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Engineering
106 Views   |  

The function  \sin ^{2}\left ( \omega t \right ) represents

  • Option 1)

    a simple harmonic motion with a period  \frac{2\pi }{\omega }

  • Option 2)

    a simple harmonic motion with a period  \frac{\pi }{\omega }

  • Option 3)

    a periodic, but not simple harmonic motion with a period \frac{2\pi }{\omega }

  • Option 4)

    a periodic, but not simple harmonic motion with a period \frac{\pi }{\omega }

 
It is a periodic motion but it is not  SHM Heance a periodic, but not simple harmonic motion with a period  is correct It is a periodic motion but it is not s.n Angular speed 2w   period 2= Option 1) a simple harmonic motion with a period  Incorrect option Option 2) a simple harmonic motion with a period  Incorrect option Option 3) a periodic, but not simple harmonic motion with a...
Engineering
102 Views   |  

If \cos ^{-1}x-\cos ^{-1}\frac{y}{2}=\alpha , then 4x^{2}-4xy\cos \alpha +y^{2} is equal to

  • Option 1)

    4

  • Option 2)

    2\sin 2\alpha

  • Option 3)

    -4\sin ^{2}\alpha

  • Option 4)

    4\sin ^{2}\alpha

 
As we learnt in  Formulae of Inverse Trigonometric Functions - - wherein     Squaring, we get   Option 1) Incorrect Option 2) Incorrect Option 3) Incorrect Option 4) Correct
Engineering
103 Views   |  

In a triangle PQR, i f \angle R= \frac{\pi }{2}.If \tan \left ( \frac{P}{2} \right )\: and \: \tan \left ( \frac{Q}{2} \right )  are the root of ax^{2}+bx+c =0,a\neq 0

  • Option 1)

    b=a+c

  • Option 2)

    b=c

  • Option 3)

    c=a+b

  • Option 4)

    a=b+c

 
As we learnt in  Trigonometric Equations - The equations involving trigonometric function of unknown angles are known as trigonometric equations. - wherein e.g.     Sum of roots  =                                   Similarly, product  of roots =      Divides (2) and (3) we get     Now (2a) - (5) gives, we get      Option 1) Incorrect Option 2) Incorrect Option 3) Correct Option 4) Incorrect
Engineering
120 Views   |  

Let f:\left ( -1,1 \right )\rightarrow B, be a function defined by f\left ( x \right )= \tan ^{-1}\left ( \frac{2x}{1-x^{2}} \right )

then f is both one­-one and onto when B is the interval

  • Option 1)

  • Option 2)

    \left ( 0,\frac{ \pi }{2} \right )

  • Option 3)

    \left ( -\frac{ \pi }{2},\frac{ \pi }{2} \right )

  • Option 4)

    \left [- \frac{ \pi }{2},\frac{ \pi }{2} \right ]

 
As we learnt in  Domains and Ranges of Inverse Trigonometric Functions - For  Domain  Range  -     Thus, range of f(x) is same as range of 2 tan-1x in the given domain. As domain is (-1,1), range of 2tan-1x will lie between Option 1) This option is incorrect Option 2) This option is incorrect Option 3) This option is correct Option 4) This option is incorrect
Engineering
110 Views   |  

The differential equation representing the family of curves y^{2}=2c(x+\sqrt{c}),\; where \; c> 0, is a parameter, is of order and degree as follows

  • Option 1)

    order 1, degree 1

  • Option 2)

    order 1, degree 2

  • Option 3)

    order 2, degree 2

  • Option 4)

    order 1, degree 3

 
As we learnt in  Degree of a Differential Equation - Degree of Highest order differential coefficient appearing in it, provided it can be expressed as a polynomial equation in derivatives - wherein Degree = 2   Degree of a Differential Equation - Degree of Highest order differential coefficient appearing in it, provided it can be expressed as a polynomial equation in derivatives -...
Engineering
107 Views   |  

If x\frac{dy}{dx}=y(\log y-\log x+1), then the solution of the equation is

  • Option 1)

    x\log \left ( \frac{y}{x} \right )=cy\;

  • Option 2)

    \; y\log \left ( \frac{x}{y} \right )=cx\;

  • Option 3)

    \; \log \left ( \frac{x}{y} \right )=cy\;

  • Option 4)

    \; \log \left ( \frac{y}{x} \right )=cx\; \;

 
As we learnt in  Homogeneous Differential Equation - Put -     Put       Option 1) Incorrect option Option 2) Incorrect option Option 3) Incorrect option Option 4) Correct option
Engineering
101 Views   |  

If the plane 2ax-3ay+4az+6=0  passes through the midpoint of the line joining the centres of the spheres x^{2}+y^{2}+z^{2}+6x-8y-2z=13  and x^{2}+y^{2}+z^{2}-10x+4y-2z=8  then a equals

  • Option 1)

    1

  • Option 2)

    –1

  • Option 3)

    2

  • Option 4)

    –2

 
As we learnt in General equation (cartesian form ) - -    Centre of Centre of Midpoint of   2ax-3ay+4az+6=0 passes through (1,1,1) 2a-3a+4a+6=0 a=-2 Option 1) 1 Incorrect Option   Option 2) –1 Incorrect Option   Option 3) 2 Incorrect Option   Option 4) –2 Correct Option  
Engineering
101 Views   |  

The distance between the line \vec{r}=2\vec{i}-2\vec{j}+3\vec{k}+\lambda (\vec{i}-\vec{j}+4\vec{k}) and the plane \vec{r}\cdot (\vec{i}+5\vec{j}+\vec{k})=5  is

  • Option 1)

    \frac{10}{3\sqrt{3}}\;

  • Option 2)

    \; \; \frac{10}{9}\;

  • Option 3)

    \; \frac{10}{3}\;

  • Option 4)

    \; \frac{3}{10}

 
As we learnt in  Angle bisector containing origin - Equation of plane bisecting the angle between the planes  containing origin is  Where       are positive   -   Distance of (2,-2,3) from x+5y+z-5=0 is Option 1) Correct Option   Option 2) Incorrect Option   Option 3) Incorrect Option   Option 4) Incorrect Option  
Engineering
109 Views   |  

If the angle \Theta between the line \frac{x+1}{1}=\frac{y-1}{2}=\frac{z-2}{2}  and the plane 2x-y+\sqrt{\lambda z}+4=0 is such that \sin \Theta =\frac{1}{3}, the value of \lambda  is

  • Option 1)

    -\frac{3}{5}\;

  • Option 2)

    \; \frac{5}{3}\;

  • Option 3)

    \; \frac{-4}{3}\;

  • Option 4)

    \frac{3}{4}

 
As we learnt in  Angle between line and Plane (vector form ) - The angle between a line and the plane is given by - wherein    Angle between line and normal to plane   Option 1) Incorrect Option   Option 2) Correct Option   Option 3) Incorrect Option   Option 4) Incorrect Option  
Engineering
88 Views   |  

The plane x+2y-z=4 cuts the sphere x^{2}+y^{2}+z^{2}-x+z-2=0  in a circle of radius

  • Option 1)

    1

  • Option 2)

    3

  • Option 3)

    \sqrt{2}

  • Option 4)

    2

 
As we learnt in  Distance of a point from plane (Cartesian form) - The length of perpendicular from to the plane is given by    -   Centre is        Perpendicular distance from centre= Option 1) 1 Correct Option Option 2) 3 Incorrect Option Option 3) Incorrect Option Option 4) 2 Incorrect Option
Engineering
105 Views   |  

The angle between the lines 2x=3y=-z  and 6x=-y=-4z  is

  • Option 1)

    90º

  • Option 2)

  • Option 3)

    30º

  • Option 4)

    45º

 
As we learnt in  Condition for perpendicularity - or -    Since Angle between the lines= Option 1) 90º Correct Option   Option 2) 0º Incorrect Option   Option 3) 30º Incorrect Option   Option 4) 45º Incorrect Option  
Engineering
119 Views   |  

The system of equations \alpha x+y+z=\alpha -1, x+\alpha y+z=\alpha -1,x+y+\alpha z=\alpha -1 has no solutions , if \alpha is

  • Option 1)

    either –2 or 1

  • Option 2)

    -2

  • Option 3)

    1

  • Option 4)

    not –2.

 

As we learnt in 

Cramer's rule for solving system of linear equations -

When \Delta =0  and \Delta _{1}=\Delta _{2}=\Delta _{3}=0 ,

then  the system of equations has infinite solutions.

- wherein

a_{1}x+b_{1}y+c_{1}z=d_{1}

a_{2}x+b_{2}y+c_{2}z=d_{2}

a_{3}x+b_{3}y+c_{3}z=d_{3}

and 

\Delta =\begin{vmatrix} a_{1} &b_{1} &c_{1} \\ a_{2} & b_{2} &c_{2} \\ a_{3}&b _{3} & c_{3} \end{vmatrix}

\Delta _{1},\Delta _{2},\Delta _{3} are obtained by replacing column 1,2,3 of \Delta by \left ( d_{1},d_{2},d_{3} \right )  column

 

 

\alpha x+ y+z= \alpha -1

x+\alpha y+z= \alpha -1

x+y+\alpha z= \alpha -1

\therefore \begin{vmatrix} \alpha &1 &1 \\ 1& \alpha &\alpha \\ 1 &1 &\alpha \end{vmatrix}= 0

\therefore \alpha(\alpha^{2}-1)- (\alpha -1)+ (1-\alpha)= 0

\therefore (\alpha-1)(\alpha^{2} + \alpha -2)= 0

(\alpha-1)(\alpha+2) (\alpha-1)= 0

(\alpha-1)^{2}(\alpha+2) = 0

\alpha= -2

 


Option 1)

either –2 or 1

Incorrect option

Option 2)

-2

Correct option

Option 3)

1

Incorrect option

Option 4)

not –2.

Incorrect option

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Engineering
156 Views   |  

If a^{2}+b^{2}+c^{2}=-2\; and

then f(x)  is a polynomial of degree

  • Option 1)

    0

  • Option 2)

    1

  • Option 3)

    2

  • Option 4)

    3

 

As we learnt in 

Property of determinant -

If each element in a row ( or column ) of a determinant is written as the sum of two or more terms then the determinant can be written as the sum of two or more determinants

- wherein

 

 f(x)= \begin{vmatrix} 1+a^{2} x&(1+b^{2})x &(1+c^{2})x\\ (1+a^{2})x&1+b^{2}x &(1+c^{2})x \\ (1+a^{2})x&(1+b^{2})x & 1+c^{2}x \end{vmatrix}   

    and a^{2} + b^{2} +c^{2}= -2

f(x)= \begin{vmatrix} 1+a^{2} x&x+b^{2}x &x+c^{2}x \\ x+a^{2}x &1+b^{2}x &x+c^{2}x \\ x+a^{2}x &x+b^{2}x & 1+c^{2}x \end{vmatrix}

c_1\rightarrow c_1+c_2+ c_3

\begin{vmatrix} 1+2x+x(a^{2}+b^{2} +c^{2}) &x+b^{2}x &x+c^{2}x \\ 1+2x+x(a^{2}+b^{2} +c^{2})&1+b^{2}x &x+c^{2}x \\ 1+2x+x(a^{2}+b^{2} +c^{2})&x+b^{2}x &1+c^{2}x \end{vmatrix}

\begin{vmatrix} 1+2x+x(-2) &x+b^{2}x &x+c^{2}x \\ 1+2x+x(-2)&1+b^{2}x &x+c^{2}x \\ 1+2x+x(-2) &x+b^{2}x &1+c^{2}x \end{vmatrix}

\begin{vmatrix} 1 &x+b^{2}x &x+c^{2}x \\ 1&1+b^{2}x &x+c^{2}x \\ 1&x+b^{2}x &1+c^{2}x \end{vmatrix}

R_1\Rightarrow R_1-R_2\ and\ R_2\Rightarrow R_2-R_3

\begin{vmatrix} 0 &x-1 &0 \\ 0 &1-x &x-1 \\ 1& x+b^{2}x &1+c^{2}x \end{vmatrix}

= -(x-1) \begin{vmatrix} 0 &x-1 \\ 1& 1+c^{2}x \end{vmatrix}

=(1-x) (1-x) = (1-x)^{2}= x^{2}-2x+1

so degree 2. 


Option 1)

0

Incorrect option

Option 2)

1

Incorrect option

Option 3)

2

Correct option

Option 4)

3

Incorrect option

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Engineering
110 Views   |  

If     then which one of the following holds for all n\geq 1 , by the principle of mathematical induction

  • Option 1)

    A^{n}=2^{n-1}A-(n-1)I\; \;

  • Option 2)

    \; A^{n}=nA-(n-1)I\; \;

  • Option 3)

    \; A^{n}=2^{n-1}A+(n-1)I\;

  • Option 4)

    \; A^{n}=nA+(n-1)I

 

As we learnt in 

Multiplication of matrices -

-

 

 A=\begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}

I=\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}

A^{n}= nA- (n-1)I

A= \begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}

A^{2}= \begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix} \begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}= \begin{bmatrix} 1 &0 \\ 2& 1 \end{bmatrix}

A^{3}= \begin{bmatrix} 1 &0 \\ 2& 1 \end{bmatrix} \begin{bmatrix} 1 &0 \\ 1& 1 \end{bmatrix}= \begin{bmatrix} 1 &0 \\ 3& 1 \end{bmatrix}

A^{n}= \begin{bmatrix} 1 &0 \\ n& 1 \end{bmatrix}

now n(A)= \begin{bmatrix} n &0 \\ n& n \end{bmatrix}

(n-1)I= (n-1)\begin{bmatrix} 1 &0 \\ 0& 1 \end{bmatrix}= \begin{bmatrix} n-1 &0 \\ 0& n-1 \end{bmatrix}

\therefore \begin{bmatrix} n &0 \\ n & n \end{bmatrix} - \begin{bmatrix} n-1 &0 \\ 0 & n-1 \end{bmatrix}= \begin{bmatrix} 1 &0 \\ n & 1 \end{bmatrix}

 


Option 1)

A^{n}=2^{n-1}A-(n-1)I\; \;

Incorrect option

Option 2)

\; A^{n}=nA-(n-1)I\; \;

Correct option

Option 3)

\; A^{n}=2^{n-1}A+(n-1)I\;

Incorrect option

Option 4)

\; A^{n}=nA+(n-1)I

Incorrect option

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Engineering
101 Views   |  

If A^{2}-A+I=0, then the inverse of A is

  • Option 1)

    A

  • Option 2)

    A+I

  • Option 3)

    I-A

  • Option 4)

    A-I

 

As we learnt in 

Property of inverse of a matrix -

Every invertible matrix possesses a unique inverse 

-

 

 A^{2}-A+I=0

A^{2}A^{-1}-AA^{-1}+IA^{-1}=0

A(AA^{-1})-I+A^{-1}=0

AI-I+A^{-1}=0

\therefore A^{-1}=I-IA

So, A^{-1}=I-IA=I-A

 

 


Option 1)

A

Incorrect option

Option 2)

A+I

Incorrect option

Option 3)

I-A

Correct option

Option 4)

A-I

Incorrect option

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Engineering
114 Views   |  

A spherical iron ball 10 cm in radius is coated with a layer of ice of uniform thickness that melts at a rate of 50 cm^{3} /min. When the thickness of ice is 5 cm, then the rate at which the thickness of ice decreases, is

  • Option 1)

    \frac{1}{18\pi }cm/min

  • Option 2)

    \frac{1}{36\pi }cm/min

  • Option 3)

    \frac{5}{6\pi }cm/min

  • Option 4)

    \frac{1}{54\pi }cm/min

 
As we learnt in  Rate Measurement - Rate of any of variable with respect to time is rate of measurement. Means according to small change in time how much other factors change is rate measurement: - wherein Where dR / dt  means Rate of change of radius.     Option 1) This option is correct Option 2) This option is incorrect Option 3) This option is incorrect Option 4) This option...
Engineering
122 Views   |  

If f is a real­-valued differentiable function satisfying  

\left | f\left ( x \right ) -f\left ( y \right )\right |\leq \left ( x-y \right )^{2},x,y \in R and f\left ( 0 \right )= 0 then f\left ( 1 \right ) equals

  • Option 1)

    1

  • Option 2)

    2

  • Option 3)

    0

  • Option 4)

    -1

 
As we learnt in  Differentiability - Let  f(x) be a real valued function defined on an open interval (a, b) and   (a, b).Then  the function  f(x) is said to be differentiable at      if -     Since Hence Option 1) Incorrect option Option 2) Incorrect option Option 3) Correct option Option 4) Incorrect option
Engineering
134 Views   |  

Suppose f\left ( x \right ) is differentiable at  x= 1 \: and\: \lim_{h\rightarrow 0}\frac{1}{h}f\left ( 1+h \right )= 5,

then {f}'\left ( 1 \right ) equals

  • Option 1)

    4

  • Option 2)

    3

  • Option 3)

    6

  • Option 4)

    5

 
As we learnt in  Differentiability - Let  f(x) be a real valued function defined on an open interval (a, b) and   (a, b).Then  the function  f(x) is said to be differentiable at      if -     Now so that must be finite as exists But can be finite only if and Therefore Option 1) Incorrect option Option 2) Incorrect option Option 3) Incorrect option Option 4) Correct option
Engineering
100 Views   |  

let P be the point (1, 0) and Q a point on the locus y^{2}=8x. The locus of mid point of PQ is

  • Option 1)

    x^{2}-4y+2=0

  • Option 2)

    x^{2}+4y+2=0

  • Option 3)

    y^{2}+4x+2=0

  • Option 4)

    y^{2}-4x+2=0

 
As we learnt in  Parametric coordinates of parabola - - wherein For the parabola.    P(1,0) and let the Parametric Coordinate of Q be  Mid point PQ is   Hence,        and  k =2t Replace (h,k) with (x,y)   Option 1) Incorrect Option 2) Incorrect Option 3) Incorrect Option 4) Correct
Engineering
120 Views   |  

The line parallel to the x-axis and passing through the intersection of the lines ax+2by+3b=0\; and\; bx-2ay-3a=0,where\; (a,b)\neq (0,0)\; is

  • Option 1)

    below the x- axis at a distance of 2/3 from it

  • Option 2)

    below the x- axis at a distance of 3/2 from it

  • Option 3)

    above  the x- axis at a distance of 2/3 from it

  • Option 4)

    above  the x- axis at a distance of 3/2 from it

 
As we learnt in  Line parallel to the y-axis - - wherein The line passes through the point (b,o).    Point of intersection of the lines Point of intersection is  line parallel to x- axis is below at a distance  . i.e.  Option 1) below the axis at a distance of 2/3 from it Incorrect option     Option 2) below the axis at a distance of 3/2 from it Correct option Option 3) above  the ...
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