Question 1 |
A function f(x), that is smooth and convex-shaped between interval \left(x_{1}, x_{u}\right) is shown in the figure. This function is observed at odd number of regularly spaced points. If the area under the function is computed numerically, then ____
the numerical value of the area obtained using
the trapezoidal rule will be less than the
actual. | |
the numerical value of the area obtained using
the trapezoidal rule will be more than the
actual. | |
the numerical value of the area obtained using
the trapezoidal rule will be exactly equal to
the actual. | |
with the given details, the numerical value of
area cannot be obtained using trapezoidal rule |
Question 1 Explanation:
Approximated function has under estimation so numerical value of the area obtained using trapezoidal rule will be less than the actual.
Question 2 |
Consider the definite integral
\int_{1}^{2}(4x^2+2x+6)dx
Let I_e be the exact value of the integral. If the same integral is estimated using Simpson's rule with 10 equal subintervals, the value is I_s . The percentage error is defined as e=100 \times (I_e-I_s)/I_e . The value of e is
\int_{1}^{2}(4x^2+2x+6)dx
Let I_e be the exact value of the integral. If the same integral is estimated using Simpson's rule with 10 equal subintervals, the value is I_s . The percentage error is defined as e=100 \times (I_e-I_s)/I_e . The value of e is
2.5 | |
3.5 | |
1.2 | |
0 |
Question 2 Explanation:
Exact value
\begin{aligned} &=\int_{1}^{2} (4x^2+2x+6)dx\\ &=\frac{4x^3}{3}+\frac{2x^2}{2}+6x\\ &=\frac{4}{3}(\pi) \times (3)+6\\ &=\frac{28}{3}+9=\frac{55}{3} \end{aligned}
Approximate value
Here f(x) is a polynomial of degree 2, so Simpsons rule gives exact value with zero error
\begin{aligned} \therefore \;\; \text{Approx value}&=\frac{55}{3}\\ \frac{I_e-I_s}{I_e}&=0\\ \therefore \;\;e=\left (\frac{I_e-I_s}{I_e} \right )\times 100&=0 \end{aligned}
\begin{aligned} &=\int_{1}^{2} (4x^2+2x+6)dx\\ &=\frac{4x^3}{3}+\frac{2x^2}{2}+6x\\ &=\frac{4}{3}(\pi) \times (3)+6\\ &=\frac{28}{3}+9=\frac{55}{3} \end{aligned}
Approximate value
Here f(x) is a polynomial of degree 2, so Simpsons rule gives exact value with zero error
\begin{aligned} \therefore \;\; \text{Approx value}&=\frac{55}{3}\\ \frac{I_e-I_s}{I_e}&=0\\ \therefore \;\;e=\left (\frac{I_e-I_s}{I_e} \right )\times 100&=0 \end{aligned}
Question 3 |
Let, f(x,y,z)=4x^2+7xy+3xz^2. The direction in which the function f(x,y,z) increases most rapidly at point P = (1,0,2) is
20\hat{i}+7\hat{j} | |
20\hat{i}+7\hat{j}+12\hat{k} | |
20\hat{i}+12\hat{j} | |
20\hat{i} |
Question 3 Explanation:
Given:
f(x,y,z)=4x^2+7xy+3xz^2
The directional derivative at point P is given by
=\triangledown f|_{point \; P}
\therefore \; \triangledown f=(8x+7y+3z^2)\hat{i}+(0+7x+0)\hat{j}+(0+0+6xz)\hat{k}
at point (1, 0, 2)
\triangledown f|_{(1,0,2)}=20\hat{i}+7\hat{j}+12\hat{k}
The directional derivative at point P is given by
=\triangledown f|_{point \; P}
\therefore \; \triangledown f=(8x+7y+3z^2)\hat{i}+(0+7x+0)\hat{j}+(0+0+6xz)\hat{k}
at point (1, 0, 2)
\triangledown f|_{(1,0,2)}=20\hat{i}+7\hat{j}+12\hat{k}
Question 4 |
Consider the following recursive iteration scheme for different values of variable P with the initial guess x_1=1:
x_{n+1}=\frac{1}{2}\left ( x_n+\frac{P}{x_n} \right ),\;\;\;n=1,2,3,4,5
For P=2,x_5 is obtained to be 1.414, rounded-off to three decimal places. For P=3,x_5 is obtained to be 1.732, rounded-off to three decimal places.
If P=10, the numerical value of x_5 is _________ . (round off to three decimal places)
x_{n+1}=\frac{1}{2}\left ( x_n+\frac{P}{x_n} \right ),\;\;\;n=1,2,3,4,5
For P=2,x_5 is obtained to be 1.414, rounded-off to three decimal places. For P=3,x_5 is obtained to be 1.732, rounded-off to three decimal places.
If P=10, the numerical value of x_5 is _________ . (round off to three decimal places)
2.155 | |
3.162 | |
1.125 | |
4.568 |
Question 4 Explanation:
x_{n+1}=\frac{1}{2}\left ( x_n+\frac{P}{x_n} \right )
Converges when x_{n+1}=x_{n}=\alpha
\begin{aligned} \alpha &=\frac{1}{2}\left ( \alpha+\frac{P}{\alpha} \right )\\ \frac{\alpha}{2}&=\frac{P}{2\alpha}\\ \alpha&=\sqrt{P} \end{aligned}
When P=2, x_5=\sqrt{2}=1.4124
When P=3, x_5=\sqrt{3}=1.732
When P=10, x_5=\sqrt{10}=3.162
Converges when x_{n+1}=x_{n}=\alpha
\begin{aligned} \alpha &=\frac{1}{2}\left ( \alpha+\frac{P}{\alpha} \right )\\ \frac{\alpha}{2}&=\frac{P}{2\alpha}\\ \alpha&=\sqrt{P} \end{aligned}
When P=2, x_5=\sqrt{2}=1.4124
When P=3, x_5=\sqrt{3}=1.732
When P=10, x_5=\sqrt{10}=3.162
Question 5 |
Numerically integrate, f(x)=10 x-20 x^{2} from lower limit a=0 to upper limit b=0.5. Use Trapezoidal rule with five equal subdivisions. The value (in units,round off to two decimal places) obtained is ____________
0.78 | |
0.65 | |
0.4 | |
0.56 |
Question 5 Explanation:
\begin{aligned} y&=10 x-20 x^{2} \\ a&=0, b=0.5, n=5 \\ \text { So, } \qquad \qquad \qquad h&=\frac{b-a}{n}=0.1 \end{aligned}
And
\begin{aligned} I &=\int_{0}^{0.5} f(x) d x=\frac{h}{2}\left[y_{0}+y_{5}+2\left(y_{1}+y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{0.1}{2}[0+0+2(0.8+1.2+1.2+0.8)] \\ &=0.40 \end{aligned}
And
\begin{aligned} I &=\int_{0}^{0.5} f(x) d x=\frac{h}{2}\left[y_{0}+y_{5}+2\left(y_{1}+y_{2}+y_{3}+y_{4}\right)\right] \\ &=\frac{0.1}{2}[0+0+2(0.8+1.2+1.2+0.8)] \\ &=0.40 \end{aligned}
There are 5 questions to complete.
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