Mathematics | ECE Electronics and Communication Engineering | TANCET Anna University Online Objective Test

TANCET Anna University ECE Electronics and Communication Engineering # Mathematics Binomial Theorem Prepare / Learn

Q1. The index of the power of x that occurs in the 5th term in the expansion of

{(2 x^{2} - \frac{1}{3 x^{2}})}^{8}

A. 12

B. 4

C. 2

D. 0

Answer : Option D
Explaination / Solution:

TANCET Anna University ECE Electronics and Communication Engineering # Mathematics Maths Sets Prepare / Learn

Q2. If A , B and C are non – empty sets , then ( A – B ) ∪( B – A ) equals :

A. ( A ∪B ) – ( A ∩ B )

B. (A∩B)−B

C. ( A ∪ B ) - B

D. ( A ∩ B ) ∪( A ∪B )

Answer : Option A
Explaination / Solution:

TANCET Anna University ECE Electronics and Communication Engineering # Mathematics Limits and Derivatives Prepare / Learn

Q3.

\frac{d}{d x} (\sin^{- 1} (\frac{2}{x^{- 1} + x}))

is equal to

A. 1−x2−−−−−√

21+x2;0<|x|<1

C. 21+x2√,|x|<1

D. 1x+x2√

Answer : Option B
Explaination / Solution:

TANCET Anna University ECE Electronics and Communication Engineering # Mathematics Linear Inequalities Prepare / Learn

Q4. Graph of the system of inequations x ≥ 0 , y ≤ 0 is

A. first quadrant

B. second quadrant

C. 4th quadrant

D. third quadrant

Answer : Option C
Explaination / Solution:

The solution region of $x \geq 0$ will be the half of XY –plane which lies on the right of y – axis , including the points on y – axis [ First and Fourth quadrants ]

The solution region of $y \leq 0$ will be the half of XY –plane which lies under x – axis , including the points on x – axis [ Third and Fourth quadrants ]

Hence the solution region of $x \geq 0, y \leq 0$ will be the the intersection of the above two regions ,which is the fourth quadrant.

TANCET Anna University ECE Electronics and Communication Engineering # Mathematics Sequences and Series Prepare / Learn

Q5.

I f \frac{a + b x}{a - b x} = \frac{b + c x}{b - c x} = \frac{c + d x}{c - d x}, x \neq 0,

then a, b, c, d are in

A. A.P.

B. H.P.

C. None of these

D. G.P.

Answer : Option D
Explaination / Solution:

TANCET Anna University ECE Electronics and Communication Engineering # Mathematics Statistics Prepare / Learn

Q6. The S.D. of the first n natural numbers is

B. n(n+1)2−−−−−√

C. n

D. n+12

Answer : Option A
Explaination / Solution:

first n natural numbers are 1,2,3.....n

we know that, $\sum x = \frac{n (n + 1)}{2}$

sum of sqaure is $\sum x^{2} = \sum (1^{2} + 2^{2} + 3^{2} + . . . . . . . . . + n^{2})$

$\sum x^{2} = \sum (1^{2} + 2^{2} + 3^{2} + . . . . . . . . . + n^{2})$

$\begin{array}{l} \sum x^{2} = \frac{n (n + 1) (2 n + 1)}{6} \\ S D = \sqrt{\frac{\sum x^{2}}{n} - {(\frac{\sum x}{n})}^{2}} \\ = \sqrt{\frac{n (n + 1) (2 n + 1)}{6 n} - {(\frac{n (n + 1)}{2 n})}^{2}} \\ = \sqrt{\frac{(n + 1) (2 n + 1)}{6} - {(\frac{(n + 1)}{2})}^{2}} \\ = \sqrt{(n + 1) [\frac{2 n + 1}{6} - \frac{(n + 1)}{4}]} \\ = \sqrt{\frac{(n + 1) (n - 1)}{12}} \\ = \sqrt{\frac{n^{2} - 1}{12}} \end{array}$