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MTH603 Quiz 3 January 2026 Set 2

Numerical Analysis PDF 40 KBUpdated 23 Oct 2026
Quiz MTH603 Quiz 3 January 2026 Set 2 PDF 40 KB
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Quiz
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MTH603 Numerical Analysis
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PDF 40 KB
Updated
23 Oct 2026
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Text from pages 2, 3, 4 of 6. Download the file for all 6 pages.

Question:

At which of the following points the Maximum value of 2nd derivative of function f(x) = -(2/x) in the interval:[1,4] exits?

A. At x=1

B. At x=4

C. At x=3 D. At x=2 Copied: 1/6/2026, 1:00:46 PM

Question:

point degree interpolation

�A�.� �y�=�f�(�x�)�=�(�x"�0�.�7�)�(�x"�0�.�9�)�(�0�.�3"�0�.�7�)�(�0�.�3"�0�.�9�)�(�0�.�5�1�8�)�+�(�x"�0�.�3�)�(�x"�0�.�9�)�(�0�.�7"�0�.�3�)�(�0�.�7 "�0�.�3�)�(�0�.�9"�0�.�7�)�(�0�.�0�6�7�)�y�=�f�(�x�)�=�(�x"�0�.�7�)�(�x"�0�.�9�)�(�0�.�3"�0�.�7�)�(�0�.�3"�0�.�9�)�(�0�.�5�1�8�)�+�(�x"�0�.�3�)�(� �(�x"�0�.�3�)�(�x"�0�.�7�)�(�0�.�9"�0�.�3�)�(�0�.�9"�0�.�7�)�(�0�.�0�6�7�)�y� �=� �f�(�x�)� �=� �\�f�r�a�c�{�{�(�x� �-� �0�.�7�)�(�x� �-� �0�.�9�)�}�}�{�{� \frac{{(x - 0.3)(x - 0.9)}}{{(0.7 - 0.3)(0.7 - 0.9)}}(0.248) + \frac{{(x - 0.3)(x - 0.7)}}{{(0.9 - 0.3)(0.9 - 0.7)}}(0.067) �B�.� �y�=�f�(�x�)�=�(�x"�0�.�7�)�(�x"�0�.�9�)�(�0�.�3"�0�.�7�)�(�0�.�3"�0�.�9�)�(�0�.�0�6�7�)�+�(�x"�0�.�3�)�(�x"�0�.�9�)�(�0�.�7"�0�.�3�)�(�0�.�7 "�0�.�3�)�(�0�.�9"�0�.�7�)�(�0�.�2�4�8�)�y�=�f�(�x�)�=�(�x"�0�.�7�)�(�x"�0�.�9�)�(�0�.�3"�0�.�7�)�(�0�.�3"�0�.�9�)�(�0�.�0�6�7�)�+�(�x"�0�.�3�)�(� �(�x"�0�.�3�)�(�x"�0�.�7�)�(�0�.�9"�0�.�3�)�(�0�.�9"�0�.�7�)�(�0�.�2�4�8�)�y� �=� �f�(�x�)� �=� �\�f�r�a�c�{�{�(�x� �-� �0�.�7�)�(�x� �-� �0�.�9�)�}�}�{�{� \frac{{(x - 0.3)(x - 0.9)}}{{(0.7 - 0.3)(0.7 - 0.9)}}(0.518) + \frac{{(x - 0.3)(x - 0.7)}}{{(0.9 - 0.3)(0.9 - 0.7)}}(0.248) �C�.� �y�=�f�(�x�)�=�(�x"�0�.�7�)�(�x"�0�.�9�)�(�0�.�3"�0�.�7�)�(�0�.�3"�0�.�9�)�(�0�.�2�4�8�)�+�(�x"�0�.�3�)�(�x"�0�.�9�)�(�0�.�7"�0�.�3�)�(�0�.�7 "�0�.�3�)�(�0�.�9"�0�.�7�)�(�0�.�5�1�8�)�y�=�f�(�x�)�=�(�x"�0�.�7�)�(�x"�0�.�9�)�(�0�.�3"�0�.�7�)�(�0�.�3"�0�.�9�)�(�0�.�2�4�8�)�+�(�x"�0�.�3�)�(� �(�x"�0�.�3�)�(�x"�0�.�7�)�(�0�.�9"�0�.�3�)�(�0�.�9"�0�.�7�)�(�0�.�5�1�8�)�y� �=� �f�(�x�)� �=� �\�f�r�a�c�{�{�(�x� �-� �0�.�7�)�(�x� �-� �0�.�9�)�}�}�{�{� \frac{{(x - 0.3)(x - 0.9)}}{{(0.7 - 0.3)(0.7 - 0.9)}}(0.067) + \frac{{(x - 0.3)(x - 0.7)}}{{(0.9 - 0.3)(0.9 - 0.7)}}(0.518) �D�.� �y�=�f�(�x�)�=�(�x"�0�.�7�)�(�x"�0�.�9�)�(�0�.�3"�0�.�7�)�(�0�.�3"�0�.�9�)�(�0�.�0�6�7�)�+�(�x"�0�.�3�)�(�x"�0�.�9�)�(�0�.�7"�0�.�3�)�(�0�.�7 "�0�.�3�)�(�0�.�9"�0�.�7�)�(�0�.�5�1�8�)�y�=�f�(�x�)�=�(�x"�0�.�7�)�(�x"�0�.�9�)�(�0�.�3"�0�.�7�)�(�0�.�3"�0�.�9�)�(�0�.�0�6�7�)�+�(�x"�0�.�3�)�(� �(�x"�0�.�3�)�(�x"�0�.�7�)�(�0�.�9"�0�.�3�)�(�0�.�9"�0�.�7�)�(�0�.�5�1�8�)�y� �=� �f�(�x�)� �=� �\�f�r�a�c�{�{�(�x� �-� �0�.�7�)�(�x� �-� �0�.�9�)�}�}�{�{� \frac{{(x - 0.3)(x - 0.9)}}{{(0.7 - 0.3)(0.7 - 0.9)}}(0.248) + \frac{{(x - 0.3)(x - 0.7)}}{{(0.9 - 0.3)(0.9 - 0.7)}}(0.518) Copied: 1/6/2026, 1:01:11 PM

Question:

Zero-th order divided difference is defined as

A. None of the given choices B. y[x0]=y1 C. y[x0]=x0

D. y[x0]=y0

Copied: 1/6/2026, 1:02:14 PM

Question:

points

A. -3

B. -1 C. 0 D. -2 Copied: 1/6/2026, 1:02:47 PM

Question:

Geometrically the definite integral of any continuous function f(x) in the interval [a,b] gives ----------.

A. Area of Trapezoid with dimension of ‘a’ and ‘b’

B. Volume with dimensions f(x), ‘a’ and ‘b’

C. Length of segment AB on real line

D. Area under f(x) on [a,b]

Copied: 1/6/2026, 1:03:09 PM

Question:

To evaluate a definite integral of tabular function f(x), piecewise quardratic approximation led to ---------

A. Trapezoidal Method

B. Simpson’s Rule

Copied: 1/6/2026, 1:03:51 PM

Question:

points zero

A. none

B. 46 C. 35 D. 42 Copied: 1/6/2026, 1:04:14 PM

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