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(1+$\omega^{2}$-$\omega$)(1-$\omega^{2}$+$\omega$) is

1. 4

2. $\omega$

3. 2

4. Zero

With the help of 2 properties of cube roots of unity, we can solve this question. Property 1,  Property 2,  Now we have to find the value of (1+-)(1-+) Using property 2,  Now use property 1, Option (1) is correct
Engineering
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Let where       Then   equals :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

Option 2

Engineering
56 Views   |

If (10)9 + 2(11)1   (10)8 + 3(11)2  (10)7 +......  +10 (11)9 = k (10)9, then k is equal to :

• Option 1)

100

• Option 2)

110

• Option 3)

• Option 4)

Use

Sum of n terms of a GP -

$S_{n}= \left\{\begin{matrix} a\frac{\left ( r^{n}-1 \right )}{r-1}, &if \: r\neq 1 \\ n\, a, & if \, r= 1 \end{matrix}\right.$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

$n\rightarrow$ number of terms

and

(10)9 + 2(11)1 (10)8 + 3(11)2 (10)7 +......  +10 (11)9 = k(10)9

Take common 109

$10^{9}\left [ 1+2\times \frac{11}{10}+3\times \left ( \frac{11}{10} \right )^{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot 10\times \left ( \frac{11}{10} \right ) ^{9}\right ]= k\left ( 10 \right )^{9}$

$\therefore k= 1+2x+3x^{2}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot 10x^{9} \: where \:\:x=\frac{11}{10}$

$kx= x+2x^{2}+3x^{3}+\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot 10x^{10}$

Subtract

$k-kx= 1+x+x^{2}+x^{3}+x^{4}\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot (-10x^{10})$

$k(1-x)= \frac{1(x^{10}-1)}{x-1}-10x^{10}$

$k\left ( 1-\frac{11}{10} \right )= \frac{\left ( \frac{11}{10} \right )^{10}-1}{\frac{11}{10}-1}-10\times \left ( \frac{11}{10} \right )^{10}$

$-\frac{k}{10}= \frac{\left ( \frac{11}{10} \right )^{10}}{\frac{1}{10}}-10-\frac{11^{10}}{10^{9}}$

$\therefore k=100$

Option 1)

100

Option 2)

110

Option 3)

Option 4)

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Engineering
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If the coefficients of x3 and x4 in the expansion of (1 + ax + bx2) (1 - 2x)18 in powers of x are both zero, then (a, b) is equal to :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we have learned Expression of Binomial Theorem -   - wherein for n  +ve integral .     coeff of   coeff of    and  and   a = 16  b = 272/3        Option 1) Option 2) Option 3) Option 4)
Engineering
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The sum of first 20 terms of the sequence 0.7 ,0.77,0.777,.........,is :

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we learnt

Sum of infinite terms of a GP -

$a+ar+ar^{2}+- - - - -= \frac{a}{1-r}\\here \left | r \right |<1$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

$S=0.7+0.77+0.777...upto\: \: 20\: \: terms$

$S=\frac{7}{9}(0.9+0.99+0.999...)$

$S=\frac{7}{9}(1-0.1+1-0.01+1-0.001...)$

$S=\frac{7}{9}(20-(\frac{1}{10}+\frac{1}{100}+...upto\: \: 20\: \: terms))$

$S=\frac{7}{9}(20-\frac{\frac{1}{10}(1-\frac{1}{10^{20}})}{(1-\frac{1}{10})})$

$S=\frac{7}{9}(20-\frac{1-10^{-20}}{9})$

$S=\frac{7}{81}(179+10^{-20})$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
134 Views   |

The graph of the function  $\dpi{100} y=f\left ( x \right )$  is symmetrical about the line $\dpi{100} x=2$ ,then

• Option 1)

$f\left ( x \right )= f\left ( -x \right )$

• Option 2)

$f\left (2+ x \right )= f\left ( 2-x \right )$

• Option 3)

$f\left ( x +2\right )= f\left ( x-2 \right )$

• Option 4)

$f\left ( x \right )=- f\left ( -x \right )$

As we learnt in

Even Function -

$f(-x)= f(x)$

- wherein

Since a graph symmetric about y-axis

means  x = 0 then it is even function and f(-x) = f(x)

$\therefore$    f(0 - x) = f(0 + x)     (b < z it is symmetric about v = 0 )

But in question it is symmetric about x = 2

then f(x - 2) = f(x + 2)

Correct option is 3.

Option 1)

$f\left ( x \right )= f\left ( -x \right )$

Option 2)

$f\left (2+ x \right )= f\left ( 2-x \right )$

Option 3)

$f\left ( x +2\right )= f\left ( x-2 \right )$

Option 4)

$f\left ( x \right )=- f\left ( -x \right )$

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Engineering
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if     and

• Option 1)

is an empty set.

• Option 2)

contains exactly one element.

• Option 3)

contains exactly two elements.

• Option 4)

contains more than two elements.

As we learnt in

FUNCTIONS -

A relation f from a set A to a set B is said to be a function if every element of set A has one and only one image in set B.

-

$f(x)+2f \left(\frac{1}{x} \right )=3x$

Put $\frac{1}{x}$ at the place of

$f\left(\frac{1}{x} \right )+2f(x)=\frac{3}{x}$                                                    (i)

$2f\left(\frac{1}{x} \right )+f(x)=3x$                                                (ii)

Multiplying (i) by 2

$2f\left(\frac{1}{x} \right )+4f(x)=\frac{6}{x}$

$\underline{2f\left(\frac{1}{x} \right )+f(x)=3x}$

$3f(x)=\frac{6}{x}-3x$

$f(x)=\frac{2}{x}-3x$

and             $f(-x)=\frac{2}{-x}+x$

$\therefore\ \; \frac{2}{x}-x=-\frac{2}{x}+x$

$\Rightarrow\ \; \frac{4}{x}-2x=0$

$\Rightarrow\ \; \frac{4-2x^{2}}{x}=0$

$\Rightarrow\ \; 4=2x^{2}$

$\Rightarrow\ \; x^{2}=2$

$x=\pm \sqrt{2}, \; x \neq 0$

Correct option is 3.

Option 1)

is an empty set.

Option 2)

contains exactly one element.

Option 3)

contains exactly two elements.

Option 4)

contains more than two elements.

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Engineering
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The real number for which the equation, $2x^2+3x+k = 0$  has two distinct real roots in

• Option 1)

does not exist.

• Option 2)

lies between 1 and 2 .

• Option 3)

lies between 2 and 3 .

• Option 4)

lies between -1 and 0 .

As we have learned

Quadratic Expression Graph when a> 0 & D > 0 -

Real and distinct roots of

$f\left ( x \right )= ax^{2}+bx+c$

& $D= b^{2}-4ac$

- wherein

$\frac{-b}{2a}=-3/4$    is the abscissa of vertex

and , it should lie in(0,1 ) but it's not true

S, no value of 'k' exists

Option 1)

does not exist.

Option 2)

lies between 1 and 2 .

Option 3)

lies between 2 and 3 .

Option 4)

lies between -1 and 0 .

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Engineering
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Let and be the roots of equation are in A.P. and   then the value of is ?

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we have learned

Sum of Roots in Quadratic Equation -

$\alpha +\beta = \frac{-b}{a}$

- wherein

$\alpha \: and\beta$ are root of quadratic equation

$ax^{2}+bx+c=0$

$a,b,c\in C$

Product of Roots in Quadratic Equation -

$\alpha \beta = \frac{c}{a}$

- wherein

$\alpha \: and\ \beta$ are roots of quadratic equation:

$ax^{2}+bx+c=0$

$a,b,c\in C$

@1449

$|\alpha -\beta | = \left | \frac{\sqrt{q^2}-4pr}{p} \right |$

$\left ( \because \left | \frac{\sqrt{D}}a{} \right | \right )$

Also $\frac{\alpha +\beta }{\alpha \beta }= 4$

$\Rightarrow \frac{-q}{r}= 4$

$\Rightarrow q = -4r ....(1)$

$= \sqrt{16(\frac{r}{p})^2-(4\frac{r}{p})}$

Also p+r =2q

$\Rightarrow p+r = -8r \Rightarrow r/p = -1/9$

$\therefore \frac{\left | \alpha -\beta \right |}{16\times 1/81+4/9}= \sqrt{\frac{52}{81}}=\frac{2\sqrt{13}}{9}$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
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If z is a complex number of unit modulus and argument  ,then arg    equals:

• Option 1)

• Option 2)

• Option 3)

• Option 4)

As we have learned

Euler's Form of a Complex number -

$z=re^{i\theta}$

- wherein

r denotes modulus of z and $\theta$ denotes argument of z.

Polar Form of a Complex Number -

$z=r(cos\theta+isin\theta)$

- wherein

r= modulus of z and $\theta$ is the argument of z

$|z| = 1$

Arg (z)= $\theta$

$\Rightarrow z = e^{i\theta }= \cos \theta + i \sin \theta$

So, $\frac{1+z}{1+z}= \frac{1+\cos \theta +i \sin \theta }{1+\cos \theta -i\sin \theta }$

$\frac{2 \cos^2\theta h+2 i\sin \theta h\cos \theta /2}{2\cos ^{2}\theta h-2i\sin \theta h\cos \theta }$

$=\frac{\cos \theta h+i\sin \theta h}{\cos \theta h-i\sin \theta h}$

$=\frac{e^{i\theta h}}{e^{-i\theta h}}= e^{i\theta }$

$\left ( \frac{1+z}{1+\bar{z}} \right )= \theta$

Option 1)

Option 2)

Option 3)

Option 4)

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Engineering
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If the equations have a common root, then a : b : c is :

• Option 1)

3 : 1 : 2

• Option 2)

1 : 2 : 3

• Option 3)

3 : 2 : 1

• Option 4)

1 : 3 : 2

As we have learned

Quadratic Expression Graph when a > 0 & D < 0 -

No Real and Equal root of

$f\left ( x \right )= ax^{2}+bx+c$

& $D= b^{2}-4ac$

- wherein

Condition for both roots common -

$\frac{a}{{a}'}=\frac{b}{{b}'}=\frac{c}{{c}'}$

- wherein

$ax^{2}+bx+c=0$ &

$a'x^{2}+b'x+c'=0$

are the 2 equations

For $x^2+2x+3=0$

Discriminant = 4-12 = -8 < 0

Both the roots are common as complex roots occur in conjugate $\therefore a:b:c= 1:2:3$

Option 1)

3 : 1 : 2

Option 2)

1 : 2 : 3

Option 3)

3 : 2 : 1

Option 4)

1 : 3 : 2

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Engineering
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If    is the adjoint of a 3 x 3  matrix A and = 4,then is equal to :

• Option 1)

0

• Option 2)

4

• Option 3)

11

• Option 4)

5

As we have learned

Property of adjoint of A -

$\left | adj A \right |=\left | A \right |^{n-1}$

- wherein

$adj A$ denotes adjoint of $A$ and  $\left |A \right |$  denotes determinant  of $A$ and $n$ is the order of the matrix

$|adj \; \; A| = |A|^{3-1}$

$\Rightarrow |adj \; \; A| =4^2=16$

$\Rightarrow 0-\alpha (-2)+3(-2)=16$

$\Rightarrow 2\alpha -6 = 16$

$\Rightarrow \alpha = 11$

Option 1)

0

Option 2)

4

Option 3)

11

Option 4)

5

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Engineering
139 Views   |

Let A and B be two sets containing 2 elements and 4 elements respectively. The number of subsets of AB having 3 or more elements is :

• Option 1)

211

• Option 2)

256

• Option 3)

220

• Option 4)

219

n(A) = 4, n(B) = 2

$n(A\times B)=8$

Number of sbsets having atlest 3 elements

$=2^{8}-\left(1+^{8}C_{1}+^{8}C_{2} \right )=219$

Option 1)

211

Incorrect

Option 2)

256

Incorrect

Option 3)

220

Incorrect

Option 4)

219

Correct

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Engineering
116 Views   |

In how many differnt ways can 3 different rings be worn in 5 fingers of a hand?

• Option 1)

3! x 5!

• Option 2)

35

• Option 3)

210

• Option 4)

180

As learnt in Number of Permutations without repetition - Arrange n objects taken r at a time equivalent to filling r places from n things.   - wherein Where      First ring can be worn in 7 ways Second in 6 ways Third in 5 ways Option 1) 3! x 5! This option is incorrect. Option 2) 35 This option is incorrect. Option 3) 210 This option is correct. Option 4) 180 This option is incorrect.
Engineering
115 Views   |

if a, b and c from G.P with common ration r ,the sum of the y cordinates of the points of intersection of the line ax+by+c=0 and the curve $x+2y^{2}=0$ is

• Option 1)

$-\frac{r}{4}$

• Option 2)

$-\frac{r}{2}$

• Option 3)

$\frac{r}{2}$

• Option 4)

$\frac{r}{4}$

As learnt in

General term of a GP -

$T_{n}= ar^{n-1}$

- wherein

$a\rightarrow$ first term

$r\rightarrow$ common ratio

And,

$ax+by+c=0$

$b=ar, c=ar^{2}$

$x+ry+r^{2}=0$

Also,

$x+2y^{2}=0$

$2y^{2}-ry-r^{2}=0$

$Sum= \frac{r}{2}$

Option 1)

$-\frac{r}{4}$

This option is incorrect.

Option 2)

$-\frac{r}{2}$

This option is incorrect.

Option 3)

$\frac{r}{2}$

This option is correct.

Option 4)

$\frac{r}{4}$

This option is incorrect.

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Engineering
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$If \begin{vmatrix} y+z & x & x \\ y & z+x &y \\ z & z & x+y\end{vmatrix} =K(xyz)$

Then K is equal to

• Option 1)

4

• Option 2)

-4

• Option 3)

0

• Option 4)

None

As learnt in concept Value of determinants of order 3 - -       Option 1) 4 This option is correct. Option 2) -4 This option is incorrect. Option 3) 0 This option is incorrect. Option 4) None This option is incorrect.
Engineering
118 Views   |

With the help of matrices , the solution of the equations  $3x+y+2z=3,\, \, 2x-3y-z=-3\, \, x+2y+z=4, is\, \, given\, \, by$

• Option 1)

x=1, y=2,z=-1

• Option 2)

x=-1,y=2,z=1

• Option 3)

x=6,y=-2,z=-1

• Option 4)

x=-1,y=-2,z=1

As learnt in Cramer's rule for solving system of linear equations - When   and  , then  the system of equations has infinite solutions. - wherein and   are obtained by replacing column 1,2,3 of  by   column     We should solve such questions analytically by cross-checking options because of the time constraint. , satisfy all these equations Option 1) x=1, y=2,z=-1 This option is...
Engineering
146 Views   |

what is the 2015th term of a sequence of natural numbers written in accending order, that does not have any perfect squares.

• Option 1)

2058

• Option 2)

2059

• Option 3)

2060

• Option 4)

2062

As learnt in Sequence - Arragement of real numbers specified in a definite order, by some assigned law. - wherein Notations - or     We get,   Thus total terms are   i.e., 2060 Option 1) 2058 This option is incorrect. Option 2) 2059 This option is incorrect. Option 3) 2060 This option is correct. Option 4) 2062 This option is incorrect.
Engineering
187 Views   |

Three pairs of dice are rolled simultaneously. In how many different ways can the sum of the numbers appearing on the top faces of the 3 dices be 9?

• Option 1)

18

• Option 2)

25

• Option 3)

22

• Option 4)

27

As learnt in concept Sum Rule of Association - Let A can occurs in m ways and B can occurs in n ways and both cannot occur simultaneously. Then A or B can occurs in (m + n) ways. - wherein A or B means at least one of them.     Sum=9 can be written in several ways. (i)    6+2+1    6 ways (ii)    4+3+2    6 ways (iii)    3+3+3    1 way (iv)    2+2+5    3 ways (v)    1+3+5    6 ways (vi)   ...
Engineering
122 Views   |

There are two parallel lines with 6 points on one line and 8 points on the other. How many quadrilaterals can be formed by joining points on the two lines?

• Option 1)

210

• Option 2)

205

• Option 3)

418

• Option 4)

420

as learnt in Rule of Geometrical Permutations - There are n points in a plane such that no three of them are in the same straight line then the number of lines that can be formed by joining is/are  and number of triangle is/are . -     To form a quadrilateral we need two points from each line.  We get,  Option 1) 210 This option is incorrect. Option 2) 205 This option is incorrect. Option...
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