Solved Question Paper : BCA I Sem , Mathematics - I 2022
1. Test for consistency for the following system and find the solution, if it exists :
2x – 3y +7z = 5
3x + y – 3z = 13
2x + 19y – 47z = 32
Answer:
To test for consistency, we can put the augmented matrix in row echelon form (REF) and see if there are any rows of zeros in the coefficient matrix with a non-zero entry in the last column (the constants). If there are no such rows, the system is consistent (has a solution). If there is at least one such row, the system is inconsistent (has no solution).
The augmented matrix for the system is:
[2 -3 -7 | 5]
[3 1 -3 | 13]
[2 19 -47 | 32]
We can perform row operations to put this matrix in REF. The first step is to use the first row to eliminate the 3 in the (2,1) position:
[2 -3 -7 | 5]
[0 10 16 | -2]
[2 19 -47 | 32]
Next, we use the first row to eliminate the 2 in the (3,1) position:
[ 2 -3 -7 | 5 ]
[ 0 10 16 | -2 ]
[ 0 25 -33 | 22 ]
Finally, we use the second row to eliminate the 25 in the (3,2) position:
[ 2 -3 -7 | 5 ]
[ 0 10 16 | -2 ]
[ 0 0 -67 | 52 ]
This is now in row echelon form. There are no rows of zeros in the coefficient matrix with a non-zero entry in the last column, so the system is consistent.
To solve the system, we can work from the bottom row up. The last equation gives us:
-67z = 52
Solving for z, we get:
z = -52/67
Substituting this into the second equation, we get:
10y + 16(-52/67) = -2
Solving for y, we get:
y = -318/335
Substituting z and y into the first equation, we get:
2x - 3(-318/335) - 7(-52/67) = 5
Solving for x, we get:
x = -2201/670
Therefore, the solution to the system is:
x = -2201/670, y = -318/335, z = -52/67
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2. Verify Cayley-Hamilton theorem for the matrix
2 -1 1
A = -1 2 -1
1 -1 2
Answer:
The Cayley-Hamilton theorem states that every square matrix satisfies its own characteristic equation. The characteristic equation of a matrix A is given by:
det(A - λI) = 0
Where I is the identity matrix of the same size as A, and det denotes the determinant.
To verify the Cayley-Hamilton theorem for the matrix A, we need to show that A satisfies its own characteristic equation. First, we find the characteristic equation of A:
det(A - λI) =
| 2-λ -1 1 |
| -1 2-λ -1 |
| 1 -1 2-λ| = (2-λ)[(2-λ)^2 - 1] - (-1)[-(2-λ) + 1] + (1)[-1 + (2-λ)(-1)]
= λ^3 - 6λ^2 + 11λ - 6
Now we need to show that A satisfies this equation. To do this, we can substitute A for λ in the characteristic equation and check that the result is the zero matrix.
A^3 - 6A^2 + 11A - 6I =
| 2 -1 1 |^3 - 6| 2 -1 1 |^2 + 11| 2 -1 1 | - 6| 1 0 0 |
| -1 2 -1 | | -1 2 -1 | | -1 2 -1 | | 0 1 0 |
| 1 -1 2 | | 1 -1 2 | | 1 -1 2 | | 0 0 1 |
Performing the matrix multiplications, we get:
| 16 -18 10 | - 6| 5 -3 1 | + 11| 2 -1 1 | - 6| 1 0 0 |
| -9 16 -9 | |-3 8 -3 | | -3 8 -3 | | 0 1 0 |
| 10 -18 16 | | 1 -3 2 | | 1 -3 2 | | 0 0 1 |
=
| 0 0 0 |
| 0 0 0 |
| 0 0 0 |
Since the result is the zero matrix, we have verified the Cayley-Hamilton theorem for A.
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3. Prove that
Lim x--->0. |x|/x
Does not exist
Answer :
To prove that the limit of |x|/x as x approaches 0 does not exist, we will show that there are two different limits that the function can approach as x approaches 0, depending on whether x approaches 0 from the left or from the right.
Consider x approaching 0 from the right (i.e., x > 0). In this case, |x| is equal to x, so the function is simply equal to 1. Therefore, as x approaches 0 from the right, the limit of the function is equal to 1:
lim x → 0+ |x|/x = lim x → 0+ 1 = 1
Now consider x approaching 0 from the left (i.e., x < 0). In this case, |x| is equal to -x, so the function becomes:
|x|/x = (-x)/x = -1
Therefore, as x approaches 0 from the left, the limit of the function is equal to -1:
lim x → 0- |x|/x = lim x → 0- (-1) = 1
Since the limits from the left and right are different, the limit of the function does not exist.
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4. Prove that the function
F(x) = sin2x / x,x≠0
1, x=0
Has a removable discontinuity at the origin
Answer :
To prove that the function F(x) = sin2x / x, x≠0, 1, x=0 has a removable discontinuity at x=0, we need to show that F(x) is not continuous at x=0 but can be made continuous at this point by redefining the function.
First, let's consider the limit of F(x) as x approaches 0. We can use L'Hopital's rule to evaluate the limit:
lim x→0 sin2x / x = lim x→0 2cos2x / 1 = 2
Therefore, the limit of F(x) as x approaches 0 exists and is equal to 2.
Now, let's consider the value of F(x) at x=0. Since sin2(0) = 0 and 0/0 is undefined, we cannot directly evaluate F(0) using the given formula. However, we can redefine the function at x=0 to make it continuous.
Specifically, we can define F(x) = sin2x / x for all x≠0, and F(0) = 2. This new definition agrees with the original definition everywhere except at x=0, and it makes F(x) continuous at x=0.
Therefore, we have shown that the function F(x) has a removable discontinuity at x=0.
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5. Verify Rolle’s theorem for the function
F(x) = x3—4x in [-2,2]
Answer :
Verify Rolle’s theorem for the function
F(x) = x3—4x in [-2,2]
To verify Rolle's theorem for the function F(x) = x^3 - 4x in the interval [-2,2], we need to show that:
F(x) is continuous on [-2,2], which means that F(x) is defined and finite for all x in [-2,2].
F(x) is differentiable on (-2,2), which means that the derivative F'(x) exists and is finite for all x in (-2,2).
F(-2) = F(2), which means that the function takes the same value at the endpoints of the interval.
There exists at least one point c in (-2,2) such that F'(c) = 0, which means that the derivative of the function is zero at some point in the interval.
F(x) is a polynomial function, which is continuous for all real numbers. Therefore, F(x) is continuous on the closed interval [-2,2].
The derivative of F(x) is F'(x) = 3x^2 - 4, which is also a polynomial function and is defined and finite for all x in (-2,2). Therefore, F(x) is differentiable on the open interval (-2,2).
F(-2) = (-2)^3 - 4(-2) = -8 + 8 = 0
F(2) = (2)^3 - 4(2) = 8 - 8 = 0
Therefore, F(-2) = F(2) = 0.
To find a point c in (-2,2) such that F'(c) = 0, we can solve the equation F'(x) = 0:
3x^2 - 4 = 0
Solving for x, we get:
x = ±√(4/3)
Since √(4/3) is positive and lies in the interval (-2,2), we have found a point c in (-2,2) such that F'(c) = 0.
Therefore, all the conditions of Rolle's theorem are satisfied for the function F(x) = x^3 - 4x in the interval [-2,2]. This means that there exists at least one point c in (-2,2) such that F'(c) = 0, which is the conclusion of Rolle's theorem.
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6. Write the statement of Leibnitz’s theorem.
If y = A epx + Bqx then prove that
d2y/dx2 –( p + q ) dy/dx + pqy = 0
Answer :
Leibniz's theorem states that the nth derivative of the product of two functions u(x) and v(x) can be expressed in terms of the derivatives of u(x) and v(x) up to the nth order. Specifically, for any positive integer n:
(d^n/dx^n)(u(x)v(x)) = ∑(k=0 to n) (n choose k) * (d^k/dx^k)(u(x)) * (d^(n-k)/dx^(n-k))(v(x))
where (n choose k) denotes the binomial coefficient.
Using Leibniz's theorem, we can prove the given equation:
y = A e^(px) + B e^(qx)
Taking the first derivative with respect to x, we get:
dy/dx = Ap e^(px) + Bq e^(qx)
Taking the second derivative with respect to x, we get:
d^2y/dx^2 = Ap^2 e^(px) + Bq^2 e^(qx)
Now, we can substitute these expressions for y, dy/dx, and d^2y/dx^2 into the given equation and simplify:
d^2y/dx^2 - (p+q) dy/dx + pq y
= (Ap^2 e^(px) + Bq^2 e^(qx)) - (p+q)(Ap e^(px) + Bq e^(qx)) + pq (A e^(px) + B e^(qx))
= A(p^2 - p^2 - pq + pq) e^(px) + B(q^2 - q^2 - pq + pq) e^(qx)
= 0
Therefore, the given equation is satisfied by the function y = A e^(px) + B e^(qx), and we have used Leibniz's theorem to prove it.
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7. Evaluate
∫0a x2 (a2-x2)3/2 dx
Answer:
Let's start by making a substitution:
Let x = a sin(θ), then dx = a cos(θ) dθ
When x = 0, θ = 0
When x = a, θ = π/2
So, the integral becomes:
∫0a x^2 (a^2-x^2)^(3/2) dx = ∫0π/2 a^4 sin^4(θ) cos(θ) dθ
We can use integration by parts to solve this integral:
Let u = sin^4(θ), then du = 4 sin^3(θ) cos(θ) dθ
and dv = a^4 cos(θ) dθ, then v = a^4 sin(θ)
Using the formula for integration by parts, we have:
∫0π/2 a^4 sin^4(θ) cos(θ) dθ = [sin^4(θ) a^4 sin(θ)]0π/2 - ∫0π/2 4 sin^3(θ) cos^2(θ) a^4 sin(θ) dθ
The first term evaluates to 0 because sin(0) = sin(π/2) = 0.
Now, let's use a trigonometric identity to simplify the integrand:
4 sin^3(θ) cos^2(θ) = 4 sin^3(θ) (1 - sin^2(θ)) = 4 sin^3(θ) - 4 sin^5(θ)
So the integral becomes:
∫0π/2 a^4 sin^4(θ) cos(θ) dθ = 4a^4 ∫0π/2 sin^3(θ) dθ - 4a^4 ∫0π/2 sin^5(θ) dθ
The first integral can be solved by making the substitution u = cos(θ), then du = -sin(θ) dθ:
∫0π/2 sin^3(θ) dθ = -∫1 0 u^2 du = 1/3
The second integral can be solved using the reduction formula:
∫ sin^n(x) dx = -1/n cos(x) sin^(n-1)(x) + (n-1)/n ∫ sin^(n-2)(x) dx
Using this formula with n = 5 and integrating by parts, we get:
∫0π/2 sin^5(θ) dθ = -1/5 cos(θ) sin^4(θ) ∣0π/2 + 4/5 ∫0π/2 sin^3(θ) dθ
The first term evaluates to 0, and the second term was solved earlier.
So, putting it all together:
∫0a x^2 (a^2-x^2)^(3/2) dx = ∫0π/2 a^4 sin^4(θ) cos(θ) dθ = 4a^4 (1/3 - 4/5 (1/3)) = a^4 (8/15)
Therefore, the final answer is:
∫0a x^2 (a^2-x^2)^(3/2) dx = a^4 (8/15)
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8. If ā=2î+2ĵ-k, b=6i-3j+2k then find the value of the Unit Vector n and θ, where
θ is the angle between a and b .
Answer:
To find the unit vector n in the direction of the angle between vectors a and b, we can use the following steps:
Find the dot product of a and b.
Find the magnitude of a and b.
Use the formula cos(θ) = a•b / (|a| |b|) to find the cosine of the angle between a and b.
Use the inverse cosine function to find the angle θ.
Finally, divide vector a by its magnitude to find the unit vector in the direction of a.
Let's use these steps to solve the problem:
The dot product of a and b is:
a•b = (2)(6) + (2)(-3) + (-1)(2) = 12 - 6 - 2 = 4
The magnitudes of a and b are:
|a| = sqrt((2)^2 + (2)^2 + (-1)^2) = sqrt(9) = 3
|b| = sqrt((6)^2 + (-3)^2 + (2)^2) = sqrt(49) = 7
The cosine of the angle θ between a and b is:
cos(θ) = a•b / (|a| |b|) = 4 / (3 * 7) = 4 / 21
The angle θ between a and b is:
θ = acos(cos(θ)) = acos(4/21) ≈ 1.186 radians
The unit vector in the direction of a is:
n = a / |a| = (2î + 2ĵ - k) / 3
Therefore, the unit vector n in the direction of the angle between a and b is:
n = (2î + 2ĵ - k) / 3
and the angle θ between a and b is approximately 1.186 radians.
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