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Given that,

AB ‖ HF and CD cuts them

∠HFC = ∠FDA (Corresponding angle)

∠FDA = 28°

∠FDA + ∠FDE + ∠EDB = 180°(Linear pair)

28° + ∠FDE + 72° = 180°

∠FDE = 80°

Red panda, Black buck.

When a particular chromosome is present in 3 copies in a diploid cell, the condition is called trisomic condition.

AB is parallel to CD, P is any point

To prove:

∠ABP + ∠BPD + ∠CDP = 360°

Construction:

Draw EF ‖ AB passing through F

Proof: Since,

AB ‖ EF and AB ‖ CD

Therefore,

EF ‖ CD (Lines parallel to the same line are parallel to each other)

∠ABP + ∠EPB = 180°(Sum of co interior angles is 180°, AB ‖ EF and BP is transversal)

∠EPD + ∠COP = 180°(Sum of co. interior angles is 180°, EF ‖ CD and DP is transversal) (i)

∠EPD + ∠CDP = 180°(Sum of co. interior angles is 180°, EF ‖ CD and DP is the transversal) (ii)

Adding (i) and (ii), we get

∠ABP + ∠EPB + ∠EPD + ∠CDP = 360°

∠ABP + ∠EPD + ∠COP = 360°

Given that,

l ‖ m

Let, l ‖ m and transversal cuts them and

∠1 = 70°

∠3 = 20°

∠4 = 30°

∠1 + ∠2 = 180°(Interior angle)

∠2 = 110° (i)

∠2 = ∠5 (Vertically opposite angle)

∠5 = 110° (ii)

∠5 + ∠3 + ∠4 = 180°(Sum of angles of a triangle is 180 o )

110° + x + 30° = 180°

x = 40°

Given that,

AB ‖ CD and PQ cuts them

EL is bisector of ∠FEB

∠LEB = ∠FEL = 35°

∠FEB = ∠LEB + ∠FEL

= 35° + 35°

= 70°

∠FEB = ∠EFC = 70°(Alternate angles)

∠EFC + ∠CFQ = 180°(Linear pair)

70° + ∠CFQ = 180°

∠CFQ = 110°

The Earth Summit was held at Rio de Janeiro (Brazil) in which representatives of more than 170 countries were present. The summit promoted Convention on Biological Diversity (CBD). India became signatory to this convention in May 1994.

The major objectives were:

i. Finding and supporting various methods to conserve biological diversity. 

ii. Use of biodiversity only up to sustainable limit. 

iii. The benefits derived from use of genetic resources should be fairly and equitably shared.

A second world summit on biological diversity was held in 2002 in Johannesburg, South Africa. In the Summit, 190 countries pledged to reduce the current rate of biodiversity loss at global, regional and local levels by 2010.

When a particular chromosome is present in a single copy in a diploid cell, the condition is called monosomic condition.

Western Ghats and North East region.

Given that,

AB ‖ CD

Let, EF be the parallel line to AB and CD which passes through P.

It can be seen from the figure that alternate angles are equal

∠ABP = ∠BPF

∠CDP = ∠DPF

∠ABP + ∠CDP = ∠BPF + ∠DPF

∠ABP + ∠CDP = ∠DPB

Hence, proved

Given that,

AB ‖ CD

Produce P to Q so that PQ ‖ AB ‖ CD

∠BAP + ∠APQ = 180°(Interior angle)

132° + ∠APQ = 180°

∠APQ = 48° (i)

∠APC = ∠APQ + ∠QPC

148° = 48° + ∠QPC [From (i)]

∠QPC = 100°

∠QPC + ∠PCD = 180°(Interior angles)

100° + ∠PCD = 180°

∠PCD = 80°

∠PCD + x = 180°(Linear pair)

80° + x = 180°

x = 100°

Given that,

AB ‖ CD and transversal cuts them

Let,

∠1 = 120° + x and

∠2 = x

∠1 = ∠3 (Alternate angles)

∠3 = 120° + x (i)

∠2 + ∠3 = 180°(Linear pair)

x + 120° + x = 180°

2x = 60°

x = 30°

Correct option is (B) \(\overrightarrow{AB}\)

It is the measure of variety in genetic information contained in the organisms.

1. Ex-situ conservation

2. a. In-vitro fertilisation: Gametes of threatened species can be fertilised for their propagation. 

b. Cryopreservation techniques: Gametes of threatened species can be preserved in viable and fertile condition for long periods.

The presence of a chromosome as a single copy in a diploid cell is termed as monosomic condition.

Since,

AB ‖ CD

And, BD cuts them

y + 2y + y + 5y = 180°(Consecutive interior angle)

9y = 180°

y = 20°

Correct option is C) \(\overline{AB}\)
 

Causes of Biodiversity Losses

i. Habitat loss and fragmentation 

• Destruction of habitat is the primary cause of extinction of species. 

• The tropical rainforests initially covered 14 per cent of the land surface of earth, but now cover only 6 per cent of land area.

ii. Over-exploitation 

• When biological system is over-exploited by man for the natural resources, it results in degradation and extinction of the resources. 

• For example, Stellar’s sea cow, passenger pigeon and many marine fishes.

iii. Alien (exotic) species invasions 

• Some alien (exotic) species when introduced unintentionally or deliberately, become invasive and cause harmful impact, resulting in extinction of the indigenous species. 

• Nile perch, a large predator fish when introduced in Lake Victoria (East Africa) caused the extinction of an ecologically unique species of Cichlid fish in the lake.

iv. Co-extinctions

• When a species becomes extinct, the plant and animal species associated with it in an obligatory manner, also become extinct. 

• For example, if the host fish species becomes extinct, all those parasites exclusively dependent on it, will also become extinct; in plant–pollinator mutualism also, extinction of one results in the extinction of the other.

Genetic diversity is the measure of variety in genetic information contained in the organisms. 

Significance of large genetic diversity are as follows:

i. Larger genetic diversity provides adaptability at the time of environmental changes and helps the species in surviving. 

ii. Larger genetic diversity is also useful in the evolution of species.

International Union for Conservation of Nature and Natural Resources.

The Red Data Book is a compilation of data on species threatened with extinction and is maintained by IUCN.

Latitudinal gradients 

a. Biodiversity increases from poles to equators, i.e., from high to low latitude. 

b. Tropics (23.5°N to 23.5°S) have more species than temperate or polar regions. 

For example, Columbia located near the equator has 1,400 species of birds while New York (41.5°N) has 105 species and Greenland (71°N) has only 56 species.

Causes of Biodiversity Losses 

There are four major causes of biodiversity loss. These are also known as ‘The Evil Quartet’.

i. Habitat loss and fragmentation 

ii. Over-exploitation 

iii. Alien (exotic) species invasions 

iv. Co-extinctions

IUCN red list is a catalogue of species and subspecies that are facing the risk of extinction.

The two uses of this list are: 

i. Provides information and develops awareness about the importance of threatened species. 

ii. Identification and documentation of endangered species.

Phenotype = Genotype + Environment 

(Trait)         (Potentiality) (opportunity)

(a) This is an exception to Mendel’s principle of dominance and can be explained by the phenomenon of ‘Incomplete dominance’. It is a phenomenon where none of the two contrasting alleles or factors are dominant. The expression of the character in a hybrid or F1 individual is intermediate or a fine mixture of expression of the two factors (pink flowers in this case from two parents with red and white flowers). This may be considered as an example of quantitative inheritance where only a single gene pair is involved. F2 phenotypic ratio is 1 : 2 : 1, similar to the genotypic ratio, in which the parental characters also reappear. 

(b) Mendel’s Laws of Inheritance 

▪ Based on his hybridisation experiments, Mendel proposed the laws of inheritance. 

▪ His theory was rediscovered by Hugo de Vries of Holland, Carl Correns of Germany and Eric von Tschermak of Austria in 1901. 

(i) Law of dominance 

▪ This law states that when two alternative forms of a trait or character (genes or alleles) are present in an organism, only one factor expresses itself in F1 progeny and is called dominant while the other that remains masked is called recessive. 

(ii) Law of segregation. 

▪ This law states that the factors or alleles of a pair segregate from each other during gamete formation, such that a gamete receives only one of the two factors. They do not show any blending.

Correct option is: C) 4

i) This is not a mathematical statement. no mathematics is involved in it.

ii) This is not a statement, as its truthness cant be determined.

iii) This is not a statement. This is an ambiguous open sentence.

iv) This is a mathematical statement.

v) This is not a mathematical statement.

Correct option is: C) Theorem

Correct option is: A) Mathematical proof

Let the two positive numbers be a and b. 

Given : 

a + b = 64 … (1) 

Also, 

a³ + b³ is minima 

Assume, 

S = a³ + b³ (from equation 1) 

S = a³ + (64 – a)³ 

⇒ \(\frac{ds}{da}\) = 3a² + 3(64 – a)² × ( - 1)

⇒ \(\frac{ds}{da}\) = 0

(condition for maxima and minima) 

⇒ 3a² + 3(64 – a)² × ( - 1) = 0 

⇒ 3a² + 3(4096 + a² – 128a) × (-1) = 0 

⇒ 3a² – 3 × 4096 - 3a² + 424a = 0 

⇒ a = 32

\(\frac{d^2s}{da^2}\) = 6a + 6(64 - a) = 384

Since, 

\(\frac{d^2s}{da^2}\)> 0 

⇒ a = 32 will give minimum value

Hence, two numbers will be 32 and 32.

p(x) = x² + x + 41 

p(0) = 0² + 0 + 41 = 41 – is a prime 

p(1) = 1² + 1 + 41 = 43 – is a prime 

p(2) = 2² + 2 + 41 = 47 – is a prime 

p(3) = 3² + 3 + 41 = 53 – is a prime

p(41) = 41² + 41 + 41 

= 41(41 + 1 + 1) 

= 41 x 43 is not a prime. 

∴ p(x) = x² + x + 41 is not a prime for all x. 

∴ The conjecture “p(x) = x² + x + 41 is a prime” is false.

Let a three digit number be xyz. 

Repeat the digit = xyzxyz 

= xyz × (1001) 

= xyz × (7 × 11 × 13) 

Hence the given conjecture is true.

Choose a number = x say 

Double it = 2x 

Add nine = 2x + 9 

Add your original number 

= 2x + 9 + x = 3x + 9 

Divide by 3 = (3x + 9) ÷ 3

= 3x/3 + 9/3 = x + 3

Add 4 ⇒ x + 3 + 4 = x + 7 

Subtract your original number = 

x + 7 – x = 7 

Your result is 7 – True.

Head = H, Tail = T 

All possible outcomes = HH, HT, TH, TT

1. Things which are equal to the same things are equal to one another. 

2. If equals are added to equals, the sums are equal. 

3. If equals are subtracted from equals, the differences are equal.

4. When a pair of parallel lines are intersected by a transversal, the pairs of corresponding angles are equal. 

5. There passes infinitely many lines through a given point.

1) 9 is a prime number – False 

This is a statement because we can judge the truthness of this sentence. Clearly it is a false statement as 9 has factors other than 1 and 9, hence it is a composite number.

2) x is less than 5 – can’t say True or False 

This is not a statement. The truthness can t be verified unless the value of x is known. Hence it is a sentence only. 

3) 3 + 5 = 8 – True 

The above sentence is a statement. It is a true statement as 5 + 3 = 8. 

4) Sum of two odd numbers is even – True 

The above sentence can be verified as a true sentence by taking examples like 3 + 5 = 8, 5 + 7 = 12 etc. Hence it is a true statement. 

5) X/2+3 = 9- can’t say True or False. 

The above sentence is not a statement. Its truthness can’t be verified without the value of x.

24 = 2³ × 3¹ 

24 has (3+1) (1 + 1) = 4 × 2 = 8 factors

[1, 2, 3, 4, 6, 8, 12 and 24] 

36 = 2² × 3² 

Number of factors = (2 + 1) (2 + 1) 

3 × 3 = 9 [ 1, 2, 3. 4, 6, 9, 12, 18 and 36] 

If N = a^p . b^q . c^r …….. where N is a natural number. 

a. b, c … are primes and p, q, r ……. are positive integers then, the number of factors of N =(p- 1) (q+ 1)(r + 1)

i) Any natural number greater than 1 can be represented in prime factorization.

ii) Two times a natural number is always even.

iii) For any x > 1; 3x + 1 > 4.

iv) For any x > 0; x³ ≥ 0.

v) In an equilateral triangle, a median is also an angle bisector.

Let x be an odd number. 

Then x = 2m + 1 

(general form of ah odd number) Squaring on both sides, 

x² = (2m + 1)² 

= 4m² + 4m +1

 = 2 (2m² + 2m) + 1 

= 2K + 1 where K = 2m² + 2 

Hence x² is also odd

111111² = 12345654321 

1111111² = 1234567654321 

(111………. n-times)² = (123 … (n- 1) n (n – 1) (n – 2) 1) 

The conjecture is true.

Line-4 : 1331 = 11³ 

Line-5 : 14641 = 11⁴ 

Line – 6 : 11⁵ 

∴ Line – n = 11^n-1 

Yes the conjecture holds good for line – 6 too.

i) False. Sum of the interior angles of a quadrilateral is 360°.

ii)True. This is true for all real numbers,

iii) True. In a rhombus, both pairs of opposite sides are parallel and hence every rhombus is a parallelogram.

iv) True. This is true for any two even numbers.

v) Ambiguous. Since square of an odd number can’t be written as sum of two odd numbers.

Correct option is: B) conjecture

i) Always false. Generally 30 days or 31 days make a month except February.

ii) Ambiguous. Makarasankranthi may fall on any day of the week.

iii) Ambiguous. Sometimes the temperature may go down to 2°C in winter.

iv) We can’t say always true. To the known fact, so far we can say this.

v) Always false, ’as dogs can never fly.

vi) Ambiguous. A leap year has 29 days for February.

Answer is : D. x = -π/2

We can go through options for this question

Option a is wrong because 0 is not included in (-π,0)

At x = -π/4 value of f(x) is -√2 = -1.41

At x = -π/3 value of f(x) is -2.

At x = -π/2 value of f(x) is -1.

∴ f(x) has max value at x = -π/2.

Which is -1.

Correct option is: B) always false