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11th Standard Maths English Medium Free Online Test 1 Mark Questions 2020 - Part Eight

11th 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 A = {(x,y) : y = ex, x∈R} and B = {(x,y) : y=e-x, x ∈ R} then n(A∩B) is

    (a)

    Infinity

    (b)

    0

    (c)

    1

    (d)

    2

  2. If A = {x / x is an integer, x2 \(\le\) 4} then elements of A are

    (a)

    A = {-1, 0, 1}

    (b)

    A = {-1, 0, 1, 2}

    (c)

    A = {0, 2, 4}

    (d)

    A = {- 2, - 1, 0, 1, 2}

  3. Solve \(\sqrt{7+6x-x^2}=x+1\)

    (a)

    (1, -3)

    (b)

    (3, -1)

    (c)

    (1, -1)

    (d)

    (3, -3)

  4. The value of log 1 is

    (a)

    1

    (b)

    0

    (c)

    \(\infty\)

    (d)

    -1

  5. If \(\Sigma n=210\) then \(\Sigma { n }^{ 2 }\)=

    (a)

    2870

    (b)

    2160

    (c)

    2970

    (d)

    none of these

  6. If a vertex of a square is at the origin and its one side lies along the line 4x + 3y - 20 = 0, then the area of the square is

    (a)

    20 sq. units

    (b)

    16 sq. units

    (c)

    25 sq. units

    (d)

    4 sq.units

  7. The vector in the direction of the vector\(\hat{i}-2\hat{j}+2\hat{k}\) that has magnitude 9 is

    (a)

    \(\hat{i}-2\hat{j}+2\hat{k}\)

    (b)

    \(\frac { \hat { i } -2\hat { j } +2\hat { k } }{ 3 } \)

    (c)

    3(\(\hat{i}-2\hat{j}+2\hat{k}\))

    (d)

    9(\(\hat{i}-2\hat{j}+2\hat{k}\))

  8. Assertion (A) : \(\overset { \rightarrow }{ a } ,\overset { \rightarrow }{ b } ,\overset { \rightarrow }{ c } \) are the position vector three collinear points then 2 \(\overset { \rightarrow }{ a }=\overset { \rightarrow }{ b } +\overset { \rightarrow }{ c } \)
    Reason (R): Collinear points, have same direction

    (a)

    Both A and R are true and R is the correct explanation of A

    (b)

    Both A and R are true and R is not a correct explantion of A

    (c)

    A is true but R is false

    (d)

    A is false but R is true

  9. If P(A)=\(\frac { 1 }{ 2 } \), P(B)=\(\frac { 1 }{ 3 } \) and P(A/B) = \(\frac { 1 }{ 4 } \), then \(P(\bar { A } \cap \bar { B } )\) =

    (a)

    \(\frac { 1 }{ 12 } \)

    (b)

    \(\frac { 3 }{ 4 } \)

    (c)

    \(\frac { 1 }{ 4 } \)

    (d)

    \(\frac { 3 }{ 16 } \)

  10. Two dice are thrown. It is known that the sum of the numbers on the dice was less than 6, the probability of getting a sum 3 is

    (a)

    \(\frac { 1 }{ 18 } \)

    (b)

    \(\frac { 5 }{ 18 } \)

    (c)

    \(\frac { 1 }{ 5 } \)

    (d)

    \(\frac { 2 }{ 5 } \)

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