k
+
1
)
(
k
+
2
)
k
k
+
1
+
1
(
k
+
1
)
(
k
+
2
)
=
k
(
k
+
2
)
+
1
(
k
+
1
)
(
k
+
2
)
(k)/(k+1)+(1)/((k+1)(k+2))=(k(k+2)+1)/((k+1)(k+2)) \frac{k}{k+1} + \frac{1}{(k+1)(k+2)} = \frac{k(k+2) + 1}{(k+1)(k+2)} k k + 1 + 1 ( k + 1 ) ( k + 2 ) = k ( k + 2 ) + 1 ( k + 1 ) ( k + 2 ) Simplify the numerator:
k
(
k
+
2
)
+
1
=
k
2
+
2
k
+
1
=
(
k
+
1
)
2
k
(
k
+
2
)
+
1
=
k
2
+
2
k
+
1
=
(
k
+
1
)
2
k(k+2)+1=k^(2)+2k+1=(k+1)^(2) k(k+2) + 1 = k^2 + 2k + 1 = (k+1)^2 k ( k + 2 ) + 1 = k 2 + 2 k + 1 = ( k + 1 ) 2 So, the expression becomes:
(
k
+
1
)
2
(
k
+
1
)
(
k
+
2
)
=
k
+
1
k
+
2
(
k
+
1
)
2
(
k
+
1
)
(
k
+
2
)
=
k
+
1
k
+
2
((k+1)^(2))/((k+1)(k+2))=(k+1)/(k+2) \frac{(k+1)^2}{(k+1)(k+2)} = \frac{k+1}{k+2} ( k + 1 ) 2 ( k + 1 ) ( k + 2 ) = k + 1 k + 2 Thus, we have shown that:
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
(
k
+
1
)
(
k
+
2
)
=
k
+
1
k
+
2
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
(
k
+
1
)
(
k
+
2
)
=
k
+
1
k
+
2
(1)/(1*2)+(1)/(2*3)+cdots+(1)/((k+1)(k+2))=(k+1)/(k+2) \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{(k+1)(k+2)} = \frac{k+1}{k+2} 1 1 ⋅ 2 + 1 2 ⋅ 3 + ⋯ + 1 ( k + 1 ) ( k + 2 ) = k + 1 k + 2 Step 4: Conclusion
By the principle of Mathematical Induction, the statement is true for all
n
≥
1
n
≥
1
n >= 1 n \geq 1 n ≥ 1 . Therefore, we have proven that:
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
n
(
n
+
1
)
=
n
n
+
1
1
1
⋅
2
+
1
2
⋅
3
+
⋯
+
1
n
(
n
+
1
)
=
n
n
+
1
(1)/(1*2)+(1)/(2*3)+cdots+(1)/(n(n+1))=(n)/(n+1) \frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} + \cdots + \frac{1}{n(n+1)} = \frac{n}{n+1} 1 1 ⋅ 2 + 1 2 ⋅ 3 + ⋯ + 1 n ( n + 1 ) = n n + 1
Question:-04
How many terms of the sequence
3
,
3
,
3
3
,
…
3
,
3
,
3
3
,
…
sqrt3,3,3sqrt3,dots \sqrt{3}, 3,3 \sqrt{3}, \ldots 3 , 3 , 3 3 , …
39
+
13
3
39
+
13
3
39+13sqrt3 39+13 \sqrt{3} 39 + 13 3
Answer:
a
=
sqrt
3
r
=
3
3
=
3
S
n
=
a
(
r
n
−
1
)
r
−
1
=
39
+
13
3
=
3
(
3
n
−
1
)
3
−
1
=
39
3
−
39
+
13
×
3
−
13
3
=
(
3
)
n
+
1
−
3
=
39
3
−
13
3
=
(
3
)
n
+
1
−
3
=
27
3
=
(
3
)
n
+
1
=
3
3
3
=
(
3
)
n
+
1
=
(
3
)
6
+
1
=
(
3
)
n
+
1
∴
n
=
6
.
a
=
sqrt
3
r
=
3
3
=
3
S
n
=
a
r
n
−
1
r
−
1
=
39
+
13
3
=
3
3
n
−
1
3
−
1
=
39
3
−
39
+
13
×
3
−
13
3
=
(
3
)
n
+
1
−
3
=
39
3
−
13
3
=
(
3
)
n
+
1
−
3
=
27
3
=
(
3
)
n
+
1
=
3
3
3
=
(
3
)
n
+
1
=
(
3
)
6
+
1
=
(
3
)
n
+
1
∴
n
=
6
.
{:[a=sqrt3],[r=(3)/(sqrt3)=sqrt3],[S_(n)=(a(r^(n)-1))/(r-1)],[=39+13sqrt3=sqrt3((sqrt3^(n)-1))/(sqrt3-1)],[=39sqrt3-39+13 xx3-13sqrt3=(sqrt3)^(n+1)-sqrt3],[=39sqrt3-13sqrt3=(sqrt3)^(n+1)-sqrt3],[=27sqrt3=(sqrt3)^(n+1)],[=3^(3)sqrt3=(sqrt3)^(n+1)],[=(sqrt3)^(6+1)=(sqrt3)^(n+1)],[:.n=6.]:} \begin{aligned}
& \mathrm{a}=\mathrm{sqrt3} \\
& \mathrm{r}=\frac{3}{\sqrt{3}}=\sqrt{3} \\
& \mathrm{~S}_{\mathrm{n}}=\frac{\mathrm{a}\left(\mathrm{r}^{\mathrm{n}}-1\right)}{\mathrm{r}-1} \\
& =39+13 \sqrt{3}=\sqrt{3} \frac{\left(\sqrt{3}^{\mathrm{n}}-1\right)}{\sqrt{3}-1} \\
& =39 \sqrt{3}-39+13 \times 3-13 \sqrt{3}=(\sqrt{3})^{\mathrm{n}+1}-\sqrt{3} \\
& =39 \sqrt{3}-13 \sqrt{3}=(\sqrt{3})^{\mathrm{n}+1}-\sqrt{3} \\
& =27 \sqrt{3}=(\sqrt{3})^{\mathrm{n}+1} \\
& =3^3 \sqrt{3}=(\sqrt{3})^{\mathrm{n}+1} \\
& =(\sqrt{3})^{6+1}=(\sqrt{3})^{\mathrm{n}+1} \\
& \therefore \mathrm{n}=6 .
\end{aligned} a = sqrt 3 r = 3 3 = 3 S n = a ( r n − 1 ) r − 1 = 39 + 13 3 = 3 ( 3 n − 1 ) 3 − 1 = 39 3 − 39 + 13 × 3 − 13 3 = ( 3 ) n + 1 − 3 = 39 3 − 13 3 = ( 3 ) n + 1 − 3 = 27 3 = ( 3 ) n + 1 = 3 3 3 = ( 3 ) n + 1 = ( 3 ) 6 + 1 = ( 3 ) n + 1 ∴ n = 6 .
Question:-05
If
y
=
a
e
m
x
+
b
e
−
m
x
y
=
a
e
m
x
+
b
e
−
m
x
y=ae^(mx)+be^(-mx) y = ae^{mx} + be^{-mx} y = a e m x + b e − m x
d
2
y
d
x
2
=
m
2
y
d
2
y
d
x
2
=
m
2
y
(d^(2)y)/(dx^(2))=m^(2)y \frac{d^2y}{dx^2} = m^2y d 2 y d x 2 = m 2 y
Answer:
We are given the function:
y
=
a
e
m
x
+
b
e
−
m
x
y
=
a
e
m
x
+
b
e
−
m
x
y=ae^(mx)+be^(-mx) y = ae^{mx} + be^{-mx} y = a e m x + b e − m x We are asked to prove that:
d
2
y
d
x
2
=
m
2
y
d
2
y
d
x
2
=
m
2
y
(d^(2)y)/(dx^(2))=m^(2)y \frac{d^2y}{dx^2} = m^2y d 2 y d x 2 = m 2 y Step 1: Find the first derivative of
y
y
y y y
Differentiate
y
=
a
e
m
x
+
b
e
−
m
x
y
=
a
e
m
x
+
b
e
−
m
x
y=ae^(mx)+be^(-mx) y = ae^{mx} + be^{-mx} y = a e m x + b e − m x with respect to
x
x
x x x :
d
y
d
x
=
d
d
x
(
a
e
m
x
)
+
d
d
x
(
b
e
−
m
x
)
d
y
d
x
=
d
d
x
(
a
e
m
x
)
+
d
d
x
(
b
e
−
m
x
)
(dy)/(dx)=(d)/(dx)(ae^(mx))+(d)/(dx)(be^(-mx)) \frac{dy}{dx} = \frac{d}{dx}(ae^{mx}) + \frac{d}{dx}(be^{-mx}) d y d x = d d x ( a e m x ) + d d x ( b e − m x ) Using the chain rule, we get:
d
y
d
x
=
a
⋅
m
e
m
x
+
b
⋅
(
−
m
)
e
−
m
x
d
y
d
x
=
a
⋅
m
e
m
x
+
b
⋅
(
−
m
)
e
−
m
x
(dy)/(dx)=a*me^(mx)+b*(-m)e^(-mx) \frac{dy}{dx} = a \cdot m e^{mx} + b \cdot (-m) e^{-mx} d y d x = a ⋅ m e m x + b ⋅ ( − m ) e − m x Simplifying:
d
y
d
x
=
a
m
e
m
x
−
b
m
e
−
m
x
d
y
d
x
=
a
m
e
m
x
−
b
m
e
−
m
x
(dy)/(dx)=ame^(mx)-bme^(-mx) \frac{dy}{dx} = am e^{mx} – bm e^{-mx} d y d x = a m e m x − b m e − m x Step 2: Find the second derivative of
y
y
y y y
Now, differentiate
d
y
d
x
=
a
m
e
m
x
−
b
m
e
−
m
x
d
y
d
x
=
a
m
e
m
x
−
b
m
e
−
m
x
(dy)/(dx)=ame^(mx)-bme^(-mx) \frac{dy}{dx} = am e^{mx} – bm e^{-mx} d y d x = a m e m x − b m e − m x with respect to
x
x
x x x :
d
2
y
d
x
2
=
d
d
x
(
a
m
e
m
x
)
−
d
d
x
(
b
m
e
−
m
x
)
d
2
y
d
x
2
=
d
d
x
(
a
m
e
m
x
)
−
d
d
x
(
b
m
e
−
m
x
)
(d^(2)y)/(dx^(2))=(d)/(dx)(ame^(mx))-(d)/(dx)(bme^(-mx)) \frac{d^2y}{dx^2} = \frac{d}{dx}(am e^{mx}) – \frac{d}{dx}(bm e^{-mx}) d 2 y d x 2 = d d x ( a m e m x ) − d d x ( b m e − m x ) Again, applying the chain rule:
d
2
y
d
x
2
=
a
m
2
e
m
x
+
b
m
2
e
−
m
x
d
2
y
d
x
2
=
a
m
2
e
m
x
+
b
m
2
e
−
m
x
(d^(2)y)/(dx^(2))=am^(2)e^(mx)+bm^(2)e^(-mx) \frac{d^2y}{dx^2} = am^2 e^{mx} + bm^2 e^{-mx} d 2 y d x 2 = a m 2 e m x + b m 2 e − m x Step 3: Express
d
2
y
d
x
2
d
2
y
d
x
2
(d^(2)y)/(dx^(2)) \frac{d^2y}{dx^2} d 2 y d x 2
y
y
y y y
Notice that:
a
e
m
x
+
b
e
−
m
x
=
y
a
e
m
x
+
b
e
−
m
x
=
y
ae^(mx)+be^(-mx)=y ae^{mx} + be^{-mx} = y a e m x + b e − m x = y Thus, we can write:
d
2
y
d
x
2
=
m
2
(
a
e
m
x
+
b
e
−
m
x
)
=
m
2
y
d
2
y
d
x
2
=
m
2
(
a
e
m
x
+
b
e
−
m
x
)
=
m
2
y
(d^(2)y)/(dx^(2))=m^(2)(ae^(mx)+be^(-mx))=m^(2)y \frac{d^2y}{dx^2} = m^2(ae^{mx} + be^{-mx}) = m^2y d 2 y d x 2 = m 2 ( a e m x + b e − m x ) = m 2 y Conclusion:
We have proven that:
d
2
y
d
x
2
=
m
2
y
d
2
y
d
x
2
=
m
2
y
(d^(2)y)/(dx^(2))=m^(2)y \frac{d^2y}{dx^2} = m^2y d 2 y d x 2 = m 2 y
Question:-06
Integrate function
f
(
x
)
=
x
(
x
+
1
)
(
2
x
−
1
)
f
(
x
)
=
x
(
x
+
1
)
(
2
x
−
1
)
f(x)=(x)/((x+1)(2x-1)) f(x) = \frac{x}{(x+1)(2x-1)} f ( x ) = x ( x + 1 ) ( 2 x − 1 )
x
x
x x x
Answer:
To integrate the function
f
(
x
)
=
x
(
x
+
1
)
(
2
x
−
1
)
f
(
x
)
=
x
(
x
+
1
)
(
2
x
−
1
)
f(x)=(x)/((x+1)(2x-1)) f(x) = \frac{x}{(x+1)(2x-1)} f ( x ) = x ( x + 1 ) ( 2 x − 1 ) , we can use partial fraction decomposition to break it down into simpler terms that are easier to integrate.
Step 1: Partial Fraction Decomposition
We want to express
x
(
x
+
1
)
(
2
x
−
1
)
x
(
x
+
1
)
(
2
x
−
1
)
(x)/((x+1)(2x-1)) \frac{x}{(x+1)(2x-1)} x ( x + 1 ) ( 2 x − 1 ) as a sum of simpler fractions:
x
(
x
+
1
)
(
2
x
−
1
)
=
A
x
+
1
+
B
2
x
−
1
x
(
x
+
1
)
(
2
x
−
1
)
=
A
x
+
1
+
B
2
x
−
1
(x)/((x+1)(2x-1))=(A)/(x+1)+(B)/(2x-1) \frac{x}{(x+1)(2x-1)} = \frac{A}{x+1} + \frac{B}{2x-1} x ( x + 1 ) ( 2 x − 1 ) = A x + 1 + B 2 x − 1 Multiplying both sides by
(
x
+
1
)
(
2
x
−
1
)
(
x
+
1
)
(
2
x
−
1
)
(x+1)(2x-1) (x+1)(2x-1) ( x + 1 ) ( 2 x − 1 ) to clear the denominator:
x
=
A
(
2
x
−
1
)
+
B
(
x
+
1
)
x
=
A
(
2
x
−
1
)
+
B
(
x
+
1
)
x=A(2x-1)+B(x+1) x = A(2x-1) + B(x+1) x = A ( 2 x − 1 ) + B ( x + 1 ) Expanding both sides:
x
=
A
(
2
x
−
1
)
+
B
(
x
+
1
)
=
2
A
x
−
A
+
B
x
+
B
x
=
A
(
2
x
−
1
)
+
B
(
x
+
1
)
=
2
A
x
−
A
+
B
x
+
B
x=A(2x-1)+B(x+1)=2Ax-A+Bx+B x = A(2x – 1) + B(x + 1) = 2Ax – A + Bx + B x = A ( 2 x − 1 ) + B ( x + 1 ) = 2 A x − A + B x + B Simplifying:
x
=
(
2
A
+
B
)
x
+
(
−
A
+
B
)
x
=
(
2
A
+
B
)
x
+
(
−
A
+
B
)
x=(2A+B)x+(-A+B) x = (2A + B)x + (-A + B) x = ( 2 A + B ) x + ( − A + B ) Step 2: Solve for
A
A
A A A
B
B
B B B
Now, compare the coefficients of
x
x
x x x and the constant terms on both sides of the equation.
Coefficient of
x
x
x x x
2
A
+
B
=
1
2
A
+
B
=
1
2A+B=1 2A + B = 1 2 A + B = 1
Constant term:
−
A
+
B
=
0
−
A
+
B
=
0
-A+B=0 -A + B = 0 − A + B = 0
We have the system of equations:
2
A
+
B
=
1
2
A
+
B
=
1
2A+B=1 2A + B = 1 2 A + B = 1
−
A
+
B
=
0
−
A
+
B
=
0
-A+B=0 -A + B = 0 − A + B = 0 Solve the second equation for
B
B
B B B :
Substitute
B
=
A
B
=
A
B=A B = A B = A into the first equation:
2
A
+
A
=
1
⟹
3
A
=
1
⟹
A
=
1
3
2
A
+
A
=
1
⟹
3
A
=
1
⟹
A
=
1
3
2A+A=1Longrightarrow3A=1LongrightarrowA=(1)/(3) 2A + A = 1 \implies 3A = 1 \implies A = \frac{1}{3} 2 A + A = 1 ⟹ 3 A = 1 ⟹ A = 1 3 Since
B
=
A
B
=
A
B=A B = A B = A , we also have
B
=
1
3
B
=
1
3
B=(1)/(3) B = \frac{1}{3} B = 1 3 .
Step 3: Rewrite the integral
Now that we have
A
=
1
3
A
=
1
3
A=(1)/(3) A = \frac{1}{3} A = 1 3 and
B
=
1
3
B
=
1
3
B=(1)/(3) B = \frac{1}{3} B = 1 3 , we can rewrite the integrand:
x
(
x
+
1
)
(
2
x
−
1
)
=
1
/
3
x
+
1
+
1
/
3
2
x
−
1
x
(
x
+
1
)
(
2
x
−
1
)
=
1
/
3
x
+
1
+
1
/
3
2
x
−
1
(x)/((x+1)(2x-1))=(1//3)/(x+1)+(1//3)/(2x-1) \frac{x}{(x+1)(2x-1)} = \frac{1/3}{x+1} + \frac{1/3}{2x-1} x ( x + 1 ) ( 2 x − 1 ) = 1 / 3 x + 1 + 1 / 3 2 x − 1 Thus, the integral becomes:
∫
x
(
x
+
1
)
(
2
x
−
1
)
d
x
=
1
3
∫
1
x
+
1
d
x
+
1
3
∫
1
2
x
−
1
d
x
∫
x
(
x
+
1
)
(
2
x
−
1
)
d
x
=
1
3
∫
1
x
+
1
d
x
+
1
3
∫
1
2
x
−
1
d
x
int(x)/((x+1)(2x-1))dx=(1)/(3)int(1)/(x+1)dx+(1)/(3)int(1)/(2x-1)dx \int \frac{x}{(x+1)(2x-1)} \, dx = \frac{1}{3} \int \frac{1}{x+1} \, dx + \frac{1}{3} \int \frac{1}{2x-1} \, dx ∫ x ( x + 1 ) ( 2 x − 1 ) d x = 1 3 ∫ 1 x + 1 d x + 1 3 ∫ 1 2 x − 1 d x Step 4: Integrate
The two integrals are straightforward:
∫
1
x
+
1
d
x
=
ln
|
x
+
1
|
∫
1
x
+
1
d
x
=
ln
|
x
+
1
|
int(1)/(x+1)dx=ln |x+1| \int \frac{1}{x+1} \, dx = \ln|x+1| ∫ 1 x + 1 d x = ln | x + 1 |
∫
1
2
x
−
1
d
x
=
1
2
ln
|
2
x
−
1
|
∫
1
2
x
−
1
d
x
=
1
2
ln
|
2
x
−
1
|
int(1)/(2x-1)dx=(1)/(2)ln |2x-1| \int \frac{1}{2x-1} \, dx = \frac{1}{2} \ln|2x-1| ∫ 1 2 x − 1 d x = 1 2 ln | 2 x − 1 | Step 5: Final Answer
Now, combine the results:
∫
x
(
x
+
1
)
(
2
x
−
1
)
d
x
=
1
3
ln
|
x
+
1
|
+
1
6
ln
|
2
x
−
1
|
+
C
∫
x
(
x
+
1
)
(
2
x
−
1
)
d
x
=
1
3
ln
|
x
+
1
|
+
1
6
ln
|
2
x
−
1
|
+
C
int(x)/((x+1)(2x-1))dx=(1)/(3)ln |x+1|+(1)/(6)ln |2x-1|+C \int \frac{x}{(x+1)(2x-1)} \, dx = \frac{1}{3} \ln|x+1| + \frac{1}{6} \ln|2x-1| + C ∫ x ( x + 1 ) ( 2 x − 1 ) d x = 1 3 ln | x + 1 | + 1 6 ln | 2 x − 1 | + C where
C
C
C C C is the constant of integration.
Question:-07
If
1
,
w
,
w
2
1
,
w
,
w
2
1,w,w^(2) 1, w, w^2 1 , w , w 2
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
0
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
0
(1+w)^(2)-(1+w)^(3)+w^(2)=0 (1+w)^2 – (1+w)^3 + w^2 = 0 ( 1 + w ) 2 − ( 1 + w ) 3 + w 2 = 0
Answer:
We are given that
1
,
w
,
w
2
1
,
w
,
w
2
1,w,w^(2) 1, w, w^2 1 , w , w 2 are cube roots of unity. This means the following properties hold for
w
w
w w w :
w
3
=
1
w
3
=
1
w^(3)=1 w^3 = 1 w 3 = 1
1
+
w
+
w
2
=
0
1
+
w
+
w
2
=
0
1+w+w^(2)=0 1 + w + w^2 = 0 1 + w + w 2 = 0
We are asked to show that:
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
0
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
0
(1+w)^(2)-(1+w)^(3)+w^(2)=0 (1 + w)^2 – (1 + w)^3 + w^2 = 0 ( 1 + w ) 2 − ( 1 + w ) 3 + w 2 = 0 Step 1: Expand
(
1
+
w
)
2
(
1
+
w
)
2
(1+w)^(2) (1 + w)^2 ( 1 + w ) 2
First, let’s expand
(
1
+
w
)
2
(
1
+
w
)
2
(1+w)^(2) (1 + w)^2 ( 1 + w ) 2 :
(
1
+
w
)
2
=
(
1
+
w
)
(
1
+
w
)
=
1
+
2
w
+
w
2
(
1
+
w
)
2
=
(
1
+
w
)
(
1
+
w
)
=
1
+
2
w
+
w
2
(1+w)^(2)=(1+w)(1+w)=1+2w+w^(2) (1 + w)^2 = (1 + w)(1 + w) = 1 + 2w + w^2 ( 1 + w ) 2 = ( 1 + w ) ( 1 + w ) = 1 + 2 w + w 2 Step 2: Expand
(
1
+
w
)
3
(
1
+
w
)
3
(1+w)^(3) (1 + w)^3 ( 1 + w ) 3
Now, expand
(
1
+
w
)
3
(
1
+
w
)
3
(1+w)^(3) (1 + w)^3 ( 1 + w ) 3 :
(
1
+
w
)
3
=
(
1
+
w
)
(
1
+
w
)
2
=
(
1
+
w
)
(
1
+
2
w
+
w
2
)
(
1
+
w
)
3
=
(
1
+
w
)
(
1
+
w
)
2
=
(
1
+
w
)
(
1
+
2
w
+
w
2
)
(1+w)^(3)=(1+w)(1+w)^(2)=(1+w)(1+2w+w^(2)) (1 + w)^3 = (1 + w)(1 + w)^2 = (1 + w)(1 + 2w + w^2) ( 1 + w ) 3 = ( 1 + w ) ( 1 + w ) 2 = ( 1 + w ) ( 1 + 2 w + w 2 ) Using distributive property:
(
1
+
w
)
(
1
+
2
w
+
w
2
)
=
1
(
1
+
2
w
+
w
2
)
+
w
(
1
+
2
w
+
w
2
)
(
1
+
w
)
(
1
+
2
w
+
w
2
)
=
1
(
1
+
2
w
+
w
2
)
+
w
(
1
+
2
w
+
w
2
)
(1+w)(1+2w+w^(2))=1(1+2w+w^(2))+w(1+2w+w^(2)) (1 + w)(1 + 2w + w^2) = 1(1 + 2w + w^2) + w(1 + 2w + w^2) ( 1 + w ) ( 1 + 2 w + w 2 ) = 1 ( 1 + 2 w + w 2 ) + w ( 1 + 2 w + w 2 ) Simplifying:
=
1
+
2
w
+
w
2
+
w
+
2
w
2
+
w
3
=
1
+
2
w
+
w
2
+
w
+
2
w
2
+
w
3
=1+2w+w^(2)+w+2w^(2)+w^(3) = 1 + 2w + w^2 + w + 2w^2 + w^3 = 1 + 2 w + w 2 + w + 2 w 2 + w 3 Since
w
3
=
1
w
3
=
1
w^(3)=1 w^3 = 1 w 3 = 1 , this becomes:
1
+
2
w
+
w
2
+
w
+
2
w
2
+
1
=
2
+
3
w
+
3
w
2
1
+
2
w
+
w
2
+
w
+
2
w
2
+
1
=
2
+
3
w
+
3
w
2
1+2w+w^(2)+w+2w^(2)+1=2+3w+3w^(2) 1 + 2w + w^2 + w + 2w^2 + 1 = 2 + 3w + 3w^2 1 + 2 w + w 2 + w + 2 w 2 + 1 = 2 + 3 w + 3 w 2 Step 3: Substitute into the original expression
Now, substitute
(
1
+
w
)
2
=
1
+
2
w
+
w
2
(
1
+
w
)
2
=
1
+
2
w
+
w
2
(1+w)^(2)=1+2w+w^(2) (1 + w)^2 = 1 + 2w + w^2 ( 1 + w ) 2 = 1 + 2 w + w 2 and
(
1
+
w
)
3
=
2
+
3
w
+
3
w
2
(
1
+
w
)
3
=
2
+
3
w
+
3
w
2
(1+w)^(3)=2+3w+3w^(2) (1 + w)^3 = 2 + 3w + 3w^2 ( 1 + w ) 3 = 2 + 3 w + 3 w 2 into the original expression:
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
(
1
+
2
w
+
w
2
)
−
(
2
+
3
w
+
3
w
2
)
+
w
2
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
(
1
+
2
w
+
w
2
)
−
(
2
+
3
w
+
3
w
2
)
+
w
2
(1+w)^(2)-(1+w)^(3)+w^(2)=(1+2w+w^(2))-(2+3w+3w^(2))+w^(2) (1 + w)^2 – (1 + w)^3 + w^2 = (1 + 2w + w^2) – (2 + 3w + 3w^2) + w^2 ( 1 + w ) 2 − ( 1 + w ) 3 + w 2 = ( 1 + 2 w + w 2 ) − ( 2 + 3 w + 3 w 2 ) + w 2 Simplify:
=
1
+
2
w
+
w
2
−
2
−
3
w
−
3
w
2
+
w
2
=
1
+
2
w
+
w
2
−
2
−
3
w
−
3
w
2
+
w
2
=1+2w+w^(2)-2-3w-3w^(2)+w^(2) = 1 + 2w + w^2 – 2 – 3w – 3w^2 + w^2 = 1 + 2 w + w 2 − 2 − 3 w − 3 w 2 + w 2 Combine like terms:
=
(
1
−
2
)
+
(
2
w
−
3
w
)
+
(
w
2
−
3
w
2
+
w
2
)
=
(
1
−
2
)
+
(
2
w
−
3
w
)
+
(
w
2
−
3
w
2
+
w
2
)
=(1-2)+(2w-3w)+(w^(2)-3w^(2)+w^(2)) = (1 – 2) + (2w – 3w) + (w^2 – 3w^2 + w^2) = ( 1 − 2 ) + ( 2 w − 3 w ) + ( w 2 − 3 w 2 + w 2 )
=
−
1
−
w
−
w
2
=
−
1
−
w
−
w
2
=-1-w-w^(2) = -1 – w – w^2 = − 1 − w − w 2 But from the property of cube roots of unity, we know:
1
+
w
+
w
2
=
0
⟹
−
(
1
+
w
+
w
2
)
=
0
1
+
w
+
w
2
=
0
⟹
−
(
1
+
w
+
w
2
)
=
0
1+w+w^(2)=0Longrightarrow-(1+w+w^(2))=0 1 + w + w^2 = 0 \implies – (1 + w + w^2) = 0 1 + w + w 2 = 0 ⟹ − ( 1 + w + w 2 ) = 0 Thus:
−
1
−
w
−
w
2
=
0
−
1
−
w
−
w
2
=
0
-1-w-w^(2)=0 -1 – w – w^2 = 0 − 1 − w − w 2 = 0 Conclusion:
We have shown that:
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
0
(
1
+
w
)
2
−
(
1
+
w
)
3
+
w
2
=
0
(1+w)^(2)-(1+w)^(3)+w^(2)=0 (1 + w)^2 – (1 + w)^3 + w^2 = 0 ( 1 + w ) 2 − ( 1 + w ) 3 + w 2 = 0
Question:-08
If
α
,
β
α
,
β
alpha,beta \alpha, \beta α , β
2
x
2
−
3
x
−
5
=
0
2
x
2
−
3
x
−
5
=
0
2x^(2)-3x-5=0 2x^2 – 3x – 5 = 0 2 x 2 − 3 x − 5 = 0
α
2
,
β
2
α
2
,
β
2
alpha^(2),beta^(2) \alpha^2, \beta^2 α 2 , β 2
Answer:
We are given that
α
α
alpha \alpha α and
β
β
beta \beta β are the roots of the quadratic equation:
2
x
2
−
3
x
−
5
=
0
2
x
2
−
3
x
−
5
=
0
2x^(2)-3x-5=0 2x^2 – 3x – 5 = 0 2 x 2 − 3 x − 5 = 0 This implies, from Vieta’s formulas, that:
α
+
β
=
−
(
−
3
)
2
=
3
2
α
+
β
=
−
(
−
3
)
2
=
3
2
alpha+beta=(-(-3))/(2)=(3)/(2) \alpha + \beta = \frac{-(-3)}{2} = \frac{3}{2} α + β = − ( − 3 ) 2 = 3 2
α
β
=
−
5
2
α
β
=
−
5
2
alpha beta=(-5)/(2) \alpha \beta = \frac{-5}{2} α β = − 5 2 We need to find a quadratic equation whose roots are
α
2
α
2
alpha^(2) \alpha^2 α 2 and
β
2
β
2
beta^(2) \beta^2 β 2 .
Step 1: Find the sum and product of
α
2
α
2
alpha^(2) \alpha^2 α 2
β
2
β
2
beta^(2) \beta^2 β 2
Sum of
α
2
α
2
alpha^(2) \alpha^2 α 2
β
2
β
2
beta^(2) \beta^2 β 2
We use the identity:
α
2
+
β
2
=
(
α
+
β
)
2
−
2
α
β
α
2
+
β
2
=
(
α
+
β
)
2
−
2
α
β
alpha^(2)+beta^(2)=(alpha+beta)^(2)-2alpha beta \alpha^2 + \beta^2 = (\alpha + \beta)^2 – 2\alpha \beta α 2 + β 2 = ( α + β ) 2 − 2 α β Substitute
α
+
β
=
3
2
α
+
β
=
3
2
alpha+beta=(3)/(2) \alpha + \beta = \frac{3}{2} α + β = 3 2 and
α
β
=
−
5
2
α
β
=
−
5
2
alpha beta=(-5)/(2) \alpha \beta = \frac{-5}{2} α β = − 5 2 :
α
2
+
β
2
=
(
3
2
)
2
−
2
×
−
5
2
α
2
+
β
2
=
3
2
2
−
2
×
−
5
2
alpha^(2)+beta^(2)=((3)/(2))^(2)-2xx(-5)/(2) \alpha^2 + \beta^2 = \left( \frac{3}{2} \right)^2 – 2 \times \frac{-5}{2} α 2 + β 2 = ( 3 2 ) 2 − 2 × − 5 2
α
2
+
β
2
=
9
4
+
5
=
9
4
+
20
4
=
29
4
α
2
+
β
2
=
9
4
+
5
=
9
4
+
20
4
=
29
4
alpha^(2)+beta^(2)=(9)/(4)+5=(9)/(4)+(20)/(4)=(29)/(4) \alpha^2 + \beta^2 = \frac{9}{4} + 5 = \frac{9}{4} + \frac{20}{4} = \frac{29}{4} α 2 + β 2 = 9 4 + 5 = 9 4 + 20 4 = 29 4 Product of
α
2
α
2
alpha^(2) \alpha^2 α 2
β
2
β
2
beta^(2) \beta^2 β 2
We use the identity:
α
2
β
2
=
(
α
β
)
2
α
2
β
2
=
(
α
β
)
2
alpha^(2)beta^(2)=(alpha beta)^(2) \alpha^2 \beta^2 = (\alpha \beta)^2 α 2 β 2 = ( α β ) 2 Substitute
α
β
=
−
5
2
α
β
=
−
5
2
alpha beta=(-5)/(2) \alpha \beta = \frac{-5}{2} α β = − 5 2 :
α
2
β
2
=
(
−
5
2
)
2
=
25
4
α
2
β
2
=
−
5
2
2
=
25
4
alpha^(2)beta^(2)=((-5)/(2))^(2)=(25)/(4) \alpha^2 \beta^2 = \left( \frac{-5}{2} \right)^2 = \frac{25}{4} α 2 β 2 = ( − 5 2 ) 2 = 25 4 The quadratic equation with roots
α
2
α
2
alpha^(2) \alpha^2 α 2 and
β
2
β
2
beta^(2) \beta^2 β 2 is given by:
x
2
−
(
α
2
+
β
2
)
x
+
α
2
β
2
=
0
x
2
−
(
α
2
+
β
2
)
x
+
α
2
β
2
=
0
x^(2)-(alpha^(2)+beta^(2))x+alpha^(2)beta^(2)=0 x^2 – (\alpha^2 + \beta^2)x + \alpha^2 \beta^2 = 0 x 2 − ( α 2 + β 2 ) x + α 2 β 2 = 0 Substitute
α
2
+
β
2
=
29
4
α
2
+
β
2
=
29
4
alpha^(2)+beta^(2)=(29)/(4) \alpha^2 + \beta^2 = \frac{29}{4} α 2 + β 2 = 29 4 and
α
2
β
2
=
25
4
α
2
β
2
=
25
4
alpha^(2)beta^(2)=(25)/(4) \alpha^2 \beta^2 = \frac{25}{4} α 2 β 2 = 25 4 :
x
2
−
29
4
x
+
25
4
=
0
x
2
−
29
4
x
+
25
4
=
0
x^(2)-(29)/(4)x+(25)/(4)=0 x^2 – \frac{29}{4}x + \frac{25}{4} = 0 x 2 − 29 4 x + 25 4 = 0 Multiply the entire equation by 4 to clear the denominators:
4
x
2
−
29
x
+
25
=
0
4
x
2
−
29
x
+
25
=
0
4x^(2)-29 x+25=0 4x^2 – 29x + 25 = 0 4 x 2 − 29 x + 25 = 0 Final Answer:
The quadratic equation whose roots are
α
2
α
2
alpha^(2) \alpha^2 α 2 and
β
2
β
2
beta^(2) \beta^2 β 2 is:
4
x
2
−
29
x
+
25
=
0
4
x
2
−
29
x
+
25
=
0
4x^(2)-29 x+25=0 4x^2 – 29x + 25 = 0 4 x 2 − 29 x + 25 = 0
Question:-09
Solve the inequality
3
5
(
x
−
2
)
≤
5
3
(
2
−
x
)
3
5
(
x
−
2
)
≤
5
3
(
2
−
x
)
(3)/(5)(x-2) <= (5)/(3)(2-x) \frac{3}{5}(x – 2) \leq \frac{5}{3}(2 – x) 3 5 ( x − 2 ) ≤ 5 3 ( 2 − x )
Answer:
To solve the inequality:
3
5
(
x
−
2
)
≤
5
3
(
2
−
x
)
3
5
(
x
−
2
)
≤
5
3
(
2
−
x
)
(3)/(5)(x-2) <= (5)/(3)(2-x) \frac{3}{5}(x – 2) \leq \frac{5}{3}(2 – x) 3 5 ( x − 2 ) ≤ 5 3 ( 2 − x ) Step 1: Eliminate the fractions
Multiply both sides of the inequality by the least common denominator (LCD) of 5 and 3, which is 15, to eliminate the fractions:
15
×
(
3
5
(
x
−
2
)
)
≤
15
×
(
5
3
(
2
−
x
)
)
15
×
3
5
(
x
−
2
)
≤
15
×
5
3
(
2
−
x
)
15 xx((3)/(5)(x-2)) <= 15 xx((5)/(3)(2-x)) 15 \times \left( \frac{3}{5}(x – 2) \right) \leq 15 \times \left( \frac{5}{3}(2 – x) \right) 15 × ( 3 5 ( x − 2 ) ) ≤ 15 × ( 5 3 ( 2 − x ) ) Simplifying both sides:
3
×
3
(
x
−
2
)
≤
5
×
5
(
2
−
x
)
3
×
3
(
x
−
2
)
≤
5
×
5
(
2
−
x
)
3xx3(x-2) <= 5xx5(2-x) 3 \times 3(x – 2) \leq 5 \times 5(2 – x) 3 × 3 ( x − 2 ) ≤ 5 × 5 ( 2 − x ) This gives:
9
(
x
−
2
)
≤
25
(
2
−
x
)
9
(
x
−
2
)
≤
25
(
2
−
x
)
9(x-2) <= 25(2-x) 9(x – 2) \leq 25(2 – x) 9 ( x − 2 ) ≤ 25 ( 2 − x ) Step 2: Expand both sides
Expand the expressions on both sides:
9
x
−
18
≤
50
−
25
x
9
x
−
18
≤
50
−
25
x
9x-18 <= 50-25 x 9x – 18 \leq 50 – 25x 9 x − 18 ≤ 50 − 25 x Step 3: Collect like terms
Move all terms involving
x
x
x x x to one side and constants to the other side. Add
25
x
25
x
25 x 25x 25 x to both sides and add
18
18
18 18 18 to both sides:
9
x
+
25
x
≤
50
+
18
9
x
+
25
x
≤
50
+
18
9x+25 x <= 50+18 9x + 25x \leq 50 + 18 9 x + 25 x ≤ 50 + 18 Simplifying:
34
x
≤
68
34
x
≤
68
34 x <= 68 34x \leq 68 34 x ≤ 68 Step 4: Solve for
x
x
x x x
Divide both sides by 34:
x
≤
68
34
x
≤
68
34
x <= (68)/(34) x \leq \frac{68}{34} x ≤ 68 34 Simplifying:
x
≤
2
x
≤
2
x <= 2 x \leq 2 x ≤ 2 Step 5: Graph the solution
The solution is
x
≤
2
x
≤
2
x <= 2 x \leq 2 x ≤ 2 . To graph this on a number line:
Final Answer:
The solution to the inequality is
x
≤
2
x
≤
2
x <= 2 x \leq 2 x ≤ 2 , and the graph is a number line with a closed circle at
x
=
2
x
=
2
x=2 x = 2 x = 2 and shading to the left of
x
=
2
x
=
2
x=2 x = 2 x = 2 .
Question:-10
If a positive number exceeds its positive square root by 12, then find the number.
Answer:
Let the positive number be
x
x
x x x , and its positive square root is
x
x
sqrtx \sqrt{x} x . The problem states that the number exceeds its square root by 12. This gives us the equation:
x
=
x
+
12
x
=
x
+
12
x=sqrtx+12 x = \sqrt{x} + 12 x = x + 12 Step 1: Isolate the square root term
Subtract
x
x
sqrtx \sqrt{x} x from both sides:
x
−
x
=
12
x
−
x
=
12
x-sqrtx=12 x – \sqrt{x} = 12 x − x = 12 Step 2: Substitute
y
=
x
y
=
x
y=sqrtx y = \sqrt{x} y = x
Let
y
=
x
y
=
x
y=sqrtx y = \sqrt{x} y = x , which means
y
2
=
x
y
2
=
x
y^(2)=x y^2 = x y 2 = x . Substituting
y
y
y y y into the equation:
y
2
−
y
=
12
y
2
−
y
=
12
y^(2)-y=12 y^2 – y = 12 y 2 − y = 12 Step 3: Rearrange into a quadratic equation
Rearrange the equation:
y
2
−
y
−
12
=
0
y
2
−
y
−
12
=
0
y^(2)-y-12=0 y^2 – y – 12 = 0 y 2 − y − 12 = 0 Step 4: Solve the quadratic equation
Now, solve the quadratic equation
y
2
−
y
−
12
=
0
y
2
−
y
−
12
=
0
y^(2)-y-12=0 y^2 – y – 12 = 0 y 2 − y − 12 = 0 using the quadratic formula. The quadratic formula is:
y
=
−
b
±
b
2
−
4
a
c
2
a
y
=
−
b
±
b
2
−
4
a
c
2
a
y=(-b+-sqrt(b^(2)-4ac))/(2a) y = \frac{-b \pm \sqrt{b^2 – 4ac}}{2a} y = − b ± b 2 − 4 a c 2 a For the equation
y
2
−
y
−
12
=
0
y
2
−
y
−
12
=
0
y^(2)-y-12=0 y^2 – y – 12 = 0 y 2 − y − 12 = 0 , we have
a
=
1
a
=
1
a=1 a = 1 a = 1 ,
b
=
−
1
b
=
−
1
b=-1 b = -1 b = − 1 , and
c
=
−
12
c
=
−
12
c=-12 c = -12 c = − 12 . Substituting these values into the quadratic formula:
y
=
−
(
−
1
)
±
(
−
1
)
2
−
4
(
1
)
(
−
12
)
2
(
1
)
y
=
−
(
−
1
)
±
(
−
1
)
2
−
4
(
1
)
(
−
12
)
2
(
1
)
y=(-(-1)+-sqrt((-1)^(2)-4(1)(-12)))/(2(1)) y = \frac{-(-1) \pm \sqrt{(-1)^2 – 4(1)(-12)}}{2(1)} y = − ( − 1 ) ± ( − 1 ) 2 − 4 ( 1 ) ( − 12 ) 2 ( 1 )
y
=
1
±
1
+
48
2
y
=
1
±
1
+
48
2
y=(1+-sqrt(1+48))/(2) y = \frac{1 \pm \sqrt{1 + 48}}{2} y = 1 ± 1 + 48 2
y
=
1
±
49
2
y
=
1
±
49
2
y=(1+-sqrt49)/(2) y = \frac{1 \pm \sqrt{49}}{2} y = 1 ± 49 2
y
=
1
±
7
2
y
=
1
±
7
2
y=(1+-7)/(2) y = \frac{1 \pm 7}{2} y = 1 ± 7 2 Thus, we have two possible solutions for
y
y
y y y :
y
=
1
+
7
2
=
8
2
=
4
y
=
1
+
7
2
=
8
2
=
4
y=(1+7)/(2)=(8)/(2)=4 y = \frac{1 + 7}{2} = \frac{8}{2} = 4 y = 1 + 7 2 = 8 2 = 4
y
=
1
−
7
2
=
−
6
2
=
−
3
y
=
1
−
7
2
=
−
6
2
=
−
3
y=(1-7)/(2)=(-6)/(2)=-3 y = \frac{1 – 7}{2} = \frac{-6}{2} = -3 y = 1 − 7 2 = − 6 2 = − 3 Step 5: Interpret the solution
Since
y
=
x
y
=
x
y=sqrtx y = \sqrt{x} y = x and
y
y
y y y must be positive, we take
y
=
4
y
=
4
y=4 y = 4 y = 4 .
Step 6: Find
x
x
x x x
Since
y
=
x
y
=
x
y=sqrtx y = \sqrt{x} y = x , we have:
x
=
4
x
=
4
sqrtx=4 \sqrt{x} = 4 x = 4 Squaring both sides:
x
=
4
2
=
16
x
=
4
2
=
16
x=4^(2)=16 x = 4^2 = 16 x = 4 2 = 16 Thus, the positive number is
x
=
16
x
=
16
x=16 x = 16 x = 16 .
Final Answer:
The positive number is
16
16
16 \boxed{16} 16 .
Question:-11
Find the area bounded by the curves
x
2
=
y
x
2
=
y
x^(2)=y x^2 = y x 2 = y
y
=
x
y
=
x
y=x y = x y = x
Answer:
We are tasked with finding the area bounded by the curves
x
2
=
y
x
2
=
y
x^(2)=y x^2 = y x 2 = y (a parabola) and
y
=
x
y
=
x
y=x y = x y = x (a straight line).
Step 1: Find the points of intersection
To find the points of intersection between the curves, set
y
=
x
y
=
x
y=x y = x y = x in the equation
x
2
=
y
x
2
=
y
x^(2)=y x^2 = y x 2 = y . Thus, we have:
x
2
=
x
x
2
=
x
x^(2)=x x^2 = x x 2 = x Rearrange this equation:
x
2
−
x
=
0
x
2
−
x
=
0
x^(2)-x=0 x^2 – x = 0 x 2 − x = 0 Factor the equation:
x
(
x
−
1
)
=
0
x
(
x
−
1
)
=
0
x(x-1)=0 x(x – 1) = 0 x ( x − 1 ) = 0 So,
x
=
0
x
=
0
x=0 x = 0 x = 0 or
x
=
1
x
=
1
x=1 x = 1 x = 1 . These are the
x
x
x x x -coordinates of the points of intersection. Substituting these values of
x
x
x x x into the equation
y
=
x
y
=
x
y=x y = x y = x gives
y
=
0
y
=
0
y=0 y = 0 y = 0 when
x
=
0
x
=
0
x=0 x = 0 x = 0 and
y
=
1
y
=
1
y=1 y = 1 y = 1 when
x
=
1
x
=
1
x=1 x = 1 x = 1 .
Thus, the points of intersection are
(
0
,
0
)
(
0
,
0
)
(0,0) (0, 0) ( 0 , 0 ) and
(
1
,
1
)
(
1
,
1
)
(1,1) (1, 1) ( 1 , 1 ) .
Step 2: Set up the integral
We are looking for the area between the curves from
x
=
0
x
=
0
x=0 x = 0 x = 0 to
x
=
1
x
=
1
x=1 x = 1 x = 1 . The area between two curves is given by the integral of the difference between the top curve and the bottom curve.
For
0
≤
x
≤
1
0
≤
x
≤
1
0 <= x <= 1 0 \leq x \leq 1 0 ≤ x ≤ 1 :
The line
y
=
x
y
=
x
y=x y = x y = x
x
2
=
y
x
2
=
y
x^(2)=y x^2 = y x 2 = y
A
=
∫
0
1
[
(
x
)
−
(
x
2
)
]
d
x
A
=
∫
0
1
[
(
x
)
−
(
x
2
)
]
d
x
A=int_(0)^(1)[(x)-(x^(2))]dx A = \int_0^1 [(x) – (x^2)] \, dx A = ∫ 0 1 [ ( x ) − ( x 2 ) ] d x Step 3: Compute the integral
Now, let’s compute the integral:
A
=
∫
0
1
(
x
−
x
2
)
d
x
A
=
∫
0
1
(
x
−
x
2
)
d
x
A=int_(0)^(1)(x-x^(2))dx A = \int_0^1 (x – x^2) \, dx A = ∫ 0 1 ( x − x 2 ) d x Break this into two integrals:
A
=
∫
0
1
x
d
x
−
∫
0
1
x
2
d
x
A
=
∫
0
1
x
d
x
−
∫
0
1
x
2
d
x
A=int_(0)^(1)xdx-int_(0)^(1)x^(2)dx A = \int_0^1 x \, dx – \int_0^1 x^2 \, dx A = ∫ 0 1 x d x − ∫ 0 1 x 2 d x Now, compute each integral:
∫
0
1
x
d
x
=
[
x
2
2
]
0
1
=
1
2
2
−
0
=
1
2
∫
0
1
x
d
x
=
x
2
2
0
1
=
1
2
2
−
0
=
1
2
int_(0)^(1)xdx=[(x^(2))/(2)]_(0)^(1)=(1^(2))/(2)-0=(1)/(2) \int_0^1 x \, dx = \left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} – 0 = \frac{1}{2} ∫ 0 1 x d x = [ x 2 2 ] 0 1 = 1 2 2 − 0 = 1 2
∫
0
1
x
2
d
x
=
[
x
3
3
]
0
1
=
1
3
3
−
0
=
1
3
∫
0
1
x
2
d
x
=
x
3
3
0
1
=
1
3
3
−
0
=
1
3
int_(0)^(1)x^(2)dx=[(x^(3))/(3)]_(0)^(1)=(1^(3))/(3)-0=(1)/(3) \int_0^1 x^2 \, dx = \left[ \frac{x^3}{3} \right]_0^1 = \frac{1^3}{3} – 0 = \frac{1}{3} ∫ 0 1 x 2 d x = [ x 3 3 ] 0 1 = 1 3 3 − 0 = 1 3
Thus, the total area is:
A
=
1
2
−
1
3
=
3
6
−
2
6
=
1
6
A
=
1
2
−
1
3
=
3
6
−
2
6
=
1
6
A=(1)/(2)-(1)/(3)=(3)/(6)-(2)/(6)=(1)/(6) A = \frac{1}{2} – \frac{1}{3} = \frac{3}{6} – \frac{2}{6} = \frac{1}{6} A = 1 2 − 1 3 = 3 6 − 2 6 = 1 6 Final Answer:
The area bounded by the curves
x
2
=
y
x
2
=
y
x^(2)=y x^2 = y x 2 = y and
y
=
x
y
=
x
y=x y = x y = x is
1
6
1
6
(1)/(6) \boxed{\frac{1}{6}} 1 6 .
Question:-12
Find the inverse of the matrix
A
=
(
1
6
4
2
4
−
1
−
1
2
5
)
A
=
1
6
4
2
4
−
1
−
1
2
5
A=([1,6,4],[2,4,-1],[-1,2,5]) A = \left(\begin{array}{ccc} 1 & 6 & 4 \\ 2 & 4 & -1 \\ -1 & 2 & 5 \end{array}\right) A = ( 1 6 4 2 4 − 1 − 1 2 5 )
Answer:
To find the inverse of matrix
A
=
(
1
6
4
2
4
−
1
−
1
2
5
)
A
=
1
6
4
2
4
−
1
−
1
2
5
A=([1,6,4],[2,4,-1],[-1,2,5]) A = \begin{pmatrix} 1 & 6 & 4 \\ 2 & 4 & -1 \\ -1 & 2 & 5 \end{pmatrix} A = ( 1 6 4 2 4 − 1 − 1 2 5 ) , we will use the formula for the inverse of a
3
×
3
3
×
3
3xx3 3 \times 3 3 × 3 matrix. The inverse of a matrix
A
A
A A A , if it exists, is given by:
A
−
1
=
1
det
(
A
)
⋅
adj
(
A
)
A
−
1
=
1
det
(
A
)
⋅
adj
(
A
)
A^(-1)=(1)/(det(A))*”adj”(A) A^{-1} = \frac{1}{\det(A)} \cdot \text{adj}(A) A − 1 = 1 det ( A ) ⋅ adj ( A ) where
det
(
A
)
det
(
A
)
det(A) \det(A) det ( A ) is the determinant of
A
A
A A A , and
adj
(
A
)
adj
(
A
)
“adj”(A) \text{adj}(A) adj ( A ) is the adjugate (transpose of the cofactor matrix) of
A
A
A A A .
Step 1: Calculate the determinant of
A
A
A A A
The determinant of matrix
A
A
A A A is calculated as:
det
(
A
)
=
1
⋅
|
4
−
1
2
5
|
−
6
⋅
|
2
−
1
−
1
5
|
+
4
⋅
|
2
4
−
1
2
|
det
(
A
)
=
1
⋅
4
−
1
2
5
−
6
⋅
2
−
1
−
1
5
+
4
⋅
2
4
−
1
2
det(A)=1*|[4,-1],[2,5]|-6*|[2,-1],[-1,5]|+4*|[2,4],[-1,2]| \det(A) = 1 \cdot \begin{vmatrix} 4 & -1 \\ 2 & 5 \end{vmatrix} – 6 \cdot \begin{vmatrix} 2 & -1 \\ -1 & 5 \end{vmatrix} + 4 \cdot \begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} det ( A ) = 1 ⋅ | 4 − 1 2 5 | − 6 ⋅ | 2 − 1 − 1 5 | + 4 ⋅ | 2 4 − 1 2 | Now, let’s calculate each of these
2
×
2
2
×
2
2xx2 2 \times 2 2 × 2 determinants:
|
4
−
1
2
5
|
=
(
4
)
(
5
)
−
(
−
1
)
(
2
)
=
20
+
2
=
22
4
−
1
2
5
=
(
4
)
(
5
)
−
(
−
1
)
(
2
)
=
20
+
2
=
22
|[4,-1],[2,5]|=(4)(5)-(-1)(2)=20+2=22 \begin{vmatrix} 4 & -1 \\ 2 & 5 \end{vmatrix} = (4)(5) – (-1)(2) = 20 + 2 = 22 | 4 − 1 2 5 | = ( 4 ) ( 5 ) − ( − 1 ) ( 2 ) = 20 + 2 = 22
|
2
−
1
−
1
5
|
=
(
2
)
(
5
)
−
(
−
1
)
(
−
1
)
=
10
−
1
=
9
2
−
1
−
1
5
=
(
2
)
(
5
)
−
(
−
1
)
(
−
1
)
=
10
−
1
=
9
|[2,-1],[-1,5]|=(2)(5)-(-1)(-1)=10-1=9 \begin{vmatrix} 2 & -1 \\ -1 & 5 \end{vmatrix} = (2)(5) – (-1)(-1) = 10 – 1 = 9 | 2 − 1 − 1 5 | = ( 2 ) ( 5 ) − ( − 1 ) ( − 1 ) = 10 − 1 = 9
|
2
4
−
1
2
|
=
(
2
)
(
2
)
−
(
4
)
(
−
1
)
=
4
+
4
=
8
2
4
−
1
2
=
(
2
)
(
2
)
−
(
4
)
(
−
1
)
=
4
+
4
=
8
|[2,4],[-1,2]|=(2)(2)-(4)(-1)=4+4=8 \begin{vmatrix} 2 & 4 \\ -1 & 2 \end{vmatrix} = (2)(2) – (4)(-1) = 4 + 4 = 8 | 2 4 − 1 2 | = ( 2 ) ( 2 ) − ( 4 ) ( − 1 ) = 4 + 4 = 8
Substitute these values into the determinant formula:
det
(
A
)
=
1
⋅
22
−
6
⋅
9
+
4
⋅
8
=
22
−
54
+
32
=
0
det
(
A
)
=
1
⋅
22
−
6
⋅
9
+
4
⋅
8
=
22
−
54
+
32
=
0
det(A)=1*22-6*9+4*8=22-54+32=0 \det(A) = 1 \cdot 22 – 6 \cdot 9 + 4 \cdot 8 = 22 – 54 + 32 = 0 det ( A ) = 1 ⋅ 22 − 6 ⋅ 9 + 4 ⋅ 8 = 22 − 54 + 32 = 0 Step 2: Conclusion
Since the determinant of
A
A
A A A is zero (
det
(
A
)
=
0
det
(
A
)
=
0
det(A)=0 \det(A) = 0 det ( A ) = 0 ), the matrix
A
A
A A A is
singular and does not have an inverse.
Final Answer:
The matrix
A
A
A A A does not have an inverse because its determinant is zero.
No inverse exists
No inverse exists
“No inverse exists” \boxed{\text{No inverse exists}} No inverse exists .
Question:-13
If
m
m
m m m
m
th
m
th
m^(“th”) m^{\text{th}} m th
n
n
n n n
n
th
n
th
n^(“th”) n^{\text{th}} n th
(
m
+
n
)
th
(
m
+
n
)
th
(m+n)^(“th”) (m+n)^{\text{th}} ( m + n ) th
Answer:
Let the arithmetic progression (A.P.) have the first term
a
a
a a a and common difference
d
d
d d d .
The general term of the A.P., which is the
k
th
k
th
k^(“th”) k^{\text{th}} k th term, is given by:
T
k
=
a
+
(
k
−
1
)
d
T
k
=
a
+
(
k
−
1
)
d
T_(k)=a+(k-1)d T_k = a + (k-1)d T k = a + ( k − 1 ) d We are given that
m
m
m m m times the
m
th
m
th
m^(“th”) m^{\text{th}} m th term is equal to
n
n
n n n times the
n
th
n
th
n^(“th”) n^{\text{th}} n th term. This can be expressed as:
m
⋅
T
m
=
n
⋅
T
n
m
⋅
T
m
=
n
⋅
T
n
m*T_(m)=n*T_(n) m \cdot T_m = n \cdot T_n m ⋅ T m = n ⋅ T n Using the formula for the
k
th
k
th
k^(“th”) k^{\text{th}} k th term, the
m
th
m
th
m^(“th”) m^{\text{th}} m th term and
n
th
n
th
n^(“th”) n^{\text{th}} n th term are:
T
m
=
a
+
(
m
−
1
)
d
T
m
=
a
+
(
m
−
1
)
d
T_(m)=a+(m-1)d T_m = a + (m-1)d T m = a + ( m − 1 ) d
T
n
=
a
+
(
n
−
1
)
d
T
n
=
a
+
(
n
−
1
)
d
T_(n)=a+(n-1)d T_n = a + (n-1)d T n = a + ( n − 1 ) d Substitute these into the given condition:
m
⋅
(
a
+
(
m
−
1
)
d
)
=
n
⋅
(
a
+
(
n
−
1
)
d
)
m
⋅
(
a
+
(
m
−
1
)
d
)
=
n
⋅
(
a
+
(
n
−
1
)
d
)
m*(a+(m-1)d)=n*(a+(n-1)d) m \cdot (a + (m-1)d) = n \cdot (a + (n-1)d) m ⋅ ( a + ( m − 1 ) d ) = n ⋅ ( a + ( n − 1 ) d ) Step 1: Expand both sides
Expand both sides of the equation:
m
⋅
a
+
m
⋅
(
m
−
1
)
⋅
d
=
n
⋅
a
+
n
⋅
(
n
−
1
)
⋅
d
m
⋅
a
+
m
⋅
(
m
−
1
)
⋅
d
=
n
⋅
a
+
n
⋅
(
n
−
1
)
⋅
d
m*a+m*(m-1)*d=n*a+n*(n-1)*d m \cdot a + m \cdot (m-1) \cdot d = n \cdot a + n \cdot (n-1) \cdot d m ⋅ a + m ⋅ ( m − 1 ) ⋅ d = n ⋅ a + n ⋅ ( n − 1 ) ⋅ d Simplifying both sides:
m
a
+
m
(
m
−
1
)
d
=
n
a
+
n
(
n
−
1
)
d
m
a
+
m
(
m
−
1
)
d
=
n
a
+
n
(
n
−
1
)
d
ma+m(m-1)d=na+n(n-1)d ma + m(m-1)d = na + n(n-1)d m a + m ( m − 1 ) d = n a + n ( n − 1 ) d Step 2: Collect like terms
Rearrange the terms to collect like terms:
m
a
−
n
a
=
n
(
n
−
1
)
d
−
m
(
m
−
1
)
d
m
a
−
n
a
=
n
(
n
−
1
)
d
−
m
(
m
−
1
)
d
ma-na=n(n-1)d-m(m-1)d ma – na = n(n-1)d – m(m-1)d m a − n a = n ( n − 1 ) d − m ( m − 1 ) d Factor out common terms:
a
(
m
−
n
)
=
d
(
n
(
n
−
1
)
−
m
(
m
−
1
)
)
a
(
m
−
n
)
=
d
(
n
(
n
−
1
)
−
m
(
m
−
1
)
)
a(m-n)=d(n(n-1)-m(m-1)) a(m-n) = d(n(n-1) – m(m-1)) a ( m − n ) = d ( n ( n − 1 ) − m ( m − 1 ) ) Step 3: Simplify the equation
Expand
n
(
n
−
1
)
n
(
n
−
1
)
n(n-1) n(n-1) n ( n − 1 ) and
m
(
m
−
1
)
m
(
m
−
1
)
m(m-1) m(m-1) m ( m − 1 ) :
a
(
m
−
n
)
=
d
(
n
2
−
n
−
(
m
2
−
m
)
)
a
(
m
−
n
)
=
d
(
n
2
−
n
−
(
m
2
−
m
)
)
a(m-n)=d(n^(2)-n-(m^(2)-m)) a(m-n) = d(n^2 – n – (m^2 – m)) a ( m − n ) = d ( n 2 − n − ( m 2 − m ) )
a
(
m
−
n
)
=
d
(
n
2
−
n
−
m
2
+
m
)
a
(
m
−
n
)
=
d
(
n
2
−
n
−
m
2
+
m
)
a(m-n)=d(n^(2)-n-m^(2)+m) a(m-n) = d(n^2 – n – m^2 + m) a ( m − n ) = d ( n 2 − n − m 2 + m ) Factor the right-hand side:
a
(
m
−
n
)
=
d
(
(
n
2
−
m
2
)
−
(
n
−
m
)
)
a
(
m
−
n
)
=
d
(
(
n
2
−
m
2
)
−
(
n
−
m
)
)
a(m-n)=d((n^(2)-m^(2))-(n-m)) a(m-n) = d((n^2 – m^2) – (n – m)) a ( m − n ) = d ( ( n 2 − m 2 ) − ( n − m ) ) Now, use the difference of squares for
n
2
−
m
2
n
2
−
m
2
n^(2)-m^(2) n^2 – m^2 n 2 − m 2 :
a
(
m
−
n
)
=
d
(
(
n
−
m
)
(
n
+
m
)
−
(
n
−
m
)
)
a
(
m
−
n
)
=
d
(
(
n
−
m
)
(
n
+
m
)
−
(
n
−
m
)
)
a(m-n)=d((n-m)(n+m)-(n-m)) a(m-n) = d((n-m)(n+m) – (n-m)) a ( m − n ) = d ( ( n − m ) ( n + m ) − ( n − m ) ) Factor out
(
n
−
m
)
(
n
−
m
)
(n-m) (n-m) ( n − m ) :
a
(
m
−
n
)
=
d
(
n
−
m
)
(
(
n
+
m
)
−
1
)
a
(
m
−
n
)
=
d
(
n
−
m
)
(
(
n
+
m
)
−
1
)
a(m-n)=d(n-m)((n+m)-1) a(m-n) = d(n-m)((n+m) – 1) a ( m − n ) = d ( n − m ) ( ( n + m ) − 1 ) Thus, we have:
a
(
m
−
n
)
=
d
(
n
−
m
)
(
n
+
m
−
1
)
a
(
m
−
n
)
=
d
(
n
−
m
)
(
n
+
m
−
1
)
a(m-n)=d(n-m)(n+m-1) a(m-n) = d(n-m)(n+m-1) a ( m − n ) = d ( n − m ) ( n + m − 1 ) Step 4: Solve for the relationship
Divide both sides by
(
m
−
n
)
(
m
−
n
)
(m-n) (m-n) ( m − n ) (assuming
m
≠
n
m
≠
n
m!=n m \neq n m ≠ n ):
a
=
d
(
n
+
m
−
1
)
a
=
d
(
n
+
m
−
1
)
a=d(n+m-1) a = d(n+m-1) a = d ( n + m − 1 ) Step 5: Find the
(
m
+
n
)
th
(
m
+
n
)
th
(m+n)^(“th”) (m+n)^{\text{th}} ( m + n ) th
Now, let’s find the
(
m
+
n
)
th
(
m
+
n
)
th
(m+n)^(“th”) (m+n)^{\text{th}} ( m + n ) th term of the A.P., which is:
T
m
+
n
=
a
+
(
m
+
n
−
1
)
d
T
m
+
n
=
a
+
(
m
+
n
−
1
)
d
T_(m+n)=a+(m+n-1)d T_{m+n} = a + (m+n-1)d T m + n = a + ( m + n − 1 ) d Substitute
a
=
d
(
n
+
m
−
1
)
a
=
d
(
n
+
m
−
1
)
a=d(n+m-1) a = d(n+m-1) a = d ( n + m − 1 ) into the equation:
T
m
+
n
=
d
(
n
+
m
−
1
)
+
(
m
+
n
−
1
)
d
T
m
+
n
=
d
(
n
+
m
−
1
)
+
(
m
+
n
−
1
)
d
T_(m+n)=d(n+m-1)+(m+n-1)d T_{m+n} = d(n+m-1) + (m+n-1)d T m + n = d ( n + m − 1 ) + ( m + n − 1 ) d Simplify:
T
m
+
n
=
d
(
n
+
m
−
1
)
+
d
(
m
+
n
−
1
)
T
m
+
n
=
d
(
n
+
m
−
1
)
+
d
(
m
+
n
−
1
)
T_(m+n)=d(n+m-1)+d(m+n-1) T_{m+n} = d(n+m-1) + d(m+n-1) T m + n = d ( n + m − 1 ) + d ( m + n − 1 )
T
m
+
n
=
2
d
(
n
+
m
−
1
)
T
m
+
n
=
2
d
(
n
+
m
−
1
)
T_(m+n)=2d(n+m-1) T_{m+n} = 2d(n+m-1) T m + n = 2 d ( n + m − 1 ) Since
a
=
d
(
n
+
m
−
1
)
a
=
d
(
n
+
m
−
1
)
a=d(n+m-1) a = d(n+m-1) a = d ( n + m − 1 ) , this implies that
T
m
+
n
=
0
T
m
+
n
=
0
T_(m+n)=0 T_{m+n} = 0 T m + n = 0 .
Conclusion:
We have shown that the
(
m
+
n
)
th
(
m
+
n
)
th
(m+n)^(“th”) (m+n)^{\text{th}} ( m + n ) th term of the A.P. is zero.
T
m
+
n
=
0
T
m
+
n
=
0
T_(m+n)=0 \boxed{T_{m+n} = 0} T m + n = 0 .
Question:-14
Show that:
i)
lim
n
→
0
|
x
|
x
lim
n
→
0
|
x
|
x
lim_(n rarr0)(|x|)/(x) \lim_{n \to 0} \frac{|x|}{x} lim n → 0 | x | x does not exist
ii)
f
(
x
)
=
|
x
|
f
(
x
)
=
|
x
|
f(x)=|x| \mathrm{f}(x) = |x| f ( x ) = | x | is continuous at
x
=
0
x
=
0
x=0 x = 0 x = 0 .
Answer:
Let’s analyze both statements one by one:
i)
lim
x
→
0
|
x
|
x
lim
x
→
0
|
x
|
x
lim_(x rarr0)(|x|)/(x) \lim_{x \to 0} \frac{|x|}{x} lim x → 0 | x | x
The expression
|
x
|
x
|
x
|
x
(|x|)/(x) \frac{|x|}{x} | x | x is defined as:
|
x
|
x
=
{
1
,
if
x
>
0
,
−
1
,
if
x
<
0.
|
x
|
x
=
1
,
if
x
>
0
,
−
1
,
if
x
<
0.
(|x|)/(x)={[1″,”,”if “x > 0″,”],[-1″,”,”if “x < 0.]:} \frac{|x|}{x} = \begin{cases}
1, & \text{if } x > 0, \\
-1, & \text{if } x < 0.
\end{cases} | x | x = { 1 , if x > 0 , − 1 , if x < 0. We now check the limit from both sides as
x
→
0
x
→
0
x rarr0 x \to 0 x → 0 :
As
x
→
0
+
x
→
0
+
x rarr0^(+) x \to 0^+ x → 0 + (approaching 0 from the right or positive side):
lim
x
→
0
+
|
x
|
x
=
lim
x
→
0
+
x
x
=
1
lim
x
→
0
+
|
x
|
x
=
lim
x
→
0
+
x
x
=
1
lim_(x rarr0^(+))(|x|)/(x)=lim_(x rarr0^(+))(x)/(x)=1 \lim_{x \to 0^+} \frac{|x|}{x} = \lim_{x \to 0^+} \frac{x}{x} = 1 lim x → 0 + | x | x = lim x → 0 + x x = 1
As
x
→
0
−
x
→
0
−
x rarr0^(-) x \to 0^- x → 0 − (approaching 0 from the left or negative side):
lim
x
→
0
−
|
x
|
x
=
lim
x
→
0
−
−
x
x
=
−
1
lim
x
→
0
−
|
x
|
x
=
lim
x
→
0
−
−
x
x
=
−
1
lim_(x rarr0^(-))(|x|)/(x)=lim_(x rarr0^(-))(-x)/(x)=-1 \lim_{x \to 0^-} \frac{|x|}{x} = \lim_{x \to 0^-} \frac{-x}{x} = -1 lim x → 0 − | x | x = lim x → 0 − − x x = − 1 Since the left-hand limit (
−
1
−
1
-1 -1 − 1 ) and the right-hand limit (
1
1
1 1 1 ) are not equal, the limit does not exist. Hence:
lim
x
→
0
|
x
|
x
does not exist.
lim
x
→
0
|
x
|
x
does not exist.
lim_(x rarr0)(|x|)/(x)” does not exist.” \lim_{x \to 0} \frac{|x|}{x} \text{ does not exist.} lim x → 0 | x | x does not exist. Conclusion for part i: true because the limit does not exist due to different values from the left and right sides of 0.
ii)
f
(
x
)
=
|
x
|
f
(
x
)
=
|
x
|
f(x)=|x| \mathrm{f}(x) = |x| f ( x ) = | x |
x
=
0
x
=
0
x=0 x = 0 x = 0
To check whether the function
f
(
x
)
=
|
x
|
f
(
x
)
=
|
x
|
f(x)=|x| f(x) = |x| f ( x ) = | x | is continuous at
x
=
0
x
=
0
x=0 x = 0 x = 0 , we need to verify the following three conditions for continuity at
x
=
0
x
=
0
x=0 x = 0 x = 0 :
f
(
0
)
f
(
0
)
f(0) f(0) f ( 0 )
lim
x
→
0
f
(
x
)
lim
x
→
0
f
(
x
)
lim_(x rarr0)f(x) \lim_{x \to 0} f(x) lim x → 0 f ( x )
lim
x
→
0
f
(
x
)
=
f
(
0
)
lim
x
→
0
f
(
x
)
=
f
(
0
)
lim_(x rarr0)f(x)=f(0) \lim_{x \to 0} f(x) = f(0) lim x → 0 f ( x ) = f ( 0 )
Let’s check each condition:
Since all three conditions for continuity are satisfied, the function
f
(
x
)
=
|
x
|
f
(
x
)
=
|
x
|
f(x)=|x| f(x) = |x| f ( x ) = | x | is continuous at
x
=
0
x
=
0
x=0 x = 0 x = 0 .
Conclusion for part ii:
The statement is
true because
f
(
x
)
=
|
x
|
f
(
x
)
=
|
x
|
f(x)=|x| f(x) = |x| f ( x ) = | x | is continuous at
x
=
0
x
=
0
x=0 x = 0 x = 0 .
Question:-15
Suriti wants to invest at most 12000 in saving certificates and National Saving Bonds. She has to invest at least 2000 in Saving certificates and at least 4000 in National Saving Bonds. If Rate of Interest on saving certificates is
8
%
8
%
8% 8\% 8 %
10
%
10
%
10% 10\% 10 %
Answer:
We are tasked with maximizing the yearly income Suriti can earn by investing in saving certificates and National Saving Bonds, subject to certain constraints.
Let:
x
x
x x x
y
y
y y y
Given Constraints:
The total investment cannot exceed 12,000:
x
+
y
≤
12000
x
+
y
≤
12000
x+y <= 12000 x + y \leq 12000 x + y ≤ 12000
The amount invested in saving certificates must be at least 2,000:
x
≥
2000
x
≥
2000
x >= 2000 x \geq 2000 x ≥ 2000
The amount invested in National Saving Bonds must be at least 4,000:
y
≥
4000
y
≥
4000
y >= 4000 y \geq 4000 y ≥ 4000
Objective:
The yearly income is determined by the interest earned from both investments. The interest from:
Saving certificates is
8
%
8
%
8% 8\% 8 %
National Saving Bonds is
10
%
10
%
10% 10\% 10 %
The total yearly income
I
(
x
,
y
)
I
(
x
,
y
)
I(x,y) I(x, y) I ( x , y ) is:
I
(
x
,
y
)
=
0.08
x
+
0.10
y
I
(
x
,
y
)
=
0.08
x
+
0.10
y
I(x,y)=0.08 x+0.10 y I(x, y) = 0.08x + 0.10y I ( x , y ) = 0.08 x + 0.10 y We aim to maximize
I
(
x
,
y
)
I
(
x
,
y
)
I(x,y) I(x, y) I ( x , y ) , subject to the constraints.
We need to maximize:
I
(
x
,
y
)
=
0.08
x
+
0.10
y
I
(
x
,
y
)
=
0.08
x
+
0.10
y
I(x,y)=0.08 x+0.10 y I(x, y) = 0.08x + 0.10y I ( x , y ) = 0.08 x + 0.10 y Subject to the following constraints:
x
+
y
≤
12000
x
+
y
≤
12000
x+y <= 12000 x + y \leq 12000 x + y ≤ 12000
x
≥
2000
x
≥
2000
x >= 2000 x \geq 2000 x ≥ 2000
y
≥
4000
y
≥
4000
y >= 4000 y \geq 4000 y ≥ 4000 Step 2: Identify the feasible region
The constraints define a feasible region in the
x
x
x x x –
y
y
y y y plane. The region is bounded by:
x
+
y
≤
12000
x
+
y
≤
12000
x+y <= 12000 x + y \leq 12000 x + y ≤ 12000
x
≥
2000
x
≥
2000
x >= 2000 x \geq 2000 x ≥ 2000
y
≥
4000
y
≥
4000
y >= 4000 y \geq 4000 y ≥ 4000
Step 3: Solve for the vertices of the feasible region
To find the vertices of the feasible region, we solve for the intersection points of the boundary lines.
Intersection of
x
=
2000
x
=
2000
x=2000 x = 2000 x = 2000
x
+
y
=
12000
x
+
y
=
12000
x+y=12000 x + y = 12000 x + y = 12000 Substituting
x
=
2000
x
=
2000
x=2000 x = 2000 x = 2000 into
x
+
y
=
12000
x
+
y
=
12000
x+y=12000 x + y = 12000 x + y = 12000 :
2000
+
y
=
12000
⟹
y
=
10000
2000
+
y
=
12000
⟹
y
=
10000
2000+y=12000Longrightarrowy=10000 2000 + y = 12000 \implies y = 10000 2000 + y = 12000 ⟹ y = 10000 So, one vertex is
(
2000
,
10000
)
(
2000
,
10000
)
(2000,10000) (2000, 10000) ( 2000 , 10000 ) .
Intersection of
y
=
4000
y
=
4000
y=4000 y = 4000 y = 4000
x
+
y
=
12000
x
+
y
=
12000
x+y=12000 x + y = 12000 x + y = 12000 Substituting
y
=
4000
y
=
4000
y=4000 y = 4000 y = 4000 into
x
+
y
=
12000
x
+
y
=
12000
x+y=12000 x + y = 12000 x + y = 12000 :
x
+
4000
=
12000
⟹
x
=
8000
x
+
4000
=
12000
⟹
x
=
8000
x+4000=12000Longrightarrowx=8000 x + 4000 = 12000 \implies x = 8000 x + 4000 = 12000 ⟹ x = 8000 So, another vertex is
(
8000
,
4000
)
(
8000
,
4000
)
(8000,4000) (8000, 4000) ( 8000 , 4000 ) .
Intersection of
x
=
2000
x
=
2000
x=2000 x = 2000 x = 2000
y
=
4000
y
=
4000
y=4000 y = 4000 y = 4000 This gives the point
(
2000
,
4000
)
(
2000
,
4000
)
(2000,4000) (2000, 4000) ( 2000 , 4000 ) .
Step 4: Calculate the objective function at each vertex
Now, evaluate
I
(
x
,
y
)
=
0.08
x
+
0.10
y
I
(
x
,
y
)
=
0.08
x
+
0.10
y
I(x,y)=0.08 x+0.10 y I(x, y) = 0.08x + 0.10y I ( x , y ) = 0.08 x + 0.10 y at each of the vertices:
At
(
2000
,
10000
)
(
2000
,
10000
)
(2000,10000) (2000, 10000) ( 2000 , 10000 ) :
I
(
2000
,
10000
)
=
0.08
(
2000
)
+
0.10
(
10000
)
=
160
+
1000
=
1160
I
(
2000
,
10000
)
=
0.08
(
2000
)
+
0.10
(
10000
)
=
160
+
1000
=
1160
I(2000,10000)=0.08(2000)+0.10(10000)=160+1000=1160 I(2000, 10000) = 0.08(2000) + 0.10(10000) = 160 + 1000 = 1160 I ( 2000 , 10000 ) = 0.08 ( 2000 ) + 0.10 ( 10000 ) = 160 + 1000 = 1160
At
(
8000
,
4000
)
(
8000
,
4000
)
(8000,4000) (8000, 4000) ( 8000 , 4000 ) :
I
(
8000
,
4000
)
=
0.08
(
8000
)
+
0.10
(
4000
)
=
640
+
400
=
1040
I
(
8000
,
4000
)
=
0.08
(
8000
)
+
0.10
(
4000
)
=
640
+
400
=
1040
I(8000,4000)=0.08(8000)+0.10(4000)=640+400=1040 I(8000, 4000) = 0.08(8000) + 0.10(4000) = 640 + 400 = 1040 I ( 8000 , 4000 ) = 0.08 ( 8000 ) + 0.10 ( 4000 ) = 640 + 400 = 1040
At
(
2000
,
4000
)
(
2000
,
4000
)
(2000,4000) (2000, 4000) ( 2000 , 4000 ) :
I
(
2000
,
4000
)
=
0.08
(
2000
)
+
0.10
(
4000
)
=
160
+
400
=
560
I
(
2000
,
4000
)
=
0.08
(
2000
)
+
0.10
(
4000
)
=
160
+
400
=
560
I(2000,4000)=0.08(2000)+0.10(4000)=160+400=560 I(2000, 4000) = 0.08(2000) + 0.10(4000) = 160 + 400 = 560 I ( 2000 , 4000 ) = 0.08 ( 2000 ) + 0.10 ( 4000 ) = 160 + 400 = 560
Step 5: Conclusion
The maximum yearly income occurs at the point
(
2000
,
10000
)
(
2000
,
10000
)
(2000,10000) (2000, 10000) ( 2000 , 10000 ) , and the maximum yearly income is:
1160
1160
1160 \boxed{1160} 1160 Final Answer:
Suriti should invest $2000 in saving certificates and $10000 in National Saving Bonds to earn the maximum yearly income of $1160.
Question:-16
A spherical balloon is being inflated at the rate of
900
cm
3
/
sec
900
cm
3
/
sec
900cm^(3)//sec 900 \, \mathrm{cm}^3/\mathrm{sec} 900 cm 3 / sec
Answer:
We are given that the volume of the spherical balloon is increasing at the rate of
900
cm
3
/
sec
900
cm
3
/
sec
900cm^(3)//sec 900 \, \mathrm{cm}^3/\mathrm{sec} 900 cm 3 / sec , and we need to find how fast the radius of the balloon is increasing when the radius is 15 cm.
The volume
V
V
V V V of a sphere with radius
r
r
r r r is given by the formula:
V
=
4
3
π
r
3
V
=
4
3
π
r
3
V=(4)/(3)pir^(3) V = \frac{4}{3} \pi r^3 V = 4 3 π r 3 Step 2: Differentiate with respect to time
We are interested in how the radius changes over time, so we differentiate both sides of the volume equation with respect to time
t
t
t t t :
d
V
d
t
=
4
π
r
2
d
r
d
t
d
V
d
t
=
4
π
r
2
d
r
d
t
(dV)/(dt)=4pir^(2)(dr)/(dt) \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} d V d t = 4 π r 2 d r d t Where:
d
V
d
t
d
V
d
t
(dV)/(dt) \frac{dV}{dt} d V d t
900
cm
3
/
sec
900
cm
3
/
sec
900cm^(3)//sec 900 \, \mathrm{cm}^3/\mathrm{sec} 900 cm 3 / sec
d
r
d
t
d
r
d
t
(dr)/(dt) \frac{dr}{dt} d r d t
Step 3: Substitute the given values
We are given that:
d
V
d
t
=
900
cm
3
/
sec
d
V
d
t
=
900
cm
3
/
sec
(dV)/(dt)=900cm^(3)//sec \frac{dV}{dt} = 900 \, \mathrm{cm}^3/\mathrm{sec} d V d t = 900 cm 3 / sec
r
=
15
cm
r
=
15
cm
r=15cm r = 15 \, \mathrm{cm} r = 15 cm
Substitute these values into the differentiated equation:
900
=
4
π
(
15
)
2
d
r
d
t
900
=
4
π
(
15
)
2
d
r
d
t
900=4pi(15)^(2)(dr)/(dt) 900 = 4 \pi (15)^2 \frac{dr}{dt} 900 = 4 π ( 15 ) 2 d r d t Step 4: Solve for
d
r
d
t
d
r
d
t
(dr)/(dt) \frac{dr}{dt} d r d t
Simplify the equation:
900
=
4
π
⋅
225
⋅
d
r
d
t
900
=
4
π
⋅
225
⋅
d
r
d
t
900=4pi*225*(dr)/(dt) 900 = 4 \pi \cdot 225 \cdot \frac{dr}{dt} 900 = 4 π ⋅ 225 ⋅ d r d t
900
=
900
π
⋅
d
r
d
t
900
=
900
π
⋅
d
r
d
t
900=900 pi*(dr)/(dt) 900 = 900 \pi \cdot \frac{dr}{dt} 900 = 900 π ⋅ d r d t Now, solve for
d
r
d
t
d
r
d
t
(dr)/(dt) \frac{dr}{dt} d r d t :
d
r
d
t
=
900
900
π
=
1
π
cm
/
sec
d
r
d
t
=
900
900
π
=
1
π
cm
/
sec
(dr)/(dt)=(900)/(900 pi)=(1)/(pi)cm//sec \frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi} \, \mathrm{cm/sec} d r d t = 900 900 π = 1 π cm / sec Final Answer:
The radius of the balloon is increasing at a rate of
1
π
cm
/
sec
1
π
cm
/
sec
(1)/(pi)cm//sec \frac{1}{\pi} \, \mathrm{cm/sec} 1 π cm / sec when the radius is 15 cm.
