Numerical Analysis MCQ Quiz - Objective Question with Answer for Numerical Analysis - Download Free PDF
Last updated on Jul 7, 2025
Latest Numerical Analysis MCQ Objective Questions
Numerical Analysis Question 1:
Let f(x) be the polynomial of degree at most 2 that interpolates the data (-1, 2), (0, 1) and (1, 2).
If g(x) is a polynomial of degree at most 3 such that f(x) + g(x) interpolates the data (-1, 2), (0, 1), (1, 2) and (2, 17)
then
Answer (Detailed Solution Below)
Numerical Analysis Question 1 Detailed Solution
Concept:
Polynomial Interpolation:
- A unique polynomial of degree ≤ n passes through (n+1) points.
- We are given that \( f(x) \) is degree ≤ 2 interpolating 3 points.
- The polynomial \( f(x) + g(x) \) is of degree ≤ 3 interpolating 4 points.
- We can compute both \( f(x) \) and \( h(x) = f(x) + g(x) \) using interpolation and then evaluate required values.
Calculation:
Given,
\( f(x) \) interpolates points: (-1, 2), (0, 1), (1, 2)
\( f(x) = x^2 + 1 \Rightarrow f(3) = 10, \, f(5) = 26 \)
From interpolation of \( f(x) + g(x) \) through: (-1, 2), (0, 1), (1, 2), (2, 17)
We obtain a cubic polynomial \( h(x) \) such that:
\( h(3) = 58, \, h(5) = 266 \)
⇒ \( g(3) = h(3) - f(3) = 58 - 10 = 48 \)
⇒ \( g(5) = h(5) - f(5) = 266 - 26 = 240 \)
Now evaluate the options:
Option 1: \( f(3) + g(3) = 10 + 48 = 58 \) Incorrect
Option 2: \( 2f(5) - g(3) = 2×26 - 48 = 52 - 48 = 4 \) Correct
Option 3: \( f(1) + g(3) = 2 + 48 = 50 \) Correct
Option 4: \( f(5) + g(3) = 26 + 48 = 74 \) Correct
∴ The correct options are: Option 2 , Option 3 and Option 4.
Numerical Analysis Question 2:
If λ ∈ ℝ and p ∈ ℝ are such that the quadrature formula
\(\rm \int_{x_0}^{x_0+h}f(x)dx\approx\lambda h(f(x_0)+f(x_0+h)+ph^3(f''(x_0)+f''(x_0+h))\)
is exact for all polynomials of degree as high as possible, then
Answer (Detailed Solution Below)
Numerical Analysis Question 2 Detailed Solution
Concept:
Quadrature Formula and Degree of Exactness:
- A quadrature formula is a numerical method to approximate the definite integral of a function over an interval.
- The degree of exactness of a quadrature formula is the highest degree of polynomial for which the formula integrates exactly.
- Given formula is: \( \int_{x_0}^{x_0+h} f(x)dx \approx \lambda h \left(f(x_0) + f(x_0+h)\right) + p h^3 \left(f''(x_0) + f''(x_0+h)\right) \)
- To determine \( \lambda \) and \( p \), compare both sides using the Taylor series expansion of \( f(x) \) around \( x_0 \).
Calculation:
Given,
Exact integral of \( f(x) \):
\( \int_{x_0}^{x_0+h} \left[ f(x_0) + f'(x_0)(x - x_0) + \frac{f''(x_0)}{2}(x - x_0)^2 + \frac{f'''(x_0)}{6}(x - x_0)^3 \right] dx \)
⇒ \( f(x_0)h + \frac{f'(x_0)h^2}{2} + \frac{f''(x_0)h^3}{6} + \frac{f'''(x_0)h^4}{24} \)
Quadrature formula expansion using Taylor series:
\( \lambda h \left[ 2f(x_0) + h f'(x_0) + \frac{h^2 f''(x_0)}{2} + \frac{h^3 f'''(x_0)}{6} \right] + p h^3 \left[ 2 f''(x_0) + h f'''(x_0) \right] \)
Compare coefficients of each term:
For \( f(x_0) \): \( 2 \lambda h = h \) ⇒ \( \lambda = \frac{1}{2} \)
For \( f'(x_0) \): \( \lambda h^2 = \frac{h^2}{2} \) ⇒ Verified for \( \lambda = \frac{1}{2} \)
For \( f''(x_0) \): \( \lambda \frac{h^3}{2} + 2 p h^3 = \frac{h^3}{6} \)
⇒ \( \frac{\lambda}{2} + 2p = \frac{1}{6} \)
For \( f'''(x_0) \): \( \lambda \frac{h^4}{6} + p h^4 = \frac{h^4}{24} \)
⇒ \( \frac{\lambda}{6} + p = \frac{1}{24} \)
Solving:
From \( \frac{\lambda}{6} + p = \frac{1}{24} \):
⇒ \( p = \frac{1}{24} - \frac{\lambda}{6} \)
Substitute into \( \frac{\lambda}{2} + 2p = \frac{1}{6} \):
⇒ \( \frac{\lambda}{2} + 2 \left( \frac{1}{24} - \frac{\lambda}{6} \right) = \frac{1}{6} \)
⇒ \( \frac{\lambda}{2} + \frac{1}{12} - \frac{\lambda}{3} = \frac{1}{6} \)
⇒ \( \frac{\lambda}{6} + \frac{1}{12} = \frac{1}{6} \)
⇒ \( \frac{\lambda}{6} = \frac{1}{12} \)
⇒ \( \lambda = \frac{1}{2} \)
Substitute back: \( p = \frac{1}{24} - \frac{1}{12} = -\frac{1}{24} \)
Verification of Options:
Option 1: \( 2\lambda + 24p \):
⇒ \( 2 \times \frac{1}{2} + 24 \times \left(-\frac{1}{24}\right) = 1 - 1 = 0 \) Correct
Option 2: \( 7\lambda - 12p \):
⇒ \( 7 \times \frac{1}{2} - 12 \times \left(-\frac{1}{24}\right) = \frac{7}{2} + \frac{1}{2} = 4 \) Correct
Option 3: \( 2\lambda + 24p \):
Already verified as 0, not -3. Incorrect
Option 4: \( 7\lambda - 12p \):
Already verified as 4, not 11. Incorrect
∴ The correct answers are 1 and 2.
Numerical Analysis Question 3:
Let f : ℝ → ℝ be such that \(\rm sup_{x\ne y}\frac{|fx)-f(y)|}{|x-y|}=L\). wjere 1 < L < ∞, Lt h : ℝ → ℝ be a differentiable function satisfying |h'(x) ≤ \(\frac{3}{4}\) for all x ∈ ℝ. For a > 0, define g(x) = af(x) + h(x) for x ∈ ℝ. Consider the sequence \(\rm \{x_k\}^\infty_{k=0}\) defined by
xk+1 = g(xk), k = 0, 1...
where x0 ∈ ℝ. The sequence \(\rm \{x_k\}^\infty_{k=0}\) converges to the solution of the equation = g(x) if
Answer (Detailed Solution Below)
Numerical Analysis Question 3 Detailed Solution
Concept:
Fixed Point Iteration Method:
- In numerical analysis, the fixed point iteration method is used to find a solution of the equation x = g(x).
- The method defines a sequence using the recursive formula: xk+1 = g(xk).
- The sequence converges to a fixed point if g(x) is a contraction mapping.
- Contraction Mapping: A function g(x) is a contraction if there exists a constant 0 < c < 1 such that |g(x) − g(y)| ≤ c × |x − y| for all x, y.
- This implies |g'(x)| < 1 for convergence.
- Given Functions:
- g(x) = αf(x) + h(x)
- With f(x) being Lipschitz with constant L i.e., |f'(x)| ≤ L
- And h(x) having derivative bound |h'(x)| ≤ 3/4
- Objective: Find the condition on α such that the iteration converges to the solution of x = g(x)
Calculation:
Given,
Lipschitz constant of f(x) = L
Bound on |h'(x)| = 3/4
Function: g(x) = αf(x) + h(x)
Derivative: g'(x) = αf'(x) + h'(x)
⇒ |g'(x)| ≤ |αf'(x)| + |h'(x)|
⇒ |g'(x)| ≤ αL + 3/4
⇒ For convergence, we need |g'(x)| < 1
⇒ αL + 3/4 < 1
⇒ αL < 1 − 3/4 = 1/4
⇒ α < 1/(4L)
∴ The sequence converges to the solution of x = g(x) if α < 1⁄(4L).
Numerical Analysis Question 4:
Let \({dy\over dx}={1\over x+y}\), x0 = 0, y(x0) = 1, h = 0.5. Using Runge-Kutta method the value of k1 and k2 are
Answer (Detailed Solution Below)
Numerical Analysis Question 4 Detailed Solution
Concept:
Let y' = f(x, y) then using Runge-Kutta method
k1 = hf(x0, y0) and
k2 = h\(f\left(x_0+\frac h2, y_0+\frac {k_1}2\right)\)
Explanation:
Given
\({dy\over dx}={1\over x+y}\), x0 = 0, y(x0) = 1, h = 0.5.
k1 = hf(x0, y0) = 0.5 × f(0, 1) = 0.5 × 1 = 0.5
k2 = h\(f\left(x_0+\frac h2, y_0+\frac {k_1}2\right)\)
= 0.5 \(f\left(0+0.25, 1+0.25\right)\)
= 0.5 × f(0.25, 1.25)
= 0.5 × \(1\over 1.50\) = 0.333
Option (2) and (3) are true.
Numerical Analysis Question 5:
Consider the Newton-Raphson method applied to approximate the square root of a positive number α. A recursion relation for the error en = xn - \(\alpha^{1/3}\) is given by
Answer (Detailed Solution Below)
Numerical Analysis Question 5 Detailed Solution
Concept:
Iterative formula for NR-method
\(x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime}\left(x_n\right)}\)
Explanation
Iterative formula for NR-method
\(x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime}\left(x_n\right)}\) --- (i)
Let f(x) = x3 - α
⇒ f'(x) = 3x2
So, putting in (i) we get
\(x_{n+1}=x_n-\frac{\left(x_n^3-\alpha\right)}{3x_n^2}\)
\(x_{n+1}=x_n-\frac{x_n}{3}+{\alpha\over 3x_n^2}\)
\(x_{n+1}=\frac23x_n+{\alpha\over 3x_n^2}\)
and given en = xn - \(\alpha^{1/3}\) ⇒ xn = en + \(\alpha^{1/3}\).
hence we get
\(e_{n+1}+\alpha^{1/3} =\frac23\left(e_{n}+\alpha^{1/3}\right)+\frac{α}{3\left(e_n+\alpha^{1/3}\right)^2}\)
\(e_{n+1}\approx {e_n^2\over e_n+2\alpha^{1/3}}\)
(2) is correct
Top Numerical Analysis MCQ Objective Questions
Which of the following values of a, b, c and d will produce a quadrature formula
\(\displaystyle\int_{−1}^1\) f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)
that has degree of precision 3?
Answer (Detailed Solution Below)
Numerical Analysis Question 6 Detailed Solution
Download Solution PDFExplanation:
Given
\(\displaystyle\int_{−1}^1\) f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)....(i)
Let f(x) = 1 then from (i)
\(\displaystyle\int_{−1}^1\)dx = a + b
⇒ 2 = a + b...(ii)
For f(x) = x ⇒ f'(x) = 1 so from (i)
\(\displaystyle\int_{−1}^1\)x dx = -a + b + c + d
0 = -a + b + c + d....(iii)
For f(x) = x2 ⇒ f'(x) = 2x so from (i)
\(\displaystyle\int_{−1}^1\)x2 dx = a + b - 2c + 2d
\(\frac23\) = a + b - 2c + 2d....(iv)
and for f(x) = x3 ⇒ f'(x) = 3x2 so from (i)
\(\displaystyle\int_{−1}^1\)x3 dx = -a + b + 3c + 3d
0 = -a + b + 3c + 3d....(v)
Multiply (iii) by 3 we get
-3a +3b + 3c + 3d = 0...(vi)
Subtract (v) and (vi) we get
2a - 2b = 0 ⇒ a = b...(vii)
Putting a = b in (ii)
2a = 2 ⇒ a = 1
Hence a = b = 1
Putting these values of a and b in (iii) and (iv) we get
c + d = 0 ...(viii) and
2 - 2c + 2d = \(\frac23\)
2c - 2d = \(\frac43\) ⇒ c - d = \(\frac23\) ...(ix)
adding (viii) and (ix) we get
2c = \(\frac23\) so c = \(\frac13\)
Hence d = - \(\frac13\)
Therefore we get a = 1, b = 1, c = \(\frac{1}{3}\), d = \(−\frac{1}{3}\)
Option (1) is correct
Let f be an infinitely differentiable real-valued function on a bounded interval I. Take n ≥ 1 interpolation points {x0, x1, ....., xn-1}. Take n additional interpolation points
xn+j = xj + ε, j = 0, 1, ....., n - 1
where ε > 0 is such that {x0, x1, ....., x2n-1} are all distinct.
Let p2n-1 be the Lagrange interpolation polynomial of degree 2n - 1 with the interpolation points {x0, x1, ....., x2n-1} for the function f.
Let q2n-1 be the Hermite interpolation polynomial of degree 2n - 1 with the interpolation points {x0, x1, ....., xn-1} for the function f. In the ε → 0 limit, the quantity
\(\sup _{x \in 1}\left|p_{2 n-1}(x)-q_{2 n-1}(x)\right|\)
Answer (Detailed Solution Below)
Numerical Analysis Question 7 Detailed Solution
Download Solution PDFThe quantity
\(\sup _{x \in 1}\left|p_{2 n-1}(x)-q_{2 n-1}(x)\right|\) converges to 0
Option (3) is correct
Consider the linear system 𝑀𝑥 = 𝑏, where 𝑀 = \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\) and b = \(\begin{bmatrix}-2\\\ 5\end{bmatrix}\).
Suppose 𝑀 = 𝐿𝑈, where 𝐿 and U are lower triangular and upper triangular square matrices, respectively. Consider the following statements:
𝑃: If each element of the main diagonal of 𝐿 is 1, then 𝑡𝑟𝑎𝑐𝑒(𝑈) = 3.
𝑄: For any choice of the initial vector 𝑥(0) , the Jacobi iterates 𝑥(𝑘) , 𝑘 = 1,2,3 … converge to the unique solution of the linear system 𝑀𝑥 = 𝑏.
Then
Answer (Detailed Solution Below)
Numerical Analysis Question 8 Detailed Solution
Download Solution PDFConcept:
A square matrix (aij) is called a diagonally dominant matrix if |aii| ≥ \(\sum_{j\neq i}|a_{ij}|\) for all i
Explanation:
For P,
M = \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\)
Let M = LU, where 𝐿 and U are lower triangular and upper triangular square matrices
Each element of the main diagonal of 𝐿 is 1
Let L = \(\begin{bmatrix}1&0\\a&1\end{bmatrix}\) and U = \(\begin{bmatrix}b&c\\0&d\end{bmatrix}\)
Then
\(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\) = \(\begin{bmatrix}1&0\\a&1\end{bmatrix}\)\(\begin{bmatrix}b&c\\0&d\end{bmatrix}\)
⇒ \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\) = \(\begin{bmatrix}b&c\\ab&ac+d\end{bmatrix}\)
Comparing both sides
b = 2, c = -1, ab = -4 and ac + d = 3
Substituting b = 2 in ab = -4 we get a = -2
Again substituting a = -2 and c = -1 in ac + d = 3 we get
(-2)(-1) + d = 3 ⇒ 2 + d = 3 ⇒ d = 1
So U = \(\begin{bmatrix}2&-1\\0&1\end{bmatrix}\)
Hence trace(U) = 1 + 2 = 3
P is TRUE
For Q,
M = \(\begin{bmatrix}2&-1\\\ -4&3\end{bmatrix}\)
M is not a diagonally dominant matrix as 3 \(\ngeq\) |-4|
Then HJacobi = D-1(L + U) where
D is diagonal matric ie..e, \(\begin{bmatrix}2&0\\\ 0&3\end{bmatrix}\) and L + U = \(\begin{bmatrix}0&-1\\\ -4&0\end{bmatrix}\)
So, D-1 = \(\begin{bmatrix}\frac12&0\\\ 0&\frac13\end{bmatrix}\)
So HJacobi = \(\begin{bmatrix}\frac12&0\\\ 0&\frac13\end{bmatrix}\)\(\begin{bmatrix}0&-1\\\ -4&0\end{bmatrix}\) = \(\begin{bmatrix}0&-\frac12\\\ -\frac43&0\end{bmatrix}\)
Hence eigenvalues re given by
λ2 - 0λ - 2/3 = 0
⇒ λ = \(\pm\sqrt{\frac23}\)
Since |λ| < 1
Hence for any choice of the initial vector 𝑥(0) , the Jacobi iterates 𝑥(𝑘) , 𝑘 = 1,2,3 … converge to the unique solution of the linear system 𝑀𝑥 = 𝑏.
Q is TRUE
both 𝑃 and 𝑄 are TRUE
(1) is correct
Numerical Analysis Question 9:
Using Euler’s method with the step size 0.05, the approximate value of the solution for the initial value problem
\(\frac{d y}{d x}=\sqrt{3 x+2 y+1}, \quad y(1)=1 \text {, }\)
at x = 1.1 (rounded off to two decimal places), is
Answer (Detailed Solution Below)
Numerical Analysis Question 9 Detailed Solution
Concept:
Euler method to solve y' = f(x, y), y(x0) = y0, with step size h is
yn+1 = yn + h.[f(xn, yn)]
Explanation:
\(\frac{d y}{d x}=\sqrt{3 x+2 y+1}, \quad y(1)=1 \text {, }\) h = 0.05,
x0 = 1, y0 = 1, h = 0.05
x1 = x0 + h = 1 + 0.05 = 1.05
x2 = x0 + 2h = 1 + 2(0.05) = 1 + 0.1 = 1.1
So we need to find y(x2) = y2
y1 = y0 + h. f(x0, y0) = 1 + 0.05 × f(1, 1) = 1.12247
y2 = y1 + h. f(x1, y1) = 1.12247 + 0.05 × f(1.05, 1.12247) = 1.248 ≈ 1.25 rounded off to two decimal places
Hence option (3) is correct
Numerical Analysis Question 10:
Consider the Newton-Raphson method applied to approximate the square root of a positive number α. A recursion relation for the error en = xn -√α is given by
Answer (Detailed Solution Below)
Numerical Analysis Question 10 Detailed Solution
Explanation
Recall: Iterative formula for NR-method
\(x_{n+1}=x_n-\frac{f\left(x_n\right)}{f^{\prime}\left(x_n\right)}\) --- (i)
Nos let x = √α ⇒ x2 - α = 0
so assume that f(x) = x2 - α
⇒ f'(x) = 2x
and given en = xn - √α ⇒xn = en + √α
so, by (i)
\(e_{n+1}+√{α} =\left(e_{n+1}+√{α}\right)-\frac{\left(e_n+√{α}\right)^2-α}{2\left(e_n+√{α}\right)}\)
\(\Rightarrow e_{n+1}+√{α} =\frac{2\left[e_n^2+α+2 e_n √{α}\right]-\left(e_n+√{α}\right)^2+α}{2\left(e_{n+}+√{α}\right)}\)
\(e_{n+1} =\frac{e_n^2+α+2 e_n \sqrtα+α}{2\left(e_n+√{α}\right)}-√{α}\)
\(=\frac{e_n^2+α+2 e_n \sqrt{α}+α-2 e_n \sqrt{α}-2 α}{2\left(e_n+\sqrt{α}\right)}\)
\(=e_n^2 / 2\left(e_n+\sqrt{α}\right)\)
(3) is correct
Numerical Analysis Question 11:
Let \(a,b,c\in \mathbb{R}\) be such that the quadrature rule \(\int_{ - 1}^1 {f(x)dx = af( - 1) + bf'(0) + cf'(1)}\) is exact for all polynomials of degree less than or equal to 2. Then a + b + c equal to
Answer (Detailed Solution Below)
Numerical Analysis Question 11 Detailed Solution
Concept:
Basic ideas of polynomial .
Calculation:
Given
\(\int_{ - 1}^1 f(x)dx\) = af(-1) + bf'(0) + cf'(1) . . . . . . (i)
is exact for all polynomials of degree less than or equal to 2
let f(x) = 1
So from equation (i), we get
\(\int_{ - 1}^1 1 dx \) = a + 0 + 0
2 = a . . . . . . . (ii)
let f(x) = x
now from equation (i) , we get
\(\int_{ - 1}^1 xdx\) = -a + b + c
b + c = 2 . . . . . . (iii)
now adding equation (ii) and (iii) , we get
a + b + c = 2 + 2
a + b + c = 4
Hence option (1) correct
Numerical Analysis Question 12:
Let \({dy\over dx}={1\over x+y}\), x0 = 0, y(x0) = 1, h = 0.5. Using Runge-Kutta method the value of k1 and k2 are
Answer (Detailed Solution Below)
Numerical Analysis Question 12 Detailed Solution
Concept:
Let y' = f(x, y) then using Runge-Kutta method
k1 = hf(x0, y0) and
k2 = h\(f\left(x_0+\frac h2, y_0+\frac {k_1}2\right)\)
Explanation:
Given
\({dy\over dx}={1\over x+y}\), x0 = 0, y(x0) = 1, h = 0.5.
k1 = hf(x0, y0) = 0.5 × f(0, 1) = 0.5 × 1 = 0.5
k2 = h\(f\left(x_0+\frac h2, y_0+\frac {k_1}2\right)\)
= 0.5 \(f\left(0+0.25, 1+0.25\right)\)
= 0.5 × f(0.25, 1.25)
= 0.5 × \(1\over 1.50\) = 0.333
Option (2) and (3) are true.
Numerical Analysis Question 13:
Let S denote the set of all 2 × 2 matrices A such that the iterative sequence generated by the Gauss-Seidel method applied to the system of linear equations \(\rm A\begin{pmatrix}x_1\\\ x_2\end{pmatrix}=\rm \begin{pmatrix}2\\\ 3\end{pmatrix}\) converges for every initial guess. Then which of the following statements are true?
Answer (Detailed Solution Below)
Numerical Analysis Question 13 Detailed Solution
We will update the solution later.
Numerical Analysis Question 14:
Let 𝑝(𝑥) = 𝑥3 − 2𝑥 + 2. If 𝑞(𝑥) is the interpolating polynomial of degree less than or equal to 4 for the data
x | -2 | -1 | 0 | 1 | 3 |
q(x) | p(-2) | p(-1) | 2.5 | p(1) | p(3) |
then the value of \(\frac{d^4q}{dx^4}\) at 𝑥 = 0 is ______.
Answer (Detailed Solution Below) 2
Numerical Analysis Question 14 Detailed Solution
Concept:
Explanation:
𝑝(𝑥) = 𝑥3 − 2𝑥 + 2
Given
x | -2 | -1 | 0 | 1 | 3 |
q(x) | p(-2) | p(-1) | 2.5 | p(1) | p(3) |
Now, p(-2) = -8 + 4 + 2 = -2; p(-1) = -1 + 2 + 2 = 3
p(1) = 1 - 2 + 2 = 1, p(3) = 27 - 6 + 2 = 23
Then the given data becomes
x | -2 | -1 | 0 | 1 | 3 |
q(x) | -2 | 3 | 2.5 | 1 | 23 |
Then using Lagrange interpolating polynomial
q(x) = \(\frac{(x+1)(x-0)(x-1)(x-3)}{(-2+1)(-2-0)(-2-1)(-2-3)}\) × (-2) + \(\frac{(x+2)(x-0)(x-1)(x-3)}{(-1+2)(-1-0)(-1-1)(-1-3)}\) × 3 + \(\frac{(x+2)(x+1)(x-1)(x-3)}{(0+2)(0+1)(0-1)(0-3)}\) × 2.5 + \(\frac{(x+2)(x+1)(x-0)(x-3)}{(1+2)(1+1)(1-0)(0-3)}\) × 1 + \(\frac{(x+2)(x+1)(x-0)(x-1)}{(3+2)(3+1)(3-0)(3-1)}\) × 23
Now, \(\frac{d^4q}{dx^4}\) at x = 0, will be 4! × coefficient of x4
So, \(\frac{d^4q}{dx^4}\) = 24(\({-2\over30}+{-3\over8}+{5\over12}+{1\over-12}+{23\over120}\))
= - \(\frac85\) - 9 + 10 - 2 + \(\frac{23}5\) = 2
Hence answer is 2
Numerical Analysis Question 15:
Which of the following values of a, b, c and d will produce a quadrature formula
\(\displaystyle\int_{−1}^1\) f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)
that has degree of precision 3?
Answer (Detailed Solution Below)
Numerical Analysis Question 15 Detailed Solution
Explanation:
Given
\(\displaystyle\int_{−1}^1\) f(x) dx ≈ af(−1) + bf(1) + cf'(−1) + df'(1)....(i)
Let f(x) = 1 then from (i)
\(\displaystyle\int_{−1}^1\)dx = a + b
⇒ 2 = a + b...(ii)
For f(x) = x ⇒ f'(x) = 1 so from (i)
\(\displaystyle\int_{−1}^1\)x dx = -a + b + c + d
0 = -a + b + c + d....(iii)
For f(x) = x2 ⇒ f'(x) = 2x so from (i)
\(\displaystyle\int_{−1}^1\)x2 dx = a + b - 2c + 2d
\(\frac23\) = a + b - 2c + 2d....(iv)
and for f(x) = x3 ⇒ f'(x) = 3x2 so from (i)
\(\displaystyle\int_{−1}^1\)x3 dx = -a + b + 3c + 3d
0 = -a + b + 3c + 3d....(v)
Multiply (iii) by 3 we get
-3a +3b + 3c + 3d = 0...(vi)
Subtract (v) and (vi) we get
2a - 2b = 0 ⇒ a = b...(vii)
Putting a = b in (ii)
2a = 2 ⇒ a = 1
Hence a = b = 1
Putting these values of a and b in (iii) and (iv) we get
c + d = 0 ...(viii) and
2 - 2c + 2d = \(\frac23\)
2c - 2d = \(\frac43\) ⇒ c - d = \(\frac23\) ...(ix)
adding (viii) and (ix) we get
2c = \(\frac23\) so c = \(\frac13\)
Hence d = - \(\frac13\)
Therefore we get a = 1, b = 1, c = \(\frac{1}{3}\), d = \(−\frac{1}{3}\)
Option (1) is correct