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Published on: 05/11/2020
12th Standard Maths Differentials and Partial Derivatives English Medium Free Online Test One Mark Questions 2020 - 2021
Download Tamil Nadu 12th Standard Maths question papers, model tests, one-mark questions, important questions, and public exam papers in PDF format. Free study materials and answer keys for TN State Board students.
Questions + Answers key
Take MCQ Maths Test1.
If u = y sin x then \(\frac { { \partial }^{ 2 }u }{ \partial x\partial y } \) = ..........
cos x
cos y
sin x
0
2.
3.
The approximate value of (627)\(\frac14\) is ................
5.002
5.003
5.005
5.004
4.
If f (x, y) = x3 + y3 - 3xy2 then \(\frac { { \partial }f }{ \partial { x } } \) at x = 2,_____________
-15
15
-9
16
5.
If f(x, y, z) = sin (xy) + sin (yz) + sin (zx) then fxx is _____________
-y sin (xy) + z2 cos (xz)
y sin (xy) - z2 cos (xz)
y sin (xy) + z2 cos (xz)
-y2 sin (xy) - z2 cos (xz)
6.
If y = x4 - 10 and if x changes from 2 to 1.99, the approximate change in y is ________
-32
-0.32
- 10
10
7.
If w (x, y, z) = x2 (y - z) + y2 (z - x) + z2(x - y), then \(\frac { { \partial }w }{ \partial x } +\frac { \partial w }{ \partial y } +\frac { \partial w }{ \partial z } \) is
xy + yz + zx
x(y + z)
y(z + x)
0
8.
The approximate change in the volume V of a cube of side x metres caused by increasing the side by 1% is
0.3xdx m3
0.03x m3
0.03x2 m3
0.03x3 m3
9.
If we measure the side of a cube to be 4 cm with an error of 0.1 cm, then the error in our calculation of the volume is
0.4 cu.cm
0.45 cu.cm
2 cu.cm
4.8 cu.cm
10.
A circular template has a radius of 10 cm. The measurement of radius has an approximate error of 0.02 cm. Then the percentage error in calculating area of this template is
0.2%
0.4%
0.04%
0.08%
1.
(a)
cos x
2.
(b)
3.
(d)
5.004
4.
(a)
-15
5.
(d)
-y2 sin (xy) - z2 cos (xz)
6.
(b)
-0.32
7.
(d)
0
8.
\({ f }_{ y }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { y }^{ 2 } } } \left( \frac { -x }{ { y }^{ 2 } } \right) =\frac { -\frac { x }{ { y }^{ 2 } } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { x }{ x }^{ 2 } } } } \)
= \(\frac { -x }{ { x }^{ 2 }+{ y }^{ 2 } } \)
\({ f }_{ x }=\frac { 1 }{ 1+\frac { { x }^{ 2 } }{ { { y }^{ 2 } } } } \left( \frac { 1 }{ y } \right) =\frac { \frac { 1 }{ y } }{ \frac { { x }^{ 2 }+{ y }^{ 2 } }{ { { y }^{ 2 } } } } \)
\(=\frac { y }{ { { x }^{ 2 }+{ y }^{ 2 } } } \)
\({ f }_{ xy }=\frac { \partial }{ \partial x } ({ f }_{ y })\)
\({ f }_{ xy }=-\left[ \frac { { (x }^{ 2 }+{ y }^{ 2 })(1)-x{ (2x) } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { x }^{ 2 }+{ y }^{ 2 }-2{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=-\left[ \frac { { y }^{ 2 }-{ x }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \right] \)
\(=\frac { x^{ 2 }-{ y }^{ 2 } }{ { ({ x }^{ 2 }+{ y }^{ 2 }) }^{ 2 } } \) ...(1)
\({ f }_{ xy }=\frac { \partial }{ \partial y } ({ f }_{ x })\)
\({ f }_{ xy }=\frac { ({ x }^{ 2 }+{ y }^{ 2 })(1)-y(2y) }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }+{ y }^{ 2 }-2{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \)
\(=\frac { { x }^{ 2 }-{ y }^{ 2 } }{ ({ x }^{ 2 }+{ { y }^{ 2 }) }^{ 2 } } \) ...(2)
∴ From (1) and (2), fxy = fyz
9.
(d)
4.8 cu.cm
10.
(b)
0.4%
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Tamilnadu Stateboard 12th Standard Subjects

Maths

Chemistry

Physics

Biology

Computer Science

Business Maths and Statistics

Economics

Commerce

Accountancy

History

Computer Applications

Biology

Computer Technology

Computer Applications

Computer Science

Business Maths and Statistics

Commerce

Economics

Maths

Chemistry

Physics

Computer Technology

History

Accountancy

Tamil

English

French
Tamilnadu Stateboard Standards