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12th Standard Maths English Medium Free Online Test One Mark Questions with Answer Key 2020 - Part Nine

12th Standard

    Reg.No. :
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Maths

Time : 00:10:00 Hrs
Total Marks : 10

    Answer all the questions

    10 x 1 = 10
  1. If ATA−1 is symmetric, then A2 =

    (a)

    A-1

    (b)

    (AT)2

    (c)

    AT

    (d)

    (A-1)2

  2. If |z| = 1, then the value of \(\frac { 1+z }{ 1+\overline { z } }\) is

    (a)

    z

    (b)

    \(\bar { z } \)

    (c)

    \(\cfrac { 1 }{ z } \)

    (d)

    1

  3. If a = 3 + i and z = 2 - 3i, then the points on the Argand diagram representing az, 3az and - az are ___________

    (a)

    Vertices of a right angled triangle

    (b)

    Vertices of an equilateral triangle

    (c)

    Vertices of an isosceles

    (d)

    Collinear

  4. If z1, z2, z3 are the vertices of a parallelogram, then the fourth vertex z4 opposite to z2 is _____

    (a)

    z1 + z2 - z2

    (b)

    z1 + z2 - z3

    (c)

    z1 + z2 - z3

    (d)

    z1 - z2 - z3

  5. The domain of the function defined by \(f(x)=\sin ^{-1} \sqrt{x-1}\) is

    (a)

    [1, 2]

    (b)

    [-1, 1]

    (c)

    [0, 1]

    (d)

    [-1, 0]

  6. The number of solutions of the equation \({ tan }^{ -1 }2x+{ tan }^{ -1 }3x=\frac { \pi }{ 4 } \) ____________

    (a)

    2

    (b)

    3

    (c)

    1

    (d)

    none

  7. The area of quadrilateral formed with foci of the hyperbolas \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1 \text { and } \frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=-1\)

    (a)

    4(a2+b2)

    (b)

    2(a2+b2)

    (c)

    a2 +b2

    (d)

    \(\frac { 1 }{ 2 } \)(a2+b2)

  8. If \(\vec { a } \) and \(\vec { b } \) are unit vectors such that \([\vec { a } ,\vec { b },\vec { a } \times \vec { b } ]=\frac { 1}{ 4 } \), then the angle between \(\vec { a } \) and \(\vec { b } \) is

    (a)

    \(\frac { \pi }{ 6 } \)

    (b)

    \(\frac { \pi }{ 4 } \)

    (c)

    \(\frac { \pi }{ 3 } \)

    (d)

    \(\frac { \pi }{ 2 } \)

  9. The tangent to the curve y2 - xy + 9 = 0 is vertical when 

    (a)

    y = 0

    (b)

    \(\\ \\ y=\pm \sqrt { 3 } \)

    (c)

    \(y=\frac { 1 }{ 2 } \)

    (d)

    \(y=\pm 3\)

  10. If \(g(x, y)=3 x^{2}-5 y+2 y^{2}, x(t)=e^{t}\) and y(t) = cos t, then \(\frac{dg}{dt}\) is equal to

    (a)

    6e2t + 5 sin t - 4 cos t sin t

    (b)

    6e2t- 5 sin t + 4 cos t sin t

    (c)

    3e2t+ 5 sin t + 4 cos t sin t

    (d)

    3e2t - 5 sin t + 4 cos t sin t

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