Your Answer : (Not Answered)

Correct Answer : A

The area is 11.5 sq. unit.

The table gives 9 equally spaced ordinates with \(h = 1\), that is 8 intervals, an even number, so Simpson's one-third rule applies:

\[ \begin{aligned} A &= \tfrac{h}{3}[(y_{1} + y_{9}) + 4(y_{2} + y_{4} + y_{6} + y_{8}) \\ &\quad + 2(y_{3} + y_{5} + y_{7})] \\ &= \tfrac{1}{3}[2.5 + 4(5.8) + 2(4.4)] \\ &= \tfrac{1}{3}(34.5) = 11.5 \end{aligned} \]

The trapezoidal rule gives \(1.25 + 10.2 = 11.45\), which also rounds to 11.5, so either standard rule leads to option A. 12.0, 10.5 and 10.2 are not produced by any rule; 10.2 is just the sum of the interior ordinates.

Your Answer : (Not Answered)

Correct Answer : D

\(\Delta^{5}u_{0} =\) -61.

Expand \(\Delta^{5} = (E - 1)^{5}\) with binomial coefficients:

\[ \begin{aligned} \Delta^{5}u_{0} &= u_{5} - 5u_{4} + 10u_{3} \\ &\quad - 10u_{2} + 5u_{1} - u_{0} \\ &= 0 - 35 + 30 - 80 + 25 - 1 \\ &= -61 \end{aligned} \]

A difference table gives the same result. Option B, \(-60\), is what you get by dropping the final \(-u_{0}\), and 60 and 24 come from sign errors in the alternating sum.

Your Answer : (Not Answered)

Correct Answer : D

The value is 2.

The shift operator acts by \(E f(x) = f(x + h)\), so \(E^{-1}f(x) = f(x - 1)\) with \(h = 1\), and \(\Delta\) commutes with E. Hence

\[ \frac{\Delta^{2}}{E}x^{2} = \Delta^{2}(x - 1)^{2} \]

The second forward difference of any quadratic \(ax^{2} + bx + c\) with step 1 is the constant \(2a\). Here \(a = 1\), so the result is 2, whatever the shift. Directly: \(\Delta x^{2} = 2x + 1\) and \(\Delta(2x + 1) = 2\).

0 would be the third difference of a quadratic, and \(\pm 1\) come from forgetting the factor 2.

Your Answer : (Not Answered)

Correct Answer : D

The exact rule is the composite trapezoidal rule by dividing [-1, 2] into 3 equal subintervals.

\(f(x) = |x| + |x - 1| + |x - 2|\) is piecewise linear on [-1, 2], with corners only at \(x = 0\) and \(x = 1\). Dividing [-1, 2] into three equal parts puts the nodes at -1, 0, 1, 2, so on each subinterval \(f\) is linear and the trapezoidal rule is exact.

Check: \(f(-1) = 6, f(0) = 3, f(1) = 2, f(2) = 3\), so the composite rule gives \(\frac{1}{2}(6 + 3) + \frac{1}{2}(3 + 2) + \frac{1}{2}(2 + 3) = 9.5\), the exact integral. The single trapezoid and single Simpson's rule ignore the corners, and the four-subinterval Simpson's rule places nodes at -0.25 and 1.25 with panels straddling the corners, where a quadratic cannot match a kink.

Your Answer : (Not Answered)

Correct Answer : B

Both P and Q are TRUE.

Write \(L = \begin{bmatrix} 1 & 0 \\ l & 1 \end{bmatrix}\) and \(U = \begin{bmatrix} u_{11} & u_{12} \\ 0 & u_{22} \end{bmatrix}\), so

\[ LU = \begin{bmatrix} u_{11} & u_{12} \\ l u_{11} & l u_{12} + u_{22} \end{bmatrix} \]

For P, matching with \(\begin{bmatrix} 0 & 5 \\ 0 & 7 \end{bmatrix}\) gives \(u_{11} = 0\), \(u_{12} = 5\), \(l \cdot 0 = 0\) and \(5l + u_{22} = 7\). The value of \(l\) is free and \(u_{22} = 7 - 5l\), so there are infinitely many factorizations.

For Q, matching with \(\begin{bmatrix} 0 & 0 \\ 2 & 5 \end{bmatrix}\) gives \(u_{11} = 0\) and then \(l u_{11} = 2\), that is \(0 = 2\): no factorization exists. Both statements hold, so every option calling one of them FALSE is wrong.

Your Answer : (Not Answered)

Correct Answer : A

S1 is TRUE and S2 is FALSE.

A fixed-point scheme \(x_{k+1} = g(x_{k})\) converges from every start in an interval if g maps the interval into itself and \(|g'| \lt 1\) there.

S1: \(g(x) = \sqrt{10/(4 + x)}\) is decreasing, with \(g(1) = \sqrt{2} \approx 1.414\) and \(g(1.5) \approx 1.348\), so it maps the interval into itself; and \(|g'(x)| = \sqrt{10}/(2(4 + x)^{3/2}) \le 0.15\). So it converges: S1 is true.

S2: \(g(x) = \tfrac{1}{2}\sqrt{10 - x^{3}}\) is also decreasing, with \(g(1) = 1.5\) and \(g(1.5) \approx 1.287\), so it maps (1, 3/2) into itself, and \(|g'(x)| = 3x^{2}/(4\sqrt{10 - x^{3}})\) is at most about 0.66 there. It converges for every start in the interval, so it never diverges and S2 is false.

Your Answer : (Not Answered)

Correct Answer : B

The sequence converges to \(\sqrt{2}\) whenever \(x_{0} \gt \sqrt{2/3}\).

The scheme is Newton's method for \(x^{2} - 2 = 0\): \(x - \frac{x^{2}-2}{2x} = \frac{x}{2} + \frac{1}{x}\). For any \(x_{0} \gt 0\), the AM-GM inequality gives

\[ \begin{aligned} x_{1} &= \frac{x_{0}}{2} + \frac{1}{x_{0}} \\ &\geq 2\sqrt{\frac{1}{2}} = \sqrt{2} \end{aligned} \]

From then on the terms stay at or above \(\sqrt{2}\) and decrease, so they converge to the positive fixed point \(\sqrt{2}\). Every \(x_{0} \gt \sqrt{2/3}\) is positive, so the statement holds.

The map is odd, so a negative start converges to \(-\sqrt{2}\). That makes 'for all \(x_{0} \neq 0\)' false, and also the (-1, 1) option, which includes negative starts. The sequence does not diverge for any non-zero start.

Your Answer : (Not Answered)

Correct Answer : A

The order of convergence is \(\frac{1 + \sqrt{5}}{2}\), about 1.618.

For the secant method near a simple root, the errors satisfy \(e_{n+1} \approx C e_{n} e_{n-1}\). Seeking \(|e_{n+1}| \approx A|e_{n}|^{p}\) and substituting gives \(p^{2} = p + 1\), whose positive root is the golden ratio:

\[ p = \frac{1 + \sqrt{5}}{2} \approx 1.618 \]

So the secant method is superlinear but slower than Newton's quadratic method. \(\frac{2}{1 + \sqrt{5}} \approx 0.618\) is the reciprocal, less than 1, which is not an order of convergence; \(\frac{1+\sqrt{5}}{3} \approx 1.08\) and \(\frac{3}{1+\sqrt{5}} \approx 0.93\) do not solve \(p^{2} = p + 1\).

Your Answer : (Not Answered)

Correct Answer : C

Both S1 and S2 are TRUE.

The Jacobi scheme for the scaled system solves each equation for its diagonal unknown using the previous values:

\[ \begin{aligned} x_{1} &= 4 - x_{2} + 2x_{3} = 4 - 1 + 2 = 5 \\ x_{2} &= 4 - 3x_{1} - 2x_{3} = 4 - 3 - 2 = -1 \\ x_{3} &= 1 - x_{1} - x_{2} = 1 - 1 - 1 = -1 \end{aligned} \]

so S1 holds. The iteration matrix is \(B = I - D^{-1}A\), whose characteristic equation works out to \(\lambda^{3} - 3\lambda - 4 = 0\). It has a real root near 2.2, so the spectral radius exceeds 1 and the iteration diverges for a general starting vector (any whose error has a component along that eigenvector). So S2 holds too.

Your Answer : (Not Answered)

Correct Answer : A

Both I and II are TRUE.

In \(A + \varepsilon I_{3}\) each diagonal entry is \(2 + \varepsilon\), and the off-diagonal entries in each row have absolute values summing to \(1 + 1 = 2\). Since \(2 + \varepsilon \gt 2\) for \(\varepsilon \gt 0\), the matrix is strictly diagonally dominant by rows.

For a strictly diagonally dominant matrix both the Jacobi and the Gauss-Seidel iterations converge for every starting vector. So I and II both hold, and the options denying either are wrong.

Your Answer : (Not Answered)

Correct Answer : A

Both P and Q are TRUE.

P: Doolittle factorisation (unit diagonal in L):

\[ L = \begin{bmatrix} 1 & 0 \\ -2 & 1 \end{bmatrix},\ U = \begin{bmatrix} 2 & -1 \\ 0 & 1 \end{bmatrix} \]

since \(l_{21} = -4/2 = -2\) and \(u_{22} = 3 - (-2)(-1) = 1\). So trace(U) = 2 + 1 = 3.

Q: the Jacobi iteration matrix is \(-D^{-1}(L+U)\) built from the off-diagonal entries:

\[ J = \begin{bmatrix} 0 & 1/2 \\ 4/3 & 0 \end{bmatrix},\quad \lambda^{2} = \frac{2}{3} \]

Its spectral radius is \(\sqrt{2/3} \approx 0.82 \lt 1\), so the Jacobi iterates converge from every starting vector. (M is not diagonally dominant, so that sufficient condition would not settle it; the spectral radius does.)

Your Answer : (Not Answered)

Correct Answer : C

After two Jacobi iterations the solution is \([2.64\ \ 1.70\ \ -1.12]^{T}\).

Solve each equation for its diagonal unknown and use only the previous iterate on the right:

\[ \begin{aligned} x &= (10 + y - z)/5 \\ y &= (12 - 2x)/4 \\ z &= (-1 - x - y)/5 \end{aligned} \]

From \((2, 3, 0)\): \(x^{(1)} = 13/5 = 2.6\), \(y^{(1)} = 8/4 = 2\), \(z^{(1)} = -6/5 = -1.2\). Then:

\[ \begin{aligned} x^{(2)} &= (10 + 2 + 1.2)/5 = 2.64 \\ y^{(2)} &= (12 - 5.2)/4 = 1.70 \\ z^{(2)} &= (-1 - 2.6 - 2)/5 = -1.12 \end{aligned} \]

All options agree on 2.64; the distractors flip the sign of \(y\) or of \(z\). \(y\) stays positive because \(12 - 2x\) is positive for \(x\) near 2.6, and \(z\) is negative because \(-1 - x - y\) is.

Your Answer : (Not Answered)

Correct Answer : C

The second-order Runge-Kutta choice is \(a = 0.25, b = 0.75, \alpha = 2/3, \beta = 2/3\).

Matching the Taylor expansion of \(y(x_{n})\) to second order requires

\[ \begin{aligned} a + b &= 1 \\ b\alpha &= \tfrac{1}{2},\quad b\beta = \tfrac{1}{2} \end{aligned} \]

For \(a = 0.25, b = 0.75, \alpha = \beta = 2/3\): \(a + b = 1\) and \(b\alpha = b\beta = 0.5\). This is Ralston's method.

\(a = b = 1\) gives \(a + b = 2\). \(a = b = 0.5, \alpha = \beta = 2\) gives \(b\alpha = 1\). \(a = b = 0.5, \alpha = 1, \beta = 2\) gives \(b\beta = 1\). Each fails a condition.

Your Answer : (Not Answered)

Correct Answer : C

Euler's method gives 1.222.

Euler's method updates \(y_{n+1} = y_{n} + h\,f(x_{n}, y_{n})\) with \(f(x, y) = x^{2} + y^{2}\) and \(h = 0.1\).

\[ \begin{aligned} y_{1} &= 1 + 0.1(0^{2} + 1^{2}) = 1.1 \\ y_{2} &= 1.1 + 0.1(0.1^{2} + 1.1^{2}) \\ &= 1.1 + 0.1(0.01 + 1.21) = 1.222 \end{aligned} \]

So \(y(0.2) \approx 1.222\). The value 1.122 adds the second increment, 0.122, to the starting value 1 instead of to 1.1; 1.110 and 1.322 do not come from the two Euler steps.

Your Answer : (Not Answered)

Correct Answer : C

Two iterations of bisection give 0.75.

Let \(f(x) = xe^{x} - 1\). Then \(f(0) = -1 \lt 0\) and \(f(1) = e - 1 \gt 0\), so a root lies in [0, 1].

\[ \begin{aligned} x_{1} &= 0.5:\ f(0.5) = 0.5e^{0.5} - 1 = -0.176 \lt 0 \\ x_{2} &= \tfrac{0.5 + 1}{2} = 0.75 \end{aligned} \]

Since \(f(0.5) \lt 0\), the root lies in [0.5, 1], and the second midpoint is 0.75. (The true root is about 0.567.) 0.25 would be the second midpoint only if \(f(0.5)\) were positive; 0.50 is the first iterate, and 0.65 is not a bisection midpoint of [0, 1].

Your Answer : (Not Answered)

Correct Answer : C

After two iterations the estimate is 10.548.

Each step multiplies by \(A\) and scales by the entry of largest magnitude.

\[ \begin{aligned} A[1, 1, 1]^{T} &= [-8, 4, 18]^{T} \\ &= 18\,[-0.444, 0.222, 1]^{T} \end{aligned} \]

\[ \begin{aligned} A[-0.444, 0.222, 1]^{T} \\ = [10.548, -1.104, -7.768]^{T} \end{aligned} \]

The first component is \(6.660 + 0.888 + 3 = 10.548\), the entry of largest magnitude, so the eigenvalue estimate after two iterations is 10.548 (with the vector rounded to three decimals, as the options assume). 7.768 is the magnitude of the third component, not the largest one, and 18 from the first iteration is not offered; 9.468 and 19.468 do not arise from the iteration.

Your Answer : (Not Answered)

Correct Answer : D

The value of \(\frac{1}{\alpha\beta}\) is 12.

Differentiate the Lagrange interpolating polynomial on the nodes -1, 0, 2 and evaluate at 0. Each coefficient is the derivative at 0 of the corresponding basis polynomial:

\[ \begin{aligned} \ell_{-1}(x) &= \frac{x(x-2)}{3},\ \ \ell_{-1}'(0) = -\frac{2}{3} \\ \ell_{0}(x) &= -\frac{(x+1)(x-2)}{2},\ \ \ell_{0}'(0) = \frac{1}{2} \\ \ell_{2}(x) &= \frac{x(x+1)}{6},\ \ \ell_{2}'(0) = \frac{1}{6} \end{aligned} \]

The first coefficient matches the given -2/3, which confirms the set-up. So \(\alpha = 1/2\) and \(\beta = 1/6\), giving \(\alpha\beta = 1/12\) and \(1/(\alpha\beta) = 12\).

Check: the coefficients must sum to 0 (a constant has zero derivative): -2/3 + 1/2 + 1/6 = 0. The options 3, 6 and 9 come from reading 1/\(\alpha\) or 1/\(\beta\) alone or from a slip in one basis polynomial.

Your Answer : (Not Answered)

Correct Answer : B

The formula gives \(y'(2) = 8\).

Here \(h = 1\) and \(x_{k} = 2\), so the forward differences start from \(y = 2\). From the values \(2, -3, 4, -5\):

\[ \begin{aligned} \Delta y &: -5,\ 7,\ -9 \\ \Delta^{2} y &: 12,\ -16 \\ \Delta^{3} y &: -28 \end{aligned} \]

Substituting the leading differences into the given formula:

\[ \begin{aligned} y'(2) &\approx -5 + \tfrac{1}{2}(12) - \tfrac{1}{4}(-28) \\ &= -5 + 6 + 7 = 8 \end{aligned} \]

The paper prints its own coefficients (\(+\frac{1}{2}, -\frac{1}{4}\)), and the answer follows them as given. Using the differences that start at \(x = 1\) instead, or dropping the third-difference term (which gives 1), leads away from 8; 4, 12 and 16 come from such slips.