Explore topic-wise InterviewSolutions in .

Correct option is D) oil

Correct answer is (a) 4

Given (-10,-4)(-8,-1)(-3,-5)

area of triangle formula

1/2 |x₁ (y₂-y₃)+x₂ (y₃-y₁)+x₃ (y₁-y₂)|

take -10 as x₁,-4 as y₁,-8 as x₂,-1 as y₂,-3 as x₃,-5 as y₃

than

1/2|-10 (-1-(-5))+(-8)(-5-(-4))+(-3)(-4-(-1))|

1/2|-10 (-1+5)+(-8)(-5+4)+(-3)(-4+1)|

1/2|-10 (+4)+(-8)(-1)+(-3)(-3)|

1/2|40+8+9|

1/2|57|

|28.5|

Concepts:

If 'p' then 'q' , denoted by p → q where p and q are the hypothesis and conclusion respectively.

p → q denotes implies also,

~ p denotes not p means negation

~ q denotes not q means negation

q → p denotes converse, which means if q then p.

the converse is not true even if the implication is true.

~ p → ~ q denotes inverse, which means if not p then not q.

the inverse is not true even if the implication is true. 

~ q → ~ p  contrapositive,

the contrapositive is true if the implication is true and vice versa.

Given:

p : I like the music

q : I give you a prize

Truth table 

Where T is true denotes positive statement and F is False denotes the negative statement,

The logical equivalence from the truth table :

Implication and contrapositive result is the same i.e, if p happens then q will happen which implies if q won't happen then p also won't happen.

Thus "If I like the music then I will give you a prize" logical equivalent statement is "If I do not give you a prize then I have not liked the music".

Hence, option 1 is the correct answer.

NOTE:

This is not an english language problem, It is about discrete mathematics proposition logic equivalence.

A farmer grows 3 crops in rotation – A, B, C.

A is grown between the months Jan‐June.

B between April – September.

C between July‐December.

Each crop takes 3 months from sowing to harvesting. At a time only one crop can be grown.

Farmer starts in the month of January by sowing A. If there is no gap between each crop, then harvesting of A is March. B is sowing in the month of April and B is harvesting in the month of June and C is sowing in the month of July and harvesting in the month of September.

So, C would be harvested in the month of September.

Hence, September is the correct answer.

Correct option is B) hand over fist

Correct option is A) barely

Some supporters of 'party Y' did not agree for the alliance with the 'party X' : False because nothing is mentioned about alliance plan of 'party Y' supporters.

There is at least one supporter of 'party Y' who knew some supporters of 'party X' as a friend : True as it is mentioned clearly in the statement given in question.

No supporters of 'party X' supported Z's campaign strategy : False as it is mentioned in the statement that some supporters of 'party X' supported Z's campaign strategy.

No supporters of 'party X' knew Z : False as it is mentioned in the statement that some supporters of 'party X' knew Z.

Hence, option 2 is the correct answer.

Correct option is A) made up

Correct option is A) What I plan to do

option (a) is correct i.e. 2a      

cuz {a+a=2a}

Put \(y=x-2 \,in \,eqquation\, of \,ellipse\)

\(3x^2+4y^2=24\) we obtain

\(3x^2+4(x-2)^2=24\)

\(=3x^2+4x^2-16x+16=24\)

\(=7x^2-16x-8=0\)

\(=x=\frac{16\pm \sqrt{256+224}}{14}\)

\(x=\frac{16\pm \sqrt{480}}{14}=\frac{8\pm2\sqrt{30}}{7}\)

\(y=\frac{8+2\sqrt{30}}{7}-2\,or\,y=\frac{8-2\sqrt{30}}{7}-2\)

\(=y=\frac{2\sqrt{30}-6}{7}\,or\,y=\frac{-2\sqrt{30}-6}{7}\)

But line y = x - 2 intersects ellipse not touching it 

Hence option (D) is corret for line y = x-2.

Let the major axis is 2a.

Focus is (3,0) lies on x-axis.

Hence, Major axis of ellipse lies on x-axis.

Given that   \(\frac{2b^2}{a}=\frac{2a}{2} ⇒\frac{2b^2}{a}=a\)

⇒ \(\)2b² = a² ⇒  \(b^2=\frac{a^2}{2}\)     ----(i)

∴  \(e =\sqrt{1-\frac{b^2}{a^2}}\) \(= \sqrt{1 - \frac{1}{2}}\)  (From (i))

 = \(e =\frac{1}{√2}\)

∴ Focus of ellipse is (ae,o) = (3,0)

⇒ ae = 3

⇒ \(a×\frac{1}{√2}=3\)

⇒ \(a=3√2\)

∴ b² = \(\frac{18}{2}\) = 9   (from (ii))

b = 3

Hence , equation of ellipse is 

\(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\)

⇒ \(\frac{x^2}{18}+\frac{y^2}{9}=1\)

 is equation of required ellipse.

High level knowledge and research related activities are Quinary sector.

Panama canal is known as ‘Gateway of Pacific’.

The practice of buying goods from other countries is known as Import.

Cotton textile industry is traditional industry of India.

Problems of Human settlement in developing countries are Agglomeration of population, Slums and Lack of drinking water. 

\(\int^{2\pi}_0[\cot^{-1}\,x]dx\)

\(= \int^{\cot 1}_0[\cot^{-1}\,x]dx\) \(+\, \int^{2\pi}_{\cot 1}[\cot^{-1}\,x]dx\)

\(= \int^{\cot 1}_0 1\,\text{dx}\) \(+\, \int^{2\pi}_{\cot 1} 0\,\text{dx}\)

(\(\because\) when \(x \in [0, \cot 1],\) \(\cot^{-1}x \in \left[\frac{\pi}{2}, 1\right]\) \(\Rightarrow [\cot^{-1} x] = 1\) and when \(x \in [\cot 1, 2\pi],\) \(\cot^{-1}x \in [1, 0]\) \(\Rightarrow [\cot^{-1} x] = 0\))

\(= [x]^{\cot 1}_0 = \cot 1 - 0 = \cot 1.\)

Correct option (C) Statement – 1 is True, Statement – 2 is False.

Explanation:

Statement–I is true, as Difference of the roots of a quadratic equation is always D , D being the discriminant of the quadratic equation and the two given equations have the same discriminant.

Statement – II is false as if two quadratic equations over reals have the same product of the coefficients, their discriminents need not be same.

Hence (c) is the correct answer. 

Given F = 1015, 

F=1015, t = 3/12 year, r = 6/100 = 0.06, BD = ? 

BD = Ftr = 1015 x 3/12 × 0.06 = 15.22

Given sin A = 3/5, cos B = 4/5

⇒ cos A = 4/5 & sin B = 3/5

∵ sin²A + cos²A = 1

(i) sin(A + B) = sinA cosB + cosA sinB 

= 3/5. 4/5 + 4/5.3/5 = 24/25

(ii) cos(A – B) = COSA COSB + sinA sinB

= 4/5.4/5 + 3/5.3/5 = 16/25 + 9/25 = 25/25 = 1

Also f(x) at x = 5 is 10; i.e, f(5) = 10 

Here lim_(x→5) f(x) = f(5) = 10.

∴ the function f(x) is continuous at x = 5

Given points are A(a, b + c), B(b, c + a) and C(c, a + b). 

We know that area of triangle whose vertices are (x₁, y₁), (x₂, y₂) and (x₃, y₃) is given by \(\frac{1}{2}\) [x₁ (y₂ − y₃) + x₂ (y₃ − y₁ ) + x₃ (y₁ − y₂)]. 

∴ Area of triangle formed by the points A, B and C 

= \(\frac{1}{2}\) [a((c + a) − (a + b)) + ((a + b) − (b+ c)) +c ((b + c)– (c + a))] 

= \(\frac{1}{2}\) [a(c – b) + b(a – c) + c(b – a)] 

= \(\frac{1}{2}\) [ac – ab + ba – bc + cb – ca] 

= \(\frac{1}{2}\) [ac – ca + ba – ab + cb – bc] 

= \(\frac{1}{2}\) [ac – ac + ba – ba + cb – cb] (∵ ac = ca & ba = ab & bc = cb) 

= \(\frac{1}{2}\) × 0 = 0. 

Area of triangle formed by the point A, B and C is zero which is condition that A, B & C are collinear. 

Hence, points A(a, b + c), B(b, c + a), and C(c, a + b) are collinear.

Option : (c)

As A is singular matrix

⇒ |A| = 0 

⇒ 2k² − 32 = 0 

⇒ k = ±4

b² -4ac = 12² -4(-9)4

Roots are real and unequal 

= 144 +144 = 228 >0

Length = √op² -r² = √(13² -5²) = 12 cm

We have function defined on such that f(x) = 2x ∀ x∊N . 

Let x₁, x₂ ∊ N such that x₁ ≠ x₂. 

⇒ 2x₁ ≠ 2x₂ (By multiplying by 2 both sides of equation. ) 

⇒ f(x₁) ≠ f(x₂). 

Hence, different elements have different images under the function . 

This implies each element of connected with different elements (even numbers) of under the function , therefore, function f(x) is one-one function. 

Let f(x) = 3 ∊ N. 

⇒ 2x = 3 ⇒ x = \(\frac{3}{2}\)∉ N . 

Hence, 3 ∊ N does not have pre-image under the function f, therefore, function f(x) is an into function. 

No odd numbers have pre-image of function . 

Hence, function f ∶ N → N such that f(x) = 2x is one-one and into function.

Let set = {1, 2, 3}.

And relation on set is defined as = {(1, 1), (2, 2), (3, 3), (1, 3), (3, 2), (1, 2)}. 

Since, if there is any (a, b) ∊ R and (b, c) ∊R for a,b,c ∊ R, then (a, c) ∊ R.

Hence, relation R is a transitive relation on set A.

2(6x -2) = (8x +4) (2x +1)

12x -4 = 10x +11

2x = 15x =15/2