cbse 12th mathematics | Online Objective Test

Q1. Let A = {1, 2, 3}, then the domain of the relation R = {(1, 1), (2, 3), (2, 1)} defined on A is

A. {1, 2}

B. {1, 3}

C. none of these.

D. {1, 2, 3}

Answer : Option A
Explaination / Solution:

Since the domain is represented by the x- co ordinate of the ordered pair ( x , y ).Therefore, domain of the given relation is { 1 , 2 }.

Integrals Prepare / Learn

Q2.

\int_{0}^{2} \sqrt{\frac{2 + x}{2 - x}} d x

is equal to

A. π+32

B. π+1

C. π+2

D. π2+1

Answer : Option C
Explaination / Solution:

Vector Algebra Prepare / Learn

Q3. Negative of a Vector

\vec{a}

is a

A. A vector whose magnitude is the same as that

\vec{a}

of but direction is 120° to that of

\vec{a}

B. A scalar whose magnitude is the same as that

\vec{a}

C. A vector whose magnitude is the same as that

\vec{a}

of but direction is opposite to that of

\vec{a}

D. A vector whose magnitude is the same as that

\vec{a}

of but direction is perpendicular to that of

\vec{a}

Answer : Option C
Explaination / Solution:

egative of

\vec{a}

is equal to -

\vec{a}

, i.e. A vector whose magnitude is the same as that of

\vec{a}

,but direction is opposite to that of

\vec{a}

Probability Prepare / Learn

Q4. If P(A) =

\frac{6}{11}

, P(B) =

\frac{5}{11}

and P(A ∪ B) =

\frac{7}{11} .

find P(A∩B)

A. 4/11

B. 4/13

C. 5/11

D. 5/17

Answer : Option A
Explaination / Solution:

Inverse Trigonometric Functions Prepare / Learn

Q5. The period of the function f(x)=cos 4x + tan 3x is

A. none of these

B. π2

C. π3

D. π

Answer : Option D
Explaination / Solution:

f(π)=(cos4π+tan3π)gives the same value as f (0) .Therefore,the period of the function is π

Continuity and Differentiability Prepare / Learn

Q6. The function f (x) = 1 + | sin x l is

A. differentiable nowhere

B. differentiable everywhere.

C. continuous nowhere

D. continuous everywhere

Answer : Option D
Explaination / Solution:

f (x) = 1 + | s i n x |

is not derivable at those x for which

x f o r w h i c h s i n x = 0, h o w e v e r, 1 + | s i n x |

is continuous everywhere (being the sum of two continuous functions)

Differential Equations Prepare / Learn

Q7. What is the order of differential equation :

\frac{d^{3} y}{d x^{3}} + \frac{d^{2} y}{d x^{2}} + {(\frac{d y}{d x})}^{2} = e^{x}

A. 2

B. 1

C. 0

D. 3

Answer : Option D
Explaination / Solution:

Order = 3.Since the third derivative is the highest derivative present in the equation. i.e.

\frac{d^{3} y}{d x^{3}}

Three Dimensional Geometry Prepare / Learn

Q8.

Cartesian equation of a line that passes through two points

(x_{1}, y_{1}, z_{1})

and

(x_{2}, y_{2}, z_{2})

A. x−x1x2−x1=y−y1y2−y1=z+z1z2−z1

B. x−x1x2−x1=y−y1y2−y1=z+z1z2−z1

\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}}

D. x+x1x2−x1=y−y1y2+y1=z−z1z2−z1

Answer : Option C
Explaination / Solution:

Cartesian equation of a line that passes through two points

(x_{1}, y_{1}, z_{1})

and

(x_{2}, y_{2}, z_{2})

is given by :

\frac{x - x_{1}}{x_{2} - x_{1}} = \frac{y - y_{1}}{y_{2} - y_{1}} = \frac{z - z_{1}}{z_{2} - z_{1}}

Application of Integrals Prepare / Learn

Q9. The area enclosed by the curve

\frac{x^{2}}{25} + \frac{y^{2}}{16} = 1

is equal to

A. 20πsq.units

B. 5πsq.units

C. none of these

D. 10πsq.units

Answer : Option A
Explaination / Solution:

The area of the standard ellipse is given by ;πab. Here , a = 5 and b = 4 Therefore , the area of curve is π(5)(4)=20π.

Matrices Prepare / Learn

Q10.

If A and B are invertible matrices of the same order, then $(A B)^{- 1}$ is equal to

A. A−1B

B. A−1B−1

C. B−1A−1

D. AB−1

Answer : Option C
Explaination / Solution:

By the property of inverse

(A B)^{- 1}

is equal to

B^{- 1} A^{- 1}

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