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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 f_{k}\left ( x \right )= \frac{1}{k}\left ( \sin ^{k}x+\cos ^{k}x \right )where x\epsilon R\: \: and\: \: k\geqslant 1      Then f_{4}(x)-f_{6}(x)  equals :

  • Option 1)

    \frac{1}{4}

  • Option 2)

    \frac{1}{12}

  • Option 3)

    \frac{1}{6}

  • Option 4)

    \frac{1}{3}

 

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)

    \frac{121}{10}

  • Option 4)

    \frac{441}{100}

 

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)

\frac{121}{10}

Option 4)

\frac{441}{100}

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Engineering
319 Views   |  

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)

    \left ( 14,\; \frac{272}{3} \right )

  • Option 2)

    \left ( 16,\; \frac{272}{3} \right )

  • Option 3)

    \left ( 16,\; \frac{251}{3} \right )

  • Option 4)

    \left ( 14,\; \frac{251}{3} \right )

 
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
282 Views   |  

The sum of first 20 terms of the sequence 0.7 ,0.77,0.777,.........,is :

  • Option 1)

    \frac{7}{9}\left ( 99+10^{-20} \right )

  • Option 2)

    \frac{7}{81}\left ( 179-10^{-20} \right )

  • Option 3)

    \frac{7}{9}\left ( 99-10^{-20} \right )

  • Option 4)

    \frac{7}{81}\left ( 179+10^{-20} \right )

 

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)

\frac{7}{9}\left ( 99+10^{-20} \right )

Option 2)

\frac{7}{81}\left ( 179-10^{-20} \right )

Option 3)

\frac{7}{9}\left ( 99-10^{-20} \right )

Option 4)

\frac{7}{81}\left ( 179+10^{-20} \right )

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Engineering
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The graph of the function  y=f\left ( x \right )  is symmetrical about the line 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

Symmetric about Y - axis

 

 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   f(x)+2f\left ( \frac{1}{x} \right )=3x,x\neq 0,  and  S=\left \{ x\, \epsilon \, R : f(x)=f(-x)\right \};then S:

  • 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
124 Views   |  

The real number k for which the equation, 2x^2+3x+k = 0  has two distinct real roots in \left [ 0,1 \right ]

 

 

  • 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
124 Views   |  

Let \alpha and \beta be the roots of equation px^{2}+qx+r=0,p\neq 0.\; if\; p,q,r are in A.P. and \frac{1}{\alpha }+\frac{1}{\beta }=4,   then the value of \left | \alpha -\beta \right | is ?

  • Option 1)

    \frac{\sqrt{34}}{9}

  • Option 2)

    \frac{2\sqrt{13}}{9}

  • Option 3)

    \frac{\sqrt{61}}{9}

  • Option 4)

    \frac{2\sqrt{17}}{9}

 

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)

\frac{\sqrt{34}}{9}

Option 2)

\frac{2\sqrt{13}}{9}

Option 3)

\frac{\sqrt{61}}{9}

Option 4)

\frac{2\sqrt{17}}{9}

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Engineering
133 Views   |  

If z is a complex number of unit modulus and argument \theta ,then arg  (\frac{1+z}{1+\bar{z}})  equals:

  • Option 1)

    \pi - \theta

  • Option 2)

    -\theta

  • Option 3)

    \dfrac{\pi}{2}-\theta

  • Option 4)

    \theta

 

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)

\pi - \theta

Option 2)

-\theta

Option 3)

\dfrac{\pi}{2}-\theta

Option 4)

\theta

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Engineering
135 Views   |  

If the equations x^{2}+2x+3=0\; and\; ax^{2}+bx+c=0,a,b,c\; \epsilon R, 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
105 Views   |  

If    is the adjoint of a 3 x 3  matrix A and \left | A \right | = 4,then \alpha 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 A\timesB 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
95 Views   |  

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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