Q&A - Ask Doubts and Get Answers

Sort by :
Clear All
Q
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 .

View More
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}

View More
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

View More
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

View More
Engineering
88 Views   |  

 If z is a complex number such that \left | z \right |\geq 2,   then the minimum value of \left | z+\frac{1}{2} \right |:  

  • Option 1)

    is strictly greater than \frac{5}{2}

  • Option 2)

    is strictly greater than \frac{3}{2}  but less than \frac{5}{2}

  • Option 3)

    is equal to  \frac{5}{2}

  • Option 4)

    lies in the interval (1, 2)

 

As we have learned

Triangle Law of Inequality in Complex Numbers -

|z_{1}-z_{2}|\geq \left || z_{1} \right |-| z_{2} \right |||

- wherein

|.| denotes modulus of z in complex numbers

 

 \left | z+\frac{1}{z} \right |= \left | z-(-\frac{1}{z}) \right |

\geq \left | |z|- (-\frac{1}{z}) \right |

= |z| - 1/z (\because |z|\geq 2)

 

\geq 2-1/2 = 3/2

 

\left | z+1/z \right |\geq 3/2

3/2 lies in the interval (1,2)

 

 

 

 

 


Option 1)

is strictly greater than \frac{5}{2}

Option 2)

is strictly greater than \frac{3}{2}  but less than \frac{5}{2}

Option 3)

is equal to  \frac{5}{2}

Option 4)

lies in the interval (1, 2)

View More
Engineering
128 Views   |  

 A  complex  number  z  is  said  to  be   unimodular if  \left | z \right |=1. Suppose z1 and z2 are complex numbers such that

     \frac{z_{1-2z_{2}}}{2-z_{1}\bar{z}_{2}} is unimodular and z2 is not unimodular.Then the point \frac{z_{1-2z_{2}}}{2-z_{1}\bar{z}_{2}}1 lies on a :

  • Option 1)

    straight line parallel to x-axis.

  • Option 2)

    straight line parallel to y-axis.

  • Option 3)

    circle of radius 2.

  • Option 4)

    circle of radius \sqrt{2}

 

As we have learned

Property of conjugate of complex number -

z\bar{z}=\left |z \right |^{2}

- wherein

  z=x+iy\bar{z}=conjugate \: of\: z   

 \left |z \right |=\sqrt{x^{2}+y^{2}}

 

 Given , \left | \frac{z_1-2z_2}{2-z_1\bar{z}_2} \right |= 1

\Rightarrow |(z_1-2z_2)|^2= |(z-z_1\bar{z}_2)|^2

\Rightarrow (z_1-2z_2)(\bar{z}_1-2\bar{z} _2)= (2-z_1\bar{z}_2)(2-\bar{z}_1z_2)

\Rightarrow (z_1)^2 (1-|z_2|^2)= 4 (1-|z_2|^2)

\Rightarrow |z_1|^2 = 4 (\because |z|\neq 1)

\Rightarrow |z_1|= 2

 

 

 

 

 

 


Option 1)

straight line parallel to x-axis.

Option 2)

straight line parallel to y-axis.

Option 3)

circle of radius 2.

Option 4)

circle of radius \sqrt{2}

View More
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

View More
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

View More
Engineering
440 Views   |  

The length of a focal chord of the parabola y2 = 4ax at a distance b from the vertex is C. Then

  • Option 1)

    2a2 = bc 

  • Option 2)

    a3 = b2c

  • Option 3)

    ac = b2

  • Option 4)

    b2c = 4a3

 
Find intersections A(x1, y1), B(x2, y2) of P & L Eliminate y from (1) & (2): m2x2 - 2am x + m2 a2 = 4ax  m2 x2-2a(m + 2) x + m2 a2 = 0                     ..............(3) x1, x2 are the roots, x1 + x2 = 2a (m+2)/m2 ;; x1 x2 = a2         .........(4) Eliminate x from (1) & (2): y = m(y2/4a) - ma my2 - 4a y-4a2 m = 0                        ..........(5) y1, y2 are the roots, y1 + y2 = 4a/m   ...
Engineering
136 Views   |  

The inverse of the function   y=\left [ 1-\left ( x-3 \right )^{4} \right ]^{\frac{1}{7}}  is

  • Option 1)

    3+\left ( 1-x^{7} \right )^{\frac{1}{4}}

  • Option 2)

    3-\left ( 1+x^{7} \right )^{\frac{1}{4}}

  • Option 3)

    3-\left ( 1-x^{7} \right )^{\frac{1}{4}}

  • Option 4)

    None of these

 
As we learnt in  Co - Domain of function - All possible outcomes for the function f(x) is known as co - domain unless not specified in question. -     Range of function - All possible values of  domain   is known as Range  -     Option 1) Option is correct Option 2) Option is incorrect Option 3) Option is incorrect Option 4) None of these Option is incorrect
Engineering
118 Views   |  

The radius of the circle passing through the points (1,2) (5,2)and (5,-2) is 

  • Option 1)

    5\sqrt{2}

  • Option 2)

    2\sqrt{5}

  • Option 3)

    3\sqrt{2}

  • Option 4)

    2\sqrt{2}

 
As we learnt in Equation of a circle - - wherein Circle with centre and radius .    On substituting the  value of  (x, y) as (1, 2), we get              ....................(1)  On substituting the  value of  (x, y) as (5, 2), we get         .........................(2) On substituting the  value of  (x, y) as (5, -2), we get        .......................(3) On subtracting  Eq. (1) - (2), we...
Engineering
104 Views   |  

The radius of the circle passing through the points (1,2),(5,2) and (5,-2) is 

  • Option 1)

    5\sqrt{2}

  • Option 2)

    2\sqrt{5}

  • Option 3)

    3\sqrt{2}

  • Option 4)

    2\sqrt{2}

 

As we learnt in

Equation of a circle -

\left ( x-h \right )^{2}+\left ( y-k \right )^{2}= r^{2}

- wherein

Circle with centre \left ( h,k \right ) and radius r.

 

 On substituting the  value of  (x, y) as (1, 2), we get 

(1-h)^{2}+(2-k)^{2}=r^{2}            ....................(1)

 On substituting the  value of  (x, y) as (5, 2), we get

(5-h)^{2}+(2-k)^{2}=r^{2}        .........................(2)

On substituting the  value of  (x, y) as (5, -2), we get

(5-h)^{2}+(-2-k)^{2}=r^{2}       .......................(3)

On subtracting  Eq. (1) - (2), we get 

h = 3

On subtacting Eq. (2) - Eq. (3), we get

k = 0

\therefore r = 2\sqrt{2}

 

 

 

On


Option 1)

5\sqrt{2}

Incorrect

Option 2)

2\sqrt{5}

Incorrect

Option 3)

3\sqrt{2}

Incorrect

Option 4)

2\sqrt{2}

Correct

View More
Engineering
102 Views   |  

\lim_{x \to b}\frac{\sqrt{x-a}-\sqrt{b-a}}{x^{2}-b^{2}}

  • Option 1)

    \frac{1}{4b\sqrt{a-b}}

  • Option 2)

    \frac{1}{4b\sqrt{b-a}}

  • Option 3)

    \frac{1}{4a\sqrt{a-b}}

  • Option 4)

    \frac{1}{b\sqrt{b-a}}

 
As we learnt in Method of Rationalisation - Rationalisation method is used when we have RADICAL SIGNS in an expression.(like  1/2,  1/3 etc) and there exists a negative sign between two terms of an algebraic expression. - wherein                 Option 1) Incorrect Option 2) Correct Option 3) Incorrect Option 4) Incorrect
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.

View More
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...
Exams
Articles
Questions