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Engineering
111 Views   |

Let A and B denote the statements

$A:\cos \alpha +\cos \beta +\cos \gamma = 0$

$B:\sin \alpha +\sin \beta +\sin \gamma = 0$

If $\cos \left (\beta -\gamma \right )+\cos \left ( \gamma -\alpha \right )+\cos \left ( \alpha -\beta \right )= -\frac{3}{2},$ then

• Option 1)

A is false and B is true

• Option 2)

Both A and B are true

• Option 3)

Both A and B are false

• Option 4)

A is true and B is false

As we learnt in   Addition Formulae - - wherein A and B are two angles.     Adding   to equation, we get              This is only true if  Hence both statements true. Option 1) A is false and B is true This option is incorrect. Option 2) Both A and B are true This option is correct. Option 3) Both A and B are false This option is incorrect. Option 4) A is true and B is false This...
Engineering
125 Views   |

If  $\vec{u},\vec{v},\vec{w}$ are non-coplanar vectors and p,q,are real number, then the equality,$\left [ 3\vec{u}\: p\vec{v}\: p\vec{w} \right ]-\left [ p\vec{v}\: \vec{w}\: q\vec{u} \right ]-\left [ 2\vec{w}\: q\vec{v}\: q\vec{u} \right ]= 0$

hold for

• Option 1)

exactly two values of  $\left ( p,q \right )$

• Option 2)

more than two but not all values of $\left ( p,q \right )$

• Option 3)

all values of  $\left ( p,q \right )$

• Option 4)

exactly one value of  $\left ( p,q \right )$

As we learnt in  Scalar Triple Product - - wherein Scalar Triple Product of three vectors .     But Option 1) exactly two values of  This option is incorrect. Option 2) more than two but not all values of This option is incorrect. Option 3) all values of  This option is incorrect. Option 4) exactly one value of  This option is correct.
Engineering
170 Views   |

The projections of a vector on the three coordinate axis are 6, -3, 2 respectively. The direction cosines  of the vector are

• Option 1)

$\frac{6}{5},\frac{-3}{5},\frac{2}{5}$

• Option 2)

$\frac{6}{7},\frac{-3}{7},\frac{2}{7}$

• Option 3)

$\frac{-6}{7},\frac{-3}{7},\frac{2}{7}$

• Option 4)

$6,-3,2$

As we learnt in  Direction Cosines of a Vector -  are the Direction Cosines. They are denoted by l, m, n,  - wherein    Projections are (6,-3,2) Magnitude = Option 1) This solution is incorrect  Option 2) This solution is correct  Option 3) This solution is incorrect  Option 4) This solution is incorrect
Engineering
104 Views   |

The differential equation which represents the family of curves $y= c_{1}e^{c_{2}x},$ where $c_{1}\: and\: c_{2}$ are arbitrary constants, is

• Option 1)

$y{}''= {y}'y$

• Option 2)

$y{y}''= {y}'$

• Option 3)

$y{y}''= \left ( {y}' \right )^{2}$

• Option 4)

${y}'= y^{2}$

As we learnt in  Formation of Differential Equations - A differential equation can be derived from its equation by the process of differentiation and other algebraical process of elimination -     Option 1) This is incorrect option Option 2) This is incorrect option Option 3) This is correct option Option 4) This is incorrect option
Engineering
99 Views   |

Let y be an implicit function of  x defined by $x^{2x}-2x^{x}\cot y-1= 0$

then ${y}'\left ( 1 \right )$ equals

• Option 1)

$1$

• Option 2)

$\log 2$

• Option 3)

$-\log 2$

• Option 4)

$0$

As we learnt in  Differential Equations - An equation involving independent variable (x), dependent variable (y) and derivative of dependent variable with respect to independent variable  - wherein eg:        Put x=1                         at    Correct optoin is 1 Option 1) Correct Option 2) Incorrect Option 3) Incorrect Option 4) Incorrect
Engineering
104 Views   |

Let the line $\dpi{100} \frac{x-2}{3}= \frac{y-1}{-5}= \frac{z+2}{2}$ lie in the plane

$\dpi{100} x+3y-\alpha z+\beta = 0$.Then $\dpi{100} \left ( \alpha ,\beta \right )$ equals to

• Option 1)

$(-6,7)$

• Option 2)

$(5,-15)$

• Option 3)

$(-5,5)$

• Option 4)

$(6,-17)$

As we learnt in  Condition for line to be lie in plane - and or   and -     (2,1,-2) lies in  Also,  Since line 1 isperpendicular to normal plane, 3-15-2k=0     k= -6   Option 1) Correct answer Option 2) Incorrect answer Option 3) Incorrect answer Option 4) Incorrect answer
Engineering
120 Views   |

Let a, b, c be such that  $b(a+c)$ $\neq$ $0$.If

then the value of $n$ is

• Option 1)

any even integer

• Option 2)

any odd integer

• Option 3)

any integer

• Option 4)

zero

As we learnt in

Value of determinants of order 3 -

-

$\begin{vmatrix} a &a+1&a-1 \\ -b & b+1 &b-1\\ c&c-1& c+1 \end{vmatrix}+\begin{vmatrix} a+1&b+1&c-1 \\ a-1& b-1 &c+1\\ \left ( -1\right )^{n+2}a& \left ( -1\right )^{n+1}b& \left ( -1\right )^{n}c \end{vmatrix}= 0$

$\begin{vmatrix} a &a+1&a-1 \\ -b & b+1 &b-1\\ c&c-1& c+1 \end{vmatrix}+\begin{vmatrix} a+1&b+1&c-1 \\ a-1& b-1 &c+1\\ \left ( -1\right )^{n}a& \left ( -1\right )^{n}b& \left ( -1\right )^{n}c \end{vmatrix}= 0$

$\begin{vmatrix} a &a+1&a-1 \\ -b & b+1 &b-1\\ c&c-1& c+1 \end{vmatrix}+(-1)^{n}\begin{vmatrix} a+1&b+1&c-1 \\ a-1& b-1 &c+1\\ a& -b& c \end{vmatrix}= 0$

$C_{1}\leftrightarrow C_{2}$

$\begin{vmatrix} a+1 &a&a-1 \\ b+1 & -b &b-1\\ c-1&c& c+1 \end{vmatrix}+(-1)^{n}\begin{vmatrix} a+1&b+1&c-1 \\ a-1& b-1 &c+1\\ a& -b& c \end{vmatrix}= 0$

$C_{2}\leftrightarrow C_{3}$

$\begin{vmatrix} a+1 &a-1&a \\ b+1 & b-1 &-b\\ c-1&c+1& c \end{vmatrix}+(-1)^{n}\begin{vmatrix} a+1&b+1&c-1 \\ a-1& b-1 &c+1\\ a& -b& c \end{vmatrix}= 0$

$\begin{vmatrix} a+1 &a-1&a \\ b+1 & b-1 &-b\\ c-1&c+1& c \end{vmatrix}\ (1+(-1)^{n})=0$

$\therefore 1+(-1)^{n}=0$

So n is any odd integer.

Option 1)

any even integer

Incorrect Option

Option 2)

any odd integer

Correct Option

Option 3)

any integer

Incorrect Option

Option 4)

zero

Incorrect Option

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Engineering
154 Views   |

Directions : Questions are Assertion- Reason type questions. Each of these questions contains two statements :

Statement- 1 (Assertion) and Statement - 2 (Reason).

Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.

Question : Let $A$ be a  2 x 2 matrix

Statement-1 : $\dpi{100} adj\, (adj\; A)=A$

Statement-2 : $\dpi{100} \left |adj\; A \right |=\left | A \right |$

• Option 1)

Statement-1 is true,Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

• Option 2)

Statement- 1 is true, Statement-2 is false

• Option 3)

Statement-1 is false, Statement-2 is true

• Option 4)

Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1

As we learnt in

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

$|adjA|= |A|^{n-1}= |A|^{2-1}= |A|$      Statement 2 is true

$adj|adjA|= |A|^{n-2}A= |A|^{0}A= A$  Statement 1 is true

Option 1)

Statement-1 is true,Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

Correct Option

Option 2)

Statement- 1 is true, Statement-2 is false

Incorrect Option

Option 3)

Statement-1 is false, Statement-2 is true

Incorrect Option

Option 4)

Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1

Incorrect Option

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Engineering
133 Views   |

Given $\dpi{100} P\left ( x \right )= x^{4}+ax^{3}+bx^{2}+cx+d$ such that x = 0 is the only real root of $\dpi{100} P\left ( x \right )= 0$.If P(-1) < P(1) then in the interval $\dpi{100} \left [ -1,1 \right ]$:

• Option 1)

P(-1)  is not minimum but P(1) is the maxmium of P

• Option 2)

P(-1)  is  minimum but P(1) is not the maxmium of P

• Option 3)

neither P(-1)  is the  minimum nor  P(1) is the maxmium of P

• Option 4)

P(-1)  is  minimum and P(1) is the maxmium of P

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 Option 1) P(-1)  is not minimum but P(1) is the maxmium of P Correct Option 2) P(-1)  is  minimum but P(1) is not the maxmium of P Incorrect Option 3) neither P(-1)  is the  minimum nor  P(1) is the...
Engineering
140 Views   |

Directions : Questions are Assertion- Reason type questions. Each of these questions contains two statements :

Statement- 1 (Assertion) and Statement - 2 (Reason).

Each of these questions also has four alternative choices, only one of which is the correct answer. You have to select the correct choice.

Question :

Statement- 1 : $\dpi{100} \sim (p\leftrightarrow \sim q)\;$ is equivalent to $\dpi{100} p\leftrightarrow q$

Statement- 2 : $\dpi{100} \sim (p\leftrightarrow \sim q)\;$   is a tautology.

• Option 1)

Statement-1 is true,Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

• Option 2)

Statement- 1 is true, Statement-2 is false

• Option 3)

Statement-1 is false, Statement-2 is true

• Option 4)

Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1

As we learnt in If and only if - -     Tautology - A statement pattern is called tautalogy, if it is always true, whatever may be the truth values of constitute statements. -     Fallacy or contradiction - A statement pattern is called a fallacy if it is always false, whatever may be the truth values of its constituent statements. -     is equivalent  to    as their truth tables are...
Engineering
108 Views   |

If P and Q are the points of intersection of the circles

$x^{2}+y^{2}+3x+7y+2p-5= 0$

and $x^{2}+y^{2}+2x+2y-p^{2}= 0$, then there is a circle passing through P, Q and (1,1)for

• Option 1)

all except one value of  p

• Option 2)

all except two values of  p

• Option 3)

exactly one value of  p

• Option 4)

all values of p

As we learnt in Family of circle - - wherein Equation of the family of circles passing through point of intersection .                                                                                                                                                                                  Equation of common chord family of circles is It passes through (1,1) So,  Option 1) all except...
Engineering
116 Views   |

The ellipse $x^{2}+4y^{2}= 4$ is inscribed in a rectangle aligned with the coordinate axes, which in turn is inscribed in another ellipse that passes through the point (4, 0). Then the equation of the ellipse is

• Option 1)

$x^{2}+12y^{2}=16$

• Option 2)

$4x^{2}+48y^{2}=48$

• Option 3)

$4x^{2}+64y^{2}=48$

• Option 4)

$x^{2}+16y^{2}=16$

As we learnt in Stanard equation -   - wherein Semi major axis Semi minor axis  Equation is   It passes through (2, 1) Solving,  Hence equation is    Option 1) Correct Option 2)   Incorrect Option 3) Incorrect Option 4) Incorrect
Engineering
646 Views   |

Three distinct points  A,B and C are given in the 2-dimensional coordinate plane such that the ratio of the distance of any one of them from the point (1, 0) to the distance from the point ( –1, 0) is equal to 1/3. Then the circumcentre of the triangle ABC is at the point

• Option 1)

$\left ( \frac{5}{4},0 \right )$

• Option 2)

$\left ( \frac{5}{2},0 \right )$

• Option 3)

$\left ( \frac{5}{3},0 \right )$

• Option 4)

$\left ( 0,0 \right )$

As we learnt in  Circumcentre - The point of intersection of the perpendicular bisectors of the sides of a triangle. - wherein The centre of the circumcircle of a triangle.    Let x be (h,k) A,B, C lie on locus of x. given that  Centre is Circumstance is    Option 1) This option is correct. Option 2) This option is incorrect. Option 3) This option is incorrect. Option 4) This option...
Engineering
142 Views   |

The lines $\dpi{100} p\left ( p^{2}+1 \right )x-y+q= 0$ and

$\dpi{100} \left ( p^{2}+1 \right )^{2}\! x+\left ( p^{2}+1 \right )y+2q= 0$ are perpendicular to a common line for .

• Option 1)

exactly one value of  p

• Option 2)

exactly two value of  p

• Option 3)

more than two value of  p

• Option 4)

No value of  p

As we learnt in  Parallel lines -     - wherein The two lines are   and    If two lines are perpendicular to same line, they must be parallel. only one value   Option 1) exactly one value of  p This option is correct. Option 2) exactly two value of  p This option is incorrect. Option 3) more than two value of  p This option is incorrect. Option 4) No value of  p This option is incorrect.
Engineering
125 Views   |

The shortest distance between the line $\dpi{100} y-x = 1$ and the curve $\dpi{100} x=y^{2}$ is

• Option 1)

$\frac{2\sqrt{3}}{8}$

• Option 2)

$\frac{3\sqrt{2}}{5}$

• Option 3)

$\frac{\sqrt{3}}{4}$

• Option 4)

$\frac{3\sqrt{2}}{8}$

As we learnt in Perpendicular distance of a point from a line -     - wherein   is the distance from the line .    Let (a2 , a) be a point on x = y2 Distance between (a2, a) and x- y+ 1= is It is minimum when So minimum distance =    Option 1) This is incorrect. Option 2) This is incorrect. Option 3) This is incorrect. Option 4) This is incorrect.
Engineering
123 Views   |

In a binomial distribution $\dpi{100} B\left ( n,p= \frac{1}{4} \right ),$ if the probability of at least one success is greater than or equal to $\dpi{100} \frac{9}{10}$,then n is greater than

• Option 1)

$\frac{1}{\log_{10}4+\log_{10}3}$

• Option 2)

$\frac{9}{\log_{10}4-\log_{10}3}$

• Option 3)

$\frac{4}{\log_{10}4-\log_{10}3}$

• Option 4)

$\frac{1}{\log_{10}4-\log_{10}3}$

As we learnt in Binomial Distribution - Let E be an event and p+q = 1 then X :          0                         1                            2         ..................     n P(x):      qn                                          pn -         n [log 4 - log 3] log 10 Thus  Option 1) this is incorrect option Option 2) this is correct option Option 3) this is incorrect option Option...
Engineering
183 Views   |

One ticket is selected at random from 50 tickets numbered 00, 01, 02, ...., 49. Then the probability that the sum of the digits on the selected ticket is 8, given that the product of these digits is zero, equals

• Option 1)

$\frac{1}{7}$

• Option 2)

$\frac{5}{14}$

• Option 3)

$\frac{1}{50}$

• Option 4)

$\frac{1}{14}$

As we learnt in  Probability of occurrence of an event - Let S be the sample space then the probability of occurrence of an event E is denoted by P(E) and it is defined as  - wherein Where n repeated experiment and E occurs r times.   If the product of the digit is zero, ticket must be one of 00, 01, 02, --- 09, 10, 20, 30, 40. Total of 14 numbers, only one with sum= 8. P(A)= Option...
Engineering
111 Views   |

For real x,let $\dpi{100} f\left ( x \right )= x^{3}+5x+1$ then

• Option 1)

$f$ is onto $R$ but not one-one

• Option 2)

$f$ is one-one and onto $R$

• Option 3)

$f$ is neither one-one nor onto $R$

• Option 4)

$f$ is one-one but not  onto $R$

As we learnt in

Onto function -

If  f:A$\rightarrow$B is such that each & every element in B is the image of atleast one element in A.Then it is Onto function.

- wherein

The range of f is equal to codomain of f.

One - One or Injective function -

A line parallel to x - axis cut the curve at most one point.

-

f(x) = x3 + 5x + 1

f'(x) = 3x2 + 5 > 0

Which is strictly increasing function so that it is one - one function.

For $x\epsilon R$      $f(x)\epsilon R$

So it is onto function.

Correct option is 2.

Option 1)

$f$ is onto $R$ but not one-one

This is an incorrect option.

Option 2)

$f$ is one-one and onto $R$

This is the correct option.

Option 3)

$f$ is neither one-one nor onto $R$

This is an incorrect option.

Option 4)

$f$ is one-one but not  onto $R$

This is an incorrect option.

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Engineering
139 Views   |

If A,B and C are three sets such that

$\dpi{100} A\cap B= A\cap C \: and \: A\cup B= A\cup C$ then

• Option 1)

$A=C$

• Option 2)

$B=C$

• Option 3)

$A\cap B= \phi$

• Option 4)

$A= B$

As we learnt in

UNION OF SETS -

Let A and B be any two sets. The union of A and B is the set which consists of all the elements of A and all the elements of B, the common elements being taken only once. The symbol ‘∪’ is used to denote the union.

- wherein

Symbolically, we write A U B = {x: x ∈ A or x ∈ B}.

INTERSECTION OF SETS -

The intersection of sets A and B is the set of all elements which are common to both A and B. The symbol ‘∩ ‘is used to denote the intersection.

- wherein

Symbolically, we write A ∩ B = {x: x ∈ A and x ∈ B}.

$A\cap B=A\cap C$  and  $A\cup B=A\cup C$ then B = C

Let $x\epsilon A\cap B$

$x\epsilon A\ \; and\ \; x\epsion B$ and $y\epsilon A$  or  $y\epsilon B$

So $y\epsilon A$  and $y\epsilon C$  and  $y\epsilon A$  or   $y\epsilon B$

Both will be equal only and only if B = C

Correct option is 2.

Option 1)

$A=C$

This is an incorrect option.

Option 2)

$B=C$

This is the correct option.

Option 3)

$A\cap B= \phi$

This is an incorrect option.

Option 4)

$A= B$

This is an incorrect option.

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Engineering
108 Views   |

Let $\dpi{100} f(x)=(x+1)^{2}-1,x\geq -1$

Statement - 1 : The set $\dpi{100} \left \{ x:f(x)=f^{-1}(x) \right \}=\left \{ 0,-1 \right \}$

Statement - 2 : $\dpi{100} f$ is a bijection.

Statement - 1 (Assertion) and Statement - 2 (Reason).

• Option 1)

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

• Option 2)

Statement- 1 is true, Statement-2 is false

• Option 3)

Statement-1 is false, Statement-2 is true

• Option 4)

Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1

As we learnt in

Bijective Function -

The function which is both one-one and onto is Bijective Function.

-

f(x) = (x + 1)2 - 1     $x\geq-1$

Let  y = (x + 1)2 - 1

$\therefore\ \; x+1=\pm\sqrt{y+1}$

$\therefore\ \; x =\pm\sqrt{y+1}-1$

Correct option is 2.

Option 1)

Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1

This is an incorrect option.

Option 2)

Statement- 1 is true, Statement-2 is false

This is the correct option.

Option 3)

Statement-1 is false, Statement-2 is true

This is an incorrect option.

Option 4)

Statement-1 is true, Statement-2 is true; Statement-2 is correct explanation for Statement-1

This is an incorrect option.

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