Explore topic-wise InterviewSolutions in Current Affairs.

Correct answer is (d) hand

1. When flowers open, their sex organs are exposed for further process of fertilization then it is chasmogamous condition. 

2. Pollinating agents can easily transfer pollen grains in such flowers for self as well as cross pollination.

3. When flowers are closed, they are self pollinated in bud conditions then this condition is cleistogamy. 

4. When some plants have both of these types of flowers it ensures pollination and fertilization leading to seed setting. When seeds are formed then perpetuation of species is achieved as new plants will germinate from it.

The pollen grains are produced in large number from versatile anthers and are dry, light in weight for their easy dispersal.

Correct answer is (b) protogyny

The ovule which has a downwardly directed micropyle is called an anatropous ovule.

In angiosperms, the diploid condition is restored by the fusion of two haploid gametes.

Human beings maintain a constant body temperature of 37°C.

• In summers: The outside temperature is much higher than our body temperature. Therefore, we sweat profusely. This results in evaporative cooling and our body temperature is brought down to normal (37°C).

• In winters: The outside temperature is much lower than our body temperature. Therefore, we start to shiver; this action (of shivering) is a kind of exercise (work) that produces heat and raises the body temperature.

Fusion of small sized, morphologically different gametes is called merogamy.

The scientific term used for water pollinated flowers is hydrophilous.

1. Endosperm, that is developed after fertilization is a nutritive tissue for developing embryo. 

2. Endosperm stores food material. 

3. In some seeds this reserved food is partially utilized by embryo for development, E.g; Castor. 

4. The endosperm remains in the seed and it is utilized further during seed germination. Hence the seed is endospermic or albuminous.

Maturation of anther and stigma at different times is called dichogamy.

1. Endospermic seed : Castor 

2. Non-endospermic seed : Bean.

Correct answer is (b) haustorium

(b) anisogamy

Correct answer is (c) Octant

The most perfect evidence of evolution is Fossils.

1. Habitat loss and fragmentation, Overexploitation, Alien species invasion and Coextinction.

2. effects of biodiversity loss

• decline in plant production
• Lowered resistance to environment

Golden Rice...

Smelting:- The oxides of less electropositive metals like Pb, Zn, Fe, Sn, Cu etc. are reduced by strongly heating them with coal or coke. Reduction of oxide with carbon at high temperature is known as smelting.

C₂H₅OH (Ethylalcohol) + K₂Cr₂O₇/H₂SO₄ → C₆H₅NH₂ (Aniline)

To put into the world something which should make the meanest, humblest citizen, a little happier or better.

This is better than all he has done before. It is surely beautiful for it makes one happy to look at it.

One can find happiness for oneself by seeking it for others.

The memories of his sister made his heartache.

Neotype Specimen is derived from non – original collection selected as the type, when original specimen is missing or destroyed.

a. am drawing 

b. makes 

c. do

Un – unimportant, unfair, unusual, uninteresting, unlucky 

Im – immature, impossible, immortal, immobile, impatient 

In – insecure, inappropriate, incorrect, indirect, incredible

Dis- disqualify, disable, disapprove, disagree, disconnect

1. The sight of the trees and fields, the deep shadows and hills beyond, the glimmering sunlight along with rustling leaves and rippling stream make the world beautiful. 

2. According to the girl if we want to give our best we should do it for the people we love. 

3. To do anything which might represent the world badly or imperfectly. 

4. Echo

dedication, glimmer, rustling, rippling, honor, sin, grief, seize.

November 13

Monday 

A memorable day! My promise to her has been fulfilled. But, I miss my little sister a lot. She is the reason for all the appreciation I am receiving today. How happy she would be if she were alive. She was the one who wished the most to see me successful. It was her words of love and encouragement that kept me going. How I wish she were there with me. In fact it is her memories that gave me the strength to complete our dream picture. I was so pleased when I saw the happiness on the face of the people who came to see my picture. All the appreciation! and happiness is dedicated to my sister.

A very warm good morning to all seated here. Today I am very happy as I stand before you. On this occasion I remember my dear little sister. She is the person behind my success. She was the one who wished the most to see me successful. It was her words of love and encouragement that kept me going. I wish she were here with me. In fact, it is her memories that inspired me to complete our dream picture. I dedicate all the appreciation and happiness to my dear sister. She still continues to inspire me. My dear friends, before I conclude I would like to tell you that you must motivate others at the same time you can receive motivation from others.

Correct Answer is : (b) Neotype

(d) Evolutionary relationships

Correct Answer is : (c) 3

(b) Artificial system

(i) Given, f : R → R

and f(x) = x²

then if x < 0 ⇒ x² > 0

x = 0 ⇒ x² = 0

x > 0 ⇒ x² > 0

So, f(x) = x² ≥ 0 ∀ x ∈ R

Hence, range of R = R⁺ ∪ {0} or {x ∈ R | 0 ≤ x < ∞}

(ii) f(x) = 4

⇒ x² = 4

⇒ x = ±2

Hence, {x : y (x) = 4} = {-2, 2}.

(iii) f(y) = -1

⇒ y² = -1

⇒ y = ±√1

⇒ {y : f(y) = -1} = Φ null set.

Answer is (C)

f(x) = e^x

Let x, y ∈ R are such that

f(x) = f(y)

⇒ e^x = e^y

⇒ x log_e e = y log_e e

⇒ x = y

f(x) = f(y)

⇒ x = y [∵ log_e e = 1]

f is one-one function ∀ x, y ∈ R

Given, A = {-2, -1, 0, 1, 2}

and R = set of real numbers

Given: f(x) = x² + 1

then f(-2) = (-2)² + 1 = 5

f(-1) = (-1)² + 1 = 2

f(0) = (0)² + 1 = 1

f(1) = (1)² + 1 = 2

f(2) = (2)² + 1 = 5

Hence, range of f = (1, 2, 5}

(i) Given f : Q → Q and f(x) = 3x + 7

where Q is set of rational numbers.

Let x₁, x₂ ∈ Q are in this way f(x₁) = f(x₂)

f(x₁) = f(x₂)

⇒ 3x₁ + 7 = 3x₂ + 7

⇒ 3x₁ = 3x₂

⇒ x₁ = x₂

⇒ f(x₁) = f(x₂)

⇒ x₁ = x₂

x₁, x₂ ∈ Q

Hence, f is one-one function.

Now, Let y ∈ Q co-domain

If possible, then let pre-image of y is x in domain Q then

f(x) = y

⇒ 3x + 7 = y

⇒ x = (y -7)/3 ∈ Q

So, every element in the co-domain of Q has pre-image in Q

Hence, f is onto function.

Thus, f is a one-one, onto function.

(ii) f : C → R : f(x + iy) = x

Given : function

f : C → R and f(x + iy) = x

Here C = set of complex numbers

R = set of real numbers

Let x + iy and x – iy (y ≠ 0) are different elements in domain C.

f(x + iy) = x and f (x – iy) = x

⇒ f(x + iy) = f(x – iy)

So, two different elements of domain R have the same image.

So, f is a many-one function.

Here Range of f = (x : x + iy ∈ C} – R Co-domain

[Range of x + iy in x ∈ R, y ∈ R and i = √-1 ]

f is a onto function.

Thus, f is a many-one, onto function.

(iii) f : R → [-1, 1], f(x) = sin x

Given: f : R → [-1, 1] and f(x) = sinx

where R is set of real numbers.

Let x₁, x₂ ∈ R

If f(x₁) = f(x₂) ⇒ sin x₁ = sin x₂

⇒ x₁ = nπ + (-1)ⁿx₂, n ∈ I (General value of sin θ)

⇒ x₁ ≠ x₂

f is not one-one function so, it is many-one function

Again, Let y ∈ Y

If possible then let pre-image of y under f is x then
f(x) = y

⇒ sin x = y

⇒ x = sin^-1y

So, -1 ≤ y ≤ 1

Many value of sin^-1y present in R, then

x ∈ R ∀ y ∈ [-1, 1]

f is onto function.

Thus, f is many-one, onto function.

(iv) f : N → Z, f(x) = |x|

Given f : N → Z and f(x) = |x|

where N = Set of natural number = {1, 2, 3, 4,…}

and Z = set of natural numbers = {0, ± 1, ± 2, ± 3, …}

Let x₁, x₂ ∈ N

If f(x₁) = f(x₂) ⇒ |x₁| = |x₂| [∵ x₁ > 0, x₂ > 0, x₁, x₂ ∈ N]

⇒ x₁ = x₂

f is one-one function.

Again  Range of f = {|x| : x ∈ N], Z ≠ Z (co-domain)

So, f is not onto function.

Hence, f is into function.

Thus, f is one-one into function.

Answer is (D)

Given: f : R → R and f(x) = x² + x

where R is a set of real numbers.

Let x₁, x₂ ∈ R are such that f(x₁) = f(x₂)

f(x₁) = f(x₂)

⇒ x₁² + x₁ = x₂² + x₂

⇒ x₁² – x₂² + x₁ – x₂ = 0

⇒ (x₁ – x₂)(x₁ + x₂) + 1 (x₁ – x₂) = 0

⇒ (x₁ – x₂) (x₁ + x₂ + 1} = 0

⇒ x₁ = x₂, x₁ = -(x₂ + 1) ∀ x₁, x₂ ∈ R

Here, element of set A relates to two elements of set B.

So, it is a many-one function.

Again, let y ∈ R (co-domain)

If possible then let pre-image of y is x in domain R.
then f(x) = y

⇒ x² + x = y

⇒ x(x + 1) = y

⇒ x = y, x = y – 1

If y < 1, then there is no real value of x.

So, pre-image of many elements of R does not exist in domain R, 

so, f is an into function.

Thus, f is many-one, into function.

Answer is (C)

f : R₀ → R⁺, f(x) = |x|

Pre-image of every positive real number is exists in domain R₀. 

So, function is onto.

(i) f : N → N, f(x) = 2x

(ii) f : R₀ → R+, f(x) = x²

(iii) f : z₀ → N, f(x) = |x|

(iv) f : Z → Z, f(x) = 2x

(v) f : R → R₁, f(x) = x²

(vi) f : Z → f(z) = -x

Answer is  (C)

f(x) = x³

Given, f : R → R and f(x) = x³

Let y ∈ R (co-domain) if possible, then let pre-image of y is x in domain R, then

f(x) = y

⇒ x³ = y

⇒ x = y^1/3 ∈ R ∀ y ∈ R

So, pre-image of each value of y is exist in domain R.

So, R is onto function.

Answer is (B)

Given: f : N → N and f(x) = 2x + 3

where N = set of natural numbers

Let x₁, x₂ ∈ N is such that f(x₁) = f(x₂)

f(x₁) = f(x₂)

⇒ 2x₁ + 3 = 2x₂ + 3

⇒ 2x₁ = 2x₂

⇒ x₁ = x₂

f(x₁) = f(x₂) ⇒ x₁ = x₂ ∀ x₁, x₂ ∈ N

f is one-one function.

Again, let y ∈ N (co-domain) if possible than let pre image of x is in domain N then f(x) = y

f(x) = y

⇒ 2x + 3 = y

⇒ x = (y - 3)/2 ∈ N

At y = 1, then x = (1 - 3)/2 = -1 ∉ N

If this way, y has many values for which x is not exist in domain A. 

So, f is into function.

Answer is (D)

Here, f₁ = {(x, y) : y = x + 1} 

= {(1, 2), (2, 3), (3, 4)}

f₂ = {(x, y) : x + y > 4} 

= (1, 4), (2, 3), (2, 4), (3, 2), (3, 3), (3, 4), (4, 1), (4, 2), (4, 3), (4, 4)}

f₃ = {(x, y) : y < x} 

= {(2, 1), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3)}

f₄ = {(x, y) : x + y = 5} 

= {(x, y) : x = 5 – y} 

= {(1, 4), (2, 3), (3, 3), (4, 1)}

It is clear that only f₄ is a function because every element of A correspoints to unique element in B.

Answer is  (A)

A = {a, b, c} 

Number of elements in A = n(A) = 3

Then, numbers of elements in A × A 

= n(A × A) = 3² = 9

So, the number of relations in A are 

= 2ⁿ – 1 = 2⁹ – 1 

= 512 – 1 = 511.

Answer is (A)

Given : N = set of natural numbers = (1, 2, 3, 4, …}

A relation R is N is defined as

xRy ⇒ “x is divisor of y” ∀ x, y ∈ N

⇒ y/x  = k ∈ N

where N is a set of natural numbers ∀ x, y ∈ N

From option ‘A’, 36/9, 36/3 , 9/3 all are natural numbers

From option ‘B’, 11/7 ∉ N, from option ‘C’, 9/6 ∉ N

From option ‘D’, 3/2 ∉ N, 

Answer is (B)

Given : Set A = {1, 2, 3} and R = {(1, 1), (2, 2), (3, 3), (1, 2), (2, 3), (1, 3)}

(i) Reflexivity: (1, 1) ∈ R, (2, 2) ∈ R, (3, 3) ∈ R

(a, a) ∈ R

R is reflexive.

(ii) Symmetricity: (1, 2) = (2, 1) ∉ R

(1, 3) = (3, 1 ) ∉ R

(2, 3) = (3, 2) ∉ R

(a, b) ∈ R ⇒ (b, a) ∉ R 

So, R is not symmetric

Given : Set A = {1, 2, 3}

Relation R in A is defined as

R = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 3), (1, 3), (3, 1) (2, 3), (3, 2)}

(i) Reflexivity : 

Here

(1, 1) ∈ R

(2, 2) ∈ R

(3, 3) ∈ R

So, ∀ a ∈ A ⇒ (a, a) ∈ R

R is not reflexive.

(ii) Symmetricity :

Here

(1, 2) ∈ R ⇔ (2, 1) ∈ R

(2, 3) ∈ R ⇔ (3, 2) ∈ R

(1, 3) ∈ R ⇔ (3, 1) ∈ R

So, (a, b) ∈ R ⇒ (b, a) ∈ R
R is symmetric relation.

(iii) Transitivity:

(1, 2) ∈ R, (2, 1) ∈ R ⇔ (1, 1) ∈ R etc.,

So, by definition of transitive relation.

R is transitive if

(a, b) ∈ R, (b, c) ∈ R

⇒ (a, c) ∈ R ∀ a, b, c ∈ A

(3) 5

If A and B are Bijective (one-one and onto) function then n (A) = n (B)