Mathematics | CSE, IT Computer Science Engineering and Information Technology | TANCET Anna University Online Objective Test

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Relations and Functions Prepare / Learn

Q1. A relation R from C(complex no.) to R(real no.) is defined by x Ry iff |x| = y. Which of the following is correct?

A. (1 + i)R2

B. iR1

C. 3R(–3)

D. (2 + 3 i)R13

Answer : Option B
Explaination / Solution:

As

x \in C

i.e. x is a complex no., then

| i | = \sqrt{0^{2} + 1^{2}} = 1

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Determinants Prepare / Learn

Q2. The only integral root of the equation det.

A. 2

B. 4

C. 3

D. 1

Answer : Option D
Explaination / Solution:

The value of determinant is 0 if any two rows or column are identical and Clearly , y = 1 satisfies it. $[C_{3} = 3 C_{1}]$

if we take common as 3 from C3 THEN C1 And C3 Becomes identical after puting y=1.

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Straight Lines Prepare / Learn

Q3. A line is equally inclined to the axis and the length of perpendicular from the origin upon the line is √2. A possible equation of the line is

A. y = x + 1

B. x + y = 2

C. y = x +

\sqrt{2}

D. y =

\sqrt{2}

x + 2 .

Answer : Option B
Explaination / Solution:

Since the line is equally inclined the slope of the line should be -1, because it makes 1350 in the positive direction of the X axis

This implies the equation of the line is y= -x +c

i.e; x+y - c =0

distance of the line from the origin is given as $\sqrt{2}$

Therefore $\sqrt{2}$ = $\frac{| c |}{\sqrt{1^{2} + 1^{2}}}$

This implies c = 2

Hence the equation of the line is x+y=2

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Mathematical Reasoning Prepare / Learn

Q4. The false statement in the following is

A. p∧∼p is a contradiction

B. ∼(∼p)↔p is a tautology

C. p∨∼p is a tautology

D. (p→q)↔(∼q→∼p) is a contradiction

Answer : Option D
Explaination / Solution:

p→q An implication statement,∼q→∼p a contrapositive statement. Implication and contrapositve have same meaning.hence they have identical values. p↔p≡T. hence the above statement will be true always.

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Integrals Prepare / Learn

Q5.

\int (\tan x + c o t x)

dx is equal to

A. none of these.

B. log | sin x + cos x|+C

C. log | cot x|+C

D. log | tan x|+C

Answer : Option D
Explaination / Solution:

[−log| cosx|+log| sin x|]=log | tan x|+C

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Vector Algebra Prepare / Learn

Q6. If a vector

\vec{r}

is expressed in component form as

\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}

then x, y and z are referred to as

A. unit scalars

B. scaling factors

C. vector components or rectangular components

D. unit vectors

Answer : Option C
Explaination / Solution:

In

\vec{r}

\vec{r} = x \hat{i} + y \hat{j} + z \hat{k}

x , y and z represents the components of

\vec{r}

along x-axis ,y-axis and z-axis respectively.

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Probability Prepare / Learn

Q7. Compute P(A|B), if P(B) = 0.5 and P (A ∩ B) = 0.32

A. P(A|B) =

\frac{16}{33}

B. P(A|B) =

\frac{16}{33}

C. P(A|B) =

\frac{15}{27}

D. P(A|B) =

\frac{16}{29}

Answer : Option A
Explaination / Solution:

We have , P(B) = 0.5 and P (A ∩ B) = 0.32

P (A / B) = \frac{P (A \cap B)}{P (B)} = \frac{0.32}{0.5} = \frac{16}{25}

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Inverse Trigonometric Functions Prepare / Learn

Q8. The range of

\tan^{- 1}

x is

A. (π,π)

B. R

C. [−π2,π2]

D. [-π2,π2]

Answer : Option D
Explaination / Solution:

the domain range of tan function are $(- \frac{π}{2}, \frac{π}{2})$ and R

But tan function is one one and onto only when we shrink range to (-1,1)

Now tan function which is one and on to has domain $(- \frac{π}{2}, \frac{π}{2})$ and range(-1,1)

hence domain and range of $\tan^{- 1}$ function is (-1,1 ) to $(- \frac{π}{2}, \frac{π}{2})$

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Continuity and Differentiability Prepare / Learn

Q9. If x + | y | = 2y , then y as a function of x is

A. not continuous at x = 0

B. not defined for all real x

C. such that dy/dx = 1/3 for x<0

D. differentiable for all x

Answer : Option C
Explaination / Solution:

TANCET Anna University CSE, IT Computer Science Engineering and Information Technology # Mathematics Conic Sections Prepare / Learn

Q10. The length of latus rectum of an ellipse is one-third of its major axis. Its eccentricity would be

A. 12√

B. 23

C. 13√

\sqrt{\frac{2}{3}}

Answer : Option D
Explaination / Solution:

$2 b^{2} / a^{2}$ = 2a/3

Hence $\frac{b^{2}}{a^{2}}$ = 1/3. Hence e2 = 1- $\frac{b^{2}}{a^{2}}$ = 1 - (1/3) = 2/3

Therefore e = $\sqrt{\frac{2}{3}}$

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